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\begin{center}

THEORY OF NUMBERS

QUADRATIC RESIDUES

\end{center}

The problem can be reduced to a study of the congruence

$$x^2\equiv a \textrm{ mod }p$$

Suppose $a\not\equiv0$ mod $p$.

If $x^2\equiv a$ is soluble $a$ is called a quadratic residue mod
$p$.

If $x^2\equiv a$ is not soluble $a$ is a quadratic non-residue mod
$p$.

\begin{description}

\item[The Legendre Symbol]
$\left(\frac{n}{p}\right)=\left\{\begin{array}{cl}+1&amp;\textrm{if
$n$ is quadratic residue mod $p$.}\\-1&amp;\textrm{if $n$ is quadratic
non-residue mod $p$}\\0&amp;\textrm{if $n\equiv0$ mod
$p$}\end{array}\right.$

\item[Theorem]
$\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)=0$

i.e. $\exists$ the same number of quadratic residues and quadratic
non-residues.

\item[Proof]
$\pm1,\pm2,\ldots\pm\frac{1}{2}(p-1)$ form an R.S.R. so $1^2\
2^2\ldots\left(\frac{p-1}{2}\right)^2$ contain all the quadratic
residues.

$x^2\equiv y^2$ mod $p\Rightarrow x\equiv \pm y$ mod $p$

and this does not occur with the above R.S.R. so these are all the
quadratic residues mod $p$ and there are $\frac{p-1}{2}$ of them.

\item[Theorem]
$\left(\frac{n}{p}\right)\equiv n^\frac{p-1}{2}$ mod $p$.

\item[Proof]
$n\equiv 0$ is trivial. If $n$ is a quadratic residue $\exists\ x$
such that $n\equiv x^2$ then

$(x^2)^\frac{p-1}{2}=x^{p-1}\equiv1$ by Fermats theorem. $\exists
\frac{p-1}{2}$ \ quadratic residues and $n^\frac{p-1}{2}\equiv1$
has at most $\frac{p-1}{2}$ solutions therefore the quadratic
residues are all the solutions of $n^\frac{p-1}{2}=1$.

Suppose $(n,\ p)=1$ then $n^{p-1}\equiv 1$ therefore


$\left(n^\frac{p-1}{2}-1\right)\left(n^\frac{p-1}{2}+1\right)\equiv0$
therefore

$n^\frac{p-1}{2}\equiv\pm1$

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

\item[Theorem]
For every pair of integers $m,\ n$ we have

$$\left(\frac{mn}{p}\right)=
\left(\frac{m}{p}\right)\left(\frac{n}{p}\right)$$

\item[Proof]
If $E=\left(\frac{mn}{p}\right)-
\left(\frac{m}{p}\right)\left(\frac{n}{p}\right)$ then
$E\equiv(mn)^\frac{p-1}{2}-m^\frac{p-1}{2}n^\frac{p-1}{2}\equiv0$.


But $|E|\leq2$ and $p\geq3$ therefore $E=0$.

\item[Gauss's Lemma]
Suppose $n\not\equiv0$ mod $p$. Let $\mu$ be the number of those
numbers $1,\ n,\ 2n\ldots\frac{p-1}{2}n$ whose remainder mod $p$
is $&gt;\frac{1}{2}p$. Then $\left(\frac{n}{p}\right)=(-1)^\mu$

\item[Proof]
Let the remainders $&gt;\frac{1}{2}p$ be $\alpha_1\ldots\alpha_\mu$.
Let those $&lt;\frac{1}{2}p$ be $\beta_1\ldots\beta_\nu$. Then
$\mu+\nu=\frac{p-1}{2}$.

Consider the $\frac{p-1}{2}$ numbers $p-\alpha_1,\
p-\alpha_2\ldots p-\alpha_\mu,\ \beta_1,\ \beta_2,\ldots\beta_nu$

$1\leq\beta\leq\frac{p-1}{2}$ and $1\leq
p-\alpha\leq\frac{p-1}{2}$

The $\beta_i$ are distinct for $k'n\equiv k''n$ mod $p\Rightarrow
k'\equiv k''$ mod $p,\ (n,\ p)=1$

Similarly the $p-\alpha_j$ are distinct.

Now $p-\alpha_j\equiv\beta_i$ mod $p\Rightarrow
\alpha_j+\beta_i\equiv p\equiv0$ mod $p$. Let $\alpha_j=un\
\beta_i=vn$

Then $(u+v)n\equiv0$ mod $p$ therefore $u+v\equiv0$ mod $p$.

