Finding Inifintely many solutions to Diophantine Equations of the form, 3 3 3 A X + B Y + B Z = CONSTANT, for A and B At Most, 10 With Detailed Explanation and Motovation By Shalosh B. Ekhad Chapter , 1 3 3 3 Infinitely many solutions to the Diophantine Equation, X + 2 Y + 2 Z = -4913 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 X + 2 Y + 2 Z = -4913 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 W + X + 2 Y + 2 Z = 0 and get the following special case W = -19, X = -17, Y = -15, Z = 21 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 17 x, Y = n y - 15 x, Z = -n y + 21 x Plugging into the euqation 3 3 3 3 W + X + 2 Y + 2 Z = 0 we get 2 2 -36 x y (3 m y - n y - 6 m x + 36 n x) = 0 dividing by , x y, we have 2 2 (216 m - 1296 n) x + (-108 m + 36 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 108 m - 36 n , y = 216 m - 1296 n Plugging them back, we get the following parametric solution to , 3 3 3 3 W + X + 2 Y + 2 Z = 0 2 2 2 2 W = -51 m - 36 m n + 19 n , X = -57 m + 36 m n + 17 n , 2 2 2 2 Y = -45 m + 6 m n - 21 n , Z = 63 m - 6 m n + 15 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 431 t ) d(i) t = --------------------- / 2 ----- 102 (t - 1726 t + 1) i = 0 infinity ----- \ i 12 t ) e(i) t = --------------- / 2 ----- t - 1726 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -51 m - 36 m n + 19 n = --- 204 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -57 m + 36 m n + 17 n , Y = -45 m + 6 m n - 21 n , 2 2 Z = 63 m - 6 m n + 15 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 11290733 t + 14207654 t - 19 ) a(i) t = ----------------------------- / 2 ----- (t - 1) (t - 2979074 t + 1) i = 0 infinity ----- 2 \ i 3 (4072853 t - 1957786 t + 5) ) b(i) t = - ------------------------------ / 2 ----- (t - 1) (t - 2979074 t + 1) i = 0 infinity ----- 2 \ i 3 (3445591 t - 5560670 t + 7) ) c(i) t = ------------------------------ / 2 ----- (t - 1) (t - 2979074 t + 1) i = 0 In Maple notation, these generating functions are (11290733*t^2+14207654*t-19)/(t-1)/(t^2-2979074*t+1) -3*(4072853*t^2-1957786*t+5)/(t-1)/(t^2-2979074*t+1) 3*(3445591*t^2-5560670*t+7)/(t-1)/(t^2-2979074*t+1) Then for all i>=0 we have 3 3 3 a(i) + 2 b(i) + 2 c(i) = -4913 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 2 3 3 3 Infinitely many solutions to the Diophantine Equation, X + 3 Y + 3 Z = -512 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 X + 3 Y + 3 Z = -512 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 W + X + 3 Y + 3 Z = 0 and get the following special case W = -19, X = -8, Y = -15, Z = 18 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 8 x, Y = n y - 15 x, Z = -n y + 18 x Plugging into the euqation 3 3 3 3 W + X + 3 Y + 3 Z = 0 we get 2 2 -27 x y (3 m y - n y - 33 m x + 33 n x) = 0 dividing by , x y, we have 2 2 (891 m - 891 n) x + (-81 m + 27 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 81 m - 27 n , y = 891 m - 891 n Plugging them back, we get the following parametric solution to , 3 3 3 3 W + X + 3 Y + 3 Z = 0 2 2 2 2 W = -24 m - 33 m n + 19 n , X = -57 m + 33 m n + 8 n , 2 2 2 2 Y = -45 m + 33 m n - 18 n , Z = 54 m - 33 m n + 15 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 755 t ) d(i) t = -------------------- / 2 ----- 48 (t - 3886 t + 1) i = 0 infinity ----- \ i 36 t ) e(i) t = --------------- / 2 ----- t - 3886 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -24 m - 33 m n + 19 n = -- 96 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -57 m + 33 m n + 8 n , Y = -45 m + 33 m n - 18 n , 2 2 Z = 54 m - 33 m n + 15 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 11483189 t + 49107422 t - 19 ) a(i) t = ----------------------------- / 2 ----- (t - 1) (t - 15100994 t + 1) i = 0 infinity ----- 2 \ i 3 (4038413 t - 679186 t + 5) ) b(i) t = - ----------------------------- / 2 ----- (t - 1) (t - 15100994 t + 1) i = 0 infinity ----- 2 \ i 18 (602185 t - 1162058 t + 1) ) c(i) t = ------------------------------ / 2 ----- (t - 1) (t - 15100994 t + 1) i = 0 In Maple notation, these generating functions are (11483189*t^2+49107422*t-19)/(t-1)/(t^2-15100994*t+1) -3*(4038413*t^2-679186*t+5)/(t-1)/(t^2-15100994*t+1) 18*(602185*t^2-1162058*t+1)/(t-1)/(t^2-15100994*t+1) Then for all i>=0 we have 3 3 3 a(i) + 3 b(i) + 3 c(i) = -512 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 3 3 3 3 Infinitely many solutions to the Diophantine Equation, X + 4 Y + 4 Z = -3375 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 X + 4 Y + 4 Z = -3375 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 W + X + 4 Y + 4 Z = 0 and get the following special case W = -17, X = -15, Y = -5, Z = 13 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 17 x, X = -m y - 15 x, Y = n y - 5 x, Z = -n y + 13 x Plugging into the euqation 3 3 3 3 W + X + 4 Y + 4 Z = 0 we get 2 2 -96 x y (m y - n y - 2 m x + 18 n x) = 0 dividing by , x y, we have (192 m - 1728 n) x - 96 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 96 (m - n) (m + n), y = 192 m - 1728 n Plugging them back, we get the following parametric solution to , 3 3 3 3 W + X + 4 Y + 4 Z = 0 2 2 2 2 W = -15 m - 18 m n + 17 n , X = -17 m + 18 m n + 15 n , 2 2 2 2 Y = -5 m + 2 m n - 13 n , Z = 13 m - 2 m n + 5 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 3079 t ) d(i) t = --------------------- / 2 ----- 30 (t - 12098 t + 1) i = 0 infinity ----- \ i 165 t ) e(i) t = ---------------- / 2 ----- t - 12098 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -15 m - 18 m n + 17 n = -- 60 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -17 m + 18 m n + 15 n , Y = -5 m + 2 m n - 13 n , 2 2 Z = 13 m - 2 m n + 5 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 480712303 t + 669271714 t - 17 ) a(i) t = ------------------------------- / 2 ----- (t - 1) (t - 146361602 t + 1) i = 0 infinity ----- 2 \ i 5 (67090321 t + 16544878 t + 1) ) b(i) t = - -------------------------------- / 2 ----- (t - 1) (t - 146361602 t + 1) i = 0 infinity ----- 2 \ i 215273533 t - 633449546 t + 13 ) c(i) t = ------------------------------- / 2 ----- (t - 1) (t - 146361602 t + 1) i = 0 In Maple notation, these generating functions are (480712303*t^2+669271714*t-17)/(t-1)/(t^2-146361602*t+1) -5*(67090321*t^2+16544878*t+1)/(t-1)/(t^2-146361602*t+1) (215273533*t^2-633449546*t+13)/(t-1)/(t^2-146361602*t+1) Then for all i>=0 we have 3 3 3 a(i) + 4 b(i) + 4 c(i) = -3375 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 4 3 3 3 Infinitely many solutions to the Diophantine Equation, X + 5 Y + 5 Z = -27 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 X + 5 Y + 5 Z = -27 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 W + X + 5 Y + 5 Z = 0 and get the following special case W = -17, X = -3, Y = -7, Z = 11 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 17 x, X = -m y - 3 x, Y = n y - 7 x, Z = -n y + 11 x Plugging into the euqation 3 3 3 3 W + X + 5 Y + 5 Z = 0 we get 2 2 -60 x y (m y - n y - 14 m x + 18 n x) = 0 dividing by , x y, we have (840 m - 1080 n) x - 60 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 60 (m - n) (m + n), y = 840 m - 1080 n Plugging them back, we get the following parametric solution to , 3 3 3 3 W + X + 5 Y + 5 Z = 0 2 2 2 2 W = -3 m - 18 m n + 17 n , X = -17 m + 18 m n + 3 n , 2 2 2 2 