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If $p$ is a prime, the discussion of the congruence $x^2 \equiv a(p)$ is fairly easy. It is solvable iff $a^{(p-1) / 2} \equiv 1(p)$. With this fact in hand a complete analysis is a simple matter. However, if the question is turned around, the problem is much more difficult. Suppose that $a$ is an integer. For which primes $p$ is the congruence $x^2 \equiv a(p)$ solvable? The answer is provided by the law of quadratic reciprocity. This law was formulated by Euler and A. M. Legendre but Gauss was the first to provide a complete proof. Gauss was extremely proud of this result. He called it the Theorema Aureum, the golden theorem.
如果 $p$ 是素数，讨论同余$x^2 \equiv a(p)$ 很简单，当且仅当 $a^{(p-1) / 2} \equiv 1(p)$ 是可解的。然而，如果把问题转过来，问题就难得多了。假设 $a$ 是一个整数。对于哪些素数 $p$，同余式 $x^2 \equiv a(p)$ 是可解的？二次互易律给出了答案。这一定律由欧拉和 A. M. 勒让德提出，但高斯是第一个提供完整证明的人。高斯对这一结果极为自豪。他称之为 Theorema Aureum。
We shall present an ingenious proof due to Eisenstein. For a somewhat more elementary and standard proof, see.
A complex number $\zeta$ is called an $n$th root of unity if $\zeta^n=1$ for some integer $n>0$. If $n$ is the least integer with this property, then $\zeta$ is called a primitive $n$th root of unity.
The $n$th roots of unity are $1, e^{2 \pi i / n}, e^{(2 \pi i / n) 2}, \ldots, e^{(2 \pi i / n)(n-1)}$. Among these the primitive $n$th roots of unity are $e^{(2 \pi i / n) k}$, where $(k, n)=1$.
If $\zeta$ is an $n$th root of unity and $m \equiv l(n)$, then $\zeta^m=\zeta^l$. If $\zeta$ is a primitive $n$th root of unity and $\zeta^m=\zeta^l$, then $m \equiv l(n)$.
If $n>0$ is odd, we have
$$
x^n-y^n=\prod_{k=0}^{n-1}\left(\zeta^k x-\zeta^{-k} y\right), \quad \text { where } \zeta=e^{2 \pi i / n}
$$
If $n$ is a positive odd integer and $f(z)=e^{2 \pi i z}-e^{-2 \pi i z}$, then
$$
\frac{f(n z)}{f(z)}=\prod_{k=1}^{(n-1) / 2} f\left(z+\frac{k}{n}\right) f\left(z-\frac{k}{n}\right)
$$
If $p$ is an odd prime, $a \in \mathbb{Z}$, and $p \nmid a$, then
$$
\prod_{l=1}^{(p-1) / 2} f\left(\frac{l a}{p}\right)=\left(\frac{a}{p}\right) \prod_{l=1}^{(p-1) / 2} f\left(\frac{l}{p}\right)
$$
Quadratic Reciprocity. Let $p$ and $q$ be distinct odd primes and $a \geq 1$ an integer. Then the following assertions are equivalent:
(a) $(p / q)(q / p)=(-1)^{((p-1) / 2)((q-1) / 2)}$.
(b) If $p \equiv \pm q(4 a), p \nmid a$, then $(a / p)=(a / q)$.
Proof. In order to show (a) implies (b) it is enough, by multiplicativity, to show that (b) holds when $a$ is prime. For $a=2$ the result follows from Proposition 5.1.3. If $a$ is an odd prime then by (a) $(a / p)=(-1)^{((p-1) / 2)((a-1) / 2)}(p / a)$. If $p \equiv q(4 a)$ then $(p / a)=(q / a)$ so that
$$
\begin{aligned}
\left(\frac{a}{p}\right) & =(-1)^{((p-1) / 2)((a-1) / 2)}\left(\frac{q}{a}\right)=(-1)^{((p-1) / 2)((a-1) / 2)}(-1)^{((q-1) / 2)((a-1) / 2)}\left(\frac{a}{q}\right) \
& =(-1)^{((a-1) / 2)((p+q-2) / 2)}\left(\frac{a}{q}\right) .
\end{aligned}
$$
But $p \equiv q(4 a)$ implies $p+q-2 \equiv 0$ (4) and the result follows. If, on the other hand $p \equiv-q(4 a)$, a similar calculation shows
$$
\left(\frac{a}{p}\right)=(-1)^{((a-1) / 2)((p+q) / 2)}\left(\frac{a}{q}\right) .
$$
Since $p+q \equiv 0$ (4) the result also holds in this case.
To show that (b) implies (a) suppose first of all that $p>q$ and $p \equiv q$ (4). The $p=q+4 a, a \geq 1$. Thus
$$
\begin{aligned}
\left(\frac{p}{q}\right) & =\left(\frac{q+4 a}{q}\right)=\left(\frac{a}{q}\right)=\left(\frac{a}{p}\right)=\left(\frac{4 a}{p}\right)=\left(\frac{p-q}{p}\right)=\left(\frac{-q}{p}\right) \
& =(-1)^{(p-1) / 2}\left(\frac{q}{p}\right) .
\end{aligned}
$$
If $p \equiv 1$ (4) then $(p / q)=(q / p)$ which gives (a). If $p \equiv 3$ (4) then $q \equiv 3$ (4) and we obtain $(p / q)=-(q / p)$ which is part (a) in that case. Finally if $p \equiv$ $-q$ (4) then, $p+q=4 a$ and
$$
\left(\frac{p}{q}\right)=\left(\frac{-q+4 a}{q}\right)=\left(\frac{a}{q}\right)=\left(\frac{a}{p}\right)=\left(\frac{4 a}{p}\right)=\left(\frac{p+q}{p}\right)=\left(\frac{q}{p}\right) .
$$
Thus $(p / q)=(q / p)$ which is the assertion of part (a) since in this case at least one of $p$ or $q$ must be congruent to 1 modulo 4 . The proof is complete.
下面是一些经典的Quadratic Reciprocity题目
