$$ \sum_{a=1}^{p-1} (\frac{a}{p})a^n \equiv{0} \pmod{p} $$
I'm supposed to show this for $p \equiv 1 \pmod 4$ with $n=1$, and for $ p\gt 5$ with $n=2$.
I already know the standard case of $n=0$. For $n \gt 0$ I assume a proof uses some information about the distribution of the QRs and NRs between $1$ and $p$ (because the higher numbers will have more weight in the sum), but I'm not sure what symmetry I should be looking for.
We do the $p\equiv 1\pmod{4}$ case.
Note that since $-1$ is a quadratic residue of $p$, we have $\left(\frac{a}{p}\right)=\left(\frac{p-a}{p}\right)$.
Note also that $p-a\equiv -a\pmod{p}$. So if $n=1$, or more generally if $n$ is odd, then $$\left(\frac{a}{p}\right)(p-a)^n+\left(\frac{a}{p}\right)a^n\equiv 0\pmod{p}.$$ The symmetry used is the left-right symmetry of the Legendre symbol in this case, and the fact that $(-a)^n=-a^n$ if $n$ is odd.