3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.063, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 1 \ j ---------------- = ) a[1][j] x 3 2 / -x - x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 2 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 2 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j ---------------- = ) a[1][j] x 3 2 / -x - x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 are the following triples {[0, 1, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.027, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 4 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j ---------------- = ) a[1][j] x 3 2 / -x - x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 are the following triples {[1, 1, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.037, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.021, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 7 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x + 1 \ j ---------------- = ) a[1][j] x 3 2 / -x - x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 are the following triples {[1, 1, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 2 x + 2 \ j ---------------- = ) a[1][j] x 3 2 / -x - x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + x[1] x[3] 2 2 3 2 + 2 x[1] x[2] - 2 x[1] x[2] x[3] - x[1] x[3] + 2 x[2] - 2 x[2] x[3] 3 + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + x[1] x[3] + 2 x[1] x[2] - 2 x[1] x[2] x[3] 2 3 2 3 - x[1] x[3] + 2 x[2] - 2 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.036, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 2 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 0, a[2][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 1 \ j ------------------ = ) a[2][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 are the following triples {[1, 0, 0], [1, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j ------------------ = ) a[1][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 2 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j ------------------ = ) a[2][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 are the following triples {[0, 1, 0], [1, 1, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.037, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 2 x + 2 \ j ------------------ = ) a[1][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 2 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 2 x + 2 \ j ------------------ = ) a[2][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 are the following triples {[2, 0, 0], [2, 0, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following, 4, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -2 x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 2 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 2 x + 2 \ j ------------------ = ) a[2][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = a[3][j - 1] + 2 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 1, a[3][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 2 \ j ------------------ = ) a[3][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = a[4][j - 1] + 2 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 2, a[4][1] = 2, a[4][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -x + 2 \ j ------------------ = ) a[4][j] x 3 2 / -x - 2 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 are the following triples {[1, 1, 1], [2, 0, 1], [2, 1, 4], [2, 2, 5]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + x[1] x[3] 2 2 3 2 + 5 x[1] x[2] - x[1] x[2] x[3] - 2 x[1] x[3] + 3 x[2] - x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + x[1] x[3] + 5 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 3 x[2] - x[2] x[3] - 2 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.037, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 By Shalosh B. Ekhad The following, 92, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -2 x - x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 3 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j ------------------ = ) a[2][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = a[3][j - 1] + 3 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 0, a[3][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 2 x + 2 \ j ------------------ = ) a[3][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = a[4][j - 1] + 3 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 2, a[4][1] = 1, a[4][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -3 x - x + 2 \ j ------------------ = ) a[4][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[5][j], for j from 0 to infinity as the solution of the recurrence a[5][j] = a[5][j - 1] + 3 a[5][j - 2] + a[5][j - 3] Subject to the initial conditions a[5][0] = 2, a[5][1] = 2, a[5][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -2 x + 2 \ j ------------------ = ) a[5][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[5][j], a[5][j + 1], a[5][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[6][j], for j from 0 to infinity as the solution of the recurrence a[6][j] = a[6][j - 1] + 3 a[6][j - 2] + a[6][j - 3] Subject to the initial conditions a[6][0] = 2, a[6][1] = 3, a[6][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -x + x + 2 \ j ------------------ = ) a[6][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[6][j], a[6][j + 1], a[6][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[7][j], for j from 0 to infinity as the solution of the recurrence a[7][j] = a[7][j - 1] + 3 a[7][j - 2] + a[7][j - 3] Subject to the initial conditions a[7][0] = 3, a[7][1] = 0, a[7][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 3 x + 3 \ j ------------------ = ) a[7][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[7][j], a[7][j + 1], a[7][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[8][j], for j from 0 to infinity as the solution of the recurrence a[8][j] = a[8][j - 1] + 3 a[8][j - 2] + a[8][j - 3] Subject to the initial conditions a[8][0] = 3, a[8][1] = 1, a[8][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 2 x + 3 \ j ------------------ = ) a[8][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[8][j], a[8][j + 1], a[8][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[9][j], for j from 0 to infinity as the solution of the recurrence a[9][j] = a[9][j - 1] + 3 a[9][j - 2] + a[9][j - 3] Subject to the initial conditions a[9][0] = 3, a[9][1] = 2, a[9][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -4 x - x + 3 \ j ------------------ = ) a[9][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[9][j], a[9][j + 1], a[9][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[10][j], for j from 0 to infinity as the solution of the recurrence a[10][j] = a[10][j - 1] + 3 a[10][j - 2] + a[10][j - 3] Subject to the initial conditions a[10][0] = 3, a[10][1] = 3, a[10][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -3 x + 3 \ j ------------------ = ) a[10][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[10][j], a[10][j + 1], a[10][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[11][j], for j from 0 to infinity as the solution of the recurrence a[11][j] = a[11][j - 1] + 3 a[11][j - 2] + a[11][j - 3] Subject to the initial conditions a[11][0] = 3, a[11][1] = 4, a[11][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -2 x + x + 3 \ j ------------------ = ) a[11][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[11][j], a[11][j + 1], a[11][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[12][j], for j from 0 to infinity as the solution of the recurrence a[12][j] = a[12][j - 1] + 3 a[12][j - 2] + a[12][j - 3] Subject to the initial conditions a[12][0] = 3, a[12][1] = 5, a[12][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -x + 2 x + 3 \ j ------------------ = ) a[12][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[12][j], a[12][j + 1], a[12][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[13][j], for j from 0 to infinity as the solution of the recurrence a[13][j] = a[13][j - 1] + 3 a[13][j - 2] + a[13][j - 3] Subject to the initial conditions a[13][0] = 4, a[13][1] = 0, a[13][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 4 x + 4 \ j ------------------ = ) a[13][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[13][j], a[13][j + 1], a[13][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[14][j], for j from 0 to infinity as the solution of the recurrence a[14][j] = a[14][j - 1] + 3 a[14][j - 2] + a[14][j - 3] Subject to the initial conditions a[14][0] = 4, a[14][1] = 1, a[14][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -7 x - 3 x + 4 \ j ------------------ = ) a[14][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[14][j], a[14][j + 1], a[14][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[15][j], for j from 0 to infinity as the solution of the recurrence a[15][j] = a[15][j - 1] + 3 a[15][j - 2] + a[15][j - 3] Subject to the initial conditions a[15][0] = 4, a[15][1] = 2, a[15][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 2 x + 4 \ j ------------------ = ) a[15][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[15][j], a[15][j + 1], a[15][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[16][j], for j from 0 to infinity as the solution of the recurrence a[16][j] = a[16][j - 1] + 3 a[16][j - 2] + a[16][j - 3] Subject to the initial conditions a[16][0] = 4, a[16][1] = 3, a[16][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -5 x - x + 4 \ j ------------------ = ) a[16][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[16][j], a[16][j + 1], a[16][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[17][j], for j from 0 to infinity as the solution of the recurrence a[17][j] = a[17][j - 1] + 3 a[17][j - 2] + a[17][j - 3] Subject to the initial conditions a[17][0] = 4, a[17][1] = 4, a[17][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -4 x + 4 \ j ------------------ = ) a[17][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[17][j], a[17][j + 1], a[17][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[18][j], for j from 0 to infinity as the solution of the recurrence a[18][j] = a[18][j - 1] + 3 a[18][j - 2] + a[18][j - 3] Subject to the initial conditions a[18][0] = 4, a[18][1] = 5, a[18][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -3 x + x + 4 \ j ------------------ = ) a[18][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[18][j], a[18][j + 1], a[18][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[19][j], for j from 0 to infinity as the solution of the recurrence a[19][j] = a[19][j - 1] + 3 a[19][j - 2] + a[19][j - 3] Subject to the initial conditions a[19][0] = 4, a[19][1] = 6, a[19][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -2 x + 2 x + 4 \ j ------------------ = ) a[19][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[19][j], a[19][j + 1], a[19][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[20][j], for j from 0 to infinity as the solution of the recurrence a[20][j] = a[20][j - 1] + 3 a[20][j - 2] + a[20][j - 3] Subject to the initial conditions a[20][0] = 4, a[20][1] = 7, a[20][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -x + 3 x + 4 \ j ------------------ = ) a[20][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[20][j], a[20][j + 1], a[20][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[21][j], for j from 0 to infinity as the solution of the recurrence a[21][j] = a[21][j - 1] + 3 a[21][j - 2] + a[21][j - 3] Subject to the initial conditions a[21][0] = 5, a[21][1] = 0, a[21][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 5 x + 5 \ j ------------------ = ) a[21][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[21][j], a[21][j + 1], a[21][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[22][j], for j from 0 to infinity as the solution of the recurrence a[22][j] = a[22][j - 1] + 3 a[22][j - 2] + a[22][j - 3] Subject to the initial conditions a[22][0] = 5, a[22][1] = 1, a[22][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -9 x - 4 x + 5 \ j ------------------ = ) a[22][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[22][j], a[22][j + 1], a[22][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[23][j], for j from 0 to infinity as the solution of the recurrence a[23][j] = a[23][j - 1] + 3 a[23][j - 2] + a[23][j - 3] Subject to the initial conditions a[23][0] = 5, a[23][1] = 2, a[23][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 3 x + 5 \ j ------------------ = ) a[23][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[23][j], a[23][j + 1], a[23][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[24][j], for j from 0 to infinity as the solution of the recurrence a[24][j] = a[24][j - 1] + 3 a[24][j - 2] + a[24][j - 3] Subject to the initial conditions a[24][0] = 5, a[24][1] = 3, a[24][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -7 x - 2 x + 5 \ j ------------------ = ) a[24][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[24][j], a[24][j + 1], a[24][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[25][j], for j from 0 to infinity as the solution of the recurrence a[25][j] = a[25][j - 1] + 3 a[25][j - 2] + a[25][j - 3] Subject to the initial conditions a[25][0] = 5, a[25][1] = 4, a[25][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -6 x - x + 5 \ j ------------------ = ) a[25][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[25][j], a[25][j + 1], a[25][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[26][j], for j from 0 to infinity as the solution of the recurrence a[26][j] = a[26][j - 1] + 3 a[26][j - 2] + a[26][j - 3] Subject to the initial conditions a[26][0] = 5, a[26][1] = 5, a[26][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -5 x + 5 \ j ------------------ = ) a[26][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[26][j], a[26][j + 1], a[26][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[27][j], for j from 0 to infinity as the solution of the recurrence a[27][j] = a[27][j - 1] + 3 a[27][j - 2] + a[27][j - 3] Subject to the initial conditions a[27][0] = 5, a[27][1] = 6, a[27][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -4 x + x + 5 \ j ------------------ = ) a[27][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[27][j], a[27][j + 1], a[27][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[28][j], for j from 0 to infinity as the solution of the recurrence a[28][j] = a[28][j - 1] + 3 a[28][j - 2] + a[28][j - 3] Subject to the initial conditions a[28][0] = 5, a[28][1] = 7, a[28][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -3 x + 2 x + 5 \ j ------------------ = ) a[28][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[28][j], a[28][j + 1], a[28][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[29][j], for j from 0 to infinity as the solution of the recurrence a[29][j] = a[29][j - 1] + 3 a[29][j - 2] + a[29][j - 3] Subject to the initial conditions a[29][0] = 6, a[29][1] = 0, a[29][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -12 x - 6 x + 6 \ j ------------------ = ) a[29][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[29][j], a[29][j + 1], a[29][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[30][j], for j from 0 to infinity as the solution of the recurrence a[30][j] = a[30][j - 1] + 3 a[30][j - 2] + a[30][j - 3] Subject to the initial conditions a[30][0] = 6, a[30][1] = 1, a[30][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -11 x - 5 x + 6 \ j ------------------ = ) a[30][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[30][j], a[30][j + 1], a[30][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[31][j], for j from 0 to infinity as the solution of the recurrence a[31][j] = a[31][j - 1] + 3 a[31][j - 2] + a[31][j - 3] Subject to the initial conditions a[31][0] = 6, a[31][1] = 2, a[31][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 4 x + 6 \ j ------------------ = ) a[31][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[31][j], a[31][j + 1], a[31][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[32][j], for j from 0 to infinity as the solution of the recurrence a[32][j] = a[32][j - 1] + 3 a[32][j - 2] + a[32][j - 3] Subject to the initial conditions a[32][0] = 6, a[32][1] = 3, a[32][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -9 x - 3 x + 6 \ j ------------------ = ) a[32][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[32][j], a[32][j + 1], a[32][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[33][j], for j from 0 to infinity as the solution of the recurrence a[33][j] = a[33][j - 1] + 3 a[33][j - 2] + a[33][j - 3] Subject to the initial conditions a[33][0] = 6, a[33][1] = 4, a[33][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 2 x + 6 \ j ------------------ = ) a[33][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[33][j], a[33][j + 1], a[33][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[34][j], for j from 0 to infinity as the solution of the recurrence a[34][j] = a[34][j - 1] + 3 a[34][j - 2] + a[34][j - 3] Subject to the initial conditions a[34][0] = 6, a[34][1] = 5, a[34][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -7 x - x + 6 \ j ------------------ = ) a[34][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[34][j], a[34][j + 1], a[34][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[35][j], for j from 0 to infinity as the solution of the recurrence a[35][j] = a[35][j - 1] + 3 a[35][j - 2] + a[35][j - 3] Subject to the initial conditions a[35][0] = 6, a[35][1] = 6, a[35][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -6 x + 6 \ j ------------------ = ) a[35][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[35][j], a[35][j + 1], a[35][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[36][j], for j from 0 to infinity as the solution of the recurrence a[36][j] = a[36][j - 1] + 3 a[36][j - 2] + a[36][j - 3] Subject to the initial conditions a[36][0] = 6, a[36][1] = 7, a[36][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -5 x + x + 6 \ j ------------------ = ) a[36][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[36][j], a[36][j + 1], a[36][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[37][j], for j from 0 to infinity as the solution of the recurrence a[37][j] = a[37][j - 1] + 3 a[37][j - 2] + a[37][j - 3] Subject to the initial conditions a[37][0] = 7, a[37][1] = 0, a[37][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -14 x - 7 x + 7 \ j ------------------ = ) a[37][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[37][j], a[37][j + 1], a[37][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[38][j], for j from 0 to infinity as the solution of the recurrence a[38][j] = a[38][j - 1] + 3 a[38][j - 2] + a[38][j - 3] Subject to the initial