Y = -7 m + 14 m n - 11 n , Z = 11 m - 14 m n + 7 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 5 t ) d(i) t = ----------------- / 2 ----- 6 (t - 46 t + 1) i = 0 infinity ----- \ i t ) e(i) t = ------------- / 2 ----- t - 46 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -3 m - 18 m n + 17 n = -- 12 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -17 m + 18 m n + 3 n , Y = -7 m + 14 m n - 11 n , 2 2 Z = 11 m - 14 m n + 7 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 223 t + 3058 t - 17 ) a(i) t = ------------------------- / 2 ----- (t - 1) (t - 2114 t + 1) i = 0 infinity ----- 2 \ i 151 t + 802 t + 7 ) b(i) t = - ------------------------- / 2 ----- (t - 1) (t - 2114 t + 1) i = 0 infinity ----- 2 \ i 107 t - 1078 t + 11 ) c(i) t = ------------------------- / 2 ----- (t - 1) (t - 2114 t + 1) i = 0 In Maple notation, these generating functions are (223*t^2+3058*t-17)/(t-1)/(t^2-2114*t+1) -(151*t^2+802*t+7)/(t-1)/(t^2-2114*t+1) (107*t^2-1078*t+11)/(t-1)/(t^2-2114*t+1) Then for all i>=0 we have 3 3 3 a(i) + 5 b(i) + 5 c(i) = -27 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 5 3 3 3 Infinitely many solutions to the Diophantine Equation, X + 6 Y + 6 Z = -1 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 X + 6 Y + 6 Z = -1 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 W + X + 6 Y + 6 Z = 0 and get the following special case W = -17, X = -1, Y = -8, Z = 11 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 17 x, X = -m y - x, Y = n y - 8 x, Z = -n y + 11 x Plugging into the euqation 3 3 3 3 W + X + 6 Y + 6 Z = 0 we get 2 2 -54 x y (m y - n y - 16 m x + 19 n x) = 0 dividing by , x y, we have (864 m - 1026 n) x - 54 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 54 (m - n) (m + n), y = 864 m - 1026 n Plugging them back, we get the following parametric solution to , 3 3 3 3 W + X + 6 Y + 6 Z = 0 2 2 2 2 2 2 W = -m - 19 m n + 17 n , X = -17 m + 19 m n + n , Y = -8 m + 16 m n - 11 n , 2 2 Z = 11 m - 16 m n + 8 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 125999 t ) d(i) t = ---------------------- / 2 ----- 2 (t - 3048190 t + 1) i = 0 infinity ----- \ i 73584 t ) e(i) t = ------------------ / 2 ----- t - 3048190 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -m - 19 m n + 17 n = -1/4 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 2 2 X = -17 m + 19 m n + n , Y = -8 m + 16 m n - 11 n , Z = 11 m - 16 m n + 8 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 104088100015 t + 4639105929058 t - 17 ) a(i) t = -------------------------------------- / 2 ----- (t - 1) (t - 9291462276098 t + 1) i = 0 infinity ----- 2 \ i 8 (8570034145 t + 137624302366 t + 1) ) b(i) t = - -------------------------------------- / 2 ----- (t - 1) (t - 9291462276098 t + 1) i = 0 infinity ----- 2 \ i 51212256491 t - 1220766948598 t + 11 ) c(i) t = ------------------------------------- / 2 ----- (t - 1) (t - 9291462276098 t + 1) i = 0 In Maple notation, these generating functions are (104088100015*t^2+4639105929058*t-17)/(t-1)/(t^2-9291462276098*t+1) -8*(8570034145*t^2+137624302366*t+1)/(t-1)/(t^2-9291462276098*t+1) (51212256491*t^2-1220766948598*t+11)/(t-1)/(t^2-9291462276098*t+1) Then for all i>=0 we have 3 3 3 a(i) + 6 b(i) + 6 c(i) = -1 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 6 3 3 3 Infinitely many solutions to the Diophantine Equation, X + 7 Y + 7 Z = -1 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 X + 7 Y + 7 Z = -1 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 W + X + 7 Y + 7 Z = 0 and get the following special case W = -17, X = -1, Y = -3, Z = 9 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 17 x, X = -m y - x, Y = n y - 3 x, Z = -n y + 9 x Plugging into the euqation 3 3 3 3 W + X + 7 Y + 7 Z = 0 we get 2 2 -18 x y (3 m y - 7 n y - 48 m x + 84 n x) = 0 dividing by , x y, we have 2 2 (864 m - 1512 n) x + (-54 m + 126 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 54 m - 126 n , y = 864 m - 1512 n Plugging them back, we get the following parametric solution to , 3 3 3 3 W + X + 7 Y + 7 Z = 0 2 2 2 2 W = -3 m - 84 m n + 119 n , X = -51 m + 84 m n + 7 n , 2 2 2 2 Y = -9 m + 48 m n - 63 n , Z = 27 m - 48 m n + 21 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 83367976481 t ) d(i) t = ---------------------------- / 2 ----- 6 (t - 1894008923602 t + 1) i = 0 infinity ----- \ i 10281386730 t ) e(i) t = ------------------------ / 2 ----- t - 1894008923602 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -3 m - 84 m n + 119 n = -- 12 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -51 m + 84 m n + 7 n , Y = -9 m + 48 m n - 63 n , 2 2 Z = 27 m - 48 m n + 21 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- \ i ) a(i) t = / ----- i = 0 2 34724901549173735884303 t + 2131843989180770888194114 t - 17 ------------------------------------------------------------- 2 (t - 1) (t - 3587269802684006672654402 t + 1) infinity ----- \ i ) b(i) t = / ----- i = 0 2 3 (6159932574516656646001 t + 313497772615147304283598 t + 1) - -------------------------------------------------------------- 2 (t - 1) (t - 3587269802684006672654402 t + 1) infinity ----- \ i ) c(i) t = / ----- i = 0 2 9 (767203393388006441841 t - 107319771789942660085042 t + 1) ------------------------------------------------------------- 2 (t - 1) (t - 3587269802684006672654402 t + 1) In Maple notation, these generating functions are (34724901549173735884303*t^2+2131843989180770888194114*t-17)/(t-1)/(t^2-\ 3587269802684006672654402*t+1) -3*(6159932574516656646001*t^2+313497772615147304283598*t+1)/(t-1)/(t^2-\ 3587269802684006672654402*t+1) 9*(767203393388006441841*t^2-107319771789942660085042*t+1)/(t-1)/(t^2-\ 3587269802684006672654402*t+1) Then for all i>=0 we have 3 3 3 a(i) + 7 b(i) + 7 c(i) = -1 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 7 3 3 3 Infinitely many solutions to the Diophantine Equation, X + 9 Y + 9 Z = -512 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 X + 9 Y + 9 Z = -512 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 W + X + 9 Y + 9 Z = 0 and get the following special case W = -19, X = -8, Y = -8, Z = 11 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 8 x, Y = n y - 8 x, Z = -n y + 11 x Plugging into the euqation 3 3 3 3 W + X + 9 Y + 9 Z = 0 we get 2 2 -81 x y (m y - n y - 11 m x + 19 n x) = 0 dividing by , x y, we have (891 m - 1539 n) x - 81 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 81 (m - n) (m + n), y = 891 m - 1539 n Plugging them back, we get the following parametric solution to , 3 3 3 3 W + X + 9 Y + 9 Z = 0 2 2 2 2 W = -8 m - 19 m n + 19 n , X = -19 m + 19 m n + 8 n , 2 2 2 2 Y = -8 m + 11 m n - 11 n , Z = 11 m - 11 m n + 8 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 5294691 t ) d(i) t = ------------------------ / 2 ----- 16 (t - 27177902 t + 1) i = 0 infinity ----- \ i 436540 t ) e(i) t = ------------------- / 2 ----- t - 27177902 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -8 m - 19 m n + 19 n = -- 32 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -19 m + 19 m n + 8 n , Y = -8 m + 11 m n - 11 n , 2 2 Z = 11 m - 11 m n + 8 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 560288964891221 t + 2421706136305598 t - 19 ) a(i) t = -------------------------------------------- / 