Find the sum of all quadratic residues modulo $p$ where $p \equiv 1 \pmod{4}$
Because $p\equiv 1\bmod 4$, we have that $-1$ is a quadratic residue modulo $p$. The product of two quadratic residues is a quadratic residue. This means that for any quadratic residue $a$, we have that $-a$ is also a quadratic residue. Thus the sum cancels to $0\bmod p$.
Because there are $\frac{p-1}{2}$ quadratic residues mod $p$, and they all occur in pairs $a$ and $p-a$, there are $\frac{p-1}{4}$ pairs each of whose sum is $p$, hence the sum of them all is $\frac{p(p-1)}{4}$.