But $1\leq u+v\leq p-1$ and so we have a contradiction.

Hence $p-\alpha_1,\ p-\alpha_2\ldots p-\alpha_\mu,\ \beta_1,\
\beta_2\ldots\beta_\nu$ is a rearrangement of
$1,2,\ldots\frac{p-1}{2}$. Therefore

\begin{eqnarray*}
(p-\alpha_1)(p-\alpha_2)\ldots(p-\alpha_\mu)
\beta_1\ldots\beta_\nu&amp;\equiv&amp;1,2\ldots\frac{p-1}{2}\ (p)\\
\textrm{therefore }
(-1)^\mu\prod_{j=1}^\frac{p-1}{2}jn&amp;\equiv&amp;\prod_{j=1}^\frac{p-1}{2}j
\textrm{ mod }p\\ \textrm{therefore }(-1)^\mu
n^\frac{n-1}{2}&amp;\equiv&amp;1\textrm{ mod }p\\ \textrm{therefore
}\left(\frac{n}{p}\right)&amp;\equiv&amp;(-1)^\mu\textrm{ mod }p\\
\textrm{therefore }\left(\frac{n}{p}\right)&amp;=&amp;(-1)^\mu\ p\geq3
\end{eqnarray*}

\item[The Law of Quadratic Reciprocity]
Suppose $p,\ q$ are distinct odd primes. Then
$\left(\frac{p}{q}\right)\left(\frac{p}{p}\right)=
(-1)^{\frac{p-1}{2}\frac{q-1}{2}}$

i.e. $\left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)$ unless
$p\equiv p\equiv -1$ mod $4$ when
$\left(\frac{p}{q}\right)=-\left(\frac{q}{p}\right)$

\item[Proof 1]
Let $p'=\frac{p-1}{2},\ q'=\frac{q-1}{2}$. Write
$s=\sum_{m=1}^{p'}\left[\frac{mq}{p}\right]$

For each $m=1,2,\ldots p'$

$$mq=p\left[\frac{mq}{p}\right]+\left\{\begin{array}{cl}\alpha&amp;\textrm{if
}&gt;\frac{1}{2}p\\\beta&amp;\textrm{if
}&lt;\frac{1}{2}p\end{array}\right.$$

Summing over $m$

$$\frac{p^2-1}{8}q=ps+\sum\alpha+\sum\beta$$

Now $\sum p-\alpha+\sum\beta=1+\ldots+p'=\frac{p^2-1}{8}$ i.e.
$\mu p-\sum\alpha+\sum\beta=\frac{p^2-1}{8}$

But $sp+\sum\alpha+\sum\beta=\frac{p^2-1}{8}q$ therefore

$$\frac{p^2-1}{8}(q-1)=p(s-\mu)+2\sum\alpha$$

$\frac{p^2-1}{8}q-1$ is even. $2\sum\alpha$ is even therefore
$s\equiv\mu$ mod 2.

By Gauss's Lemma $\left(\frac{q}{p}\right)=(-1)^\mu$ therefore
$\left(\frac{q}{p}\right)=(-1)^s$

Write $t=\sum_{m=1}^{q'}\left[\frac{mp}{q}\right]$, then
$\left(\frac{p}{q}\right)=(-1)^t$

So S.T.P $s+t=p'q'$

Consider the set of all numbers $qx-py,\ x=1,2,\ldots p'\
y=1,2,\ldots q'$

This set contains $p'q'$ numbers. No element in this set is zero.

The number of positive numbers in this set is

$\sum_{x=1}^{p'}\left[\frac{qx}{p}\right]=s$

The number of negative numbers in this set is

$\sum_{y=1}^q\left[\frac{py}{q}\right]=t$

Hence the result.

\item[Proof 2]
Write $e(\alpha)=e^{2\pi i\alpha}$

Suppose $k$ is a natural number and $a$ is an integer. We define
$S(a,k)=\sum_{x=1}^ke\left(\frac{ax^2}{k}\right)$. This is called
a Gaussian sum.

\item[Theorem A](Proof postponed)

If $k$ is odd

$$S(1,k)=\left\{\begin{array}{cl}k^\frac{1}{"}&amp;\textrm{ if
$k\equiv1$ mod 4}\\ik^\frac{1}{2}&amp;\textrm{if $k\equiv-1$ mod
4}\end{array}\right.$$

$S(1,k)=\frac{1}{2}(1+i)(1-i^{3k})k^\frac{1}{2}$.

