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QUESTION

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\item[(i)]
State the Law of Quadratic Reciprocity.

\item[(ii)]
Use (i) to evaluate the Legendre symbol $\left(\frac{3}{p}\right)$
when $p$ is an odd prime. More precisely, show that

$$\left(\frac{3}{p}\right)=\left\{\begin{array}{cl}1&\textrm{if
$p\equiv \pm1$ (modulo 12)},\\-1&\textrm{if $p\equiv \pm5$ (modulo
12).}\end{array}\right.$$

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ANSWER

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\item[(i)]
The Law of Quadratic Reciprocity states that, if $p$ and $q$ are
distinct odd primes then

$$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)(q-1)/4}$$

\item[(ii)]
Therefore

$$\left(\frac{3}{p}\right)\left(\frac{p}{3}\right)=(-1)^{\frac{(p-1)}{2}}$$

and

$$\left(\frac{p}{3}\right)=\left\{\begin{array}{cl}1&\textrm{ if
$p\equiv1$ (modulo 3),}\\-1&\textrm{ if $p\equiv-1$ (modulo
3).}\end{array}\right.$$

Also $p=3k\pm1$ can only happen if $k=2s$, since $p$ is odd. Hence
we write $p=6s\pm1$. The possibilities modulo 12 for $p$ are
$12t-1, 12+1, 12t+5, 12+7$ which we shall deal with case by case.

If $p=12-1$ then

$$\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)
(-1)^{\frac{(p-1)}{2}}=(-1)(-1)^{(6t-1)}=1.$$

If $p=12t+1$ then

$$\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)
(-1)^{\frac{(p-1)}{2}}=(-1)^{(6t)}=1.$$

If $p=12t+5$ then

$$\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)
(-1)^{\frac{(p-1)}{2}}=(-1)^{(6t+3)}=-1,$$

as required.

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