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\centerline
{
\bf Solutions to Attendance Quiz \# 20 for Dr. Z.'s Number Theory
Course for Nov. 14, 2013
}
\bigskip

{\bf 1.} By `brute force', Find the set of quadratic residues, and the set of quadratic non-residues of $p=13$.

{\bf Sol. to 1}:
$$
QR(13)= \{ 0^2, 1^2, 2^2, 3^2, 4^2, 5^2, 6^2 \} \,\, mod \,\, 13
$$
$$
=\{ 0, 1, 4, 9, 16, 25, 36 \} \,\, mod \,\, 13
$$
$$
=\{ 0, 1, 4, 9, 3, 12, 10 \} 
$$
$$
=\{ 0, 1, 3, 4, 9, 10, 12 \} \quad .
$$
{\bf Ans. to 1}: The set of quadratic residues modulo $13$ is
$\{ 0, 1, 3, 4, 9, 10, 12 \}$ .



{\bf 2.}  Using the Quadratic Reciprocity Law and Rules 1-3, find (no credit for other methods) 
$$
\left ( \frac{7}{17} \right ) 
$$

{\bf Sol. to 2}: By Rule $4$ {\bf the Quadratic Reciprocity Law}
$$
\left ( \frac{7}{17} \right )\left ( \frac{17}{7} \right )=
(-1)^{(7-1)(17-1)/4}=(-1)^{(6)(16)/4}=(-1)^{24}=1 \quad,
$$
so
$$
\left ( \frac{7}{17} \right ) =
\left ( \frac{17}{7} \right ) = \left ( \frac{3}{7} \right )  \quad ,
$$
since $17 \,\, mod 7=3$.

Moving right along, using Rule 4 (QRL) once again
$$
\left ( \frac{3}{7} \right )
\left ( \frac{7}{3} \right )=
(-1)^{(3-1)(7-1)/4}=(-1)^{(2)(6)/4}=(-1)^{3}=-1 \quad .
$$
So
$$
\left ( \frac{3}{7} \right )= -\left ( \frac{7}{3} \right ) \quad .
$$
But since $7 \,\, mod \,\, 3=1$, 
$$
\left ( \frac{7}{3} \right ) =\left ( \frac{1}{3} \right ) = 1 
\quad.
$$

(Of course $\left ( \frac{1}{p} \right )$ is {\bf always} $1$, since $1^2=1$.

Going back we have,
$$
\left ( \frac{3}{7} \right ) = -1 \quad ,
$$
and finally,
$$
\left ( \frac{7}{17} \right ) = -1 \quad .
$$

{\bf Ans. to 2}:  $\left ( \frac{7}{17} \right ) =-1$,
in other words $7$ is a {\bf quadratic NON-residue} of $17$.

\end




