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I should show $n^{5}-1= (n-1) \left( n^{4}+n^{3}+n^{2}+n+1\right)$ prime. So, how?

Proof trying. We know $\left( n^{4}+n^{3}+n^{2}+n+1\right)$ is odd.

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    If $n > 2$, then $n^5-1$ is divisible by $(n-1) > 2$, so $n^5-1$ would not be prime. Therefore, for $n^5-1$ to be prime, $n=2$ (which we see that $31$ is divisible by $2$). – user2825632 Oct 10 '16 at 01:54
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    That factorization of $n^5-1$ is all you need to know. What is $n-1$ if $n>2$? The latter factor is an even bigger number, so the product can't possibly be prime. – Matt Samuel Oct 10 '16 at 01:54
  • You can try proving the contrapositive of the statement. If "n" is not equal to two then $n^{5} -1$ will not be prime. This is easy to prove due the way you factored $n^{5} -1$. – user262291 Oct 10 '16 at 01:56
  • Why $\left( n-1\right) >2$? –  Oct 11 '16 at 04:40
  • @user2825632 Why $(n-1)>2$, why not $(n-1)>1$? –  Oct 12 '16 at 00:53
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    You're right - it should have been $(n-1) \ge 2$. – user2825632 Oct 12 '16 at 01:33

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$n^m-1$ is divisible by $n-1$ for all natural numbers $m, n\ge 2$. Proceed from there.

Oscar Lanzi
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