A generalization of Hurwitz's theorem about prime numbers

Question

Can you prove or disprove a generalization of Hurwitz's theorem about prime numbers given below?

Theorem.(Hurwitz) Let $F_n(x)$ denote an irreducible factor of degree $\phi(n)$ of $x^n-1$. Then if there exists an integer $q$ such that $F_{p-1}(q)$ is divisible by $p$, $p$ is a prime.

Conjecture.(Generalization of Hurwitz's theorem) Let $F_n^{(a)}(x)$ denote an irreducible factor of degree $\phi(n)$ of $x^n-a^n$ , where $a$ is a positive integer. Let $p$ be a natural number greater than one such that $\operatorname{gcd}(a,p)=1$ . If there exists an integer $q$ such that $F_{p-1}^{(a)}(q) \equiv 0 \pmod{p} $ then $p$ is a prime.

I was searching for counterexample using this PARI/GP code.

This question arose from this answer by Gerry Myerson .

Answer 1

I believe this mostly trivial — since $\gcd(a,p)=1$ then you can 'divide by $a$' modulo $p$. Writing $_aF_n(x)$ for your factor of $x^n-a^n$ and using the existing notation $F_n$ for a factor of $x^n-1$, we have $_aF_n(x)=a^j F_n(x/a)$ where $j$ is the degree of the specific $F_n$. The condition that $_aF_{p-1}(q)\equiv 0$ is then exactly the condition that $F_{p-1}(q/a)\equiv 0\pmod p$.

Answer 2

If the $k-1$-th cyclotomic polynomial $\Phi_{k-1}(x)$ has a root $\xi$ in $\Bbb{Z}/k\Bbb{Z}$ then let a prime $p| k$, the roots of $\Phi_k(x)\bmod p$ lie in the algebraic closure of $\Bbb{F}_p$ and have order $k-1$. One of those roots is $\xi\bmod p$ so that $k-1 | p-1$ ie. $p=k$.

Next we have the factorizations in irreducible $$x^{k-1}-1=\prod_j g_j(x)\in \Bbb{Z}[x]$$ $$x^{k-1}-a^{k-1}=a^{k-1} ((x/a)^{k-1}-1)= \prod_j a^{\deg(g_j)}g_j(x/a)\in \Bbb{Z}[x]$$ So that $F_{k-1}^{(a)}(x) = a^{\phi(k-1)} \Phi_{k-1}(x/a)$ and $F_{k-1}(x)$ has a root in $\Bbb{Z}/k\Bbb{Z}$ iff $\Phi_{k-1}(x)$ does.