Proof of Quadratic Reciprocity

It is now straightforward to deduce the quadratic reciprocity law.

Proof. [First Proof of Theorem 4.1.7] 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 $p=4a+q$ ,

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

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 4.3.4 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 4.3.4 implies that $\left(\frac{a}{q}\right) = \left(\frac{a}{p}\right)$ . Since $(-1)^{\frac{p-1}{2}\cdot \frac{q-1}{2}}=1$ , the proof is complete. $\qedsymbol$