conditions a[38][0] = 7, a[38][1] = 1, a[38][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -13 x - 6 x + 7 \ j ------------------ = ) a[38][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[38][j], a[38][j + 1], a[38][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[39][j], for j from 0 to infinity as the solution of the recurrence a[39][j] = a[39][j - 1] + 3 a[39][j - 2] + a[39][j - 3] Subject to the initial conditions a[39][0] = 7, a[39][1] = 2, a[39][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -12 x - 5 x + 7 \ j ------------------ = ) a[39][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[39][j], a[39][j + 1], a[39][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[40][j], for j from 0 to infinity as the solution of the recurrence a[40][j] = a[40][j - 1] + 3 a[40][j - 2] + a[40][j - 3] Subject to the initial conditions a[40][0] = 7, a[40][1] = 3, a[40][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -11 x - 4 x + 7 \ j ------------------ = ) a[40][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[40][j], a[40][j + 1], a[40][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[41][j], for j from 0 to infinity as the solution of the recurrence a[41][j] = a[41][j - 1] + 3 a[41][j - 2] + a[41][j - 3] Subject to the initial conditions a[41][0] = 7, a[41][1] = 4, a[41][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 3 x + 7 \ j ------------------ = ) a[41][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[41][j], a[41][j + 1], a[41][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[42][j], for j from 0 to infinity as the solution of the recurrence a[42][j] = a[42][j - 1] + 3 a[42][j - 2] + a[42][j - 3] Subject to the initial conditions a[42][0] = 7, a[42][1] = 5, a[42][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -9 x - 2 x + 7 \ j ------------------ = ) a[42][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[42][j], a[42][j + 1], a[42][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[43][j], for j from 0 to infinity as the solution of the recurrence a[43][j] = a[43][j - 1] + 3 a[43][j - 2] + a[43][j - 3] Subject to the initial conditions a[43][0] = 7, a[43][1] = 6, a[43][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -8 x - x + 7 \ j ------------------ = ) a[43][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[43][j], a[43][j + 1], a[43][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[44][j], for j from 0 to infinity as the solution of the recurrence a[44][j] = a[44][j - 1] + 3 a[44][j - 2] + a[44][j - 3] Subject to the initial conditions a[44][0] = 8, a[44][1] = 0, a[44][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -16 x - 8 x + 8 \ j ------------------ = ) a[44][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[44][j], a[44][j + 1], a[44][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[45][j], for j from 0 to infinity as the solution of the recurrence a[45][j] = a[45][j - 1] + 3 a[45][j - 2] + a[45][j - 3] Subject to the initial conditions a[45][0] = 8, a[45][1] = 1, a[45][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -15 x - 7 x + 8 \ j ------------------ = ) a[45][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[45][j], a[45][j + 1], a[45][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[46][j], for j from 0 to infinity as the solution of the recurrence a[46][j] = a[46][j - 1] + 3 a[46][j - 2] + a[46][j - 3] Subject to the initial conditions a[46][0] = 8, a[46][1] = 2, a[46][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -14 x - 6 x + 8 \ j ------------------ = ) a[46][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[46][j], a[46][j + 1], a[46][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[47][j], for j from 0 to infinity as the solution of the recurrence a[47][j] = a[47][j - 1] + 3 a[47][j - 2] + a[47][j - 3] Subject to the initial conditions a[47][0] = 8, a[47][1] = 3, a[47][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -13 x - 5 x + 8 \ j ------------------ = ) a[47][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[47][j], a[47][j + 1], a[47][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[48][j], for j from 0 to infinity as the solution of the recurrence a[48][j] = a[48][j - 1] + 3 a[48][j - 2] + a[48][j - 3] Subject to the initial conditions a[48][0] = 8, a[48][1] = 4, a[48][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -12 x - 4 x + 8 \ j ------------------ = ) a[48][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[48][j], a[48][j + 1], a[48][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[49][j], for j from 0 to infinity as the solution of the recurrence a[49][j] = a[49][j - 1] + 3 a[49][j - 2] + a[49][j - 3] Subject to the initial conditions a[49][0] = 8, a[49][1] = 5, a[49][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -11 x - 3 x + 8 \ j ------------------ = ) a[49][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[49][j], a[49][j + 1], a[49][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[50][j], for j from 0 to infinity as the solution of the recurrence a[50][j] = a[50][j - 1] + 3 a[50][j - 2] + a[50][j - 3] Subject to the initial conditions a[50][0] = 8, a[50][1] = 6, a[50][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 2 x + 8 \ j ------------------ = ) a[50][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[50][j], a[50][j + 1], a[50][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[51][j], for j from 0 to infinity as the solution of the recurrence a[51][j] = a[51][j - 1] + 3 a[51][j - 2] + a[51][j - 3] Subject to the initial conditions a[51][0] = 9, a[51][1] = 0, a[51][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -18 x - 9 x + 9 \ j ------------------ = ) a[51][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[51][j], a[51][j + 1], a[51][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[52][j], for j from 0 to infinity as the solution of the recurrence a[52][j] = a[52][j - 1] + 3 a[52][j - 2] + a[52][j - 3] Subject to the initial conditions a[52][0] = 9, a[52][1] = 1, a[52][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -17 x - 8 x + 9 \ j ------------------ = ) a[52][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[52][j], a[52][j + 1], a[52][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[53][j], for j from 0 to infinity as the solution of the recurrence a[53][j] = a[53][j - 1] + 3 a[53][j - 2] + a[53][j - 3] Subject to the initial conditions a[53][0] = 9, a[53][1] = 2, a[53][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -16 x - 7 x + 9 \ j ------------------ = ) a[53][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[53][j], a[53][j + 1], a[53][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[54][j], for j from 0 to infinity as the solution of the recurrence a[54][j] = a[54][j - 1] + 3 a[54][j - 2] + a[54][j - 3] Subject to the initial conditions a[54][0] = 9, a[54][1] = 3, a[54][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -15 x - 6 x + 9 \ j ------------------ = ) a[54][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[54][j], a[54][j + 1], a[54][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[55][j], for j from 0 to infinity as the solution of the recurrence a[55][j] = a[55][j - 1] + 3 a[55][j - 2] + a[55][j - 3] Subject to the initial conditions a[55][0] = 9, a[55][1] = 4, a[55][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -14 x - 5 x + 9 \ j ------------------ = ) a[55][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[55][j], a[55][j + 1], a[55][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[56][j], for j from 0 to infinity as the solution of the recurrence a[56][j] = a[56][j - 1] + 3 a[56][j - 2] + a[56][j - 3] Subject to the initial conditions a[56][0] = 9, a[56][1] = 5, a[56][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -13 x - 4 x + 9 \ j ------------------ = ) a[56][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[56][j], a[56][j + 1], a[56][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[57][j], for j from 0 to infinity as the solution of the recurrence a[57][j] = a[57][j - 1] + 3 a[57][j - 2] + a[57][j - 3] Subject to the initial conditions a[57][0] = 10, a[57][1] = 0, a[57][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -20 x - 10 x + 10 \ j ------------------ = ) a[57][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[57][j], a[57][j + 1], a[57][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[58][j], for j from 0 to infinity as the solution of the recurrence a[58][j] = a[58][j - 1] + 3 a[58][j - 2] + a[58][j - 3] Subject to the initial conditions a[58][0] = 10, a[58][1] = 1, a[58][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -19 x - 9 x + 10 \ j ------------------ = ) a[58][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[58][j], a[58][j + 1], a[58][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[59][j], for j from 0 to infinity as the solution of the recurrence a[59][j] = a[59][j - 1] + 3 a[59][j - 2] + a[59][j - 3] Subject to the initial conditions a[59][0] = 10, a[59][1] = 2, a[59][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -18 x - 8 x + 10 \ j ------------------ = ) a[59][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[59][j], a[59][j + 1], a[59][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[60][j], for j from 0 to infinity as the solution of the recurrence a[60][j] = a[60][j - 1] + 3 a[60][j - 2] + a[60][j - 3] Subject to the initial conditions a[60][0] = 10, a[60][1] = 3, a[60][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -17 x - 7 x + 10 \ j ------------------ = ) a[60][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[60][j], a[60][j + 1], a[60][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[61][j], for j from 0 to infinity as the solution of the recurrence a[61][j] = a[61][j - 1] + 3 a[61][j - 2] + a[61][j - 3] Subject to the initial conditions a[61][0] = 10, a[61][1] = 4, a[61][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -16 x - 6 x + 10 \ j ------------------ = ) a[61][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[61][j], a[61][j + 1], a[61][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[62][j], for j from 0 to infinity as the solution of the recurrence a[62][j] = a[62][j - 1] + 3 a[62][j - 2] + a[62][j - 3] Subject to the initial conditions a[62][0] = 10, a[62][1] = 5, a[62][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -15 x - 5 x + 10 \ j ------------------ = ) a[62][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[62][j], a[62][j + 1], a[62][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[63][j], for j from 0 to infinity as the solution of the recurrence a[63][j] = a[63][j - 1] + 3 a[63][j - 2] + a[63][j - 3] Subject to the initial conditions a[63][0] = 11, a[63][1] = 0, a[63][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -22 x - 11 x + 11 \ j ------------------ = ) a[63][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[63][j], a[63][j + 1], a[63][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[64][j], for j from 0 to infinity as the solution of the recurrence a[64][j] = a[64][j - 1] + 3 a[64][j - 2] + a[64][j - 3] Subject to the initial conditions a[64][0] = 11, a[64][1] = 1, a[64][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -21 x - 10 x + 11 \ j ------------------ = ) a[64][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[64][j], a[64][j + 1], a[64][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[65][j], for j from 0 to infinity as the solution of the recurrence a[65][j] = a[65][j - 1] + 3 a[65][j - 2] + a[65][j - 3] Subject to the initial conditions a[65][0] = 11, a[65][1] = 2, a[65][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -20 x - 9 x + 11 \ j ------------------ = ) a[65][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[65][j], a[65][j + 1], a[65][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[66][j], for j from 0 to infinity as the solution of the recurrence a[66][j] = a[66][j - 1] + 3 a[66][j - 2] + a[66][j - 3] Subject to the initial conditions a[66][0] = 11, a[66][1] = 3, a[66][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -19 x - 8 x + 11 \ j ------------------ = ) a[66][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[66][j], a[66][j + 1], a[66][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[67][j], for j from 0 to infinity as the solution of the recurrence a[67][j] = a[67][j - 1] + 3 a[67][j - 2] + a[67][j - 3] Subject to the initial conditions a[67][0] = 11, a[67][1] = 4, a[67][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -18 x - 7 x + 11 \ j ------------------ = ) a[67][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[67][j], a[67][j + 1], a[67][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[68][j], for j from 0 to infinity as the solution of the recurrence a[68][j] = a[68][j - 1] + 3 a[68][j - 2] + a[68][j - 3] Subject to the initial conditions a[68][0] = 12, a[68][1] = 0, a[68][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -24 x - 12 x + 12 \ j ------------------ = ) a[68][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[68][j], a[68][j + 1], a[68][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[69][j], for j from 0 to infinity as the solution of the recurrence a[69][j] = a[69][j - 1] + 3 a[69][j - 2] + a[69][j - 3] Subject to the initial conditions a[69][0] = 12, a[69][1] = 1, a[69][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -23 x - 11 x + 12 \ j ------------------ = ) a[69][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[69][j], a[69][j + 1], a[69][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[70][j], for j from 0 to infinity as the solution of the recurrence a[70][j] = a[70][j - 1] + 3 a[70][j - 2] + a[70][j - 3] Subject to the initial conditions a[70][0] = 12, a[70][1] = 2, a[70][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -22 x - 10 x + 12 \ j ------------------ = ) a[70][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[70][j], a[70][j + 1], a[70][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[71][j], for j from 0 to infinity as the solution of the recurrence a[71][j] = a[71][j - 1] + 3 a[71][j - 2] + a[71][j - 3] Subject to the initial conditions a[71][0] = 12, a[71][1] = 3, a[71][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -21 x - 9 x + 12 \ j ------------------ = ) a[71][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[71][j], a[71][j + 1], a[71][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[72][j], for j from 0 to infinity as the solution of the recurrence a[72][j] = a[72][j - 1] + 3 a[72][j - 2] + a[72][j - 3] Subject to the initial conditions a[72][0] = 12, a[72][1] = 4, a[72][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -20 x - 8 x + 12 \ j ------------------ = ) a[72][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[72][j], a[72][j + 1], a[72][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[73][j], for j from 0 to infinity as the solution of the recurrence a[73][j] = a[73][j - 1] + 3 a[73][j - 2] + a[73][j - 3] Subject to the initial conditions a[73][0] = 13, a[73][1] = 0, a[73][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -26 x - 13 x + 13 \ j ------------------ = ) a[73][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[73][j], a[73][j + 1], a[73][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[74][j], for j from 0 to infinity as the solution of the recurrence a[74][j] = a[74][j - 1] + 3 a[74][j - 2] + a[74][j - 3] Subject to the initial conditions a[74][0] = 13, a[74][1] = 1, a[74][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -25 x - 12 x + 13 \ j ------------------ = ) a[74][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[74][j], a[74][j + 1], a[74][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[75][j], for j from 0 to infinity as the solution of the recurrence a[75][j] = a[75][j - 1] + 3 a[75][j - 2] + a[75][j - 3] Subject to the initial conditions a[75][0] = 13, a[75][1] = 2, a[75][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -24 x - 11 x + 13 \ j ------------------ = ) a[75][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[75][j], a[75][j + 1], a[75][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[76][j], for j from 0 to infinity as the solution of the recurrence a[76][j] = a[76][j - 1] + 3 a[76][j - 2] + a[76][j - 3] Subject to the initial conditions a[76][0] = 13, a[76][1] = 3, a[76][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -23 x - 10 x + 13 \ j ------------------ = ) a[76][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[76][j], a[76][j + 1], a[76][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[77][j], for j from 0 to infinity as the solution of the recurrence a[77][j] = a[77][j - 1] + 3 a[77][j - 2] + a[77][j - 3] Subject to the initial conditions a[77][0] = 14, a[77][1] = 0, a[77][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -28 x - 14 x + 14 \ j ------------------ = ) a[77][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[77][j], a[77][j + 1], a[77][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[78][j], for j from 0 to infinity as the solution of the recurrence a[78][j] = a[78][j - 1] + 3 a[78][j - 2] + a[78][j - 3] Subject to the initial conditions a[78][0] = 14, a[78][1] = 1, a[78][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -27 x - 13 x + 14 \ j ------------------ = ) a[78][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[78][j], a[78][j + 1], a[78][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[79][j], for j from 0 to infinity as the solution of the recurrence a[79][j] = a[79][j - 1] + 3 a[79][j - 2] + a[79][j - 3] Subject to the initial conditions a[79][0] = 14, a[79][1] = 2, a[79][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -26 x - 12 x + 14 \ j ------------------ = ) a[79][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[79][j], a[79][j + 1], a[79][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[80][j], for j from 0 to infinity as the solution of the recurrence a[80][j] = a[80][j - 1] + 3 a[80][j - 2] + a[80][j - 3] Subject to the initial conditions a[80][0] = 14, a[80][1] = 3, a[80][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -25 x - 11 x + 14 \ j ------------------ = ) a[80][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[80][j], a[80][j + 1], a[80][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[81][j], for j from 0 to infinity as the solution of the recurrence a[81][j] = a[81][j - 1] + 3 a[81][j - 2] + a[81][j - 3] Subject to the initial conditions a[81][0] = 15, a[81][1] = 0, a[81][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -30 x - 15 x + 15 \ j ------------------ = ) a[81][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[81][j], a[81][j + 