If $p, q$ are primes and $p$ does not divide $n-1$ and $p$ divides $n^q-1$. How to prove that $q|p-1$.
What we can see is that $p|(n^{q-1}-n^{q-2}+\cdots -1)$. Now how do I prove that $p|q-1$
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1can you use Fermat's little theorem? – J. W. Tanner Jul 01 '19 at 20:09
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What can you say about the order of $n$ is a certain group of a relevant order? – Jyrki Lahtonen Jul 01 '19 at 20:13
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1Note you have switched $p$ and $q$ around in the end part of your question where you ask "Now how do I prove that $p|q-1$". – John Omielan Jul 01 '19 at 20:20