Show that $x^4 +1 | x^{p^2-1}-1$, for prime $p>3$

Question

I'm reading through a solution in my lecture notes that says that $x^4 +1 | x^{p^2-1}-1$, for prime $p>3$.

I've tried proving this myself but I'm struggling and not sure if it's true or if it's a mistake.

I know that $8 | p^2-1$ so setting $p^2 -1 = 8k$ I can get $$x^{p^2-1}-1 = x^{8k} -1 = (x^{4k})^2 -1 = (x^{4k}+1)(x^{4k}-1)$$

but this clearly isn't what I want. I'm wondering if there's a similar sort of manipulation I can do that will enable me to get $x^4+1$ as a factor, but I can't spot it. Any help you could offer would be appreciated.

Answer 1

$8|p^2-1$ and hence $x^4+1$ divides $(x^4+1)(x^4-1)=x^8-1$, which divides $x^{p^2-1}-1$, since if $m|n$ in $\mathbb{N}_{>0}$ then $x^{m}-1|x^{n}-1$.

Answer 2

$p$ is an odd number. $p^2 - 1$ is divisible by $8$

Your right hand expression is divisible by $x^8-1$