I am considering, for an odd prime $p,$ the polynomial $f_p(x)=(x^p-1)/(x-1),$ which is known to be irreducible. I am wondering whether the following claim is true: If for some integer $a>1$ and positive $d>1$ we have that $d$ divides $f_p(a),$ then $d$ is of the form $p^ku$ where $k \ge 0$ and $u \equiv 1$ mod $p.$
My question generalizes another question on this site, which was the case $p=7.$ I could not prove that case. I did a lot of trials of the $p=7$ version and several other small primes $p.$ I did not come up with a particular $a$ in those cases wherein my claim did not hold.
Any information about this claim would be appreciated.
Here is a link to the related question which mine generalizes.
We usually write $\Phi_p$ instead of $f_p$ for this polynomial, which is more commonly known as the '$p$'th cyclotomic polynomial.'
Let $p,q$ be primes and let $a\in\mathbb{Z}$ such that $q\mid \Phi_p(a)$. This means that $\overline{a}\in(\mathbb{Z}/q\mathbb{Z})^\times$ has multiplicative order $p$. Hence, $p\mid \#(\mathbb{Z}/q\mathbb{Z})^\times=q-1$. In other words, $q\equiv 1\pmod p$.
It immediately follows that if $u,a\in\mathbb{Z}$ such that $p\nmid u$ and $u\mid \Phi_p(a)$, then $u\equiv 1\pmod p$, because every prime dividing $u$ is congruent to $1$ modulo $p$.
Now, can you show that $p^2\nmid \Phi_p(a)$ for all $a\in\mathbb{Z}$ and all $p>2$ prime? This lets you conclude not only your result, but also that $k\in\{0,1 \}$.