TITLE: $x^m-1 \nmid f(x)$ in $\mathbb{Z}/p\mathbb{Z}[x]$ where $f(x)=(x+1)((x+1)^{2m}+(x+1)^{m}+1)$
QUESTION [8 upvotes]: Problem: Suppose that $p$ is a prime. Suppose that there is $m \in \mathbb{N}$ such that $p=1+3m$. Define:
$$f(x)=(x+1)((x+1)^{2m}+(x+1)^{m}+1) \in \mathbb{Z}/p\mathbb{Z}[x]$$
I would like to prove that if $p \neq 7$ and $p \neq 13$ then $x^m-1 \nmid f(x)$ in $\mathbb{Z}/p\mathbb{Z}[x]$.
Attempt: I observed that $f(x-1)(x^m-1)=x^p-x$. Then I observed that $x^p-x$ has all its roots in $\mathbb{Z}/p\mathbb{Z}$ and that its roots are all distinct. I would like to prove that for $m \geq 6$ then $x^m-1$ and $((x-1)^m-1)$ share at least a root. This is sufficient because then we would have that $x^p-x$ has at least one multiple roots (I mean that it is not simple).

REPLY [3 votes]: An element  $a\in{\bf Z}/p{\bf Z}$ is a root of $x^m-1$ if and only is it is a non-zero cube. If it is also a root of $(x-1)^m-1$, then $a+1$ is also a non-zero cube. This leads to a ${\bf Z}/p{\bf Z}$-rational point $P$ on the  projective genus $1$ curve $x^3=y^3+z^3$
with the property that $x,y,z\not=0$. Since there are nine points with $x,y,z=0$, Hasse's theorem implies that $P$ exists whenever $p+1-2\sqrt{p}>9$. In other words, when $p>16$.