The Quadratic Reciprocity Law

With the lemma in hand, it is straightforward to deduce the quadratic reciprocity law.

Theorem 3.1 (Gauss)   Suppose that $p$ and $q$ are distinct odd primes. Then

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

Proof. First suppose that $p\equiv q\pmod{4}$. By swapping $p$ and $q$ if necessary, we may assume that $p>q$, and write $p-q=4a$. Since $p=4a+q$,

$$\left(\frac{p}{q}\right) = \left(\frac{4a+q}{q}\right) = \left(\frac{4a}{q}\right) = \left(\frac{a}{q}\right),$$

and

$$\left(\frac{q}{p}\right) = \left(\frac{p-4a}{p}\right) = \left(\frac{-4a}{p}\right) = \left(\frac{-1}{p}\right)\cdot \left(\frac{a}{p}\right).$$

Proposition 2.2 implies that $\left(\frac{a}{q}\right) = \left(\frac{a}{p}\right)$, since $p\equiv q\pmod{4a}$. Thus

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

where the last equality is because $\frac{p-1}{2}$ is even if and only if $\frac{q-1}{2}$ is even.

Next suppose that $p\not\equiv q\pmod{4}$, so $p\equiv -q\pmod{4}$. Write $p+q=4a$. We have

$$\left(\frac{p}{q}\right) = \left(\frac{4a-q}{q}\right) = \left(\frac{a}{q}\right),$$    and $$\quad \left(\frac{q}{p}\right) = \left(\frac{4a-p}{p}\right) = \left(\frac{a}{p}\right).$$

Since $p\equiv -q\pmod{4a}$, Proposition 2.2 implies that $\left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)$. Since $(-1)^{\frac{p-1}{2}\cdot \frac{q-1}{2}}=1$, the proof is complete. $\qedsymbol$