Prove if, $2^n - 1$ is prime, then $n$ is prime.

Question

Prove, when $n$ is a positive integer, if $2^n - 1$ is prime, then $n$ is prime.

I did read some sort of proving on the web, but I could not understand it...

Any help?

And if possible, could the explanations be very clear?

Thank you guys :)

So, I've found this from somewhere:

Theorem. If $2^n−1$ is prime then $n$ is prime.

Proof. Suppose that $2^n−1$ is prime, and write $n=st $ where $s,t $ are positive integers. Since $x^s−1=(x−1)(x^{s−1}+x^{s−2}+⋯+1)$ , we can substitute $x=2^t $ to see that $2^t−1$ is a factor of $2^n−1$. Since $2^n−1$ is prime there are only two possibilities, $2^t−1=1$ or $2^t−1=2^n−1 $. Therefore $t=1 $ or $t=n$. We have shown that the only possible factorisations of $n$ are $n×1$ and $1×n$. Hence, $n$ is prime.

However, I have a few questions:

1. how do you know $x^s−1=(x−1)(x^{s−1}+x^{s−2}+⋯+1) $
2. why is$ 2^t−1 $a factor of $2^n−1$
3. how do you determine $2^t−1=1$or$2^t−1=2^n−1$from 'Since $2^n−1$ is prime'
4. 'the only possible factorisations of $n$ are$ n×1$ and $1×n$. ' how?
5. how did 'Hence, $n$ is prime' come out as result?

Answer 1

Hint: First, you'd need to know that $(a-b)\mid (a^n - b^n)$ for all $n\in\mathbb{N}$. Once you do that, consider the contrapositive (if $n$ is not prime, then $2^n - 1$ is not prime).

Answer 2

If you know the formula for the sum of the terms of a geometric progression find the sum of first $q-1$ terms of a G.P. with first term $1$ and common ratio $2^p$, and remember that the sum is an integer.