2 ----- (t - 1) (t - 738638357121602 t + 1) i = 0 infinity ----- 2 \ i 8 (44263820187601 t + 17479943410798 t + 1) ) b(i) t = - -------------------------------------------- / 2 ----- (t - 1) (t - 738638357121602 t + 1) i = 0 infinity ----- 2 \ i 291856232068451 t - 785806340855662 t + 11 ) c(i) t = ------------------------------------------- / 2 ----- (t - 1) (t - 738638357121602 t + 1) i = 0 In Maple notation, these generating functions are (560288964891221*t^2+2421706136305598*t-19)/(t-1)/(t^2-738638357121602*t+1) -8*(44263820187601*t^2+17479943410798*t+1)/(t-1)/(t^2-738638357121602*t+1) (291856232068451*t^2-785806340855662*t+11)/(t-1)/(t^2-738638357121602*t+1) Then for all i>=0 we have 3 3 3 a(i) + 9 b(i) + 9 c(i) = -512 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 8 Infinitely many solutions to the Diophantine Equation, 3 3 3 X + 10 Y + 10 Z = -1331 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 X + 10 Y + 10 Z = -1331 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 W + X + 10 Y + 10 Z = 0 and get the following special case W = -19, X = -11, Y = -8, Z = 11 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 11 x, Y = n y - 8 x, Z = -n y + 11 x Plugging into the euqation 3 3 3 3 W + X + 10 Y + 10 Z = 0 we get 2 2 -90 x y (m y - n y - 8 m x + 19 n x) = 0 dividing by , x y, we have (720 m - 1710 n) x - 90 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 90 (m - n) (m + n), y = 720 m - 1710 n Plugging them back, we get the following parametric solution to , 3 3 3 3 W + X + 10 Y + 10 Z = 0 2 2 2 2 W = -11 m - 19 m n + 19 n , X = -19 m + 19 m n + 11 n , 2 2 2 2 Y = -8 m + 8 m n - 11 n , Z = 11 m - 8 m n + 8 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 1167019 t ) d(i) t = ----------------------- / 2 ----- 22 (t - 5177198 t + 1) i = 0 infinity ----- \ i 74820 t ) e(i) t = ------------------ / 2 ----- t - 5177198 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -11 m - 19 m n + 19 n = -- 44 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -19 m + 19 m n + 11 n , Y = -8 m + 8 m n - 11 n , 2 2 Z = 11 m - 8 m n + 8 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 40425430057181 t + 108101565579638 t - 19 ) a(i) t = ------------------------------------------ / 2 ----- (t - 1) (t - 26803379131202 t + 1) i = 0 infinity ----- 2 \ i 8 (3166463953801 t - 395437915802 t + 1) ) b(i) t = - ----------------------------------------- / 2 ----- (t - 1) (t - 26803379131202 t + 1) i = 0 infinity ----- 2 \ i 11 (1935378965881 t - 3950670629882 t + 1) ) c(i) t = ------------------------------------------- / 2 ----- (t - 1) (t - 26803379131202 t + 1) i = 0 In Maple notation, these generating functions are (40425430057181*t^2+108101565579638*t-19)/(t-1)/(t^2-26803379131202*t+1) -8*(3166463953801*t^2-395437915802*t+1)/(t-1)/(t^2-26803379131202*t+1) 11*(1935378965881*t^2-3950670629882*t+1)/(t-1)/(t^2-26803379131202*t+1) Then for all i>=0 we have 3 3 3 a(i) + 10 b(i) + 10 c(i) = -1331 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 9 Infinitely many solutions to the Diophantine Equation, 3 3 3 2 X + 3 Y + 3 Z = -1458 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 2 X + 3 Y + 3 Z = -1458 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 2 W + 2 X + 3 Y + 3 Z = 0 and get the following special case W = -15, X = -9, Y = -2, Z = 14 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 15 x, X = -m y - 9 x, Y = n y - 2 x, Z = -n y + 14 x Plugging into the euqation 3 3 3 3 2 W + 2 X + 3 Y + 3 Z = 0 we get 2 2 -36 x y (4 m y - 3 n y - 24 m x + 48 n x) = 0 dividing by , x y, we have 2 2 (864 m - 1728 n) x + (-144 m + 108 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 144 m - 108 n , y = 864 m - 1728 n Plugging them back, we get the following parametric solution to , 3 3 3 3 2 W + 2 X + 3 Y + 3 Z = 0 2 2 2 2 W = -36 m - 48 m n + 45 n , X = -60 m + 48 m n + 27 n , 2 2 2 2 Y = -8 m + 24 m n - 42 n , Z = 56 m - 24 m n + 6 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 861703129 t ) d(i) t = -------------------------- / 2 ----- 72 (t - 3532638098 t + 1) i = 0 infinity ----- \ i 18846165 t ) e(i) t = --------------------- / 2 ----- t - 3532638098 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -36 m - 48 m n + 45 n = --- 144 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -60 m + 48 m n + 27 n , Y = -8 m + 24 m n - 42 n , 2 2 Z = 56 m - 24 m n + 6 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 15 (1021439247291680255 t + 3274793056974913346 t - 1) ) a(i) t = ------------------------------------------------------- / 2 ----- (t - 1) (t - 12479531931441057602 t + 1) i = 0 infinity ----- 2 \ i 2 (6901258307495764681 t + 15193650685875288118 t + 1) ) b(i) t = - ------------------------------------------------------- / 2 ----- (t - 1) (t - 12479531931441057602 t + 1) i = 0 infinity ----- 2 \ i 2 (3070861130151963727 t - 25165770123523016534 t + 7) ) c(i) t = ------------------------------------------------------- / 2 ----- (t - 1) (t - 12479531931441057602 t + 1) i = 0 In Maple notation, these generating functions are 15*(1021439247291680255*t^2+3274793056974913346*t-1)/(t-1)/(t^2-\ 12479531931441057602*t+1) -2*(6901258307495764681*t^2+15193650685875288118*t+1)/(t-1)/(t^2-\ 12479531931441057602*t+1) 2*(3070861130151963727*t^2-25165770123523016534*t+7)/(t-1)/(t^2-\ 12479531931441057602*t+1) Then for all i>=0 we have 3 3 3 2 a(i) + 3 b(i) + 3 c(i) = -1458 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 10 Infinitely many solutions to the Diophantine Equation, 3 3 3 2 X + 5 Y + 5 Z = -1024 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 2 X + 5 Y + 5 Z = -1024 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 2 W + 2 X + 5 Y + 5 Z = 0 and get the following special case W = -17, X = -8, Y = -3, Z = 13 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 17 x, X = -m y - 8 x, Y = n y - 3 x, Z = -n y + 13 x Plugging into the euqation 3 3 3 3 2 W + 2 X + 5 Y + 5 Z = 0 we get 2 2 -150 x y (m y - n y - 9 m x + 16 n x) = 0 dividing by , x y, we have (1350 m - 2400 n) x - 150 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 150 (m - n) (m + n), y = 1350 m - 2400 n Plugging them back, we get the following parametric solution to , 3 3 3 3 2 W + 2 X + 5 Y + 5 Z = 0 2 2 2 2 W = -8 m - 16 m n + 17 n , X = -17 m + 16 m n + 8 n , 2 2 2 2 Y = -3 m + 9 m n - 13 n , Z = 13 m - 9 m n + 3 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 8513 t ) d(i) t = --------------------- / 2 ----- 16 (t - 39202 t + 1) i = 0 infinity ----- \ i 693 t ) e(i) t = ---------------- / 2 ----- t - 39202 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -8 m - 16 m n + 17 n = -- 32 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 2 2 X = -17 m + 16 m n + 8 n , Y = -3 m + 9 m n - 13 n , Z = 13 m - 9 m n + 3 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 1261814383 t + 5653771234 t - 17 ) a(i) t = --------------------------------- / 2 ----- (t - 1) (t - 1536796802 t + 1) i = 0 infinity ----- 2 \ i 3 (322050961 t + 958613038 t + 1) ) b(i) t = - ---------------------------------- / 2 ----- (t - 1) (t - 1536796802 t + 1) i = 0 infinity ----- 2 \ i 461427133 t - 4303419146 t + 13 ) c(i) t = -------------------------------- / 2 ----- (t - 1) (t - 