\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION

\begin{description}

\item[(i)]
Prove that if $p$ is an odd prime, then
$\left(\frac{-2}{p}\right)=\left\{\begin{array}{cl}1&\textrm{ if
$p=1$ or 3 mod 8}\\-1&\textrm{if $p=5$ or 7 mod
8}\end{array}\right.$

\item[(ii)]
Prove that if $p$ is an odd prime$>3$, then
$\left(\frac{-}{p}\right)=\left\{\begin{array}{cl}1&\textrm{ if
$p=1$ mod 6}\\-1&\textrm{if $p=5$ mod 6}\end{array}\right.$

\item[(iii)]
Describe (in terms of congruence modulo a suitable $n$) all primes
$p$ for which $\left(\frac{3}{p}\right)=1$.

\end{description}



ANSWER

\begin{description}

\item[(i)]
$\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)
\left(\frac{2}{p}\right)$.

We know
$\left(\frac{-1}{p}\right)=\left\{\begin{array}{cl}1&\textrm{if
$p\equiv1$ mod 4}\\-1&\textrm{if $p\equiv3$ mod
4}\end{array}\right.$ (th.1.7)

and $\left(\frac{2}{p}\right)=\left\{\begin{array}{cl}1&\textrm{
if $p\equiv\pm1$ mod 8}\\-1&\textrm{if $p\equiv\pm3$ mod
8}\end{array}\right.$ (th.7.3).

Noting that $p\equiv1$ mod 4$\Leftrightarrow p\equiv1$ or 5 mod 8,
i.e. $p\equiv1$ or $-3$ mod 8, and that $p\equiv3$ mod
4$\Leftrightarrow p\equiv3$ or 7 mod 8, i.e. $p\equiv3$ or $-1$
mod 8, we may put these together to deduce
$\left(\frac{-2}{p}\right)=\left\{\begin{array}{cl}1&\textrm{if
$p\equiv$ or 3 mod 8}\\-1&\textrm{if $p\equiv -1$ or $-3$ mod
8}\\& \textrm{(i.e.7 or 5 mod 8) }\end{array}\right.$

\item[(ii)]
$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)
\left(\frac{3}{p}\right)$.

As before
$\left(\frac{-1}{p}\right)=\left\{\begin{array}{cl}1&\textrm{if
$p\equiv1$ mod 4}\\-1&\textrm{if $p\equiv3$ mod
4}\end{array}\right.$

Thus if $p\equiv 1$ mod 4,
$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)
\left(\frac{3}{p}\right)=1.\left(\frac{3}{p}\right)=
\left(\frac{p}{3}\right)$ by quadratic reciprocity, and if
$p\equiv 3$ mod 4,
$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)
\left(\frac{3}{p}\right)\left(\frac{3}{p}\right)=-1.
\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)$ again by
quadratic reciprocity, as here $p\equiv3$ mod 4 and $3\equiv 3$
mod 4. Thus in all cases
$\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)$.

Thus
$\left(\frac{-3}{p}\right)=\left\{\begin{array}{cl}1&\textrm{if
$p\equiv1$ mod 6}\\-1&\textrm{if $p\equiv2$ mod
3}\end{array}\right.$ and as $p$ is prime, and $p\neq3$, we know
$p\equiv1$ or 5 mod 6, with the congruence class $p\equiv1$ mod 6
covering all primes $\equiv1$ mod 3, and the congruence class
$p\equiv5$ mod 6 covering all primes $\equiv2$ mod 3.

Hence
$\left(\frac{-3}{p}\right)=\left\{\begin{array}{cl}1&\textrm{ if
$p\equiv1$ mod 6}\\-1&\textrm{if $p\equiv 5$ mod
6}\end{array}\right.$ as required.

\item[(iii)]
By quadratic reciprocity, as $3\equiv3$ mod 4,
$\left(\frac{3}{p}\right)=\left\{\begin{array}{cl}
\left(\frac{p}{3}\right)& \textrm{if $p\equiv1$ mod
4}\\-\left(\frac{p}{3}\right)&\textrm{ if $p\equiv3$ mod
4}\end{array}\right.$

Now $\left(\frac{p}{3}\right)=\left\{\begin{array}{cl}1&\textrm{if
$p\equiv1$ mod 3}\\-1&\textrm{if $p\equiv2$ mod 3}\\0&\textrm{if
}p\equiv3\end{array}\right.$

Hence

$\left(\frac{3}{p}\right)\\=\left\{\begin{array}{cl}1&\textrm{if
$p\equiv1$ mod 4 and $p\equiv1$ mod 3 or $p\equiv3$ mod 4 and
$p\equiv2$ mod 3}\\ -1&\textrm{if $p\equiv1$ mod 4 and $p\equiv2$
mod 3 or $p\equiv3$ mod 4 and $p\equiv1$ mod 3}\end{array}\right.$

Now the Chinese Remainder Theorem tells us that the simultaneous
congruences $p\equiv a$ mod 4 and $p\equiv b$ mod 3 have a unique
solution mod 12. Thus expressing the results modulo 12 gives
$\left(\frac{3}{p}\right)=\left\{\begin{array}{cl}1&\textrm{if
$p\equiv1$ mod 12 or $p\equiv11$ mod 12}\\-1&\textrm{if $p\equiv5$
mod 12 or $p\equiv 7$ mod 12}\end{array}\right.$ or
$\left(\frac{3}{p}\right)=\left\{\begin{array}{cl}1&\textrm{if
$p\equiv\pm1$ mod 12}\\-1&\textrm{if $p\equiv\pm5$
mod12}\end{array}\right.$

\end{description}





\end{document}