1], a[81][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[82][j], for j from 0 to infinity as the solution of the recurrence a[82][j] = a[82][j - 1] + 3 a[82][j - 2] + a[82][j - 3] Subject to the initial conditions a[82][0] = 15, a[82][1] = 1, a[82][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -29 x - 14 x + 15 \ j ------------------ = ) a[82][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[82][j], a[82][j + 1], a[82][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[83][j], for j from 0 to infinity as the solution of the recurrence a[83][j] = a[83][j - 1] + 3 a[83][j - 2] + a[83][j - 3] Subject to the initial conditions a[83][0] = 15, a[83][1] = 2, a[83][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -28 x - 13 x + 15 \ j ------------------ = ) a[83][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[83][j], a[83][j + 1], a[83][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[84][j], for j from 0 to infinity as the solution of the recurrence a[84][j] = a[84][j - 1] + 3 a[84][j - 2] + a[84][j - 3] Subject to the initial conditions a[84][0] = 16, a[84][1] = 0, a[84][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -32 x - 16 x + 16 \ j ------------------ = ) a[84][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[84][j], a[84][j + 1], a[84][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[85][j], for j from 0 to infinity as the solution of the recurrence a[85][j] = a[85][j - 1] + 3 a[85][j - 2] + a[85][j - 3] Subject to the initial conditions a[85][0] = 16, a[85][1] = 1, a[85][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -31 x - 15 x + 16 \ j ------------------ = ) a[85][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[85][j], a[85][j + 1], a[85][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[86][j], for j from 0 to infinity as the solution of the recurrence a[86][j] = a[86][j - 1] + 3 a[86][j - 2] + a[86][j - 3] Subject to the initial conditions a[86][0] = 16, a[86][1] = 2, a[86][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -30 x - 14 x + 16 \ j ------------------ = ) a[86][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[86][j], a[86][j + 1], a[86][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[87][j], for j from 0 to infinity as the solution of the recurrence a[87][j] = a[87][j - 1] + 3 a[87][j - 2] + a[87][j - 3] Subject to the initial conditions a[87][0] = 17, a[87][1] = 0, a[87][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -34 x - 17 x + 17 \ j ------------------ = ) a[87][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[87][j], a[87][j + 1], a[87][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[88][j], for j from 0 to infinity as the solution of the recurrence a[88][j] = a[88][j - 1] + 3 a[88][j - 2] + a[88][j - 3] Subject to the initial conditions a[88][0] = 17, a[88][1] = 1, a[88][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -33 x - 16 x + 17 \ j ------------------ = ) a[88][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[88][j], a[88][j + 1], a[88][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[89][j], for j from 0 to infinity as the solution of the recurrence a[89][j] = a[89][j - 1] + 3 a[89][j - 2] + a[89][j - 3] Subject to the initial conditions a[89][0] = 18, a[89][1] = 0, a[89][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -36 x - 18 x + 18 \ j ------------------ = ) a[89][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[89][j], a[89][j + 1], a[89][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[90][j], for j from 0 to infinity as the solution of the recurrence a[90][j] = a[90][j - 1] + 3 a[90][j - 2] + a[90][j - 3] Subject to the initial conditions a[90][0] = 18, a[90][1] = 1, a[90][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -35 x - 17 x + 18 \ j ------------------ = ) a[90][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[90][j], a[90][j + 1], a[90][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[91][j], for j from 0 to infinity as the solution of the recurrence a[91][j] = a[91][j - 1] + 3 a[91][j - 2] + a[91][j - 3] Subject to the initial conditions a[91][0] = 19, a[91][1] = 0, a[91][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -38 x - 19 x + 19 \ j ------------------ = ) a[91][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[91][j], a[91][j + 1], a[91][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[92][j], for j from 0 to infinity as the solution of the recurrence a[92][j] = a[92][j - 1] + 3 a[92][j - 2] + a[92][j - 3] Subject to the initial conditions a[92][0] = 20, a[92][1] = 0, a[92][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -40 x - 20 x + 20 \ j ------------------ = ) a[92][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[92][j], a[92][j + 1], a[92][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 are the following triples {[1, 0, 1], [1, 1, 3], [2, 0, 2], [2, 1, 4], [2, 2, 6], [2, 3, 8], [3, 0, 3], [3, 1, 5], [3, 2, 7], [3, 3, 9], [3, 4, 11], [3, 5, 13], [4, 0, 4], [4, 1, 6], [4, 2, 8], [4, 3, 10], [4, 4, 12], [4, 5, 14], [4, 6, 16], [4, 7, 18], [5, 0, 5], [5, 1, 7], [5, 2, 9], [5, 3, 11], [5, 4, 13], [5, 5, 15], [5, 6, 17], [5, 7, 19], [6, 0, 6], [6, 1, 8], [6, 2, 10], [6, 3, 12], [6, 4, 14], [6, 5, 16], [6, 6, 18], [6, 7, 20], [7, 0, 7], [7, 1, 9], [7, 2, 11], [7, 3, 13], [7, 4, 15], [7, 5, 17], [7, 6, 19], [8, 0, 8], [8, 1, 10], [8, 2, 12], [8, 3, 14], [8, 4, 16], [8, 5, 18], [8, 6, 20], [9, 0, 9], [9, 1, 11], [9, 2, 13], [9, 3, 15], [9, 4, 17], [9, 5, 19], [10, 0, 10], [10, 1, 12], [10, 2, 14], [10, 3, 16], [10, 4, 18], [10, 5, 20], [11, 0, 11], [11, 1, 13], [11, 2, 15], [11, 3, 17], [11, 4, 19], [12, 0, 12], [12, 1, 14], [12, 2, 16], [12, 3, 18], [12, 4, 20], [13, 0, 13], [13, 1, 15], [13, 2, 17], [13, 3, 19], [14, 0, 14], [14, 1, 16], [14, 2, 18], [14, 3, 20], [15, 0, 15], [15, 1, 17], [15, 2, 19], [16, 0, 16], [16, 1, 18], [16, 2, 20], [17, 0, 17], [17, 1, 19], [18, 0, 18], [18, 1, 20], [19, 0, 19], [20, 0, 20]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.036, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -3 x - x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 3 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 2 \ j ------------------ = ) a[1][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 are the following triples {[2, 0, 5]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.035, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 4 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j ------------------ = ) a[1][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 are the following triples {[0, 1, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 7 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -2 x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 7 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 3 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 2 x + 2 \ j ------------------ = ) a[2][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 are the following triples {[1, 1, 2], [2, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 8 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 2 x + 2 \ j ------------------ = ) a[1][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 3 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 4, a[2][1] = 3, a[2][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 4 \ j ------------------ = ) a[2][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 are the following triples {[2, 0, 0], [4, 3, 14]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.028, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 9 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 2, a[1][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x + x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - 3 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 are the following triples {[1, 2, 4]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + x[1] x[3] 2 2 3 2 2 + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] - 2 x[2] x[3] - 2 x[2] x[3] 3 + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 6 x[1] x[2] + x[1] x[3] + 10 x[1] x[2] - 3 x[1] x[3] + 4 x[2] 2 2 3 - 2 x[2] x[3] - 2 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following, 3, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -2 x + 1 \ j ------------------ = ) a[2][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 5, a[3][1] = 1, a[3][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -15 x - 4 x + 5 \ j ------------------ = ) a[3][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 are the following triples {[1, 0, 0], [1, 1, 3], [5, 1, 6]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.038, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following, 7, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j ------------------ = ) a[1][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 2, a[2][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -x + x + 1 \ j ------------------ = ) a[2][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 0, a[3][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -7 x - 2 x + 2 \ j ------------------ = ) a[3][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = a[4][j - 1] + 4 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 3, a[4][1] = 0, a[4][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 3 x + 3 \ j ------------------ = ) a[4][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[5][j], for j from 0 to infinity as the solution of the recurrence a[5][j] = a[5][j - 1] + 4 a[5][j - 2] + a[5][j - 3] Subject to the initial conditions a[5][0] = 3, a[5][1] = 0, a[5][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -x - 3 x + 3 \ j ------------------ = ) a[5][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[5][j], a[5][j + 1], a[5][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[6][j], for j from 0 to infinity as the solution of the recurrence a[6][j] = a[6][j - 1] + 4 a[6][j - 2] + a[6][j - 3] Subject to the initial conditions a[6][0] = 3, a[6][1] = 5, a[6][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -3 x + 2 x + 3 \ j ------------------ = ) a[6][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[6][j], a[6][j + 1], a[6][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[7][j], for j from 0 to infinity as the solution of the recurrence a[7][j] = a[7][j - 1] + 4 a[7][j - 2] + a[7][j - 3] Subject to the initial conditions a[7][0] = 4, a[7][1] = 3, a[7][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -9 x - x + 4 \ j ------------------ = ) a[7][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[7][j], a[7][j + 1], a[7][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 are the following triples {[0, 1, 0], [1, 2, 5], [2, 0, 1], [3, 0, 2], [3, 0, 11], [3, 5, 14], [4, 3, 10] } Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 3, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 2 x + 2 \ j ------------------ = ) a[1][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 2, a[2][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -4 x + 2 \ j ------------------ = ) a[2][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 10, a[3][1] = 2, a[3][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -30 x - 8 x + 10 \ j ------------------ = ) a[3][j] x 3 2 / -x - 4 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 are the following triples {[2, 0, 0], [2, 2, 6], [10, 2, 12]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.038, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + x[1] x[3] 2 2 3 2 + 17 x[1] x[2] + x[1] x[2] x[3] - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + x[1] x[3] + 17 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 5 x[2] - 3 x[2] x[3] - 2 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -5 x - x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 2 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 5, a[1][1] = 0, a[1][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -22 x - 5 x + 5 \ j ------------------ = ) a[1][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 are the following triples {[5, 0, 3]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -9 x - 2 x + 2 \ j ------------------ = ) a[1][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 are the following triples {[2, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -3 x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 are the following triples {[1, 1, 3]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 6 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j ------------------ = ) a[1][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 6 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 10, a[2][1] = 3, a[2][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -39 x - 7 x + 10 \ j ------------------ = ) a[2][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 are the following triples {[0, 1, 0], [10, 3, 14]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 4, a[1][1] = 0, a[1][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -x - 4 x + 4 \ j ------------------ = ) a[1][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 are the following triples {[4, 0, 19]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.025, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 2 x + 2 \ j ------------------ = ) a[1][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 4, a[1][1] = 1, a[1][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -16 x - 3 x + 4 \ j ------------------ = ) a[1][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 4, a[2][1] = 4, a[2][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -11 x + 4 \ j ------------------ = ) a[2][j] x 3 2 / -x - 5 x - x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 are the following triples {[4, 1, 5], [4, 4, 13]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + x[1] x[3] 2 2 3 2 + 26 x[1] x[2] + 2 x[1] x[2] x[3] - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] 2 3 - 2 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + x[1] x[3] + 26 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 6 x[2] - 4 x[2] x[3] - 2 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x + x \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 0, a[2][1] = 1, a[2][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j ------------------ = ) a[2][j] x 3 2 / -x - x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 are the following triples {[0, 1, 0], [0, 1, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.038, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.022, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 are the following triples {[1, 1, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 4 x + 2 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.041, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following, 3, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -3 x - x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -2 x - x + 1 \ j ------------------ = ) a[2][j] x 3 2 / -x - x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 2 a[3][j - 1] + a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 1, a[3][1] = 2, a[3][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j ------------------ = ) a[3][j] x 3 2 / -x - x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 are the following triples {[1, 1, 0], [1, 1, 1], [1, 2, 4]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 3 x[1] x[2] - x[1] x[2] x[3] - x[1] x[3] + 3 x[2] + 3 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 2 x[1] x[3] + 3 x[1] x[2] - x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 3 x[2] + 3 x[2] x[3] - 4 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 2 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 2 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 2 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 are the following triples {[1, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 2 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 2 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 are the following triples {[0, 1, 1], [1, 1, 3]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 are the following triples {[0, 1, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 2, a[1][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 2 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 4 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 2 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 are the following triples {[1, 2, 5], [2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 10 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -2 x - x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + x[1] x[2] x[3] - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 2 x[1] x[3] + 6 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 5 x[2] + 2 x[2] x[3] - 4 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 2 x[1] < 0 are the following triples {[1, 1, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following, 4, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 2 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 3 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 0, a[2][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 2 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 2 a[3][j - 1] + 3 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 1, a[3][1] = 0, a[3][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 1 \ j -------------------- = ) a[3][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 2 a[4][j - 1] + 3 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 1, a[4][1] = 1, a[4][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 1 \ j -------------------- = ) a[4][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 0], [1, 0, 1], [1, 0, 2], [1, 1, 4]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following, 4, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 3 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 2, a[2][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 2 a[3][j - 1] + 3 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 0, a[3][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 4 x + 2 \ j -------------------- = ) a[3][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 2 a[4][j - 1] + 3 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 2, a[4][1] = 3, a[4][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 2 \ j -------------------- = ) a[4][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 are the following triples {[0, 1, 1], [1, 2, 6], [2, 0, 3], [2, 3, 11]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad The following, 4, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 3 