Question: > 
> Disprove that if $n$ is an odd positive integer greater than $1$ then $2^n-1$ is prime.
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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 $2^n+1=2^{2x+1}-1$

$=6(2^n)-1$

Answer: 
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-positiv*ing*" 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