Prove that $\frac{2^a+1}{2^b-1}$ is not an integer 
Let $a$ and $b$ be positive integers with $a>b>2$. Prove that $\frac{2^a+1}{2^b-1}$ is not an integer.

This is equivalent to showing there always exists some power of a prime $p$ such that $2^a+1 \not \equiv 0 \pmod{p^a}$ but $2^b-1 \equiv 0 \pmod{p^a}$. How do we prove the statement from this or is there an easier way?

if $b$ divides $a$ then $2^b - 1$ divides $2^a-1$.
We can use the Euclidean algroithm to find $q,r$ such that $a = qb + r$ with $r < b$.
$2^a+1  = (2^{qb})(2^r) + 2^r - 2^r + 1 = (2^{qb} - 1)(2^r) + 2^r + 1$
$2^b-1$ divides $(2^{qb} - 1)2^r$ leaving a remainder $2^r+1$
If $r<b$ and $b>1,  2^r + 1 < 2^b - 1$  and $2^b - 1$ cannot divide $2^r + 1$
