How do Ι show if $a^n-1$ is prime for $n >1$ then $a=2$?

Question

It is well known that $a^n-1$ is prime Mersann formula for some pimes $p$ , I would like to show this implication if $a^n-1$ is prime for $n >1$ with $a$ is a positive integer then $a=2$ ?

Edit: I edited the question as it is related to the precedent question

Note : I have tried to show that for $a$ is odd it's obvious no integer $a$ satisfy $(a^n-1)$ which means it's not prime for an odd integer $a$

I know , in my question i assume that if a^n-1 is prime number than a =2 — zeraoulia rafik, Mar 28 '16 at 17:57
But $3^1 - 1$ is a counterexample to "if $a^n-1$ is prime then $a = 2$" unless you put additional restrictions on $n$. — , Mar 28 '16 at 17:59
thanks for that i forget to add exeception for n=1 , now it's fixed — zeraoulia rafik, Mar 28 '16 at 18:03

score 5 · Accepted Answer · answered Mar 28 '16 at 17:45

5

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

So, as long as $n>1$,you get two terms.

answered Mar 28 '16 at 17:45

N. S.

132,525

score 1 · Answer 2 · answered Mar 28 '16 at 17:48

1

Hint let $a\neq 2$ then we can do $a^n-1^n=(a-1)(a^{n-1}+....+)$ now if $a\neq 2$ then the subtraction a-1 becomes even as 2 is largest even prime which gives odd sum when 1 is subtracted and all other primes give even numbers. Tgus $a=2$

answered Mar 28 '16 at 17:48

Archis Welankar

15,910

How do Ι show if $a^n-1$ is prime for $n >1$ then $a=2$?

2 Answers2