Prime Divisors of Cyclotomic Polynomials

Theorem
Let $n>0$ be a positive integer.

Let $\Phi_n(x)$ denote the $n$th cyclotomic polynomial.

Let $a\in\Z$ be an integer such that $\Phi_n(a)\neq0$.

Let $p$ be a prime divisor of $\Phi_n(a)$ (which is guaranteed to be an integer by Cyclotomic Polynomial has Integer Coefficients).

Then $p\equiv1\pmod n$ or $p\mid n$.

Proof
Let $k$ be the order of $a$ modulo $p$. By Element to Power of Multiple of Order is Identity, $k\mid p-1$. If $k=n$, the result follows.

If $k<n$, by Product of Cyclotomic Polynomials, there exists $d\mid k$ such that $p\mid\Phi_d(a)$.

Consequently, $a$ is a double root of $\Phi_d\Phi_n$ modulo $p$.

Again by Product of Cyclotomic Polynomials, $a$ is a double root of $x^n-1$ modulo $p$. Thus, by Double Root of Polynomial is Root of Derivative, $a$ is a root of the derivative of $x^n-1$ modulo $p$, which is the constant polynomial $n$.

Thus $n\equiv0\pmod p$, for a nonzero constant polynomial has no roots.