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -2 x - x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 2 a[3][j - 1] + 3 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 1, a[3][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 3 x + 2 \ j -------------------- = ) a[3][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 2 a[4][j - 1] + 3 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 2, a[4][1] = 4, a[4][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -x + 2 \ j -------------------- = ) a[4][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 are the following triples {[0, 1, 0], [1, 1, 3], [2, 1, 6], [2, 4, 13]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 4, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 4 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 3 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 4 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 2 a[3][j - 1] + 3 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 0, a[3][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 4 x + 2 \ j -------------------- = ) a[3][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 2 a[4][j - 1] + 3 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 2, a[4][1] = 2, a[4][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 2 x + 2 \ j -------------------- = ) a[4][j] x 3 2 / -x - 3 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 are the following triples {[2, 0, 0], [2, 0, 2], [2, 0, 4], [2, 2, 8]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.040, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 11 x[1] x[2] + 3 x[1] x[2] x[3] - 3 x[1] x[3] + 7 x[2] + x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 2 x[1] x[3] + 11 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 7 x[2] + x[2] x[3] - 4 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 By Shalosh B. Ekhad The following, 72, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 2 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -2 x - x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 2 a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 1, a[3][1] = 2, a[3][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j -------------------- = ) a[3][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 2 a[4][j - 1] + 4 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 2, a[4][1] = 0, a[4][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 4 x + 2 \ j -------------------- = ) a[4][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[5][j], for j from 0 to infinity as the solution of the recurrence a[5][j] = 2 a[5][j - 1] + 4 a[5][j - 2] + a[5][j - 3] Subject to the initial conditions a[5][0] = 2, a[5][1] = 1, a[5][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 3 x + 2 \ j -------------------- = ) a[5][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[5][j], a[5][j + 1], a[5][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[6][j], for j from 0 to infinity as the solution of the recurrence a[6][j] = 2 a[6][j - 1] + 4 a[6][j - 2] + a[6][j - 3] Subject to the initial conditions a[6][0] = 2, a[6][1] = 2, a[6][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 2 x + 2 \ j -------------------- = ) a[6][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[6][j], a[6][j + 1], a[6][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[7][j], for j from 0 to infinity as the solution of the recurrence a[7][j] = 2 a[7][j - 1] + 4 a[7][j - 2] + a[7][j - 3] Subject to the initial conditions a[7][0] = 2, a[7][1] = 3, a[7][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -3 x - x + 2 \ j -------------------- = ) a[7][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[7][j], a[7][j + 1], a[7][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[8][j], for j from 0 to infinity as the solution of the recurrence a[8][j] = 2 a[8][j - 1] + 4 a[8][j - 2] + a[8][j - 3] Subject to the initial conditions a[8][0] = 2, a[8][1] = 4, a[8][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -2 x + 2 \ j -------------------- = ) a[8][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[8][j], a[8][j + 1], a[8][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[9][j], for j from 0 to infinity as the solution of the recurrence a[9][j] = 2 a[9][j - 1] + 4 a[9][j - 2] + a[9][j - 3] Subject to the initial conditions a[9][0] = 2, a[9][1] = 5, a[9][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -x + x + 2 \ j -------------------- = ) a[9][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[9][j], a[9][j + 1], a[9][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[10][j], for j from 0 to infinity as the solution of the recurrence a[10][j] = 2 a[10][j - 1] + 4 a[10][j - 2] + a[10][j - 3] Subject to the initial conditions a[10][0] = 3, a[10][1] = 0, a[10][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -9 x - 6 x + 3 \ j -------------------- = ) a[10][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[10][j], a[10][j + 1], a[10][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[11][j], for j from 0 to infinity as the solution of the recurrence a[11][j] = 2 a[11][j - 1] + 4 a[11][j - 2] + a[11][j - 3] Subject to the initial conditions a[11][0] = 3, a[11][1] = 1, a[11][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 5 x + 3 \ j -------------------- = ) a[11][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[11][j], a[11][j + 1], a[11][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[12][j], for j from 0 to infinity as the solution of the recurrence a[12][j] = 2 a[12][j - 1] + 4 a[12][j - 2] + a[12][j - 3] Subject to the initial conditions a[12][0] = 3, a[12][1] = 2, a[12][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -7 x - 4 x + 3 \ j -------------------- = ) a[12][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[12][j], a[12][j + 1], a[12][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[13][j], for j from 0 to infinity as the solution of the recurrence a[13][j] = 2 a[13][j - 1] + 4 a[13][j - 2] + a[13][j - 3] Subject to the initial conditions a[13][0] = 3, a[13][1] = 3, a[13][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 3 x + 3 \ j -------------------- = ) a[13][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[13][j], a[13][j + 1], a[13][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[14][j], for j from 0 to infinity as the solution of the recurrence a[14][j] = 2 a[14][j - 1] + 4 a[14][j - 2] + a[14][j - 3] Subject to the initial conditions a[14][0] = 3, a[14][1] = 4, a[14][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 2 x + 3 \ j -------------------- = ) a[14][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[14][j], a[14][j + 1], a[14][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[15][j], for j from 0 to infinity as the solution of the recurrence a[15][j] = 2 a[15][j - 1] + 4 a[15][j - 2] + a[15][j - 3] Subject to the initial conditions a[15][0] = 3, a[15][1] = 5, a[15][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -4 x - x + 3 \ j -------------------- = ) a[15][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[15][j], a[15][j + 1], a[15][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[16][j], for j from 0 to infinity as the solution of the recurrence a[16][j] = 2 a[16][j - 1] + 4 a[16][j - 2] + a[16][j - 3] Subject to the initial conditions a[16][0] = 4, a[16][1] = 0, a[16][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -12 x - 8 x + 4 \ j -------------------- = ) a[16][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[16][j], a[16][j + 1], a[16][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[17][j], for j from 0 to infinity as the solution of the recurrence a[17][j] = 2 a[17][j - 1] + 4 a[17][j - 2] + a[17][j - 3] Subject to the initial conditions a[17][0] = 4, a[17][1] = 1, a[17][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -11 x - 7 x + 4 \ j -------------------- = ) a[17][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[17][j], a[17][j + 1], a[17][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[18][j], for j from 0 to infinity as the solution of the recurrence a[18][j] = 2 a[18][j - 1] + 4 a[18][j - 2] + a[18][j - 3] Subject to the initial conditions a[18][0] = 4, a[18][1] = 2, a[18][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 6 x + 4 \ j -------------------- = ) a[18][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[18][j], a[18][j + 1], a[18][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[19][j], for j from 0 to infinity as the solution of the recurrence a[19][j] = 2 a[19][j - 1] + 4 a[19][j - 2] + a[19][j - 3] Subject to the initial conditions a[19][0] = 4, a[19][1] = 3, a[19][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -9 x - 5 x + 4 \ j -------------------- = ) a[19][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[19][j], a[19][j + 1], a[19][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[20][j], for j from 0 to infinity as the solution of the recurrence a[20][j] = 2 a[20][j - 1] + 4 a[20][j - 2] + a[20][j - 3] Subject to the initial conditions a[20][0] = 4, a[20][1] = 4, a[20][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 4 x + 4 \ j -------------------- = ) a[20][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[20][j], a[20][j + 1], a[20][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[21][j], for j from 0 to infinity as the solution of the recurrence a[21][j] = 2 a[21][j - 1] + 4 a[21][j - 2] + a[21][j - 3] Subject to the initial conditions a[21][0] = 4, a[21][1] = 5, a[21][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -7 x - 3 x + 4 \ j -------------------- = ) a[21][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[21][j], a[21][j + 1], a[21][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[22][j], for j from 0 to infinity as the solution of the recurrence a[22][j] = 2 a[22][j - 1] + 4 a[22][j - 2] + a[22][j - 3] Subject to the initial conditions a[22][0] = 5, a[22][1] = 0, a[22][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -15 x - 10 x + 5 \ j -------------------- = ) a[22][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[22][j], a[22][j + 1], a[22][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[23][j], for j from 0 to infinity as the solution of the recurrence a[23][j] = 2 a[23][j - 1] + 4 a[23][j - 2] + a[23][j - 3] Subject to the initial conditions a[23][0] = 5, a[23][1] = 1, a[23][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -14 x - 9 x + 5 \ j -------------------- = ) a[23][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[23][j], a[23][j + 1], a[23][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[24][j], for j from 0 to infinity as the solution of the recurrence a[24][j] = 2 a[24][j - 1] + 4 a[24][j - 2] + a[24][j - 3] Subject to the initial conditions a[24][0] = 5, a[24][1] = 2, a[24][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -13 x - 8 x + 5 \ j -------------------- = ) a[24][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[24][j], a[24][j + 1], a[24][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[25][j], for j from 0 to infinity as the solution of the recurrence a[25][j] = 2 a[25][j - 1] + 4 a[25][j - 2] + a[25][j - 3] Subject to the initial conditions a[25][0] = 5, a[25][1] = 3, a[25][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -12 x - 7 x + 5 \ j -------------------- = ) a[25][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[25][j], a[25][j + 1], a[25][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[26][j], for j from 0 to infinity as the solution of the recurrence a[26][j] = 2 a[26][j - 1] + 4 a[26][j - 2] + a[26][j - 3] Subject to the initial conditions a[26][0] = 5, a[26][1] = 4, a[26][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -11 x - 6 x + 5 \ j -------------------- = ) a[26][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[26][j], a[26][j + 1], a[26][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[27][j], for j from 0 to infinity as the solution of the recurrence a[27][j] = 2 a[27][j - 1] + 4 a[27][j - 2] + a[27][j - 3] Subject to the initial conditions a[27][0] = 5, a[27][1] = 5, a[27][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 5 x + 5 \ j -------------------- = ) a[27][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[27][j], a[27][j + 1], a[27][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[28][j], for j from 0 to infinity as the solution of the recurrence a[28][j] = 2 a[28][j - 1] + 4 a[28][j - 2] + a[28][j - 3] Subject to the initial conditions a[28][0] = 6, a[28][1] = 0, a[28][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -18 x - 12 x + 6 \ j -------------------- = ) a[28][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[28][j], a[28][j + 1], a[28][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[29][j], for j from 0 to infinity as the solution of the recurrence a[29][j] = 2 a[29][j - 1] + 4 a[29][j - 2] + a[29][j - 3] Subject to the initial conditions a[29][0] = 6, a[29][1] = 1, a[29][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -17 x - 11 x + 6 \ j -------------------- = ) a[29][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[29][j], a[29][j + 1], a[29][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[30][j], for j from 0 to infinity as the solution of the recurrence a[30][j] = 2 a[30][j - 1] + 4 a[30][j - 2] + a[30][j - 3] Subject to the initial conditions a[30][0] = 6, a[30][1] = 2, a[30][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -16 x - 10 x + 6 \ j -------------------- = ) a[30][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[30][j], a[30][j + 1], a[30][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[31][j], for j from 0 to infinity as the solution of the recurrence a[31][j] = 2 a[31][j - 1] + 4 a[31][j - 2] + a[31][j - 3] Subject to the initial conditions a[31][0] = 6, a[31][1] = 3, a[31][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -15 x - 9 x + 6 \ j -------------------- = ) a[31][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[31][j], a[31][j + 1], a[31][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[32][j], for j from 0 to infinity as the solution of the recurrence a[32][j] = 2 a[32][j - 1] + 4 a[32][j - 2] + a[32][j - 3] Subject to the initial conditions a[32][0] = 6, a[32][1] = 4, a[32][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -14 x - 8 x + 6 \ j -------------------- = ) a[32][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[32][j], a[32][j + 1], a[32][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[33][j], for j from 0 to infinity as the solution of the recurrence a[33][j] = 2 a[33][j - 1] + 4 a[33][j - 2] + a[33][j - 3] Subject to the initial conditions a[33][0] = 7, a[33][1] = 0, a[33][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -21 x - 14 x + 7 \ j -------------------- = ) a[33][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[33][j], a[33][j + 1], a[33][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[34][j], for j from 0 to infinity as the solution of the recurrence a[34][j] = 2 a[34][j - 1] + 4 a[34][j - 2] + a[34][j - 3] Subject to the initial conditions a[34][0] = 7, a[34][1] = 1, a[34][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -20 x - 13 x + 7 \ j -------------------- = ) a[34][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[34][j], a[34][j + 1], a[34][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[35][j], for j from 0 to infinity as the solution of the recurrence a[35][j] = 2 a[35][j - 1] + 4 a[35][j - 2] + a[35][j - 3] Subject to the initial conditions a[35][0] = 7, a[35][1] = 2, a[35][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -19 x - 12 x + 7 \ j -------------------- = ) a[35][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[35][j], a[35][j + 1], a[35][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[36][j], for j from 0 to infinity as the solution of the recurrence a[36][j] = 2 a[36][j - 1] + 4 a[36][j - 2] + a[36][j - 3] Subject to the initial conditions a[36][0] = 7, a[36][1] = 3, a[36][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -18 x - 11 x + 7 \ j -------------------- = ) a[36][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[36][j], a[36][j + 1], a[36][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[37][j], for j from 0 to infinity as the solution of the recurrence a[37][j] = 2 a[37][j - 1] + 4 a[37][j - 2] + a[37][j - 3] Subject to the initial conditions a[37][0] = 7, a[37][1] = 4, a[37][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -17 x - 10 x + 7 \ j -------------------- = ) a[37][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[37][j], a[37][j + 1], a[37][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[38][j], for j from 0 to infinity as the solution of the recurrence a[38][j] = 2 a[38][j - 1] + 4 a[38][j - 2] + a[38][j - 3] Subject to the initial conditions a[38][0] = 8, a[38][1] = 0, a[38][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -24 x - 16 x + 8 \ j -------------------- = ) a[38][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[38][j], a[38][j + 1], a[38][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[39][j], for j from 0 to infinity as the solution of the recurrence a[39][j] = 2 a[39][j - 1] + 4 a[39][j - 2] + a[39][j - 3] Subject to the initial conditions a[39][0] = 8, a[39][1] = 1, a[39][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -23 x - 15 x + 8 \ j -------------------- = ) a[39][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[39][j], a[39][j + 1], a[39][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[40][j], for j from 0 to infinity as the solution of the recurrence a[40][j] = 2 a[40][j - 1] + 4 a[40][j - 2] + a[40][j - 3] Subject to the initial conditions a[40][0] = 8, a[40][1] = 2, a[40][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -22 x - 14 x + 8 \ j -------------------- = ) a[40][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[40][j], a[40][j + 1], a[40][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[41][j], for j from 0 to infinity as the solution of the recurrence a[41][j] = 2 a[41][j - 1] + 4 a[41][j - 2] + a[41][j - 3] Subject to the initial conditions a[41][0] = 8, a[41][1] = 3, a[41][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -21 x - 13 x + 8 \ j -------------------- = ) a[41][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[41][j], a[41][j + 1], a[41][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[42][j], for j from 0 to infinity as the solution of the recurrence a[42][j] = 2 a[42][j - 1] + 4 a[42][j - 2] + a[42][j - 3] Subject to the initial conditions a[42][0] = 8, a[42][1] = 4, a[42][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -20 x - 12 x + 8 \ j -------------------- = ) a[42][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[42][j], a[42][j + 1], a[42][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[43][j], for j from 0 to infinity as the solution of the recurrence a[43][j] = 2 a[43][j - 1] + 4 a[43][j - 2] + a[43][j - 3] Subject to the initial conditions a[43][0] = 9, a[43][1] = 0, a[43][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -27 x - 18 x + 9 \ j -------------------- = ) a[43][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[43][j], a[43][j + 1], a[43][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[44][j], for j from 0 to infinity as the solution of the recurrence a[44][j] = 2 a[44][j - 1] + 4 a[44][j - 2] + a[44][j - 3] Subject to the initial conditions a[44][0] = 9, a[44][1] = 1, a[44][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -26 x - 17 x + 9 \ j -------------------- = ) a[44][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[44][j], a[44][j + 1], a[44][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[45][j], for j from 0 to infinity as the solution of the recurrence a[45][j] = 2 a[45][j - 1] + 4 a[45][j - 2] + a[45][j - 3] Subject to the initial conditions a[45][0] = 9, a[45][1] = 2, a[45][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -25 x - 16 x + 9 \ j -------------------- = ) a[45][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[45][j], a[45][j + 1], a[45][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[46][j], for j from 0 to infinity as the solution of the recurrence a[46][j] = 2 a[46][j - 1] + 4 a[46][j - 2] + a[46][j - 3] Subject to the initial conditions a[46][0] = 9, a[46][1] = 3, a[46][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -24 x - 15 x + 9 \ j -------------------- = ) a[46][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[46][j], a[46][j + 1], a[46][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[47][j], for j from 0 to infinity as the solution of the recurrence a[47][j] = 2 a[47][j - 1] + 4 a[47][j - 2] + a[47][j - 3] Subject to the initial conditions a[47][0] = 10, a[47][1] = 0, a[47][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -30 x - 20 x + 10 \ j -------------------- = ) a[47][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[47][j], a[47][j + 1], a[47][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[48][j], for j from 0 to infinity as the solution of the recurrence a[48][j] = 2 a[48][j - 1] + 4 a[48][j - 2] + a[48][j - 3] Subject to the initial conditions a[48][0] = 10, a[48][1] = 1, a[48][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -29 x - 19 x + 10 \ j -------------------- = ) a[48][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[48][j], a[48][j + 1], a[48][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[49][j], for j from 0 to infinity as the solution of the recurrence a[49][j] = 2 a[49][j - 1] + 4 a[49][j - 2] + a[49][j - 3] Subject to the initial conditions a[49][0] = 10, a[49][1] = 2, a[49][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -28 x - 18 x + 10 \ j -------------------- = ) a[49][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[49][j], a[49][j + 1], a[49][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[50][j], for j from 0 to infinity as the solution of the recurrence a[50][j] = 2 a[50][j - 1] + 4 a[50][j - 2] + a[50][j - 3] Subject to the initial conditions a[50][0] = 10, a[50][1] = 3, a[50][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -27 x - 17 x + 10 \ j -------------------- = ) a[50][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[50][j], a[50][j + 1], a[50][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[51][j], for j from 0 to infinity as the solution of the recurrence a[51][j] = 2 a[51][j - 1] + 4 a[51][j - 2] + a[51][j - 3] Subject to the initial conditions a[51][0] = 11, a[51][1] = 0, a[51][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -33 x - 22 x + 11 \ j -------------------- = ) a[51][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[51][j], a[51][j + 1], a[51][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[52][j], for j from 0 to infinity as the solution of the recurrence a[52][j] = 2 a[52][j - 1] + 4 a[52][j - 2] + a[52][j - 3] Subject to the initial conditions a[52][0] = 11, a[52][1] = 1, a[52][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -32 x - 21 x + 11 \ j -------------------- = ) a[52][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[52][j], a[52][j + 1], a[52][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[53][j], for j from 0 to infinity as the solution of the recurrence a[53][j] = 2 a[53][j - 1] + 4 a[53][j - 2] + a[53][j - 3] Subject to the initial conditions a[53][0] = 11, a[53][1] = 2, a[53][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -31 x - 20 x + 11 \ j -------------------- = ) a[53][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[53][j], a[53][j + 1], a[53][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[54][j], for j from 0 to infinity as the solution of the recurrence a[54][j] = 2 a[54][j - 1] + 4 a[54][j - 2] + a[54][j - 3] Subject to the initial conditions a[54][0] = 11, a[54][1] = 3, a[54][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -30 x - 19 x + 11 \ j -------------------- = ) a[54][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[54][j], a[54][j + 1], a[54][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[55][j], for j from 0 to infinity as the solution of the recurrence a[55][j] = 2 a[55][j - 1] + 4 a[55][j - 2] + a[55][j - 3] Subject to the initial conditions a[55][0] = 12, a[55][1] = 0, a[55][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -36 x - 24 x + 12 \ j -------------------- = ) a[55][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[55][j], a[55][j + 1], a[55][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[56][j], for j from 0 to infinity as the solution of the recurrence a[56][j] = 2 a[56][j - 1] + 4 a[56][j - 2] + a[56][j - 3] Subject to the initial conditions a[56][0] = 12, a[56][1] = 1, a[56][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -35 x - 23 x + 12 \ j -------------------- = ) a[56][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[56][j], a[56][j + 1], a[56][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[57][j], for j from 0 to infinity as the solution of the recurrence a[57][j] = 2 a[57][j - 1] + 4 a[57][j - 2] + a[57][j - 3] Subject to the initial conditions a[57][0] = 12, a[57][1] = 2, a[57][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -34 x - 22 x + 12 \ j -------------------- = ) a[57][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[57][j], a[57][j + 1], a[57][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[58][j], for j from 0 to infinity as the solution of the recurrence a[58][j] = 2 a[58][j - 1] + 4 a[58][j - 2] + a[58][j - 3] Subject to the initial conditions a[58][0] = 13, a[58][1] = 0, a[58][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -39 x - 26 x + 13 \ j -------------------- = ) a[58][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[58][j], a[58][j + 1], a[58][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[59][j], for j from 0 to infinity as the solution of the recurrence a[59][j] = 2 a[59][j - 1] + 4 a[59][j - 2] + a[59][j - 3] Subject to the initial conditions a[59][0] = 13, a[59][1] = 1, a[59][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -38 x - 25 x + 13 \ j -------------------- = ) a[59][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[59][j], a[59][j + 1], a[59][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[60][j], for j from 0 to infinity as the solution of the recurrence a[60][j] = 2 a[60][j - 1] + 4 a[60][j - 2] + a[60][j - 3] Subject to the initial conditions a[60][0] = 13, a[60][1] = 2, a[60][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -37 x - 24 x + 13 \ j -------------------- = ) a[60][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[60][j], a[60][j + 1], a[60][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[61][j], for j from 0 to infinity as the solution of the recurrence a[61][j] = 2 a[61][j - 1] + 4 a[61][j - 2] + a[61][j - 3] Subject to the initial conditions a[61][0] = 14, a[61][1] = 0, a[61][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -42 x - 28 x + 14 \ j -------------------- = ) a[61][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[61][j], a[61][j + 1], a[61][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[62][j], for j from 0 to infinity as the solution of the recurrence a[62][j] = 2 a[62][j - 1] + 4 a[62][j - 2] + a[62][j - 3] Subject to the initial conditions a[62][0] = 14, a[62][1] = 1, a[62][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -41 x - 27 x + 14 \ j -------------------- = ) a[62][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[62][j], a[62][j + 1], a[62][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[63][j], for j from 0 to infinity as the solution of the recurrence a[63][j] = 2 a[63][j - 1] + 4 a[63][j - 2] + a[63][j - 3] Subject to the initial conditions a[63][0] = 14, a[63][1] = 2, a[63][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -40 x - 26 x + 14 \ j -------------------- = ) a[63][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[63][j], a[63][j + 1], a[63][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[64][j], for j from 0 to infinity as the solution of the recurrence a[64][j] = 2 a[64][j - 1] + 4 a[64][j - 2] + a[64][j - 3] Subject to the initial conditions a[64][0] = 15, a[64][1] = 0, a[64][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -45 x - 30 x + 15 \ j -------------------- = ) a[64][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[64][j], a[64][j + 1], a[64][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[65][j], for j from 0 to infinity as the solution of the recurrence a[65][j] = 2 a[65][j - 1] + 4 a[65][j - 2] + a[65][j - 3] Subject to the initial conditions a[65][0] = 15, a[65][1] = 1, a[65][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -44 x - 29 x + 15 \ j -------------------- = ) a[65][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[65][j], a[65][j + 1], a[65][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[66][j], for j from 0 to infinity as the solution of the recurrence a[66][j] = 2 a[66][j - 1] + 4 a[66][j - 2] + a[66][j - 3] Subject to the initial conditions a[66][0] = 16, a[66][1] = 0, a[66][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -48 x - 32 x + 16 \ j -------------------- = ) a[66][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[66][j], a[66][j + 1], a[66][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[67][j], for j from 0 to infinity as the solution of the recurrence a[67][j] = 2 a[67][j - 1] + 4 a[67][j - 2] + a[67][j - 3] Subject to the initial conditions a[67][0] = 16, a[67][1] = 1, a[67][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -47 x - 31 x + 16 \ j -------------------- = ) a[67][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[67][j], a[67][j + 1], a[67][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[68][j], for j from 0 to infinity as the solution of the recurrence a[68][j] = 2 a[68][j - 1] + 4 a[68][j - 2] + a[68][j - 3] Subject to the initial conditions a[68][0] = 17, a[68][1] = 0, a[68][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -51 x - 34 x + 17 \ j -------------------- = ) a[68][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[68][j], a[68][j + 1], a[68][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[69][j], for j from 0 to infinity as the solution of the recurrence a[69][j] = 2 a[69][j - 1] + 4 a[69][j - 2] + a[69][j - 3] Subject to the initial conditions a[69][0] = 17, a[69][1] = 1, a[69][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -50 x - 33 x + 17 \ j -------------------- = ) a[69][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[69][j], a[69][j + 1], a[69][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[70][j], for j from 0 to infinity as the solution of the recurrence a[70][j] = 2 a[70][j - 1] + 4 a[70][j - 2] + a[70][j - 3] Subject to the initial conditions a[70][0] = 18, a[70][1] = 0, a[70][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -54 x - 36 x + 18 \ j -------------------- = ) a[70][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[70][j], a[70][j + 1], a[70][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[71][j], for j from 0 to infinity as the solution of the recurrence a[71][j] = 2 a[71][j - 1] + 4 a[71][j - 2] + a[71][j - 3] Subject to the initial conditions a[71][0] = 19, a[71][1] = 0, a[71][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -57 x - 38 x + 19 \ j -------------------- = ) a[71][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[71][j], a[71][j + 1], a[71][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 ----------------------------------------------------- Define the sequence , a[72][j], for j from 0 to infinity as the solution of the recurrence a[72][j] = 2 a[72][j - 1] + 4 a[72][j - 2] + a[72][j - 3] Subject to the initial conditions a[72][0] = 20, a[72][1] = 0, a[72][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -60 x - 40 x + 20 \ j -------------------- = ) a[72][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[72][j], a[72][j + 1], a[72][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 are the following triples {[1, 0, 1], [1, 1, 4], [1, 2, 7], [2, 0, 2], [2, 1, 5], [2, 2, 8], [2, 3, 11], [2, 4, 14], [2, 5, 17], [3, 0, 3], [3, 1, 6], [3, 2, 9], [3, 3, 12], [3, 4, 15], [3, 5, 18], [4, 0, 4], [4, 1, 7], [4, 2, 10], [4, 3, 13], [4, 4, 16], [4, 5, 19], [5, 0, 5], [5, 1, 8], [5, 2, 11], [5, 3, 14], [5, 4, 17], [5, 5, 20], [6, 0, 6], [6, 1, 9], [6, 2, 12], [6, 3, 15], [6, 4, 18], [7, 0, 7], [7, 1, 10], [7, 2, 13], [7, 3, 16], [7, 4, 19], [8, 0, 8], [8, 1, 11], [8, 2, 14], [8, 3, 17], [8, 4, 20], [9, 0, 9], [9, 1, 12], [9, 2, 15], [9, 3, 18], [10, 0, 10], [10, 1, 13], [10, 2, 16], [10, 3, 19], [11, 0, 11], [11, 1, 14], [11, 2, 17], [11, 3, 20], [12, 0, 12], [12, 1, 15], [12, 2, 18], [13, 0, 13], [13, 1, 16], [13, 2, 19], [14, 0, 14], [14, 1, 17], [14, 2, 20], [15, 0, 15], [15, 1, 18], [16, 0, 16], [16, 1, 19], [17, 0, 17], [17, 1, 20], [18, 0, 18], [19, 0, 19], [20, 0, 20]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.035, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 2 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.037, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 4 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 1, a[1][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -x - 3 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 are the following triples {[2, 1, 9]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 6 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 6 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 are the following triples {[0, 1, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 7 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 3, a[1][1] = 0, a[1][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 6 x + 3 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 are the following triples {[3, 0, 10]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 4 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 9 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 9 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -7 x - 4 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 are the following triples {[0, 1, 0], [2, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 18 x[1] x[2] + 5 x[1] x[2] x[3] - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] 3 + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 2 x[1] x[3] + 18 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 3 - 4 x[1] x[3] + 9 x[2] - 4 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following, 3, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 2 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -3 x - x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 2 a[3][j - 1] + 5 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 9, a[3][1] = 0, a[3][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -38 x - 18 x + 9 \ j -------------------- = ) a[3][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 are the following triples {[1, 0, 0], [1, 1, 4], [9, 0, 7]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad The following, 6, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 2, a[2][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -2 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 2 a[3][j - 1] + 5 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 0, a[3][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -9 x - 4 x + 2 \ j -------------------- = ) a[3][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 2 a[4][j - 1] + 5 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 3, a[4][1] = 2, a[4][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -x - 4 x + 3 \ j -------------------- = ) a[4][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[5][j], for j from 0 to infinity as the solution of the recurrence a[5][j] = 2 a[5][j - 1] + 5 a[5][j - 2] + a[5][j - 3] Subject to the initial conditions a[5][0] = 4, a[5][1] = 0, a[5][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -17 x - 8 x + 4 \ j -------------------- = ) a[5][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[5][j], a[5][j + 1], a[5][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[6][j], for j from 0 to infinity as the solution of the recurrence a[6][j] = 2 a[6][j - 1] + 5 a[6][j - 2] + a[6][j - 3] Subject to the initial conditions a[6][0] = 7, a[6][1] = 2, a[6][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -27 x - 12 x + 7 \ j -------------------- = ) a[6][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[6][j], a[6][j + 1], a[6][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 are the following triples {[0, 1, 1], [1, 2, 7], [2, 0, 1], [3, 2, 18], [4, 0, 3], [7, 2, 12]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.040, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 3, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 2 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 4 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 2 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 2, a[2][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 2 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 2 a[3][j - 1] + 5 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 18, a[3][1] = 0, a[3][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -76 x - 36 x + 18 \ j -------------------- = ) a[3][j] x 3 2 / -x - 5 x - 2 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 are the following triples {[2, 0, 0], [2, 2, 8], [18, 0, 