<s>[INST] Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$ 
Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$

I tried assuming it was equal to $\frac p q $.
I get $2^nq^2-q^2 = p^2 $ 
But I don't see where to go from there.
 [/INST] To continue your idea, rather than to restart and do one of the other correct answers.
$2^nq^2 - q^2 = p^2$.
Assume $p = 1$
The $2^nq^2 - q^2 = 1$.  Then $2^n - 1 = 1/q^2$ is an integer.  So $q=1$.  So $2^n - 1 = 1$.  So $2^n = 2$ so $n = 1 \not > 1$.
Assume $p \ne 1$.
Let $k$ be a prime factor of $p$.  Then either $k|q^2$ which isn't possible as we assumed (or should have assumed) $p/q$ is in lowest terms, or $k|2^n - 1$. 
This is true for all prime factors so $p^2|2^n -1$. 
So $q^2\frac{2^n - 1}{p^2} = 1$ and $\frac{2^n - 1}{p^2} \in \mathbb Z$ so $\frac{2^n - 1}{p^2} = 1/q^2 \in \mathbb Z$ so $q = \pm 1$ and
$2^n - 1 = p^2$.
$p$ is clearly odd so let $p = 2p' + 1$
$2^n = 4p'^2 + 4p' + 2$
$4 \not \mid 4p'^2 + 4p +2 = 2^n$ so $n < 2$.  So $n \le 1$.  So $n \not > 1$.
</s>