\item[Theorem B]

\begin{description}

\item[(i)]
If $p$ is an odd prime and $(a,\ p)=1$ then
$S(a,p)=\left(\frac{a}{p}\right)S(1,p)$

\item[(ii)]
If $(k_1,k_2)=1$ then $S(a,k_1k_2)=S(ak_1,k_2)S(ak_2,k_1)$

\end{description}

\item[Proof]

\begin{description}

\item[(i)]
$$S(a,p)=\sum_{x=1}^pe\left(\frac{ax^2}{p}\right)=\sum_{n=1}^pc(n)e\left(\frac{an}{p}\right)$$

where $c(n)$ is the number of solutions of $x^2\equiv n$ mod $p$.

When $n=p\exists 1$ solution and $1+\left(\frac{n}{p}\right)=1$.

When $(n,p)=1$ if $n$ is a quadratic residue $\exists$ 2 solutions
and $1+\left(\frac{n}{p}\right)=2$

When $(n,p)=1$ if $n$ is a quadratic non-residue $\exists$ no
solutions and $1+\left(\frac{n}{p}\right)=0$. Therefore

\begin{eqnarray*}
S(a,p)&amp;=&amp;\sum_{n=1}^p\left(1+\left(\frac{n}{p}\right)\right)e\left(\frac{an}{p}\right)\\
&amp;=&amp;\sum_{n=1}^p\left(\frac{n}{p}\right)e\left(\frac{an}{p}\right)\\
&amp;=&amp;\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)e\left(\frac{an}{p}\right)
\end{eqnarray*}

$\left[\left(\frac{p}{p}\right)=0\right]$

For
$\sum_{n=1}^pe\left(\frac{an}{p}\right)=\sum_{n=1}^pz^n=x\left(\frac{1-z^p}{1-z}\right)=0$
since $z=e^\frac{2\pi ia}{p}\neq1$ since $(a,p)=1$.

Write $an=m$

$\left(\frac{n}{p}\right)=\left(\frac{a^2n}{p}\right)=\left(\frac{a}{p}\right)
\left(\frac{an}{p}\right)=\left(\frac{a}{p}\right)\left(\frac{m}{p}\right)$

If $(a,p)=1$ an $n$ runs through a R.S.R. then $an=m$ runs through
an R.S.R. Therefore

$$\sum_{n=1}^{p-1}\frac{n}{p}e\left(\frac{an}{p}\right)=\sum_{m=1}^{p-1}
\left(\frac{a}{p}\right)\left(\frac{m}{p}\right)e\left(\frac{m}{p}\right)$$

therefore

$$S(a,p)=\left(\frac{a}{p}\right)\sum_{m=1}^{p-1}\left(\frac{m}{p}\right)e
\left(\frac{m}{p}\right)$$

In particular for $a=1$,

$S(1,p)=i\sum_{m=1}^{p-1}\left(\frac{m}{p}\right)e\left(\frac{m}{p}\right)$

Therefore $S(a,p)=\left(\frac{a}{p}\right)S(1,p)$

\item[(ii)]
Suppose $(k_1\ k_2)=1$

$S(a,k_1k_2)=\sum_{x=1}^{k_1k_2}e\left(\frac{ax^2}{k_1k_2}\right)$

Suppose $u$ runs through C.S.R. mod $k_2$ and $v$ runs through
C.S.R. mod $k_1$. Then $k_1u+k_2v$ runs through C.S.R. mod
$k_1k_2$. Therefore

\begin{eqnarray*}
S(a,k_1k_2)&amp;=&amp;\sum_{v=1}^{k_1}\sum_{u=1}^{k_2}e
\left(\frac{a(k_1u+k_2v)^2}{k_1k_2}\right)\\
&amp;=&amp;\sum_{v=1}^{k_1}\sum_{u=1}^{k_2}e\left(\frac{ak_1u^2}{k_2}\right)e
\left(\frac{ak_2v^2}{k_1}\right)\\ &amp;=&amp;S(ak_2,k_1)S(ak_1,k_2)
\end{eqnarray*}

\end{description}

Suppose now that $p,\ q$ are distinct odd primes.

