I know how to prove that, if $2^n-1$ is prime and $n>1$, then $n$ is prime.
But how do we prove that, if $a^n-1$ is prime and $n>1$, then $a$ must equal 2?
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1If you thought about the proof of the result you know how to prove, that might give you a clue! – almagest Sep 11 '14 at 14:49
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Corollary of this – Caleb Stanford Sep 13 '16 at 09:52
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If we say $P(a)=a^n-1$ we have that $P(1)=0$ and by Polynomial remainder theorem we get that $$a-1\mid a^n-1$$
kingW3
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Well, I know your main idea of proving the result asked above. But, what is $P(1)$? – Idonknow Sep 11 '14 at 15:07
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