Prove, that if
(m,n) = 1 // m and n are two different primes
then
(2^m -1, 2^mn -1/2^m -1) = 1
We have $\displaystyle \tfrac{2^{m\cdot n} - 1}{2^m - 1} = \left(2^m\right)^0 + .. + \left(2^m\right)^{n-1}\equiv n (\bmod. 2^m - 1)$
So basically we have to show that $\displaystyle n$ doesn't divide $\displaystyle 2^m - 1$ (since $\displaystyle n$ is prime)
Now this cannot always happen since we may choose, say (3, 7).