Second supplement to quadratic reciprocity

Statement

Let $p$ be an odd prime. Then $$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8},$$ where the left side is a Legendre symbol.

Remarks

So $2$ is a quadratic residue modulo $p$ if $p \equiv \pm 1 \pmod 8$, and a nonresidue if $p \equiv \pm 3 \pmod 8$. For example, $3^2 \equiv 2 \pmod 7$, but $2$ is not a square modulo $5$. One proof uses Gauss's lemma with $a = 2$.

History

Euler proved the second supplement.

Explanations

L1 human

Whether $2$ is a square modulo $p$ depends only on the remainder of $p$ on division by $8$. The remainders $1$ and $7$ give a square, and the remainders $3$ and $5$ do not.