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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

Gang men
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  • 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. – Wojowu Nov 20 '22 at 08:58
  • @Wojowu Could you give me some tips beacuse I have tried the induction but I failed. – Gang men Nov 20 '22 at 09:06
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    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 $\Phi_n(a)\neq 0$. – Wojowu Nov 20 '22 at 09:37
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    I didn't write it in the cleanest possible way, but this should help you. – Jyrki Lahtonen Nov 20 '22 at 09:42
  • @Wojowu Thanks you so much and I wonder the roots are whether congruence roots or the normal roots? – Gang men Nov 20 '22 at 09:51
  • $\Phi_{p-1}$ divides $x^{p-1}-1$ and it is coprime with each $x^d-1,d| p-1,d< p-1$. – reuns Nov 20 '22 at 09:51
  • @JyrkiLahtonen Thanks for comments but I don't know why $a$ is also a roots of $\Phi_\ell(x)$ when $\ell\mid p-1$ – Gang men Nov 20 '22 at 10:02
  • @reuns Thanks for your comments and could you give me more details about the coprime between two phynomial? – Gang men Nov 20 '22 at 10:03
  • My argument in the linked thread was not clearly written at all. I tried to edit it. May be somebody else said it better :-) – Jyrki Lahtonen Nov 20 '22 at 10:09
  • @JyrkiLahtonen It is such kind of you and I am reallly appreciated! – Gang men Nov 20 '22 at 10:11

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