bump!!!!!!! me spending so much time on this my classmates are laughing at me :V All I know is $$\sum_{k=1}^{p-1} \left\{ \frac{k^{2n}}{p} - \frac{1}{2} \right\} = \sum_{k=1}^{p-1} \left\{ \frac{k^{2}}{p} - \frac{1}{2} \right\}$$which is obvious. But idk where to continue from here. Subtracting $\frac{p-1}{2}$ from both sides we "only" need to prove the sum of all quadratic residues is equal to $p$ times the number of quadratic residues larger than $\frac{p}{2}$, which seems to me is even more abstract ·_· I know this is probably ez so just give a hint?

@above, thanks for giving source to this, here is my solution: I will continue with your work, notice that $$\left\{ \frac{k^{2}}{p} - \frac{1}{2} \right\}=\frac{1}{2}+\frac{k^{2}}{p} - \left\lfloor\frac{k^{2}}{p}+\frac{1}{2}\right\rfloor=\frac{1}{2}+\frac{k^{2}}{p}-\left\lfloor\frac{2k^{2}}{p}\right\rfloor+\left\lfloor\frac{k^{2}}{p}\right\rfloor=\frac 12+\left\{ \frac{2k^{2}}{p} \right\}-\left\{ \frac{k^{2}}{p} \right\},$$now we only need $$\sum_{k=1}^{p-1}\left\{ \frac{2k^{2}}{p} \right\}=\sum_{k=1}^{p-1}\left\{ \frac{k^{2}}{p}\right\}\iff\sum_{k=1}^{\frac{p-1}2}\left\{ \frac{2k^{2}}{p} \right\}=\sum_{k=1}^{\frac{p-1}2}\left\{ \frac{k^{2}}{p}\right\}.$$But by $p\equiv 7\pmod 8$ we get $\left(\frac 2p\right)=(-1)^{\frac{p^2-1}8}=1,$ so we are done.$\Box$