Prime Divisors of Cyclotomic Polynomials

Theorem
Let $n \ge 1$ be a positive integer.

Let $\Phi_n \left({x}\right)$ denote the $n$th cyclotomic polynomial.

Let $a \in \Z$ be an integer such that $\Phi_n \left({a}\right) \ne 0$.

Let $p$ be a prime divisor of $\Phi_n \left({a}\right)$.

Then $p \equiv 1 \pmod n$ or $p \mathrel \backslash n$.

Proof
Let $k$ be the order of $a$ modulo $p$.

By Element to Power of Multiple of Order is Identity, $k \mathrel \backslash p - 1$.

If $k = n$, the result follows.

Let $k < n$.

Then by Product of Cyclotomic Polynomials, there exists $d \mathrel \backslash k$ such that $p \mathrel \backslash \Phi_d \left({a}\right)$.

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 $n \equiv 0 \pmod p$, for a nonzero constant polynomial has no roots.

Also see

 * Cyclotomic Polynomial has Integer Coefficients