1536796802 t + 1) i = 0 In Maple notation, these generating functions are (1261814383*t^2+5653771234*t-17)/(t-1)/(t^2-1536796802*t+1) -3*(322050961*t^2+958613038*t+1)/(t-1)/(t^2-1536796802*t+1) (461427133*t^2-4303419146*t+13)/(t-1)/(t^2-1536796802*t+1) Then for all i>=0 we have 3 3 3 2 a(i) + 5 b(i) + 5 c(i) = -1024 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 11 Infinitely many solutions to the Diophantine Equation, 3 3 3 2 X + 7 Y + 7 Z = -1458 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 2 X + 7 Y + 7 Z = -1458 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 2 W + 2 X + 7 Y + 7 Z = 0 and get the following special case W = -19, X = -9, Y = -18, Z = 20 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 9 x, Y = n y - 18 x, Z = -n y + 20 x Plugging into the euqation 3 3 3 3 2 W + 2 X + 7 Y + 7 Z = 0 we get 2 2 -42 x y (4 m y - n y - 40 m x + 38 n x) = 0 dividing by , x y, we have (1680 m - 1596 n) x - 42 (2 m - n) (2 m + n) y = 0 Obviously the following values of x and y will make it zero x = 42 (2 m - n) (2 m + n), y = 1680 m - 1596 n Plugging them back, we get the following parametric solution to , 3 3 3 3 2 W + 2 X + 7 Y + 7 Z = 0 2 2 2 2 W = -36 m - 38 m n + 19 n , X = -76 m + 38 m n + 9 n , 2 2 2 2 Y = -72 m + 40 m n - 20 n , Z = 80 m - 40 m n + 18 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 188063 t ) d(i) t = ---------------------- / 2 ----- 72 (t - 912382 t + 1) i = 0 infinity ----- \ i 7056 t ) e(i) t = ----------------- / 2 ----- t - 912382 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -36 m - 38 m n + 19 n = --- 144 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -76 m + 38 m n + 9 n , Y = -72 m + 40 m n - 20 n , 2 2 Z = 80 m - 40 m n + 18 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 816380216045 t + 2933188570406 t - 19 ) a(i) t = -------------------------------------- / 2 ----- (t - 1) (t - 832440913922 t + 1) i = 0 infinity ----- 2 \ i 18 (53982266689 t - 31677629762 t + 1) ) b(i) t = - --------------------------------------- / 2 ----- (t - 1) (t - 832440913922 t + 1) i = 0 infinity ----- 2 \ i 4 (228341981957 t - 328712848138 t + 5) ) c(i) t = ---------------------------------------- / 2 ----- (t - 1) (t - 832440913922 t + 1) i = 0 In Maple notation, these generating functions are (816380216045*t^2+2933188570406*t-19)/(t-1)/(t^2-832440913922*t+1) -18*(53982266689*t^2-31677629762*t+1)/(t-1)/(t^2-832440913922*t+1) 4*(228341981957*t^2-328712848138*t+5)/(t-1)/(t^2-832440913922*t+1) Then for all i>=0 we have 3 3 3 2 a(i) + 7 b(i) + 7 c(i) = -1458 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 12 Infinitely many solutions to the Diophantine Equation, 3 3 3 2 X + 9 Y + 9 Z = -686 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 2 X + 9 Y + 9 Z = -686 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 2 W + 2 X + 9 Y + 9 Z = 0 and get the following special case W = -20, X = -7, Y = -7, Z = 13 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 20 x, X = -m y - 7 x, Y = n y - 7 x, Z = -n y + 13 x Plugging into the euqation 3 3 3 3 2 W + 2 X + 9 Y + 9 Z = 0 we get 2 2 -162 x y (m y - n y - 13 m x + 20 n x) = 0 dividing by , x y, we have (2106 m - 3240 n) x - 162 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 162 (m - n) (m + n), y = 2106 m - 3240 n Plugging them back, we get the following parametric solution to , 3 3 3 3 2 W + 2 X + 9 Y + 9 Z = 0 2 2 2 2 W = -7 m - 20 m n + 20 n , X = -20 m + 20 m n + 7 n , 2 2 2 2 Y = -7 m + 13 m n - 13 n , Z = 13 m - 13 m n + 7 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 11 t ) d(i) t = ------------------ / 2 ----- 14 (t - 62 t + 1) i = 0 infinity ----- \ i t ) e(i) t = ------------- / 2 ----- t - 62 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -7 m - 20 m n + 20 n = -- 28 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -20 m + 20 m n + 7 n , Y = -7 m + 13 m n - 13 n , 2 2 Z = 13 m - 13 m n + 7 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 4 (508 t + 2983 t - 5) ) a(i) t = ------------------------- / 2 ----- (t - 1) (t - 3842 t + 1) i = 0 infinity ----- 2 \ i 7 (199 t + 448 t + 1) ) b(i) t = - ------------------------- / 2 ----- (t - 1) (t - 3842 t + 1) i = 0 infinity ----- 2 \ i 943 t - 5492 t + 13 ) c(i) t = ------------------------- / 2 ----- (t - 1) (t - 3842 t + 1) i = 0 In Maple notation, these generating functions are 4*(508*t^2+2983*t-5)/(t-1)/(t^2-3842*t+1) -7*(199*t^2+448*t+1)/(t-1)/(t^2-3842*t+1) (943*t^2-5492*t+13)/(t-1)/(t^2-3842*t+1) Then for all i>=0 we have 3 3 3 2 a(i) + 9 b(i) + 9 c(i) = -686 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 13 Infinitely many solutions to the Diophantine Equation, 3 3 3 3 X + 4 Y + 4 Z = -3993 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 3 X + 4 Y + 4 Z = -3993 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 3 W + 3 X + 4 Y + 4 Z = 0 and get the following special case W = -13, X = -11, Y = -9, Z = 15 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 13 x, X = -m y - 11 x, Y = n y - 9 x, Z = -n y + 15 x Plugging into the euqation 3 3 3 3 3 W + 3 X + 4 Y + 4 Z = 0 we get 2 2 -72 x y (3 m y - n y - 6 m x + 24 n x) = 0 dividing by , x y, we have 2 2 (432 m - 1728 n) x + (-216 m + 72 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 216 m - 72 n , y = 432 m - 1728 n Plugging them back, we get the following parametric solution to , 3 3 3 3 3 W + 3 X + 4 Y + 4 Z = 0 2 2 2 2 W = -33 m - 24 m n + 13 n , X = -39 m + 24 m n + 11 n , 2 2 2 2 Y = -27 m + 6 m n - 15 n , Z = 45 m - 6 m n + 9 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 191 t ) d(i) t = ------------------- / 2 ----- 66 (t - 766 t + 1) i = 0 infinity ----- \ i 8 t ) e(i) t = -------------- / 2 ----- t - 766 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -33 m - 24 m n + 13 n = --- 132 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -39 m + 24 m n + 11 n , Y = -27 m + 6 m n - 15 n , 2 2 Z = 45 m - 6 m n + 9 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 1354739 t + 1923098 t - 13 ) a(i) t = --------------------------- / 2 ----- (t - 1) (t - 586754 t + 1) i = 0 infinity ----- 2 \ i 3 (506851 t - 101350 t + 3) ) b(i) t = - ---------------------------- / 2 ----- (t - 1) (t - 586754 t + 1) i = 0 infinity ----- 2 \ i 3 (393957 t - 799466 t + 5) ) c(i) t = ---------------------------- / 2 ----- (t - 1) (t - 586754 t + 1) i = 0 In Maple notation, these generating functions are (1354739*t^2+1923098*t-13)/(t-1)/(t^2-586754*t+1) -3*(506851*t^2-101350*t+3)/(t-1)/(t^2-586754*t+1) 3*(393957*t^2-799466*t+5)/(t-1)/(t^2-586754*t+1) Then for all i>=0 we have 3 3 3 3 a(i) + 4 b(i) + 4 c(i) = -3993 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 14 3 3 3 Infinitely many solutions to the Diophantine Equation, 3 X + 5 Y + 5 Z = -3 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 3 X + 5 Y + 5 Z = -3 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 3 W + 3 X + 5 Y + 5 Z = 0 and get the following special case W = -14, X = -1, Y = -12, Z = 15 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 14 x, X = -m y - x, Y = n y - 12 x, Z = -n y + 15 x Plugging into the euqation 3 3 3 3 3 W + 3 X + 5 Y + 5 Z = 0 we get 2 2 -45 x y (3 m y - n y - 39 m x + 27 n x) = 0 dividing by , x y, we have 2 2 (1755 m - 1215 