# proof of quadratic reciprocity rule

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

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

 $\displaystyle R_{0}$ $\displaystyle=$ $\displaystyle\{(x,y)\in P|x>0,y>0\}$ $\displaystyle R_{1}$ $\displaystyle=$ $\displaystyle\{(x,y)\in P|x<0,y\geq 0\}$ $\displaystyle R_{2}$ $\displaystyle=$ $\displaystyle\{(x,y)\in P|x\geq 0,y<0\}$

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 $1\leq k\leq b$ and $0\leq l\leq a$. Thus

 $N_{0}+N_{1}=ab+b-(b-v)+u$

i.e.

 $\displaystyle N_{0}+N_{1}$ $\displaystyle=$ $\displaystyle ab+u+v.$ (1)

Swapping $p$ and $q$, we have likewise

 $\displaystyle N_{0}+N_{2}$ $\displaystyle=$ $\displaystyle ab+u+v.$ (2)

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

 $\displaystyle N_{1}+N_{2}$ $\displaystyle=$ $\displaystyle ab+u+v.$ (3)

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 http://www.rzuser.uni-heidelberg.de/ hb3/fchrono.htmlLemmermeyer .

Title proof of quadratic reciprocity rule ProofOfQuadraticReciprocityRule 2013-03-22 13:16:15 2013-03-22 13:16:15 mathcam (2727) mathcam (2727) 12 mathcam (2727) Proof msc 11A15