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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.}  Using the Quadratic Reciprocity Law and Rules 1-3, find (no credit for other methods) 
$$
\left ( \frac{17}{23} \right ) 
$$

{\bf Sol. to 1}: By Rule $4$ {\bf the Quadratic Reciprocity Law}
$$
\left ( \frac{17}{23} \right )\left ( \frac{23}{17} \right )=
(-1)^{(17-1)(23-1)/4}=(-1)^{(16)(24)/4}=1 \quad,
$$
so
$$
\left ( \frac{17}{23} \right ) =
\left ( \frac{23}{17} \right ) = \left ( \frac{6}{17} \right )  \quad ,
$$
since $23 \,\, mod 17=6$.


Using rule (1)
$$
\left( \frac{6}{17} \right)
= \left( \frac{2}{17} \right) \cdot \left( \frac{3}{17} \right)    \quad .
$$

By  Rule (3)
$$
\left ( \frac{2}{17} \right ) = (-1)^{(17^2-1)/8}=(-1)^{36}=1 \quad .
$$

Again by Rule (4)
$$
\left ( \frac{3}{17} \right )
\left ( \frac{17}{3} \right )=
(-1)^{(17-1)(3-1)/4}=1^8=1 \quad .
$$
Hence
$$
\left ( \frac{3}{17} \right )= \left ( \frac{17}{3} \right )=\left ( \frac{2}{3} \right )=-1
$$
(since obviously $2$ is not a quadratic residue mod $3$, the only one is $1$. You can also use Rule (4) again, if you wish.
So
Going back we have,
$$
\left ( \frac{17}{23} \right )=
\left ( \frac{23}{17} \right ) = \left ( \frac{6}{17} \right )=
\left ( \frac{2}{17} \right )\left ( \frac{3}{17} \right )=
1 \cdot (-1)=-1
$$


{\bf Ans. to 1}:  $\left ( \frac{17}{23} \right ) =-1$,
in other words $17$ is a {\bf quadratic NON-residue} of $23$.

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