<s>[INST] Reducibility of $x^4+1$ in prime fields In Dummit and Foote's "Abstract Algebra" there is a proof for the reducibility of $x^4+1$ in which it's stated:

For odd primes p, $p^2\equiv1\pmod 8$. Hence $x^8-1\mid x^{p^2-1}-1 $.

Why does $x^8-1\mid x^{p^2-1}-1$ follow?
 [/INST] As a side answer, not to the original question, you can get this result fairly explicitly.
If $-1=a^2$, for some $a$, then $x^4+1=(x^2+a)(x^2-a)$.
If $2=a^2$ for some $a$, then $x^4+1=(x^2+ax+1)(x^2-ax+1)$.
If $-2=a^2$ for some $a$, then $x^4+1=(x^2+ax-1)(x^2-ax-1)$.
But if $-1$ and $2$ are not squares, modulo $p$, then $-2$ is a a square, modulo $p$.
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