Prime Factors of Cyclotomic Polynomials I'm trying to show that if $q$ is a prime and $f_{q}(x)$ is the $q$-th cyclotomic polynomial, then all prime divisors of $f_{q}(a)$ for some fixed $a \neq 1$ either satisfy $p \equiv 1\, \text{mod}\; q$ or $p = q$. 
Clearly $f_{q}(a) \equiv 1\, \text{mod}\; q$, because $1 + a + \cdots + a^{q-1} \equiv 2 + a + \cdots + a^{q-2} \equiv 1\, \text{mod}\; q$. However, I can't quite get from here to the conclusion. Any tips or suggestions would be appreciated.

Suppose $1+x+x^2+ \ldots + x^{q-1} \equiv 0 \pmod p$ for some integer $x$. 
If $x \equiv 1 \pmod p$, then $q \equiv 0 \pmod p$, and because $p$ and $q$ are prime, $p=q$.
If $x \neq 1 \pmod p$, then $p$ divides $(1+x+x^2+ \ldots + x^{q-1})(x-1) = x^q - 1$, which means that $x$ is a nontrivial $q$th root of unity modulo $p$. Fermat's little theorem says that $x^{p-1} \equiv x^q \equiv 1 \pmod p$. Since $x$ is not $1$ and $q$ is prime, this implies that $q$ is the order of $x$ in the group of units mod $p$, and that $p-1$ is a multiple of $q$. So $p \equiv 1 \pmod q$
