I want to show that if $2^n-1$ is prime, then $n$ is prime.
I could not see a way to a direct proof based on the definition of a prime number. In that case, we could only write $2^n-1$ as $1 \cdot (2^n-1)$, and I don't see how that would help.
I think the best way to go about this proof is to use the contrapositive of the statement, which I believe is "if $n$ is not prime, then $2^n-1$ is not prime". So, if $n$ is not prime, it is composite and can be factored into the form $ab$. We plug that into $2^n-1$ to get $2^{ab}-1$. Now, if either $a$ or $b$ is even, we could use the laws of exponents and the difference of two squares to get a factored form and therefore prove that $2^n-1$ is composite. The problem in this approach is showing that if both $a$ and $b$ are odd, how the resulting expression can factor. We write this symbolically as $2^{(2k+1)(2m+1)}-1$ for some natural numbers $k$ and $m$. I don't know how to factor this. If anyone sees how to factor this, that would be awesome. Or, if anyone sees a flaw in my approach, that would also be greatly appreciated.
Finally, I ask that y'all not directly give me the answer, but rather a powerful hint. I am really trying to learn on my own, and I feel that the struggle will eventually pay off. Thanks in advance.
