TITLE: Zolotarev's Lemma and Quadratic Reciprocity
QUESTION [14 upvotes]: The law of quadratic reciprocity is unquestionably one of the most famous results of mathematics. Carl Gauss, often called the "Prince of Mathematicians", referred to it as "The Golden Theorem". He published six proofs of it in his lifetime. To date over 200 proofs of this result have been found. The single most frustrating thing about this theorem is that there are no easy proofs of it, at least when measured relative to the simplicity of the statement and the mathematics it involves. For someone like myself, who prides themselves on being able to find very slick proofs, it can drive you insane. As an undergraduate, when confronted with the lattice-point proof in an introductory number theory class, I refused to learn it. I thought to myself there's no way I need to go through all that just to prove something so simple. There must be an easier way. Zolotarev's Proof of the law has been to date the simplest, and quite frankly most elegant, proof that I can find. The crucial step involves equating the value of the Legendre symbol with the signature of the permutation on $\mathbb{Z}_q$ induced by left multiplication. It can take a little bit longer going through it the first time, but winds up being one of those proofs you can just remember without needing to re-reference it. I had a difficult time finding the result from a single source, at least in a satisfactory form, and had to compile different results from different sources. I thought others might similarly struggle, and so I've typed it below as a resource for them.

REPLY [4 votes]: I would like to add another proof of the Zolotarev's lemma. Consider $a\in\mathbb{Z}_p^{\times}$ and permutation $\tau_a\colon x\mapsto ax$.
Note that for any permutation $\pi\in S(\mathbb{Z}_p)$ we have (here, we represent $\mathbb{Z}_p$ as set $\{0,1,\ldots,p-1\}$)
$$
\text{sgn}~\pi=(-1)^{\text{inv}~\pi}=\prod_{0\leq i<j\leq p-1}\frac{\pi(i)-\pi(j)}{i-j},
$$
where $\text{inv}~\pi$ is a number of inversions. Hence, 
$$
\text{sgn}~\tau_a=\prod_{0\leq i<j\leq p-1}\frac{\tau_a(i)-\tau_a(j)}{i-j}.
$$
Now, note that $\tau_a(i)\equiv a\cdot i\pmod p$, so
$$
\text{sgn}~\tau_a\equiv\prod_{0\leq i<j\leq p-1}a=a^{p(p-1)/2}\equiv a^{(p-1)/2}\pmod p
$$
due to Fermat's Little Theorem. Finally, by Euler's criterion we have $\left(\dfrac{a}{p}\right)\equiv a^{\frac{p-1}{2}}\pmod p$. Thus,
$$
\text{sgn}~\tau_a\equiv\left(\dfrac{a}{p}\right)\pmod p,
$$
which reduces to (because $\text{sgn}~\pi\in\{-1,1\}$ and $p>2$)
$$
\text{sgn}~\tau_a=\left(\dfrac{a}{p}\right),
$$
as desired.