Prime Divisors of Cyclotomic Polynomials

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Theorem

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

Let $\map {\Phi_n} x$ denote the $n$th cyclotomic polynomial.

Let $a \in \Z$ be an integer such that $\map {\Phi_n} a \ne 0$.

Let $p$ be a prime divisor of $\map {\Phi_n} a$.


Then $p \equiv 1 \pmod n$ or $p \divides n$.


Proof

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

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

If $k = n$, the result follows.


Let $k < n$.

Then by Product of Cyclotomic Polynomials, there exists $d \divides k$ such that $p \divides \map {\Phi_d} a$.

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

$\blacksquare$


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