Summation with Legendre Symbol $\sum_{a=1}^{p-1}\big(\frac{a}{p}\big)a=0$

Question

I am trying to show the following: if $p$ is prime that is $1$ modulo 4, then $$\sum_{a=1}^{p-1}\bigg(\frac{a}{p}\bigg)a=0$$ I know I need to use the fact that for an odd prime p, $\frac{p−1}{2}$ of the numbers $1\leq k \leq p−1$ are quadratic residues, but I do not know how $p$ being $1$ modulo $4$ comes into play. Do I need to use Fermat's theorem on sums of two squares which tells me that $p$ is the sum of two squares? Any help is appreciated.

Answer 1

If $S$ is the set of quadradic residues, then this sum is equal to $$-\sum_{a = 1}^{p-1} a + 2\sum_{a \in S} a = 2\sum_{a \in S} a - \frac{p(p-1)}{2}.$$

Since $p \equiv 1 \mod 4$, $-1$ is a quadradic residue, so each quadradic residue appears in a pair $a$ and $ p -a$. There are $\frac{p-1}{4}$ total pairs and each pair sums to $p$, so this gives $$2\cdot p\frac{(p-1)}{4} - \frac{p(p-1)}{2} = 0.$$