n) x + (-135 m + 45 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 135 m - 45 n , y = 1755 m - 1215 n Plugging them back, we get the following parametric solution to , 3 3 3 3 3 W + 3 X + 5 Y + 5 Z = 0 2 2 2 2 W = -3 m - 27 m n + 14 n , X = -42 m + 27 m n + n , 2 2 2 2 Y = -36 m + 39 m n - 15 n , Z = 45 m - 39 m n + 12 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 59 t ) d(i) t = ------------------- / 2 ----- 6 (t - 1198 t + 1) i = 0 infinity ----- \ i 20 t ) e(i) t = --------------- / 2 ----- t - 1198 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -3 m - 27 m n + 14 n = -- 12 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -42 m + 27 m n + n , Y = -36 m + 39 m n - 15 n , 2 2 Z = 45 m - 39 m n + 12 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 2 (9893 t + 352514 t - 7) ) a(i) t = ---------------------------- / 2 ----- (t - 1) (t - 1435202 t + 1) i = 0 infinity ----- 2 \ i 12 (1811 t + 16188 t + 1) ) b(i) t = - ---------------------------- / 2 ----- (t - 1) (t - 1435202 t + 1) i = 0 infinity ----- 2 \ i 15 (1185 t - 15586 t + 1) ) c(i) t = ---------------------------- / 2 ----- (t - 1) (t - 1435202 t + 1) i = 0 In Maple notation, these generating functions are 2*(9893*t^2+352514*t-7)/(t-1)/(t^2-1435202*t+1) -12*(1811*t^2+16188*t+1)/(t-1)/(t^2-1435202*t+1) 15*(1185*t^2-15586*t+1)/(t-1)/(t^2-1435202*t+1) Then for all i>=0 we have 3 3 3 3 a(i) + 5 b(i) + 5 c(i) = -3 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 15 3 3 3 Infinitely many solutions to the Diophantine Equation, 3 X + 7 Y + 7 Z = -3 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 3 X + 7 Y + 7 Z = -3 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 3 W + 3 X + 7 Y + 7 Z = 0 and get the following special case W = -20, X = -1, Y = -18, Z = 21 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 20 x, X = -m y - x, Y = n y - 18 x, Z = -n y + 21 x Plugging into the euqation 3 3 3 3 3 W + 3 X + 7 Y + 7 Z = 0 we get 2 2 -63 x y (3 m y - n y - 57 m x + 39 n x) = 0 dividing by , x y, we have 2 2 (3591 m - 2457 n) x + (-189 m + 63 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 189 m - 63 n , y = 3591 m - 2457 n Plugging them back, we get the following parametric solution to , 3 3 3 3 3 W + 3 X + 7 Y + 7 Z = 0 2 2 2 2 W = -3 m - 39 m n + 20 n , X = -60 m + 39 m n + n , 2 2 2 2 Y = -54 m + 57 m n - 21 n , Z = 63 m - 57 m n + 18 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 83 t ) d(i) t = ------------------- / 2 ----- 6 (t - 2350 t + 1) i = 0 infinity ----- \ i 28 t ) e(i) t = --------------- / 2 ----- t - 2350 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -3 m - 39 m n + 20 n = -- 12 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -60 m + 39 m n + n , Y = -54 m + 57 m n - 21 n , 2 2 Z = 63 m - 57 m n + 18 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 20 (2645 t + 136124 t - 1) ) a(i) t = ---------------------------- / 2 ----- (t - 1) (t - 5522498 t + 1) i = 0 infinity ----- 2 \ i 6 (9439 t + 89342 t + 3) ) b(i) t = - ---------------------------- / 2 ----- (t - 1) (t - 5522498 t + 1) i = 0 infinity ----- 2 \ i 21 (2337 t - 30562 t + 1) ) c(i) t = ---------------------------- / 2 ----- (t - 1) (t - 5522498 t + 1) i = 0 In Maple notation, these generating functions are 20*(2645*t^2+136124*t-1)/(t-1)/(t^2-5522498*t+1) -6*(9439*t^2+89342*t+3)/(t-1)/(t^2-5522498*t+1) 21*(2337*t^2-30562*t+1)/(t-1)/(t^2-5522498*t+1) Then for all i>=0 we have 3 3 3 3 a(i) + 7 b(i) + 7 c(i) = -3 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 16 Infinitely many solutions to the Diophantine Equation, 3 3 3 3 X + 8 Y + 8 Z = -375 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 3 X + 8 Y + 8 Z = -375 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 3 W + 3 X + 8 Y + 8 Z = 0 and get the following special case W = -19, X = -5, Y = -5, Z = 14 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 5 x, Y = n y - 5 x, Z = -n y + 14 x Plugging into the euqation 3 3 3 3 3 W + 3 X + 8 Y + 8 Z = 0 we get 2 2 -216 x y (m y - n y - 14 m x + 19 n x) = 0 dividing by , x y, we have (3024 m - 4104 n) x - 216 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 216 (m - n) (m + n), y = 3024 m - 4104 n Plugging them back, we get the following parametric solution to , 3 3 3 3 3 W + 3 X + 8 Y + 8 Z = 0 2 2 2 2 W = -5 m - 19 m n + 19 n , X = -19 m + 19 m n + 5 n , 2 2 2 2 Y = -5 m + 14 m n - 14 n , Z = 14 m - 14 m n + 5 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 2220643 t ) d(i) t = ------------------------ / 2 ----- 10 (t - 14705390 t + 1) i = 0 infinity ----- \ i 270108 t ) e(i) t = ------------------- / 2 ----- t - 14705390 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -5 m - 19 m n + 19 n = -- 20 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -19 m + 19 m n + 5 n , Y = -5 m + 14 m n - 14 n , 2 2 Z = 14 m - 14 m n + 5 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 56749867990829 t + 542967618287270 t - 19 ) a(i) t = ------------------------------------------ / 2 ----- (t - 1) (t - 216248495052098 t + 1) i = 0 infinity ----- 2 \ i 5 (8564811894937 t + 54471186662758 t + 1) ) b(i) t = - ------------------------------------------- / 2 ----- (t - 1) (t - 216248495052098 t + 1) i = 0 infinity ----- 2 \ i 2 (10771429489063 t - 168361425883310 t + 7) ) c(i) t = --------------------------------------------- / 2 ----- (t - 1) (t - 216248495052098 t + 1) i = 0 In Maple notation, these generating functions are (56749867990829*t^2+542967618287270*t-19)/(t-1)/(t^2-216248495052098*t+1) -5*(8564811894937*t^2+54471186662758*t+1)/(t-1)/(t^2-216248495052098*t+1) 2*(10771429489063*t^2-168361425883310*t+7)/(t-1)/(t^2-216248495052098*t+1) Then for all i>=0 we have 3 3 3 3 a(i) + 8 b(i) + 8 c(i) = -375 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 17 Infinitely many solutions to the Diophantine Equation, 3 3 3 3 X + 10 Y + 10 Z = -3993 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 3 X + 10 Y + 10 Z = -3993 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 3 W + 3 X + 10 Y + 10 Z = 0 and get the following special case W = -19, X = -11, Y = -15, Z = 18 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 11 x, Y = n y - 15 x, Z = -n y + 18 x Plugging into the euqation 3 3 3 3 3 W + 3 X + 10 Y + 10 Z = 0 we get 2 2 -90 x y (3 m y - n y - 24 m x + 33 n x) = 0 dividing by , x y, we have 2 2 (2160 m - 2970 n) x + (-270 m + 90 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 270 m - 90 n , y = 2160 m - 2970 n Plugging them back, we get the following parametric solution to , 3 3 3 3 3 W + 3 X + 10 Y + 10 Z = 0 2 2 2 2 W = -33 m - 33 m n + 19 n , X = -57 m + 33 m n + 11 n , 2 2 2 2 Y = -45 m + 24 m n - 18 n , Z = 54 m - 24 m n + 15 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 1079 t ) d(i) t = -------------------- / 2 ----- 66 (t - 4798 t + 1) i = 0 infinity ----- \ i 40 t ) e(i) t = --------------- / 2 ----- t - 4798 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -33 m - 33 m n + 19 n = --- 132 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -57 m + 33 m n + 11 n , Y = -45 m + 24 m n - 18 n , 2 2 Z = 54 m - 24 m n + 15 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 34768781 t + 92162438 t - 19 ) a(i) t = ----------------------------- / 2 ----- (t - 1) (t - 23020802 t + 1) i = 0 infinity ----- 2 \ i 15 (2432849 t - 1165650 t + 1) ) b(i) t = - ------------------------------- / 2 ----- (t - 1) (t - 23020802 t + 1) i = 0 infinity ----- 2 \ i 6 (5502643 t - 8670646 t + 3) ) c(i) t = ------------------------------ / 2 ----- (t - 1) (t - 23020802 t + 1) i = 0 In Maple notation, these generating functions are (34768781*t^2+92162438*t-19)/(t-1)/(t^2-23020802*t+1) -15*(2432849*t^2-1165650*t+1)/(t-1)/(t^2-23020802*t+1) 6*(5502643*t^2-8670646*t+3)/(t-1)/(t^2-23020802*t+1) Then for all i>=0 we have 3 3 3 3 a(i) + 10 b(i) + 10 c(i) = -3993 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 18 Infinitely many solutions to the Diophantine Equation, 3 3 3 4 X + 5 Y + 5 Z = -5324 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 4 X + 5 Y + 5 Z = -5324 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 4 W + 4 X + 5 Y + 5 Z = 0 and get the following special case W = -19, X = -11, Y = -16, Z = 22 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 11 x, Y = n y - 16 x, Z = -n y + 22 x Plugging into the euqation 3 3 3 3 4 W + 4 X + 5 Y + 5 Z = 0 we get 2 2 -90 x y (4 m y - n y - 32 m x + 38 n x) = 0 dividing by , x y, we have (2880 m - 3420 n) x - 90 (2 m - n) (2 m + n) y = 0 Obviously the following values of x and y will make it zero x = 90 (2 m - n) (2 m + n), y = 2880 m - 3420 n Plugging them back, we get the following parametric solution to , 3 3 3 3 4 W + 4 X + 5 Y + 5 Z = 0 2 2 2 2 W = -44 m - 38 m n + 19 n , X = -76 m + 38 m n + 11 n , 2 2 2 2 Y = -64 m + 32 m n - 22 n , Z = 88 m - 32 m n + 16 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 1167019 t ) d(i) t = ----------------------- / 2 ----- 88 (t - 5177198 t + 1) i = 0 infinity ----- \ i 37410 t ) e(i) t = ------------------ / 2 ----- t - 5177198 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -44 m - 38 m n + 19 n = --- 176 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -76 m + 38 m n + 11 n , Y = -64 m + 32 m n - 22 n , 2 2 Z = 88 m - 32 m n + 16 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 40425430057181 t + 108101565579638 t - 19 ) a(i) t = ------------------------------------------ / 2 ----- (t - 1) (t - 26803379131202 t + 1) i = 0 infinity ----- 2 \ i 16 (3166463953801 t - 395437915802 t + 1) ) b(i) t = - ------------------------------------------ / 2 ----- (t - 1) (t - 26803379131202 t + 1) i = 0 infinity ----- 2 \ i 22 (1935378965881 t - 3950670629882 t + 1) ) c(i) t = ------------------------------------------- / 2 ----- (t - 1) (t - 26803379131202 t + 1) i = 0 In Maple notation, these generating functions are (40425430057181*t^2+108101565579638*t-19)/(t-1)/(t^2-26803379131202*t+1) -16*(3166463953801*t^2-395437915802*t+1)/(t-1)/(t^2-26803379131202*t+1) 22*(1935378965881*t^2-3950670629882*t+1)/(t-1)/(t^2-26803379131202*t+1) Then for all i>=0 we have 3 3 3 4 a(i) + 5 b(i) + 5 c(i) = -5324 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 19 Infinitely many solutions to the Diophantine Equation, 3 3 3 4 X + 9 Y + 9 Z = -8788 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 4 X + 9 Y + 9 Z = -8788 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 4 W + 4 X + 9 Y + 9 Z = 0 and get the following special case W = -14, X = -13, Y = -1, Z = 13 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 14 x, X = -m y - 13 x, Y = n y - x, Z = -n y + 13 x Plugging into the euqation 3 3 3 3 4 W + 4 X + 9 Y + 9 Z = 0 we get 2 2 -324 x y (m y - n y - m x + 14 n x) = 0 dividing by , x y, we have (324 m - 4536 n) x - 324 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 324 (m - n) (m + n), y = 324 m - 4536 n Plugging them back, we get the following parametric solution to , 3 3 3 3 4 W + 4 X + 9 Y + 9 Z = 0 2 2 2 2 2 2 W = -13 m - 14 m n + 14 n , X = -14 m + 14 m n + 13 n , Y = -m + m n - 13 n , 2 2 Z = 13 m - m n + n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 6231 t ) d(i) t = --------------------- / 2 ----- 26 (t - 23102 t + 1) i = 0 infinity ----- \ i 380 t ) e(i) t = ---------------- / 2 ----- t - 23102 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -13 m - 14 m n + 14 n = -- 52 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 2 2 X = -14 m + 14 m n + 13 n , Y = -m + m n - 13 n , Z = 13 m - m n + n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 2 (793652033 t + 1211197574 t - 7) ) a(i) t = ----------------------------------- / 2 ----- (t - 1) (t - 533702402 t + 1) i = 0 infinity ----- 2 \ i 1246250281 t + 1186600918 t + 1 ) b(i) t = - -------------------------------- / 2 ----- (t - 1) (t - 533702402 t + 1) i = 0 infinity ----- 2 \ i 13 (41598601 t - 228741002 t + 1) ) c(i) t = ---------------------------------- / 2 ----- (t - 1) (t - 533702402 t + 1) i = 0 In Maple notation, these generating functions are 2*(793652033*t^2+1211197574*t-7)/(t-1)/(t^2-533702402*t+1) -(1246250281*t^2+1186600918*t+1)/(t-1)/(t^2-533702402*t+1) 13*(41598601*t^2-228741002*t+1)/(t-1)/(t^2-533702402*t+1) Then for all i>=0 we have 3 3 3 4 a(i) + 9 b(i) + 9 c(i) = -8788 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 20 3 3 3 Infinitely many solutions to the Diophantine Equation, 5 X + 6 Y + 6 Z = -5 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 5 X + 6 Y + 6 Z = -5 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 5 W + 5 X + 6 Y + 6 Z = 0 and get the following special case W = -17, X = -1, Y = -1, Z = 16 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 17 x, X = -m y - x, Y = n y - x, Z = -n y + 16 x Plugging into the euqation 3 3 3 3 5 W + 5 X + 6 Y + 6 Z = 0 we get 2 2 -270 x y (m y - n y - 16 m x + 17 n x) = 0 dividing by , x y, we have (4320 m - 4590 n) x - 270 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 270 (m - n) (m + n), y = 4320 m - 4590 n Plugging them back, we get the following parametric solution to , 3 3 3 3 5 W + 5 X + 6 Y + 6 Z = 0 2 2 2 2 2 2 W = -m - 17 m n + 17 n , X = -17 m + 17 m n + n , Y = -m + 16 m n - 16 n , 2 2 Z = 16 m - 16 m n + n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 341 t ) d(i) t = ------------------- / 2 ----- 2 (t - 6802 t + 1) i = 0 infinity ----- \ i 180 t ) e(i) t = --------------- / 2 ----- t - 6802 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -m - 17 m n + 17 n = -1/4 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 2 2 X = -17 m + 17 m n + n , Y = -m + 16 m n - 16 n , Z = 16 m - 16 m n + n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 239743 t + 37473874 t - 17 ) a(i) t = ----------------------------- / 2 ----- (t - 1) (t - 46267202 t + 1) i = 0 infinity ----- 2 \ i 225721 t + 34766278 t + 1 ) b(i) t = - ----------------------------- / 2 ----- (t - 1) (t - 46267202 t + 1) i = 0 infinity ----- 2 \ i 16 (1621 t - 2188622 t + 1) ) c(i) t = ----------------------------- / 2 ----- (t - 1) (t - 46267202 t + 1) i = 0 In Maple notation, these generating functions are (239743*t^2+37473874*t-17)/(t-1)/(t^2-46267202*t+1) -(225721*t^2+34766278*t+1)/(t-1)/(t^2-46267202*t+1) 16*(1621*t^2-2188622*t+1)/(t-1)/(t^2-46267202*t+1) Then for all i>=0 we have 3 3 3 5 a(i) + 6 b(i) + 6 c(i) = -5 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 21 3 3 3 Infinitely many solutions to the Diophantine Equation, 5 X + 7 Y + 7 Z = -40 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 