For $1\leq a\leq p-1$ and $5\leq p$, show that $$\sum_{(a/p)=1} a \equiv 0 \pmod{p}$$ where $(a/p)$ is the Legende symbol.
I know that there are as many quadratic residues as quadratic nonresidues, but I have no idea of how they are distributed.
Hints or complete solutions would be appreciated.
Note that, quadratic residues are precisely members of the set, $\{1^2,2^2,\dots,\left(\frac{p-1}{2}\right)^2\}$, where I assume the rest of the arithmetic to be done modulo $p$.
Now, $$ \sum_{k=1}^{\frac{p-1}{2}}k^2 \equiv \frac{\frac{p-1}{2}\cdot \frac{p+1}{2}p}{6} \pmod{p}. $$ Since $p\neq 2,3$, the result follows immediately.