Quote: Let $p$ be a prime number of the form $4k+1$ and $\frac{m}{n}$ is an irreducible fraction such that $$\sum_{a=2}^{p-2} \frac{1}{a^{(p-1)/2}+a^{(p+1)/2}}=\frac{m}{n}.$$Prove that $p|m+n$.

The idea is to group every residue class with its inverse. Note that it is enough to show that the sum is congruent to $-1 \pmod{p}.$ Consider everything in the finite field $\mathbb{F}_p$ $$\sum_{a=2}^{p-2} \frac{1}{a^{\frac{p-1}{2}}+a^{\frac{p+1}{2}}} = \frac{1}{2}\sum_{a=2}^{p-2} \left(\frac{1}{a^{\frac{p-1}{2}}+a^{\frac{p+1}{2}}}+\frac{1}{a^{\frac{1-p}{2}}+a^{-\frac{p+1}{2}}}\right)=$$$$\frac{1}{2} \sum_{a=2}^{p-2} \frac{a^{\frac{1-p}{2}}+a^{-\frac{p+1}{2}}+a^{\frac{p-1}{2}}+a^{\frac{p+1}{2}}}{\left(a^{\frac{p-1}{2}}+a^{\frac{p+1}{2}}\right) \left(a^{\frac{1-p}{2}}+a^{-\frac{p+1}{2}}\right)}=$$$$\frac{1}{2} \sum_{a=2}^{p-2} \frac{a^{-\frac{p+1}{2}} \left(1+a+a^p+a^{p+1} \right)}{2+a+a^{-1}}=\frac{1}{2}\sum_{a=2}^{p-2} \frac{a^{-\frac{p+1}{2}} \left(1+a+a+a^2 \right)}{a^{-1}\left(a^2+2a+1\right)}=\frac{1}{2} \sum_{a=2}^{p-2} a^{-\frac{p-1}{2}}=\frac{1}{2}\sum_{a=2}^{p-2} a^{\frac{p-1}{2}}=-1$$The last inequality follow due to Euler's criterion and because both $-1$ and $1$ are quadratic residues modulo $p$.