Since for prime $p$, $p^2-1$ is divisible by $8,$ $x^{p^2-1}-1$ is divisible by $x^8-1$.

Question

I was reading Dummit and Foote, and in there, there is a line that goes like this:

Since for prime $p$, $p^2-1$ is divisible by $8,$ $x^{p^2-1}-1$ is divisible by $x^8-1$.

I get that $p^2-1$ is divisible by $8.$ But how does that fact imply that $x^{p^2-1}-1$ is divisible by $x^8-1$?

Answer 1

If $$p^2-1=8k$$ then

$$x^{p^2-1}-1=x^{8k}-1$$

Let $y=x^8$, we have $y^k-1=(y-1)\sum_{r=0}^{k-1}y^r$. That is $y-1$ divides $y^k-1$.

That is $x^8-1$ dividis $x^{8k}-1=x^{p^2-1}-1$.

Answer 2

$$x^{ab}-1=(x^a-1)(x^{a(b-1)}+x^{a(b-2)}+x^{a(b-3)}+\cdots+x^{a}+x^0)$$