Disprove that if $n$ is an odd positive integer greater than $1$ then $2^n-1$ is prime.
So my approach to this is as follows but I have no idea where to go from here or whether this is the right approach.
Let $n=2x+1$ where $x\in\Bbb Z^+$
Then $2^n+1=2^{2x+1}-1$
$=6(2^n)-1$
On
In general:
If $n$ is composite, then $2^n-1$ is composite. The prove is not hard to follow up. since $n$ is composite, $n = km$. This implies that $2^{km}-1 = (2^k-1)(1+2^k+2^{2k}+2^{3k}...+2^{km-1})$.
If $2^n - 1$ is prime, then $n$ is prime. The prove can be deduced from the previous part by "counter-positiving" this statement. "NOTE THAT THE CONVERSE IS NOT TRUE".
A quick counter example to disprove, take $n$ to be $9$, then $ 2^9-1 = 511 = 7*73$ which is divisible by $2^3 - 1 = 7$ and $1+2^3+2^6 = 73$.
Take $$2^{11}-1=23\cdot 89$$ and this number is not prime