5 X + 7 Y + 7 Z = -40 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 5 W + 5 X + 7 Y + 7 Z = 0 and get the following special case W = -19, X = -2, Y = -2, Z = 17 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 2 x, Y = n y - 2 x, Z = -n y + 17 x Plugging into the euqation 3 3 3 3 5 W + 5 X + 7 Y + 7 Z = 0 we get 2 2 -315 x y (m y - n y - 17 m x + 19 n x) = 0 dividing by , x y, we have (5355 m - 5985 n) x - 315 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 315 (m - n) (m + n), y = 5355 m - 5985 n Plugging them back, we get the following parametric solution to , 3 3 3 3 5 W + 5 X + 7 Y + 7 Z = 0 2 2 2 2 W = -2 m - 19 m n + 19 n , X = -19 m + 19 m n + 2 n , 2 2 2 2 Y = -2 m + 17 m n - 17 n , Z = 17 m - 17 m n + 2 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 2218971 t ) d(i) t = ----------------------- / 2 ----- 4 (t - 27542702 t + 1) i = 0 infinity ----- \ i 608020 t ) e(i) t = ------------------- / 2 ----- t - 27542702 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -2 m - 19 m n + 19 n = -1/8 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -19 m + 19 m n + 2 n , Y = -2 m + 17 m n - 17 n , 2 2 Z = 17 m - 17 m n + 2 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 20814797378741 t + 1070505124442078 t - 19 ) a(i) t = ------------------------------------------- / 2 ----- (t - 1) (t - 758600433460802 t + 1) i = 0 infinity ----- 2 \ i 2 (9329366460961 t + 456477917243038 t + 1) ) b(i) t = - -------------------------------------------- / 2 ----- (t - 1) (t - 758600433460802 t + 1) i = 0 infinity ----- 2 \ i 3791020508537 t - 935405587916554 t + 17 ) c(i) t = ----------------------------------------- / 2 ----- (t - 1) (t - 758600433460802 t + 1) i = 0 In Maple notation, these generating functions are (20814797378741*t^2+1070505124442078*t-19)/(t-1)/(t^2-758600433460802*t+1) -2*(9329366460961*t^2+456477917243038*t+1)/(t-1)/(t^2-758600433460802*t+1) (3791020508537*t^2-935405587916554*t+17)/(t-1)/(t^2-758600433460802*t+1) Then for all i>=0 we have 3 3 3 5 a(i) + 7 b(i) + 7 c(i) = -40 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 22 Infinitely many solutions to the Diophantine Equation, 3 3 3 5 X + 9 Y + 9 Z = -2560 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 5 X + 9 Y + 9 Z = -2560 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 5 W + 5 X + 9 Y + 9 Z = 0 and get the following special case W = -19, X = -8, Y = -1, Z = 16 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 8 x, Y = n y - x, Z = -n y + 16 x Plugging into the euqation 3 3 3 3 5 W + 5 X + 9 Y + 9 Z = 0 we get 2 2 -405 x y (m y - n y - 11 m x + 17 n x) = 0 dividing by , x y, we have (4455 m - 6885 n) x - 405 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 405 (m - n) (m + n), y = 4455 m - 6885 n Plugging them back, we get the following parametric solution to , 3 3 3 3 5 W + 5 X + 9 Y + 9 Z = 0 2 2 2 2 W = -8 m - 17 m n + 19 n , X = -19 m + 17 m n + 8 n , 2 2 2 2 Y = -m + 11 m n - 16 n , Z = 16 m - 11 m n + n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 259 t ) d(i) t = -------------------- / 2 ----- 16 (t - 1198 t + 1) i = 0 infinity ----- \ i 20 t ) e(i) t = --------------- / 2 ----- t - 1198 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -8 m - 17 m n + 19 n = -- 32 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 2 2 X = -19 m + 17 m n + 8 n , Y = -m + 11 m n - 16 n , Z = 16 m - 11 m n + n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 953621 t + 6227198 t - 19 ) a(i) t = ---------------------------- / 2 ----- (t - 1) (t - 1435202 t + 1) i = 0 infinity ----- 2 \ i 793801 t + 4390198 t + 1 ) b(i) t = - ---------------------------- / 2 ----- (t - 1) (t - 1435202 t + 1) i = 0 infinity ----- 2 \ i 16 (16501 t - 340502 t + 1) ) c(i) t = ---------------------------- / 2 ----- (t - 1) (t - 1435202 t + 1) i = 0 In Maple notation, these generating functions are (953621*t^2+6227198*t-19)/(t-1)/(t^2-1435202*t+1) -(793801*t^2+4390198*t+1)/(t-1)/(t^2-1435202*t+1) 16*(16501*t^2-340502*t+1)/(t-1)/(t^2-1435202*t+1) Then for all i>=0 we have 3 3 3 5 a(i) + 9 b(i) + 9 c(i) = -2560 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 23 3 3 3 Infinitely many solutions to the Diophantine Equation, 6 X + 7 Y + 7 Z = -6 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 6 X + 7 Y + 7 Z = -6 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 6 W + 6 X + 7 Y + 7 Z = 0 and get the following special case W = -20, X = -1, Y = -1, Z = 19 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 20 x, X = -m y - x, Y = n y - x, Z = -n y + 19 x Plugging into the euqation 3 3 3 3 6 W + 6 X + 7 Y + 7 Z = 0 we get 2 2 -378 x y (m y - n y - 19 m x + 20 n x) = 0 dividing by , x y, we have (7182 m - 7560 n) x - 378 (m - n) (m + n) y = 0 Obviously the following values of x and y will make it zero x = 378 (m - n) (m + n), y = 7182 m - 7560 n Plugging them back, we get the following parametric solution to , 3 3 3 3 6 W + 6 X + 7 Y + 7 Z = 0 2 2 2 2 2 2 W = -m - 20 m n + 20 n , X = -20 m + 20 m n + n , Y = -m + 19 m n - 19 n , 2 2 Z = 19 m - 19 m n + n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 21 t ) d(i) t = ------------------ / 2 ----- 2 (t - 482 t + 1) i = 0 infinity ----- \ i 11 t ) e(i) t = -------------- / 2 ----- t - 482 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -m - 20 m n + 20 n = -1/4 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 2 2 X = -20 m + 20 m n + n , Y = -m + 19 m n - 19 n , Z = 19 m - 19 m n + n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 4 (226 t + 48421 t - 5) ) a(i) t = --------------------------- / 2 ----- (t - 1) (t - 232322 t + 1) i = 0 infinity ----- 2 \ i 859 t + 182092 t + 1 ) b(i) t = - --------------------------- / 2 ----- (t - 1) (t - 232322 t + 1) i = 0 infinity ----- 2 \ i 85 t - 183056 t + 19 ) c(i) t = --------------------------- / 2 ----- (t - 1) (t - 232322 t + 1) i = 0 In Maple notation, these generating functions are 4*(226*t^2+48421*t-5)/(t-1)/(t^2-232322*t+1) -(859*t^2+182092*t+1)/(t-1)/(t^2-232322*t+1) (85*t^2-183056*t+19)/(t-1)/(t^2-232322*t+1) Then for all i>=0 we have 3 3 3 6 a(i) + 7 b(i) + 7 c(i) = -6 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 24 Infinitely many solutions to the Diophantine Equation, 3 3 3 7 X + 9 Y + 9 Z = -2401 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 7 X + 9 Y + 9 Z = -2401 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 7 W + 7 X + 9 Y + 9 Z = 0 and get the following special case W = -17, X = -7, Y = -2, Z = 16 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 17 x, X = -m y - 7 x, Y = n y - 2 x, Z = -n y + 16 x Plugging into the euqation 3 3 3 3 7 W + 7 X + 9 Y + 9 Z = 0 we get 2 2 -126 x y (4 m y - 3 n y - 40 m x + 54 n x) = 0 dividing by , x y, we have 2 2 (5040 m - 6804 n) x + (-504 m + 378 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 504 m - 378 n , y = 5040 m - 6804 n Plugging them back, we get the following parametric solution to , 3 3 3 3 7 W + 7 X + 9 Y + 9 Z = 0 2 2 2 2 W = -28 m - 54 m n + 51 n , X = -68 m + 54 m n + 21 n , 2 2 2 2 Y = -8 m + 40 m n - 48 n , Z = 64 m - 40 m n + 6 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 34387799 t ) d(i) t = ------------------------- / 2 ----- 56 (t - 164279998 t + 1) i = 0 