14]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 2 x[1] x[3] 2 2 3 2 + 27 x[1] x[2] + 7 x[1] x[2] x[3] - 5 x[1] x[3] + 11 x[2] - x[2] x[3] 2 3 - 4 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 2 x[1] x[3] + 27 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 11 x[2] - x[2] x[3] - 4 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 2 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 2 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x - 3 x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.038, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 4 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -3 x + x \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 4 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 3 a[2][j - 1] + a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 0, a[2][1] = 1, a[2][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j ------------------ = ) a[2][j] x 3 2 / -x - x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 are the following triples {[0, 1, 0], [0, 1, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 7 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -2 x + x \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 are the following triples {[0, 1, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 8 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 3 a[2][j - 1] + a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 6 x + 2 \ j ------------------ = ) a[2][j] x 3 2 / -x - x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 are the following triples {[1, 1, 3], [2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 2 + 4 x[1] x[2] - x[1] x[3] + 4 x[2] + 8 x[2] x[3] - 6 x[2] x[3] 3 + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 2 3 x[1] + 2 x[1] x[2] + 3 x[1] x[3] + 4 x[1] x[2] - x[1] x[3] + 4 x[2] 2 2 3 + 8 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 3 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 3 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.025, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.038, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x - 3 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 are the following triples {[1, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 are the following triples {[0, 1, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -3 x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 3 a[2][j - 1] + 2 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 2 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 are the following triples {[0, 1, 0], [1, 1, 4]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.040, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 6 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -2 x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 3 a[2][j - 1] + 2 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 2, a[2][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 2 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 are the following triples {[0, 1, 1], [1, 2, 7]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 3 x[1] x[3] 2 2 3 2 + 7 x[1] x[2] + 3 x[1] x[2] x[3] - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] 2 3 - 6 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 3 x[1] x[3] + 7 x[1] x[2] + 3 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 7 x[2] + 7 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 3 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 3 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x - 3 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 are the following triples {[1, 1, 5]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 2, a[1][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 are the following triples {[1, 2, 8]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 are the following triples {[0, 1, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 6 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 2 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 are the following triples {[1, 1, 4]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -3 x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 12 x[1] x[2] + 6 x[1] x[2] x[3] - 3 x[1] x[3] + 10 x[2] 2 2 3 + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 3 x[1] x[3] + 12 x[1] x[2] + 6 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 10 x[2] + 6 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 3 x[1] < 0 are the following triples {[0, 1, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following, 7, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 3 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 3 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 0, a[2][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 3 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 3 a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 1, a[3][1] = 0, a[3][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x - 3 x + 1 \ j -------------------- = ) a[3][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 3 a[4][j - 1] + 4 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 1, a[4][1] = 1, a[4][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 2 x + 1 \ j -------------------- = ) a[4][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[5][j], for j from 0 to infinity as the solution of the recurrence a[5][j] = 3 a[5][j - 1] + 4 a[5][j - 2] + a[5][j - 3] Subject to the initial conditions a[5][0] = 2, a[5][1] = 1, a[5][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 5 x + 2 \ j -------------------- = ) a[5][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[5][j], a[5][j + 1], a[5][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[6][j], for j from 0 to infinity as the solution of the recurrence a[6][j] = 3 a[6][j - 1] + 4 a[6][j - 2] + a[6][j - 3] Subject to the initial conditions a[6][0] = 3, a[6][1] = 1, a[6][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -7 x - 8 x + 3 \ j -------------------- = ) a[6][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[6][j], a[6][j + 1], a[6][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[7][j], for j from 0 to infinity as the solution of the recurrence a[7][j] = 3 a[7][j - 1] + 4 a[7][j - 2] + a[7][j - 3] Subject to the initial conditions a[7][0] = 9, a[7][1] = 0, a[7][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -23 x - 27 x + 9 \ j -------------------- = ) a[7][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[7][j], a[7][j + 1], a[7][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 are the following triples {[1, 0, 0], [1, 0, 1], [1, 0, 3], [1, 1, 5], [2, 1, 9], [3, 1, 8], [9, 0, 13]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.025, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad The following, 4, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 3 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 3, a[2][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 3 a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 3, a[3][1] = 0, a[3][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 9 x + 3 \ j -------------------- = ) a[3][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 3 a[4][j - 1] + 4 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 3, a[4][1] = 4, a[4][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 5 x + 3 \ j -------------------- = ) a[4][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 are the following triples {[0, 1, 2], [1, 3, 12], [3, 0, 4], [3, 4, 19]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 6, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 6 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 3 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 6 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 3 a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 0, a[3][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 6 x + 2 \ j -------------------- = ) a[3][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 3 a[4][j - 1] + 4 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 2, a[4][1] = 2, a[4][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 4 x + 2 \ j -------------------- = ) a[4][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[5][j], for j from 0 to infinity as the solution of the recurrence a[5][j] = 3 a[5][j - 1] + 4 a[5][j - 2] + a[5][j - 3] Subject to the initial conditions a[5][0] = 4, a[5][1] = 2, a[5][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 10 x + 4 \ j -------------------- = ) a[5][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[5][j], a[5][j + 1], a[5][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[6][j], for j from 0 to infinity as the solution of the recurrence a[6][j] = 3 a[6][j - 1] + 4 a[6][j - 2] + a[6][j - 3] Subject to the initial conditions a[6][0] = 6, a[6][1] = 2, a[6][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -14 x - 16 x + 6 \ j -------------------- = ) a[6][j] x 3 2 / -x - 4 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[6][j], a[6][j + 1], a[6][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 are the following triples {[2, 0, 0], [2, 0, 2], [2, 0, 6], [2, 2, 10], [4, 2, 18], [6, 2, 16]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 19 x[1] x[2] + 9 x[1] x[2] x[3] - 4 x[1] x[3] + 13 x[2] 2 2 3 + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 3 x[1] x[3] + 19 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 13 x[2] + 5 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad The following, 59, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 3 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 3 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 2 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 3 a[3][j - 1] + 5 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 1, a[3][1] = 2, a[3][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -2 x - x + 1 \ j -------------------- = ) a[3][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 3 a[4][j - 1] + 5 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 1, a[4][1] = 3, a[4][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j -------------------- = ) a[4][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[5][j], for j from 0 to infinity as the solution of the recurrence a[5][j] = 3 a[5][j - 1] + 5 a[5][j - 2] + a[5][j - 3] Subject to the initial conditions a[5][0] = 2, a[5][1] = 0, a[5][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 6 x + 2 \ j -------------------- = ) a[5][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[5][j], a[5][j + 1], a[5][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[6][j], for j from 0 to infinity as the solution of the recurrence a[6][j] = 3 a[6][j - 1] + 5 a[6][j - 2] + a[6][j - 3] Subject to the initial conditions a[6][0] = 2, a[6][1] = 1, a[6][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -7 x - 5 x + 2 \ j -------------------- = ) a[6][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[6][j], a[6][j + 1], a[6][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[7][j], for j from 0 to infinity as the solution of the recurrence a[7][j] = 3 a[7][j - 1] + 5 a[7][j - 2] + a[7][j - 3] Subject to the initial conditions a[7][0] = 2, a[7][1] = 2, a[7][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 4 x + 2 \ j -------------------- = ) a[7][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[7][j], a[7][j + 1], a[7][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[8][j], for j from 0 to infinity as the solution of the recurrence a[8][j] = 3 a[8][j - 1] + 5 a[8][j - 2] + a[8][j - 3] Subject to the initial conditions a[8][0] = 2, a[8][1] = 3, a[8][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 3 x + 2 \ j -------------------- = ) a[8][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[8][j], a[8][j + 1], a[8][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[9][j], for j from 0 to infinity as the solution of the recurrence a[9][j] = 3 a[9][j - 1] + 5 a[9][j - 2] + a[9][j - 3] Subject to the initial conditions a[9][0] = 2, a[9][1] = 4, a[9][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 2 x + 2 \ j -------------------- = ) a[9][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[9][j], a[9][j + 1], a[9][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[10][j], for j from 0 to infinity as the solution of the recurrence a[10][j] = 3 a[10][j - 1] + 5 a[10][j - 2] + a[10][j - 3] Subject to the initial conditions a[10][0] = 3, a[10][1] = 0, a[10][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -12 x - 9 x + 3 \ j -------------------- = ) a[10][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[10][j], a[10][j + 1], a[10][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[11][j], for j from 0 to infinity as the solution of the recurrence a[11][j] = 3 a[11][j - 1] + 5 a[11][j - 2] + a[11][j - 3] Subject to the initial conditions a[11][0] = 3, a[11][1] = 1, a[11][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -11 x - 8 x + 3 \ j -------------------- = ) a[11][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[11][j], a[11][j + 1], a[11][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[12][j], for j from 0 to infinity as the solution of the recurrence a[12][j] = 3 a[12][j - 1] + 5 a[12][j - 2] + a[12][j - 3] Subject to the initial conditions a[12][0] = 3, a[12][1] = 2, a[12][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 7 x + 3 \ j -------------------- = ) a[12][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[12][j], a[12][j + 1], a[12][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[13][j], for j from 0 to infinity as the solution of the recurrence a[13][j] = 3 a[13][j - 1] + 5 a[13][j - 2] + a[13][j - 3] Subject to the initial conditions a[13][0] = 3, a[13][1] = 3, a[13][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -9 x - 6 x + 3 \ j -------------------- = ) a[13][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[13][j], a[13][j + 1], a[13][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[14][j], for j from 0 to infinity as the solution of the recurrence a[14][j] = 3 a[14][j - 1] + 5 a[14][j - 2] + a[14][j - 3] Subject to the initial conditions a[14][0] = 3, a[14][1] = 4, a[14][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 5 x + 3 \ j -------------------- = ) a[14][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[14][j], a[14][j + 1], a[14][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[15][j], for j from 0 to infinity as the solution of the recurrence a[15][j] = 3 a[15][j - 1] + 5 a[15][j - 2] + a[15][j - 3] Subject to the initial conditions a[15][0] = 4, a[15][1] = 0, a[15][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -16 x - 12 x + 4 \ j -------------------- = ) a[15][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[15][j], a[15][j + 1], a[15][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[16][j], for j from 0 to infinity as the solution of the recurrence a[16][j] = 3 a[16][j - 1] + 5 a[16][j - 2] + a[16][j - 3] Subject to the initial conditions a[16][0] = 4, a[16][1] = 1, a[16][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -15 x - 11 x + 4 \ j -------------------- = ) a[16][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[16][j], a[16][j + 1], a[16][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[17][j], for j from 0 to infinity as the solution of the recurrence a[17][j] = 3 a[17][j - 1] + 5 a[17][j - 2] + a[17][j - 3] Subject to the initial conditions a[17][0] = 4, a[17][1] = 2, a[17][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -14 x - 10 x + 4 \ j -------------------- = ) a[17][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[17][j], a[17][j + 1], a[17][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[18][j], for j from 0 to infinity as the solution of the recurrence a[18][j] = 3 a[18][j - 1] + 5 a[18][j - 2] + a[18][j - 3] Subject to the initial conditions a[18][0] = 4, a[18][1] = 3, a[18][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -13 x - 9 x + 4 \ j -------------------- = ) a[18][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[18][j], a[18][j + 1], a[18][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[19][j], for j from 0 to infinity as the solution of the recurrence a[19][j] = 3 a[19][j - 1] + 5 a[19][j - 2] + a[19][j - 3] Subject to the initial conditions a[19][0] = 4, a[19][1] = 4, a[19][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -12 x - 8 x + 4 \ j -------------------- = ) a[19][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[19][j], a[19][j + 1], a[19][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[20][j], for j from 0 to infinity as the solution of the recurrence a[20][j] = 3 a[20][j - 1] + 5 a[20][j - 2] + a[20][j - 3] Subject to the initial conditions a[20][0] = 5, a[20][1] = 0, a[20][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -20 x - 15 x + 5 \ j -------------------- = ) a[20][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[20][j], a[20][j + 1], a[20][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[21][j], for j from 0 to infinity as the solution of the recurrence a[21][j] = 3 a[21][j - 1] + 5 a[21][j - 2] + a[21][j - 3] Subject to the initial conditions a[21][0] = 5, a[21][1] = 1, a[21][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -19 x - 14 x + 5 \ j -------------------- = ) a[21][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[21][j], a[21][j + 1], a[21][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[22][j], for j from 0 to infinity as the solution of the recurrence a[22][j] = 3 a[22][j - 1] + 5 a[22][j - 2] + a[22][j - 3] Subject to the initial conditions a[22][0] = 5, a[22][1] = 2, a[22][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -18 x - 13 x + 5 \ j -------------------- = ) a[22][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[22][j], a[22][j + 1], a[22][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[23][j], for j from 0 to infinity as the solution of the recurrence a[23][j] = 3 a[23][j - 1] + 5 a[23][j - 2] + a[23][j - 3] Subject to the initial conditions a[23][0] = 5, a[23][1] = 3, a[23][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -17 x - 12 x + 5 \ j -------------------- = ) a[23][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[23][j], a[23][j + 1], a[23][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[24][j], for j from 0 to infinity as the solution of the recurrence a[24][j] = 3 a[24][j - 1] + 5 a[24][j - 2] + a[24][j - 3] Subject to the initial conditions a[24][0] = 6, a[24][1] = 0, a[24][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -24 x - 18 x + 6 \ j -------------------- = ) a[24][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[24][j], a[24][j + 1], a[24][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[25][j], for j from 0 to infinity as the solution of the recurrence a[25][j] = 3 a[25][j - 1] + 5 a[25][j - 2] + a[25][j - 3] Subject to the initial conditions a[25][0] = 6, a[25][1] = 1, a[25][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -23 x - 17 x + 6 \ j -------------------- = ) a[25][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[25][j], a[25][j + 1], a[25][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[26][j], for j from 0 to infinity as the solution of the recurrence a[26][j] = 3 a[26][j - 1] + 5 a[26][j - 2] + a[26][j - 3] Subject to the initial conditions a[26][0] = 6, a[26][1] = 2, a[26][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -22 x - 16 x + 6 \ j -------------------- = ) a[26][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[26][j], a[26][j + 1], a[26][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[27][j], for j from 0 to infinity as the solution of the recurrence a[27][j] = 3 a[27][j - 1] + 5 a[27][j - 2] + a[27][j - 3] Subject to the initial conditions a[27][0] = 6, a[27][1] = 3, a[27][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -21 x - 15 x + 6 \ j -------------------- = ) a[27][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[27][j], a[27][j + 1], a[27][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[28][j], for j from 0 to infinity as the solution of the recurrence a[28][j] = 3 a[28][j - 1] + 5 a[28][j - 2] + a[28][j - 3] Subject to the initial conditions a[28][0] = 7, a[28][1] = 0, a[28][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -28 x - 21 x + 7 \ j -------------------- = ) a[28][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[28][j], a[28][j + 1], a[28][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[29][j], for j from 0 to infinity as the solution of the recurrence a[29][j] = 3 a[29][j - 1] + 5 a[29][j - 2] + a[29][j - 3] Subject to the initial conditions a[29][0] = 7, a[29][1] = 1, a[29][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -27 x - 20 x + 7 \ j -------------------- = ) a[29][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[29][j], a[29][j + 1], a[29][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[30][j], for j from 0 to infinity as the solution of the recurrence a[30][j] = 3 a[30][j - 1] + 5 a[30][j - 2] + a[30][j - 3] Subject to the initial conditions a[30][0] = 7, a[30][1] = 2, a[30][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -26 x - 19 x + 7 \ j -------------------- = ) a[30][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[30][j], a[30][j + 1], a[30][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[31][j], for j from 0 to infinity as the solution of the recurrence a[31][j] = 3 a[31][j - 1] + 5 a[31][j - 2] + a[31][j - 3] Subject to the initial conditions a[31][0] = 7, a[31][1] = 3, a[31][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -25 x - 18 x + 7 \ j -------------------- = ) a[31][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[31][j], a[31][j + 1], a[31][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[32][j], for j from 0 to infinity as the solution of the recurrence a[32][j] = 3 a[32][j - 1] + 5 a[32][j - 2] + a[32][j - 3] Subject to the initial conditions a[32][0] = 8, a[32][1] = 0, a[32][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -32 x - 24 x + 8 \ j -------------------- = ) a[32][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[32][j], a[32][j + 1], a[32][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[33][j], for j from 0 to infinity as the solution of the recurrence a[33][j] = 3 a[33][j - 1] + 5 a[33][j - 2] + a[33][j - 3] Subject to the initial conditions a[33][0] = 8, a[33][1] = 1, a[33][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -31 x - 23 x + 8 \ j -------------------- = ) a[33][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[33][j], a[33][j + 1], a[33][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[34][j], for j from 0 to infinity as the solution of the recurrence a[34][j] = 3 a[34][j - 1] + 5 a[34][j - 2] + a[34][j - 3] Subject to the initial conditions a[34][0] = 8, a[34][1] = 2, a[34][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -30 x - 22 x + 8 \ j -------------------- = ) a[34][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[34][j], a[34][j + 1], a[34][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[35][j], for j from 0 to infinity as the solution of the recurrence a[35][j] = 3 a[35][j - 1] + 5 a[35][j - 2] + a[35][j - 3] Subject to the initial conditions a[35][0] = 8, a[35][1] = 3, a[35][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -29 x - 21 x + 8 \ j -------------------- = ) a[35][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[35][j], a[35][j + 1], a[35][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[36][j], for j from 0 to infinity as the solution of the recurrence a[36][j] = 3 a[36][j - 1] + 5 a[36][j - 2] + a[36][j - 3] Subject to the initial conditions a[36][0] = 9, a[36][1] = 0, a[36][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -36 x - 27 x + 9 \ j -------------------- = ) a[36][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[36][j], a[36][j + 1], a[36][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[37][j], for j from 0 to infinity as the solution of the recurrence a[37][j] = 3 a[37][j - 1] + 5 a[37][j - 2] + a[37][j - 3] Subject to the initial conditions a[37][0] = 9, a[37][1] = 1, a[37][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -35 x - 26 x + 9 \ j -------------------- = ) a[37][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[37][j], a[37][j + 1], a[37][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[38][j], for j from 0 to infinity as the solution of the recurrence a[38][j] = 3 a[38][j - 1] + 5 a[38][j - 2] + a[38][j - 3] Subject to the initial conditions a[38][0] = 9, a[38][1] = 2, a[38][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -34 x - 25 x + 9 \ j -------------------- = ) a[38][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[38][j], a[38][j + 1], a[38][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[39][j], for j from 0 to infinity as the solution of the recurrence a[39][j] = 3 a[39][j - 1] + 5 a[39][j - 2] + a[39][j - 3] Subject to the initial conditions a[39][0] = 10, a[39][1] = 0, a[39][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -40 x - 30 x + 10 \ j -------------------- = ) a[39][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[39][j], a[39][j + 1], a[39][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[40][j], for j from 0 to infinity as the solution of the recurrence a[40][j] = 3 a[40][j - 1] + 5 a[40][j - 2] + a[40][j - 3] Subject to the initial conditions a[40][0] = 10, a[40][1] = 1, a[40][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -39 x - 29 x + 10 \ j -------------------- = ) a[40][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[40][j], a[40][j + 1], a[40][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[41][j], for j from 0 to infinity as the solution of the recurrence a[41][j] = 3 a[41][j - 1] + 5 a[41][j - 2] + a[41][j - 3] Subject to the initial conditions a[41][0] = 10, a[41][1] = 2, a[41][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -38 x - 28 x + 10 \ j -------------------- = ) a[41][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[41][j], a[41][j + 1], a[41][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[42][j], for j from 0 to infinity as the solution of the recurrence a[42][j] = 3 a[42][j - 1] + 5 a[42][j - 2] + a[42][j - 3] Subject to the initial conditions a[42][0] = 11, a[42][1] = 0, a[42][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -44 x - 33 x + 11 \ j -------------------- = ) a[42][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[42][j], a[42][j + 1], a[42][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[43][j], for j from 0 to infinity as the solution of the recurrence a[43][j] = 3 a[43][j - 1] + 5 a[43][j - 2] + a[43][j - 3] Subject to the initial conditions a[43][0] = 11, a[43][1] = 1, a[43][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -43 x - 32 x + 11 \ j -------------------- = ) a[43][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[43][j], a[43][j + 1], a[43][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[44][j], for j from 0 to infinity as the solution of the recurrence a[44][j] = 3 a[44][j - 1] + 5 a[44][j - 2] + a[44][j - 3] Subject to the initial conditions a[44][0] = 11, a[44][1] = 2, a[44][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -42 x - 31 x + 11 \ j -------------------- = ) a[44][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[44][j], a[44][j + 1], a[44][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[45][j], for j from 0 to infinity as the solution of the recurrence a[45][j] = 3 a[45][j - 1] + 5 a[45][j - 2] + a[45][j - 3] Subject to the initial conditions a[45][0] = 12, a[45][1] = 0, a[45][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -48 x - 36 x + 12 \ j -------------------- = ) a[45][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[45][j], a[45][j + 1], a[45][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[46][j], for j from 0 to infinity as the solution of the recurrence a[46][j] = 3 a[46][j - 1] + 5 a[46][j - 2] + a[46][j - 3] Subject to the initial conditions a[46][0] = 12, a[46][1] = 1, a[46][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -47 x - 35 x + 12 \ j -------------------- = ) a[46][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[46][j], a[46][j + 1], a[46][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[47][j], for j from 0 to infinity as the solution of the recurrence a[47][j] = 3 a[47][j - 1] + 5 a[47][j - 2] + a[47][j - 3] Subject to the initial conditions a[47][0] = 12, a[47][1] = 2, a[47][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -46 x - 34 x + 12 \ j -------------------- = ) a[47][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[47][j], a[47][j + 1], a[47][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[48][j], for j from 0 to infinity as the solution of the recurrence a[48][j] = 3 a[48][j - 1] + 5 a[48][j - 2] + a[48][j - 3] Subject to the initial conditions a[48][0] = 13, a[48][1] = 0, a[48][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -52 x - 39 x + 13 \ j -------------------- = ) a[48][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[48][j], a[48][j + 1], a[48][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[49][j], for j from 0 to infinity as the solution of the recurrence a[49][j] = 3 a[49][j - 1] + 5 a[49][j - 2] + a[49][j - 3] Subject to the initial conditions a[49][0] = 13, a[49][1] = 1, a[49][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -51 x - 38 x + 13 \ j -------------------- = ) a[49][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[49][j], a[49][j + 1], a[49][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[50][j], for j from 0 to infinity as the solution of the recurrence a[50][j] = 3 a[50][j - 1] + 5 a[50][j - 2] + a[50][j - 3] Subject to the initial conditions a[50][0] = 14, a[50][1] = 0, a[50][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -56 x - 42 x + 14 \ j -------------------- = ) a[50][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[50][j], a[50][j + 1], a[50][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[51][j], for j from 0 to infinity as the solution of the recurrence a[51][j] = 3 a[51][j - 1] + 5 a[51][j - 2] + a[51][j - 3] Subject to the initial conditions a[51][0] = 14, a[51][1] = 1, a[51][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -55 x - 41 x + 14 \ j -------------------- = ) a[51][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[51][j], a[51][j + 1], a[51][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[52][j], for j from 0 to infinity as the solution of the recurrence a[52][j] = 3 a[52][j - 1] + 5 a[52][j - 2] + a[52][j - 3] Subject to the initial conditions a[52][0] = 15, a[52][1] = 0, a[52][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -60 x - 45 x + 15 \ j -------------------- = ) a[52][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[52][j], a[52][j + 1], a[52][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[53][j], for j from 0 to infinity as the solution of the recurrence a[53][j] = 3 a[53][j - 1] + 5 a[53][j - 2] + a[53][j - 3] Subject to the initial conditions a[53][0] = 15, a[53][1] = 1, a[53][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -59 x - 44 x + 15 \ j -------------------- = ) a[53][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[53][j], a[53][j + 1], a[53][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[54][j], for j from 0 to infinity as the solution of the recurrence a[54][j] = 3 a[54][j - 1] + 5 a[54][j - 2] + a[54][j - 3] Subject to the initial conditions a[54][0] = 16, a[54][1] = 0, a[54][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -64 x - 48 x + 16 \ j -------------------- = ) a[54][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[54][j], a[54][j + 1], a[54][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[55][j], for j from 0 to infinity as the solution of the recurrence a[55][j] = 3 a[55][j - 1] + 5 a[55][j - 2] + a[55][j - 3] Subject to the initial conditions a[55][0] = 16, a[55][1] = 1, a[55][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -63 x - 47 x + 16 \ j -------------------- = ) a[55][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[55][j], a[55][j + 1], a[55][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[56][j], for j from 0 to infinity as the solution of the recurrence a[56][j] = 3 a[56][j - 1] + 5 a[56][j - 2] + a[56][j - 3] Subject to the initial conditions a[56][0] = 17, a[56][1] = 0, a[56][2] = 17 Or equivalently in terms of the generating function infinity 2 ----- -68 x - 51 x + 17 \ j -------------------- = ) a[56][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[56][j], a[56][j + 1], a[56][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[57][j], for j from 0 to infinity as the solution of the recurrence a[57][j] = 3 a[57][j - 1] + 5 a[57][j - 2] + a[57][j - 3] Subject to the initial conditions a[57][0] = 18, a[57][1] = 0, a[57][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -72 x - 54 x + 18 \ j -------------------- = ) a[57][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[57][j], a[57][j + 1], a[57][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[58][j], for j from 0 to infinity as the solution of the recurrence a[58][j] = 3 a[58][j - 1] + 5 a[58][j - 2] + a[58][j - 3] Subject to the initial conditions a[58][0] = 19, a[58][1] = 0, a[58][2] = 19 Or equivalently in terms of the generating function infinity 2 ----- -76 x - 57 x + 19 \ j -------------------- = ) a[58][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[58][j], a[58][j + 1], a[58][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 ----------------------------------------------------- Define the sequence , a[59][j], for j from 0 to infinity as the solution of the recurrence a[59][j] = 3 a[59][j - 1] + 5 a[59][j - 2] + a[59][j - 3] Subject to the initial conditions a[59][0] = 20, a[59][1] = 0, a[59][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -80 x - 60 x + 20 \ j -------------------- = ) a[59][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[59][j], a[59][j + 1], a[59][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 are the following triples {[1, 0, 1], [1, 1, 5], [1, 2, 9], [1, 3, 13], [2, 0, 2], [2, 1, 6], [2, 2, 10], [2, 3, 14], [2, 4, 18], [3, 0, 3], [3, 1, 7], [3, 2, 11], [3, 3, 15], [3, 4, 19], [4, 0, 4], [4, 1, 8], [4, 2, 12], [4, 3, 16], [4, 4, 20], [5, 0, 5], [5, 1, 9], [5, 2, 13], [5, 3, 17], [6, 0, 6], [6, 1, 10], [6, 2, 14], [6, 3, 18], [7, 0, 7], [7, 1, 11], [7, 2, 15], [7, 3, 19], [8, 0, 8], [8, 1, 12], [8, 2, 16], [8, 3, 20], [9, 0, 9], [9, 1, 13], [9, 2, 17], [10, 0, 10], [10, 1, 14], [10, 2, 18], [11, 0, 11], [11, 1, 15], [11, 2, 19], [12, 0, 12], [12, 1, 16], [12, 2, 20], [13, 0, 13], [13, 1, 17], [14, 0, 14], [14, 1, 18], [15, 0, 15], [15, 1, 19], [16, 0, 16], [16, 1, 20], [17, 0, 17], [18, 0, 18], [19, 0, 19], [20, 0, 20]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.034, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 3 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.037, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 2, a[1][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -x - 4 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 are the following triples {[2, 2, 15]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.038, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 3 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 6 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 are the following triples {[0, 1, 2], [2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 3 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 3, a[1][1] = 1, a[1][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 8 x + 3 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 3 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 are the following triples {[3, 1, 16]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 3 x[1] x[3] 2 2 3 + 28 x[1] x[2] + 12 x[1] x[2] x[3] - 5 x[1] x[3] + 16 x[2] 2 2 3 + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 3 x[1] x[3] + 28 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 16 x[2] + 4 x[2] x[3] - 6 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 3 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 3 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x - 4 x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x + x \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 4 a[2][j - 1] + a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 0, a[2][1] = 1, a[2][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j ------------------ = ) a[2][j] x 3 2 / -x - x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 are the following triples {[0, 1, 0], [0, 1, 3]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.040, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 8 x + 2 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 5 x[1] x[2] + x[1] x[2] x[3] - x[1] x[3] + 5 x[2] + 15 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 4 x[1] x[3] + 5 x[1] x[2] + x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 5 x[2] + 15 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 4 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 are the following triples {[1, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 are the following triples {[0, 1, 3]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 8 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 