Applying theorem B with $a=1$ we have

\begin{eqnarray*}
S(1,pq)&amp;=&amp;S(p,q)S(q,p)\\ S(1,pq)&amp;=&amp;\left(\frac{p}{q}\right)S(1,q)
\left(\frac{q}{p}\right)S(1,p)\\
\varepsilon_{pq}(pq)^\frac{1}{2}&amp;=&amp;\left(\frac{p}{q}\right)
\varepsilon_qq^\frac{1}{2}
\left(\frac{q}{p}\right)\varepsilon_pp^\frac{1}{2}
\end{eqnarray*}

where $\varepsilon_k=\left\{\begin{array}{cl}1&amp;\textrm{if
$k\equiv1$ mod 4}\\i&amp;\textrm{if $k\equiv -1$ mod
4}\end{array}\right.$

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

In connection with theorem A it is easy to prove the result to
within a $\pm$ sign as we shall now see.

If $(2a,k)=1$

\begin{eqnarray*}
|S(a,k)|^2&amp;=&amp;S(a,k)S(-a,k)\\
&amp;=&amp;\sum_{x=1}^ke\left(\frac{ax^2}{k}\right)\sum_{y=1}^ke
\left(\frac{-a(x+y)^2}{k}\right)\textrm{ sum over C.S.R.}\\
&amp;=&amp;\sum_{x=1}^k\sum_{y=1}^ke\left(-\frac{2axy}{k}\right)e
\left(-\frac{ay^2}{k}\right)\\
&amp;=&amp;\sum{y=1}^ke\left(-\frac{ay^2}{k}\right)\sum_{x=1}^ke
\left(\frac{-2axy}{k}\right)\\ &amp;=&amp;k
\end{eqnarray*}

The inner sum $=k$ if $y=k$ and 0 otherwise.

\begin{eqnarray*}
S(1,p).S(-1,p)&amp;=&amp;p\\
\left(\frac{-1}{p}\right)\left\{S(1,p)\right\}^2&amp;=&amp;p\\
S(1,p)^2&amp;=&amp;(\varepsilon_pp^\frac{1}{2})^2\\ \textrm{ therefore
}S(1,p)&amp;=&amp;\pm\varepsilon_pp^\frac{1}{2}
\end{eqnarray*}

\item[Corollary to Gauss's Lemma]
$$\left(\frac{2}{o}\right)=(-1)^\frac{p^2-1}{8}=
\left\{\begin{array}{cl}+1&amp;\textrm{if}p\equiv\pm1\
(8)\\-1&amp;\textrm{if}p\equiv\pm3\ (8)\end{array}\right.$$

\item[Jacobi Symbol]
We define, for $m$ odd

$$\left(\frac{n}{m}\right)=\left\{\begin{array}{cl}1&amp;\textrm{if
}m=\pm1\\ \left(\frac{n}{p_1}\right)^{\alpha_1}
\left(\frac{n}{p_2}\right)^{\alpha_2}\ldots
\left(\frac{n}{p_r}\right)^{\alpha_r}&amp;\textrm{if
}m=p_1^{\alpha_1}\ldots p_r^{\alpha_r}\end{array}\right.$$

Then
$\left(\frac{n+km}{m}\right)=\frac{n}{m}\left(\frac{n}{m_1m_2}\right)
=\left(\frac{n}{m_1}\right)\left(\frac{n}{m_2}\right)$

$\left(\frac{n_1n_2}{m}\right)=\left(\frac{n_1}{m}\right)
\left(\frac{n_2}{m}\right)$

If $x^2\equiv n$ mod $m$ then $\left(\frac{n}{m}\right)=+1$ but
the converse is not true.

\item[Theorem]

\begin{description}

\item[(i)]
$\left(\frac{-1}{m}\right)=(-1)^\frac{m-1}{2}\ m&gt;0\ m $ odd.

\item[(ii)]
$\left(\frac{2}{m}\right)=(-1)^\frac{m^2-1}{8}\ m$ odd.

\item[(iii)]
$\left(\frac{n}{m}\right)\left(\frac{m}{n}\right)=(-1)^{\frac{m-1}{2}\frac{n-1}{2}}$

$m,\ n$ odd but not both negative and $(m,\ n)=1$.

