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?
 A: Assume that $2^b-1$ is a divisor of $2^a+1$ and write $a$ as $kb+r$ with $0\leq r<b$. Then:
$$2^a+1\equiv (2^b)^k\cdot 2^r+1 \equiv 2^r+1 \pmod{(2^b-1)} $$
but since $r<b$ and $b>2$, $2^r+1$ is too small to be $\equiv 0\pmod{2^b-1}$.
A: Note for this type of question is always useful to ask what how many times does the denominator easily go into the numerator, here we have $2^a/2^b = 2^{a-b}$, thus $$\frac{2^a + 1}{2^b -1} = \frac{2^a + 1 - 2^{a-b} (2^b - 1)}{2^b - 1} + 2^{a-b} = \frac{2^{a-b} + 1}{2^b - 1} + 2^{a-b}$$
Thus if $a > b$, then 
$$\frac{2^a + 1}{2^b -1} = \frac{2^{a-b} + 1}{2^b - 1} + \mbox{some integer}$$
And this jumps out to me as something that can be repeated.
It is easy to show that similarly for any $x$ with $x > b$, $$\frac{2^{x} + 1}{2^b - 1} = \frac{2^{x-b} + 1}{2^b - 1} + \mbox{some integer}$$
Thus we can repeat this process, "subtracting" $b$ from the power in the numerator and get
$$\frac{2^a + 1}{2^b -1} = \frac{2^{r} + 1}{2^b - 1} + \mbox{some integer}$$
where $r < b$.
The rest is just inequalities to show that this can never be equal to an integer.
Now as $r < b$, thus $r \leq b-1$, thus $2^r \leq 2^{b-1}$.
Also $2^{b-1} < 2^b - 2$, thus $2^r + 1 < 2^b + 1$, thus $$0< \frac{2^{r} + 1}{2^b - 1} < 1$$
Thus $$\frac{2^a + 1}{2^b -1} = \mbox{non integer} + \mbox{some integer}$$
A: 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 $2^r + 1$
A: If $b$ is not a power of $2$, there is some odd prime $p$ dividing $b$.  Then 
$2^b-1$ is divisible by $2^p-1$.  If $q$ is a prime dividing $2^p-1$, then
the order of $2$ mod $q$, i.e. in the multiplicative group mod $q$, is $p$.  But then $2^a \ne -1 \mod q$.  
This leaves the case where $b$ is a power of $2$, say $b = 2^k$, $k \ge 2$.  Now $2^b - 1$ 
 is divisible by both $3$ and $5$.  Now $2^a+1$ is divisible by $3$ iff $a$ is odd, but divisible by $5$ iff $a \equiv 2 \mod 4$.
