# Prove that the roots of cyclotomic polynomial $\Phi_{p-1}(x) \equiv 0 (mod~p)$ are exactly the primitive roots mod p

$$p$$ is a prime, and $$\Phi_{p-1}(x)$$ denote the cyclotomic polynomial of order $$p-1$$. And I want to show the following:

$$g$$ is a solution of the congruence $$\Phi_{p-1}(x) \equiv 0 (mod~p)$$ if and only if $$g$$ is a primitive root (mod p)

that is:

$$\Phi_{p-1}(g) \equiv 0 (mod~p) \iff g$$ is a primitive root (mod p)

Here is some properties about cyclotomic polynomial:

$${\textstyle \prod_{d|n}^{}}\Phi_{d}(x) = x^n-1\tag{1}$$

$$\Phi_{n}(x) = {\textstyle \prod_{d|n}^{}}(x^d-1)^{\mu(n/d)}\tag{2}$$

$$\mu(x)$$ is the Möbius inversion formula

• In general, when $p\nmid n$, the roots of $\Phi_n$ in $\mathbb F_p$ are going to be precisely the elements of order $n$ in $\mathbb F_p^\times$. This shouldn't be too hard for you to prove by induction. Nov 20, 2022 at 8:58
• @Wojowu Could you give me some tips beacuse I have tried the induction but I failed. Nov 20, 2022 at 9:06
• Firstly show that $x^n-1$ has no double roots. After you do that, note that if a root $a$ has order $d<n$, then $d\mid n$, and so $\Phi_d(a)=0$, and as there are no double roots, this implies $\Phi_n(a)\neq 0$. Nov 20, 2022 at 9:37
• I didn't write it in the cleanest possible way, but this should help you. Nov 20, 2022 at 9:42
• @Wojowu Thanks you so much and I wonder the roots are whether congruence roots or the normal roots? Nov 20, 2022 at 9:51