\end{description}

\item[Proof]
Suppose $n=\prod p;\ m=\prod q\ p\neq q$. Suppose first that $m&gt;0\
n&gt;0$

\begin{description}

\item[(i)]
$\left(\frac{-1}{m}\right)\prod_q\left(\frac{-1}{q}\right)=
(-1)^{\sum\frac{q-1}{2}}=(-1)^{Am}$

\item[(ii)]
$\left(\frac{2}{m}\right)=\prod_q\left(\frac{2}{q}\right)=
(-1)^{\sum\frac{q^2-1}{8}}=(-1)^{Bm}$

\item[(iii)]
$\left(\frac{n}{m}\right)\left(\frac{m}{n}\right)=\prod_{p,\
q}\left\{\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)\right\}
=(-1)^{\sum\frac{p-1}{2}\sum\frac{q-1}{2}}=(-1)^{AnAm}$

So R.T.P. $A_m\equiv\frac{m-1}{2}$ mod $2,\
B_m\equiv\frac{m^2-1}{8}$ mod 2.

Now $r,\ s$ odd $\Rightarrow (r-1)(s-1)\equiv0$ mod 4.

So $\frac{r-1}{2}+\frac{s-1}{2}\equiv\frac{rs-1}{2}$ mod 2 and by
induction

$\frac{r_1-1}{2}+\ldots+\frac{r_v-1}{2}\equiv\frac{r_1\ldots
r_v-1}{2}$ mod 2

Also $(r^2-1)(s^2-1)\equiv 0$ mod 64 and so

$\frac{r^2-1}{8}+\frac{s^2}{8}\equiv\frac{r^2s^2-1}{8}$ mod 8

and by induction

$\frac{r_1^2-1}{8}+\ldots+\frac{r_v^2-1}{8}=\frac{r_1^2\ldots
r_v^2-1}{8}$

Thus $A_m\equiv\frac{m-1}{2}$ mod 2 and $B_m\equiv\frac{m^2-1}{8}$
mod 8 and so mod 2.

(i) and (ii) are unaffected by our assumption that $m&gt;0\ n&gt;0$.

(iii) Suppose $m&gt;0\ n&lt;0$.

Write $n=-n'$

$$\left(\frac{n}{m}\right)\left(\frac{m}{n}\right)=
\left(-\frac{n'}{m}\right)\left(\frac{m}{n'}\right)=
\left(-\frac{1}{m}\right)\left(\frac{n'}{m}\right)
\left(\frac{m}{n'}\right)=(-1)^{\frac{m-1}{2}+ \frac{n'-1}{2}.
\frac{m-1}{2}}$$

Now $\left(\frac{m-1}{2}\right)+\left(\frac{n'-1}{2}\right)
\left(\frac{m-1}{2}\right)
=\left(\frac{m-1}{2}\right)\left(\frac{-n+1}{2}\right)=
\left(\frac{m-1}{2}\right) \left(\frac{n-1}{2}\right)$ mod 2.

We can use the Jacobi symbol to evaluate Legendre symbols by this
theorem.

e.g.$\left(\frac{31}{103}\right)=-\left(\frac{103}{31}\right)=
-\left(-\frac{21}{31}\right)=\left(\frac{31}{-21}\right)=
\left(\frac{31}{21}\right)=
\left(\frac{-11}{-21}\right)=\left(\frac{21}{11}\right)=
\left(\frac{-1}{11}\right)=-1$

\end{description}

\item[Exercise]
(1) $p$ an odd prime. Consider $1,2,3\ldots$. Pick the least
quadratic non-residue $q$ mod $p$. Prove $q=O(p^\frac{1}{2})$

It is known that $q=O(p^\alpha)\
\alpha&gt;\frac{1}{4}e^{-\frac{1}{2}}$

It is conjectured that $q=O(p^\varepsilon)$

[Hint $q$ must be prime. $\exists$ a multiple of $q$ such that
$p&lt;mq&lt;p+q$. What about $m$]

(2) $1,2,\ldots p-1\ \varepsilon_1=\pm1\ \varepsilon_2=\pm1$

For how many $n$ among $1,2,\ldots p-2$ is
$\left(\frac{n}{p}\right)=\varepsilon_1\
\left(\frac{n+1}{p}\right)=\varepsilon_2$

Suppose the answer is $\psi(\varepsilon_1,\varepsilon_2)$

$4\psi(\varepsilon_1,\varepsilon_2)=\sum_{n=1}^{p-2}
\left(1+\varepsilon_1\left(\frac{n}{p}\right)\right)
\left(1+\varepsilon_2\left(\frac{n+1}{p}\right)\right)$

\end{description}

\end{document}
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