Prove that if $a^n-1$ is prime, then $a=2$ and $n$ is prime.

Question

I just started a semester with Number Theory, and I don't know how I am expected to be dealing with such questions. I am really frustrated. I would really appreciate your help.

And $n \geq 2$.Take $n=1$ and $a=8$ for example. – abnry Mar 14 '15 at 18:26 — abnry, Mar 14 '15 at 18:26

score 8 · Accepted Answer · answered Mar 14 '15 at 17:57

8

Hint:$a^n-1= (a-1)(a^{n-1}+\cdots+a+1)$.

answered Mar 14 '15 at 17:57

Tim Raczkowski

20,046

Thank you for you hint! Any suggestion as for how to arrive at $a=2$ or $n$ is prime? Should I start with $a=2$? – Meitar Mar 14 '15 at 18:02
1

Plug in a few numbers,including $a=2$ and see what happens. – Tim Raczkowski Mar 14 '15 at 18:03
@Meitar Hint: try to take $a\to a^m$ in the formula above – Winther Mar 14 '15 at 18:03
Oh, do you mean that with the form you brought it to, I can see that $a^n-1$ is divisible not only by itself and by 1? – Meitar Mar 14 '15 at 18:08
No, if both factors are $>1$, then $a$ is composite. – Tim Raczkowski Mar 14 '15 at 18:10
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How is it contradictory to what I asked? – Meitar Mar 14 '15 at 18:13
Oh. I misread it. Sorry. – Tim Raczkowski Mar 14 '15 at 18:14

Prove that if $a^n-1$ is prime, then $a=2$ and $n$ is prime.

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