Sum of squares of the quadratic nonresidues modulo $p$ is divisible by $p$ Let $p$ be a prime number with $p > 5$. Prove that the sum of the squares of the quadratic nonresidues modulo $p$ is divisible by $p$.
My idea is to use the fact that any quadratic residue is congruent modulo $p$ to one integer in the set $\{1^2,2^2,\ldots,\left( \frac{p-1}{2} \right)^2 \}$. And none of the quadratic residues are congruent to each other. So the quadratic nonresidues must all be in the set $\{ \left( \frac{p+1}{2} \right)^2 , \left( \frac{p+3}{2}\right)^2 \ldots, (p-1)^2\}$. So the sum of the quadratic nonresidues must be given by $$\sum_{k=1}^{p-1} k^2 - \sum_{k=1}^{\frac{p-1}{2}}k^2 = \frac{(p-1)p(2p-1)}{6} - \frac{\left( \frac{p-1}{2} \right) \left( \frac{p+1}{2} \right) p}{6} = \frac{p}{6} \left( (p-1)(2p-1) - \frac{1}{4}(p-1)(p+1) \right).$$
Is this the correct approach? Prove that the term in parenthesis is an integer divisible by $6$?
 A: Using primitive roots finishes the problem quickly: Let $g$ be a primitive root, so that the sum of the squares of the quadratic non-residues is just 
$$\sum_{i=1}^{\frac{p-1}{2}}{(g^{2i-1})^2} \equiv g^2\sum_{i=0}^{\frac{p-3}{2}}{g^{4i}} \equiv g^2\frac{1-(g^4)^{\frac{p-1}{2}}}{1-g^4} \equiv \frac{g^2(1-(g^{p-1})^2)}{1-g^4} \equiv 0 \pmod{p}$$ 
where since $p>5$, $p \nmid 1-g^4$ so the manipulations above are valid. 
If you don't want to appeal to primitive roots, then something similar to what you tried can also be done. What we want is to take the sum of the squares of all non-zero elements and subtract the squares of the quadratic residues. So we get
\begin{align}
&\sum_{k=1}^{p-1}{k^2}-\sum_{k=1}^{\frac{p-1}{2}}{(k^2)^2} \\
&\equiv \frac{(p-1)p(2p-1)}{6}-\frac{(\frac{p-1}{2})(\frac{p+1}{2})(p)(3(\frac{p-1}{2})^2+3(\frac{p-1}{2})-1)}{30} \pmod{p}\\
& \equiv p[\frac{(p-1)(2p-1)}{6}-\frac{(\frac{p-1}{2})(\frac{p+1}{2})(3(\frac{p-1}{2})^2+3(\frac{p-1}{2})-1)}{30}] \pmod{p}\\
& \equiv 0 \pmod{p}
\end{align}
where we may treat $\frac{1}{30}=30^{-1}$ as an element in $\mathbb{Z}_p$ since $p>5$.
