Prove $a^n - 1$ is composite

Question

I need to prove that $a^n - 1$ is a composite number when $a>2$ and $n\ge2$ ($a,n$ are integers). I have tried a few things, but couldn't make it.

The question as originally posed made no sense. I am pleased that @Dave made some sense of it. — Hans Hüttel, Nov 23 '17 at 20:22

score 1 · Answer 1 · answered Nov 23 '17 at 20:19

1

Use the factorization $$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+1)$$

answered Nov 23 '17 at 20:19

Dave

Can you explain why this factorization helps me? – oron cohen Nov 24 '17 at 01:45
What does it mean for a number to be composite? – Dave Nov 24 '17 at 01:54
I Understood how to do it, composite is not primary number. Thank you. – oron cohen Nov 24 '17 at 09:38

score 1 · Answer 2 · answered Nov 23 '17 at 20:19

1

Note that $x^m-1=(x-1)(x^{m-1}+\dots 1)$.

answered Nov 23 '17 at 20:19

Andres Mejia

2 Answers2