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$p$ 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

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  • $\begingroup$ 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. $\endgroup$
    – Wojowu
    Commented Nov 20, 2022 at 8:58
  • $\begingroup$ @Wojowu Could you give me some tips beacuse I have tried the induction but I failed. $\endgroup$
    – Gang men
    Commented Nov 20, 2022 at 9:06
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    $\begingroup$ 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 $\Phi_n(a)\neq 0$. $\endgroup$
    – Wojowu
    Commented Nov 20, 2022 at 9:37
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    $\begingroup$ I didn't write it in the cleanest possible way, but this should help you. $\endgroup$ Commented Nov 20, 2022 at 9:42
  • $\begingroup$ @Wojowu Thanks you so much and I wonder the roots are whether congruence roots or the normal roots? $\endgroup$
    – Gang men
    Commented Nov 20, 2022 at 9:51

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