Problem - 4184
Let $p$ be an odd prime divisor of integer $(n^4 + 1)$. Show that $p\equiv 1\pmod{8}$.
Given $p$ divides $(n^4+1)$, we have $n^4\equiv -1\pmod{p}$ and $n^8\equiv 1\pmod{p}$.
It is clear that $p$ and $n$ are co-prime, therefore by Euler's theorem, we have $n^{\varphi(p)}\equiv n^{p-1}\equiv 1\pmod{p}$. Therefore $8\mid (p-1)$ or $p\equiv 1\pmod{8}$.