are the following triples {[0, 1, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 4 x[1] x[3] 2 2 3 2 + 8 x[1] x[2] + 5 x[1] x[2] x[3] - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] 2 3 - 8 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 4 x[1] x[3] + 8 x[1] x[2] + 5 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 9 x[2] + 14 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 are the following triples {[0, 1, 3]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 8 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -x - 3 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 4 a[2][j - 1] + 3 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 2, a[2][2] = 10 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 3 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 are the following triples {[1, 1, 6], [1, 2, 10]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 13 x[1] x[2] + 9 x[1] x[2] x[3] - 3 x[1] x[3] + 13 x[2] 2 2 3 + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 4 x[1] x[3] + 13 x[1] x[2] + 9 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 13 x[2] + 13 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 4 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 0, a[2][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 4 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 are the following triples {[1, 0, 0], [1, 0, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 4 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 2, a[2][2] = 11 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 are the following triples {[1, 0, 1], [1, 2, 11]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad The following, 4, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 4 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 3 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 4 a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 1, a[3][1] = 1, a[3][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -x - 3 x + 1 \ j -------------------- = ) a[3][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 4 a[4][j - 1] + 4 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 1, a[4][1] = 3, a[4][2] = 15 Or equivalently in terms of the generating function infinity 2 ----- -x - x + 1 \ j -------------------- = ) a[4][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 are the following triples {[1, 0, 3], [1, 1, 6], [1, 1, 7], [1, 3, 15]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 3, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 4 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 8 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 4 a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 0, a[3][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 8 x + 2 \ j -------------------- = ) a[3][j] x 3 2 / -x - 4 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 are the following triples {[0, 1, 3], [2, 0, 0], [2, 0, 4]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 20 x[1] x[2] + 13 x[1] x[2] x[3] - 4 x[1] x[3] + 17 x[2] 2 2 3 + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 4 x[1] x[3] + 20 x[1] x[2] + 13 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 17 x[2] + 12 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.040, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following, 6, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 4 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 0, a[2][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 4 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 4 a[3][j - 1] + 5 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 1, a[3][1] = 0, a[3][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x - 4 x + 1 \ j -------------------- = ) a[3][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 4 a[4][j - 1] + 5 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 1, a[4][1] = 1, a[4][2] = 6 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 3 x + 1 \ j -------------------- = ) a[4][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[5][j], for j from 0 to infinity as the solution of the recurrence a[5][j] = 4 a[5][j - 1] + 5 a[5][j - 2] + a[5][j - 3] Subject to the initial conditions a[5][0] = 3, a[5][1] = 1, a[5][2] = 16 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 11 x + 3 \ j -------------------- = ) a[5][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[5][j], a[5][j + 1], a[5][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[6][j], for j from 0 to infinity as the solution of the recurrence a[6][j] = 4 a[6][j - 1] + 5 a[6][j - 2] + a[6][j - 3] Subject to the initial conditions a[6][0] = 7, a[6][1] = 0, a[6][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -26 x - 28 x + 7 \ j -------------------- = ) a[6][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[6][j], a[6][j + 1], a[6][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 are the following triples {[1, 0, 0], [1, 0, 1], [1, 0, 4], [1, 1, 6], [3, 1, 16], [7, 0, 9]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 5, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 8 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 4 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 8 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 4 a[3][j - 1] + 5 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 2, a[3][1] = 0, a[3][2] = 8 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 8 x + 2 \ j -------------------- = ) a[3][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[4][j], for j from 0 to infinity as the solution of the recurrence a[4][j] = 4 a[4][j - 1] + 5 a[4][j - 2] + a[4][j - 3] Subject to the initial conditions a[4][0] = 2, a[4][1] = 2, a[4][2] = 12 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 6 x + 2 \ j -------------------- = ) a[4][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[4][j], a[4][j + 1], a[4][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[5][j], for j from 0 to infinity as the solution of the recurrence a[5][j] = 4 a[5][j - 1] + 5 a[5][j - 2] + a[5][j - 3] Subject to the initial conditions a[5][0] = 14, a[5][1] = 0, a[5][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -52 x - 56 x + 14 \ j -------------------- = ) a[5][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[5][j], a[5][j + 1], a[5][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 are the following triples {[2, 0, 0], [2, 0, 2], [2, 0, 8], [2, 2, 12], [14, 0, 18]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following, 3, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 4 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 4 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 4, a[2][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 4 a[3][j - 1] + 5 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 4, a[3][1] = 0, a[3][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -15 x - 16 x + 4 \ j -------------------- = ) a[3][j] x 3 2 / -x - 5 x - 4 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 are the following triples {[0, 1, 3], [1, 4, 20], [4, 0, 5]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 4 x[1] x[3] 2 2 3 + 29 x[1] x[2] + 17 x[1] x[2] x[3] - 5 x[1] x[3] + 21 x[2] 2 2 3 + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 4 x[1] x[3] + 29 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 21 x[2] + 11 x[2] x[3] - 8 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 4 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 4 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -x - 5 x + 1 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -5 x + x \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 6 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 0, a[2][1] = 1, a[2][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j ------------------ = ) a[2][j] x 3 2 / -x - x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 are the following triples {[0, 1, 0], [0, 1, 4]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 10 x + 2 \ j ------------------ = ) a[1][j] x 3 2 / -x - x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 2 x[1] x[2] + 5 x[1] x[3] 2 2 3 2 + 6 x[1] x[2] + 2 x[1] x[2] x[3] - x[1] x[3] + 6 x[2] + 24 x[2] x[3] 2 3 - 10 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 2 x[1] x[2] + 5 x[1] x[3] + 6 x[1] x[2] + 2 x[1] x[2] x[3] 2 3 2 2 3 - x[1] x[3] + 6 x[2] + 24 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 5 x[2] - x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 are the following triples {[1, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 are the following triples {[0, 1, 4]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 2 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 10 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 2 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 4 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 9 x[1] x[2] + 7 x[1] x[2] x[3] - 2 x[1] x[3] + 11 x[2] 2 2 3 + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 4 x[1] x[2] + 5 x[1] x[3] + 9 x[1] x[2] + 7 x[1] x[2] x[3] 2 3 2 2 3 - 2 x[1] x[3] + 11 x[2] + 23 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 2 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 2 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 1]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 are the following triples {[1, 0, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 3 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 3 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 3 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 10 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 3 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 are the following triples {[0, 1, 4], [2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 6 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 14 x[1] x[2] + 12 x[1] x[2] x[3] - 3 x[1] x[3] + 16 x[2] 2 2 3 + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 6 x[1] x[2] + 5 x[1] x[3] + 14 x[1] x[2] + 12 x[1] x[2] x[3] 2 3 2 2 3 - 3 x[1] x[3] + 16 x[2] + 22 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 3 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 3 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 are the following triples {[1, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.038, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 0, a[2][2] = 2 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 5 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 are the following triples {[1, 0, 1], [1, 0, 2]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 3, a[2][2] = 18 Or equivalently in terms of the generating function infinity 2 ----- -x - 2 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 are the following triples {[1, 0, 3], [1, 3, 18]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following set of triples of non-negative integers, [x[1], x[2], x[3]], \ are the only solutions (with non-negative integers) of the diophatine eq\ uation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -8 x - 10 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 are the following triples {[2, 0, 0]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad The following, 3, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 4 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 4 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 4 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -2 x - 4 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 4 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 5 a[3][j - 1] + 4 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 1, a[3][1] = 2, a[3][2] = 13 Or equivalently in terms of the generating function infinity 2 ----- -x - 3 x + 1 \ j -------------------- = ) a[3][j] x 3 2 / -x - 4 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 are the following triples {[0, 1, 4], [1, 1, 7], [1, 2, 13]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 8 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 21 x[1] x[2] + 17 x[1] x[2] x[3] - 4 x[1] x[3] + 21 x[2] 2 2 3 + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 8 x[1] x[2] + 5 x[1] x[3] + 21 x[1] x[2] + 17 x[1] x[2] x[3] 2 3 2 2 3 - 4 x[1] x[3] + 21 x[2] + 21 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 4 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 4 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 0 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.039, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 10 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 1 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 are the following triples {[1, 0, 0], [2, 0, 7]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 1 Or equivalently in terms of the generating function infinity 2 ----- -4 x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 3, a[2][1] = 0, a[2][2] = 5 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 15 x + 3 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 2 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 are the following triples {[1, 0, 1], [3, 0, 5]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 3 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 1, a[1][2] = 7 Or equivalently in terms of the generating function infinity 2 ----- -3 x - 4 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 1, a[2][1] = 1, a[2][2] = 9 Or equivalently in terms of the generating function infinity 2 ----- -x - 4 x + 1 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 4 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 are the following triples {[1, 1, 7], [1, 1, 9]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 By Shalosh B. Ekhad The following, 3, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 1, a[1][1] = 0, a[1][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x - 5 x + 1 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 0, a[2][2] = 3 Or equivalently in terms of the generating function infinity 2 ----- -7 x - 10 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 ----------------------------------------------------- Define the sequence , a[3][j], for j from 0 to infinity as the solution of the recurrence a[3][j] = 5 a[3][j - 1] + 5 a[3][j - 2] + a[3][j - 3] Subject to the initial conditions a[3][0] = 5, a[3][1] = 1, a[3][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -16 x - 24 x + 5 \ j -------------------- = ) a[3][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[3][j], a[3][j + 1], a[3][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 5 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 are the following triples {[1, 0, 4], [2, 0, 3], [5, 1, 14]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.025, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 6 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.038, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 7 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 2, a[1][1] = 0, a[1][2] = 0 Or equivalently in terms of the generating function infinity 2 ----- -10 x - 10 x + 2 \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 4, a[2][1] = 0, a[2][2] = 14 Or equivalently in terms of the generating function infinity 2 ----- -6 x - 20 x + 4 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 8 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 are the following triples {[2, 0, 0], [4, 0, 14]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.023, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 By Shalosh B. Ekhad There are no solutions in non-negative triples [x[1],x[2],x[3]] of the Dioph\ antine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 9 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that there do not exist such non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 hence there are no solutions at all. QED. ----------------------------------------- This ends thie article that took, 0.024, to create. ------------------------------ 3 2 2 On Solutions of the Diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 By Shalosh B. Ekhad The following, 2, sets of triples of non-negative integers, [x[1], x[2], x[3]], are the only solutions (with non-negative integers) of the diophatine \ equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 ----------------------------------------------------- Define the sequence , a[1][j], for j from 0 to infinity as the solution of the recurrence a[1][j] = 5 a[1][j - 1] + 5 a[1][j - 2] + a[1][j - 3] Subject to the initial conditions a[1][0] = 0, a[1][1] = 1, a[1][2] = 4 Or equivalently in terms of the generating function infinity 2 ----- -x + x \ j -------------------- = ) a[1][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[1][j], a[1][j + 1], a[1][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 ----------------------------------------------------- Define the sequence , a[2][j], for j from 0 to infinity as the solution of the recurrence a[2][j] = 5 a[2][j - 1] + 5 a[2][j - 2] + a[2][j - 3] Subject to the initial conditions a[2][0] = 2, a[2][1] = 3, a[2][2] = 20 Or equivalently in terms of the generating function infinity 2 ----- -5 x - 7 x + 2 \ j -------------------- = ) a[2][j] x 3 2 / -x - 5 x - 5 x + 1 ----- j = 0 Then the following triples of non-negative solutions [x[1], x[2], x[3]] = [a[2][j], a[2][j + 1], a[2][j + 2]] for j from 0 to infinity 3 2 2 are solutions of the diophantine equation, x[1] + 10 x[1] x[2] + 5 x[1] x[3] 2 2 3 + 30 x[1] x[2] + 22 x[1] x[2] x[3] - 5 x[1] x[3] + 26 x[2] 2 2 3 + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 --------------------------------------- Proof: It is readily seen that if [x[1],x[2],x[3]] is a solution of the diop\ hantine equation 3 2 2 2 x[1] + 10 x[1] x[2] + 5 x[1] x[3] + 30 x[1] x[2] + 22 x[1] x[2] x[3] 2 3 2 2 3 - 5 x[1] x[3] + 26 x[2] + 20 x[2] x[3] - 10 x[2] x[3] + x[3] = 10 Then so is the triple [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] So the mapping [x[1],x[2],x[3]] -> , [x[3] - 5 x[2] - 5 x[1], x[1], x[2]] maps solution triples to other solution triples If you start with any solution triple with non-negative entries sooner or la\ ter you would have the scenario that for the FIRST time, the first compo\ nent, x[1], is negative It is readily seen that the only such triples of non-negative integers [x[1]\ ,x[2],x[3]] that satisfy the above diophantine equatiion and in additio\ n satisfy the inequality x[3] - 5 x[2] - 5 x[1] < 0 are the following triples {[0, 1, 4], [2, 3, 20]} Going forward gives the above proposed solutions QED. ----------------------------------------- This ends thie article that took, 0.024, to create. --------------------------- This took, 7.353, seconds.