The quadratic reciprocity law is:

Theorem: (Gauss) Let $p$ and $q$ be distinct odd primes, and write $p=2a+1$ and $q=2b+1$ . Then $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{ab}$ .

($\left(\frac{v}{w}\right)$ is the Legendre symbol.)

Proof: Let $R$ be the subset $[-a,a] \times [-b,b]$ of ${\mathbb Z} \times {\mathbb Z}$ . Let $S$ be the interval $$[-(pq-1)/2, (pq-1)/2]$$ of ${\mathbb Z}$ . By the Chinese remainder theorem, there exists a unique bijection $f: S\to R$ such that, for any $s\in S$ , if we write $f(s)=(x,y)$ , then $ x\equiv s \pmod p $ and $ y\equiv s \pmod q$ . Let $P$ be the subset of $R$ consisting of the values of $f$ on $[1, (pq-1)/2 ]$ . $P$ contains, say, $u$ elements of the form $(x,0)$ such that $x<0$ , and $v$ elements of the form $(0,y)$ with $y<0$ . Intending to apply Gauss's lemma, we seek some kind of comparison between $u$ and $v$ .

We define three subsets of $P$ by \begin{eqnarray*} R_{0} & = & \{(x,y) \in P | x > 0, y > 0 \} \\ R_{1} & = & \{(x,y) \in P | x < 0, y \ge 0 \} \\ R_{2} & = & \{(x,y) \in P | x \ge 0, y < 0 \} \end{eqnarray*}and we let $N_{i}$ be the cardinal of $R_{i}$ for each $i$ .

$P$ has $ab+b$ elements in the region $y>0$ , namely $f(m)$ for all $m$ of the form $k+lq$ with $1 \le k \le b$ and $0 \le l \le a$ . Thus $$N_{0}+N_{1} = ab + b - (b-v) + u$$ i.e. \begin{eqnarray} N_{0}+N_{1} & = & ab+u+v. \end{eqnarray}Swapping $p$ and $q$ , we have likewise \begin{eqnarray} N_{0}+N_{2} & = & ab+u+v. \end{eqnarray} Furthermore, for any $s \in S$ , if $f(s)=(x,y)$ then $f(-s)=(-x,-y)$ . It follows that for any $(x,y) \in R$ other than $(0,0)$ , either $(x,y)$ or $(-x,-y)$ is in $P$ , but not both. Therefore \begin{eqnarray} N_{1}+N_{2} & = & ab+u+v. \end{eqnarray}Adding (1), (2), and (3) gives us $$0 \equiv ab + u + v \pmod 2$$ so $$(-1)^{ab}=(-1)^{u}(-1)^{v}$$ which, in view of Gauss's lemma, is the desired conclusion.

For a bibliography of the more than 200 known proofs of the QRL, see Lemmermeyer .