Show that if $p \equiv 1 \pmod 4$ then $a^{\frac{p-1}{2}}$ where $\gcd(a, p) = 1$ is a quadratic residue of $p$.
We first note that $\gcd(a,p)=1$ means that $p \not \mid a$.
We need to show there exists an $x$ such that $x^2 \equiv a^{\frac{p-1}{2}} \pmod p$, where $p=4k + 1$ for some integer $k$.
Consider $x \equiv a^{\frac{p-1}{4}}$,
$$ x^2 \equiv a^{\frac{p-1}{2}}$$
We need to confirm that $a^{\frac{p-1}{4}}$ is an integer before we conclude it is a square root of $a^{\frac{p-1}{2}}$.
$$ a^{\frac{p-1}{4}}= a^{\frac{4k+1 -1}{4}} = a^{k} $$
Since both $a$ and $k$ are integers, so is $a^k=a^{\frac{p-1}{4}}$.
And so $a^{\frac{p-1}{2}}$ is a quadratic residue of $p$ where $p \equiv 1 \pmod 4$.