How can I prove that the value of $\varphi(p^n-1)$ (where $p$ is prime and $n$ is some positive integer) is some multiple of $n$? The purpose of this is to prove that $n$ divides $\varphi(p^n-1)$.
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Related: Prove that $n^2\mid\varphi(a^n-1)$ for $n$ even – Bart Michels Apr 11 '15 at 09:28
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Consider the group of units $\Bbb Z/D\Bbb Z^\times$ where $D=p^n-1$, $p$ a prime.
What is the order of $p$?
Pedro
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the order is n, I have proved that. Now I must prove that n divides ϕ(p^n-1), and I figured the best way to do so is to prove that phi(p^n-1) is a multiple of n, but I'm not exactly sure how... – malxmusician212 Nov 04 '13 at 00:54
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@MalcolmLazarow Don't you have a theorem that says that if $f$ is the order of $g\in G$, $G$ a group, then $g^k=1\iff f\mid k$? – Pedro Nov 04 '13 at 00:56
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@Omnitic Not really. From the division algorithm and the definition of order. – Pedro Nov 04 '13 at 00:58
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