Prove that $-1$ is a quadratic residue of an odd prime $p$ if and only if $p \equiv 1 \pmod 4$.
Let's remind ourselves of Euler’s Criterion (7.5).
Let $p$ be an odd prime such that $p \not \mid a$. Then $a$ is a quadratic residue of $p$ if and only if $a^{\frac{p−1}{2}} \equiv 1 \pmod p$.
We write the odd prime $p$ as $p=4k + t$ where $t$ can only be one of 1 or 3, and $k$ is some integer.
Since $p \not \mid -1$, Euler's Criterion then tells us that $-1$ is a quadratic residue of $p$ if and only if
$$ (-1)^{\frac{p−1}{2}} \equiv (-1)^{\frac{4k + t −1}{2}} \equiv 1 \pmod p $$
If $t=3$, that is $p \equiv 3 \pmod 4$, then
$$ (-1)^{\frac{4k + t −1}{2}} \equiv (-1)^{\frac{4k + 2}{2}} \equiv (-1)^{2k + 1} \equiv -1 \not \equiv 1 \pmod p$$
If $t=1$, that is $p \equiv 1 \pmod 4$, then
$$ (-1)^{\frac{4k + t −1}{2}} \equiv (-1)^{\frac{4k + 0}{2}} \equiv (-1)^{2k} \equiv 1 \pmod p$$
So $-1$ is a quadratic residue of $p$ if and only if $p \equiv 1 \pmod 4$ means $-1$ is a quadratic residue of $p$.