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I'm trying to show that if $q$ is a prime and $f_{q}(x)$ is the $q$-th cyclotomic polynomial, then all prime divisors of $f_{q}(a)$ for some fixed $a \neq 1$ either satisfy $p \equiv 1\, \text{mod}\; q$ or $p = q$.

Clearly $f_{q}(a) \equiv 1\, \text{mod}\; q$, because $1 + a + \cdots + a^{q-1} \equiv 2 + a + \cdots + a^{q-2} \equiv 1\, \text{mod}\; q$. However, I can't quite get from here to the conclusion. Any tips or suggestions would be appreciated.

Ben Perez
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  • Not sure what you're asking for. Are you looking for the primes $p$ such that the reduced $q$-th cyclotomic polynomial splits into linear factors? – PITTALUGA Oct 25 '13 at 08:23
  • I edited the post to more clearly reflect what I'm asking for. I'm actually looking at the prime factors of the integer $f_{q}(a)$. – Ben Perez Oct 25 '13 at 10:31

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Suppose $1+x+x^2+ \ldots + x^{q-1} \equiv 0 \pmod p$ for some integer $x$. If $x \equiv 1 \pmod p$, then $q \equiv 0 \pmod p$, and because $p$ and $q$ are prime, $p=q$.

If $x \neq 1 \pmod p$, then $p$ divides $(1+x+x^2+ \ldots + x^{q-1})(x-1) = x^q - 1$, which means that $x$ is a nontrivial $q$th root of unity modulo $p$. Fermat's little theorem says that $x^{p-1} \equiv x^q \equiv 1 \pmod p$. Since $x$ is not $1$ and $q$ is prime, this implies that $q$ is the order of $x$ in the group of units mod $p$, and that $p-1$ is a multiple of $q$. So $p \equiv 1 \pmod q$

mercio
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