The latex outputs look goofy in my browser. I'm not sure whats up with that so I'm abandoning it.

Suppose that p=2^k-1 is prime. Prove that either k=2 or k is odd.

Proof;

Assume p is prime and n,p,k are integers.

p=2^k-1

assume k=2 then p=3 which is prime. this concludes the first case.

assume k is even. k=2n then,

p=2^{2n}-1=(2^n+1)(2^n-1)

Since k=2n>2, n>1 and both of the factors of p, (2^n+1) and (2^n-1) are greater then 1 for all n

Since p is a prime it has two factors p and 1

since neither factor of p can be 1, p can not be a prime if k is even.

This is a contradiction with the premise that p is a prime, by reductio ad absurdum we conclude that it is not the case that k is even.

Since k is a natural number it can only be even or odd. And as k is not even it must be odd.