<s>[INST] Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$ Problem 

Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$

My attempt was,
Since $p$ divides $n^4 + 1 \implies n^4 + 1 \equiv 0 \pmod{p} \Leftrightarrow n^4 \equiv -1 \pmod{p}$.
It follows that $(n^2)^2 \equiv -1 \pmod{p}$, which implies $-1$ is quadratic residue modulo $p$. Hence $p \equiv 1 \pmod{4} \Leftrightarrow p \equiv 1 \pmod{8}$.
Am I in the right track? 
Thanks,
 [/INST] As you have noticed, $p \equiv 1 \bmod 4$. Then $1 \equiv n^{p-1} = (n^4)^{(p-1)/4} \equiv (-1)^{(p-1)/4} \bmod p$. This means that $(p-1)/4$ is even, i.e., $p\equiv 1 \bmod 8$.
By induction, this argument generalizes to: if an odd prime $p$ divides a number of the form $n^k+1$, where $k$ is a power of $2$, then $p \equiv 1 \bmod {2k}$.
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