First supplement to quadratic reciprocity

Statement

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

Remarks

So $\left(\frac{-1}{p}\right) = 1$ if $p \equiv 1 \pmod 4$, and $\left(\frac{-1}{p}\right) = -1$ if $p \equiv 3 \pmod 4$. The proof is Euler's criterion with $a = -1$. For example, $2^2 = 4 \equiv -1 \pmod 5$, but $-1$ is not a square modulo $7$.

Explanations

L1 human

The number $-1$ is a square modulo an odd prime $p$ exactly when $p$ leaves remainder $1$ on division by $4$. So $-1$ is a square modulo $5$, $13$ and $17$, but not modulo $3$, $7$ and $11$.