infinity ----- \ i 884300 t ) e(i) t = -------------------- / 2 ----- t - 164279998 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -28 m - 54 m n + 51 n = --- 112 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -68 m + 54 m n + 21 n , Y = -8 m + 40 m n - 48 n , 2 2 Z = 64 m - 40 m n + 6 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 15761076105974383 t + 98008574351145634 t - 17 ) a(i) t = ----------------------------------------------- / 2 ----- (t - 1) (t - 26987917742880002 t + 1) i = 0 infinity ----- 2 \ i 2 (7381821932308401 t + 36759751455211598 t + 1) ) b(i) t = - ------------------------------------------------- / 2 ----- (t - 1) (t - 26987917742880002 t + 1) i = 0 infinity ----- 2 \ i 16 (348105175174901 t - 5865801848614902 t + 1) ) c(i) t = ------------------------------------------------ / 2 ----- (t - 1) (t - 26987917742880002 t + 1) i = 0 In Maple notation, these generating functions are (15761076105974383*t^2+98008574351145634*t-17)/(t-1)/(t^2-26987917742880002*t+1 ) -2*(7381821932308401*t^2+36759751455211598*t+1)/(t-1)/(t^2-26987917742880002*t+ 1) 16*(348105175174901*t^2-5865801848614902*t+1)/(t-1)/(t^2-26987917742880002*t+1) Then for all i>=0 we have 3 3 3 7 a(i) + 9 b(i) + 9 c(i) = -2401 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 25 Infinitely many solutions to the Diophantine Equation, 3 3 3 7 X + 10 Y + 10 Z = -19208 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 7 X + 10 Y + 10 Z = -19208 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 7 W + 7 X + 10 Y + 10 Z = 0 and get the following special case W = -16, X = -14, Y = -5, Z = 17 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 16 x, X = -m y - 14 x, Y = n y - 5 x, Z = -n y + 17 x Plugging into the euqation 3 3 3 3 7 W + 7 X + 10 Y + 10 Z = 0 we get 2 2 -90 x y (7 m y - 4 n y - 14 m x + 88 n x) = 0 dividing by , x y, we have 2 2 (1260 m - 7920 n) x + (-630 m + 360 n ) y = 0 Obviously the following values of x and y will make it zero 2 2 x = 630 m - 360 n , y = 1260 m - 7920 n Plugging them back, we get the following parametric solution to , 3 3 3 3 7 W + 7 X + 10 Y + 10 Z = 0 2 2 2 2 W = -98 m - 88 m n + 64 n , X = -112 m + 88 m n + 56 n , 2 2 2 2 Y = -35 m + 14 m n - 68 n , Z = 119 m - 14 m n + 20 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 195088566359961 t ) d(i) t = -------------------------------- / 2 ----- 196 (t - 758600433460802 t + 1) i = 0 infinity ----- \ i 2093314208755 t ) e(i) t = -------------------------- / 2 ----- t - 758600433460802 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -98 m - 88 m n + 64 n = --- 392 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -112 m + 88 m n + 56 n , Y = -35 m + 14 m n - 68 n , 2 2 Z = 119 m - 14 m n + 20 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- \ i ) a(i) t = 16 / ----- i = 0 2 (108999999881628831799465176149 t + 160438225705249922036887561052 t - 1) / 2 / ((t - 1) (t - 575474617646916682666870483202 t + 1)) / infinity ----- \ i ) b(i) t = - 5 / ----- i = 0 2 (333098224740507730807713676729 t + 161608025517368013940999545670 t + 1) / 2 / ((t - 1) (t - 575474617646916682666870483202 t + 1)) / infinity ----- \ i ) c(i) t = ( / ----- i = 0 2 967891124460114130521991256297 t - 3441422375749492854265557368314 t + 17) / 2 / ((t - 1) (t - 575474617646916682666870483202 t + 1)) / In Maple notation, these generating functions are 16*(108999999881628831799465176149*t^2+160438225705249922036887561052*t-1)/(t-1 )/(t^2-575474617646916682666870483202*t+1) -5*(333098224740507730807713676729*t^2+161608025517368013940999545670*t+1)/(t-1 )/(t^2-575474617646916682666870483202*t+1) (967891124460114130521991256297*t^2-3441422375749492854265557368314*t+17)/(t-1) /(t^2-575474617646916682666870483202*t+1) Then for all i>=0 we have 3 3 3 7 a(i) + 10 b(i) + 10 c(i) = -19208 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- Chapter , 26 Infinitely many solutions to the Diophantine Equation, 3 3 3 8 X + 9 Y + 9 Z = -4096 And Explaining How They Were Discovered By Shalosh B. Ekhad We are interested in finding infinitely many solutions of the cubic diophant\ ine equation 3 3 3 8 X + 9 Y + 9 Z = -4096 We first consider the diophantine equation in FOUR variables, W,X,Y,Z 3 3 3 3 8 W + 8 X + 9 Y + 9 Z = 0 and get the following special case W = -19, X = -8, Y = -16, Z = 22 Obviously, the following doubly-infinite family is also a solution W = m, X = -m, Y = n, Z = -n For any integers m and n Using the Eri Jabotinsky trick writing W = m y - 19 x, X = -m y - 8 x, Y = n y - 16 x, Z = -n y + 22 x Plugging into the euqation 3 3 3 3 8 W + 8 X + 9 Y + 9 Z = 0 we get 2 2 -162 x y (4 m y - n y - 44 m x + 38 n x) = 0 dividing by , x y, we have (7128 m - 6156 n) x - 162 (2 m - n) (2 m + n) y = 0 Obviously the following values of x and y will make it zero x = 162 (2 m - n) (2 m + n), y = 7128 m - 6156 n Plugging them back, we get the following parametric solution to , 3 3 3 3 8 W + 8 X + 9 Y + 9 Z = 0 2 2 2 2 W = -32 m - 38 m n + 19 n , X = -76 m + 38 m n + 8 n , 2 2 2 2 Y = -64 m + 44 m n - 22 n , Z = 88 m - 44 m n + 16 n Check! We now need a Lemma Lemma: Let d(i), e(i) be defined in terms of the following generating functi\ ons infinity ----- \ i -1 + 5294691 t ) d(i) t = ------------------------ / 2 ----- 64 (t - 27177902 t + 1) i = 0 infinity ----- \ i 218270 t ) e(i) t = ------------------- / 2 ----- t - 27177902 t + 1 i = 0 Then m=d(i), n=e(i) satisfy the quadratic diophantine equation 2 2 -1 -32 m - 38 m n + 19 n = --- 128 The proof is routine, since everything is in the C-finite ansatz, and in fac\ t checing it for the first three cases suffices, but it was discovered using the algorithm for solving qudaratic diophanine e\ quations Now that we have C-finite representations for m and n, we can plug them into 2 2 2 2 X = -76 m + 38 m n + 8 n , Y = -64 m + 44 m n - 22 n , 2 2 Z = 88 m - 44 m n + 16 n and get C-finite representations for X,Y,Z, leading to the following theorem\ Theorem: Let a(i), b(i), c(i) be defined in terms of the following generatin\ g functions infinity ----- 2 \ i 560288964891221 t + 2421706136305598 t - 19 ) a(i) t = -------------------------------------------- / 2 ----- (t - 1) (t - 738638357121602 t + 1) i = 0 infinity ----- 2 \ i 16 (44263820187601 t + 17479943410798 t + 1) ) b(i) t = - --------------------------------------------- / 2 ----- (t - 1) (t - 738638357121602 t + 1) i = 0 infinity ----- 2 \ i 2 (291856232068451 t - 785806340855662 t + 11) ) c(i) t = ----------------------------------------------- / 2 ----- (t - 1) (t - 738638357121602 t + 1) i = 0 In Maple notation, these generating functions are (560288964891221*t^2+2421706136305598*t-19)/(t-1)/(t^2-738638357121602*t+1) -16*(44263820187601*t^2+17479943410798*t+1)/(t-1)/(t^2-738638357121602*t+1) 2*(291856232068451*t^2-785806340855662*t+11)/(t-1)/(t^2-738638357121602*t+1) Then for all i>=0 we have 3 3 3 8 a(i) + 9 b(i) + 9 c(i) = -4096 Proof: Since everything is in the C-finite ansatz, checking it for i from 0 \ to 10 suffices, which we did and you are welcome to check. -------------------------------- ----------------------------------------------------- This ends this fascinating article that contained , 26, chapters and took, 44.503, seconds to generate