In the proof of reducibility of $x^4+1$ over $F_p$ (which is stated as a corollary of the structure theorem of the finite field $F_{p^n}$), the following implication is used in the Algebra by Dummit and Foote:
Assume now that $p$ is odd. Then $p^2-1$ is divisible by $8$ since $p$ is congruent mod $8$ to $1,3,5,7$ and all of these square to $1$ mod $8$. Hence $x^{p^2-1}-1$ is divisible by $x^8-1$.
Can anyone help me with the implication "hence" here?
On
I believe another way is to go through algebra. To show $m|n \Rightarrow x^m-1|x^n-1$ consider $\zeta_n$ the $n^{th}$ root of unity (1). The group of unity has order $\phi (n)$ and $\phi(m)$ respectively, where $\phi$ is the Euler function. These are cyclic groups, and in particular, you can find $\phi(m) | \phi(n)$ so that you can realize the roots of unity of $m$ as a unique subgroup of the group of the roots of unity of $n$.Thus the roots of $x^m-1$ are also roots of $x^n-1$ in their respective splitting field we see that $x^m-1|x^n-1$. Your question follows. I realize this is a bit over the top but wtv.
If $n$ divides $m$ then $x^n-1$ divides $x^m-1$. Indeed, call $y=x^n$, $d=m/n$. Then we must show that $y-1$ divides $y^d-1$, but this is true because $1$ is a root of $p(y)=y^d-1$