<s>[INST] 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?
 [/INST] 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 $$\frac{2^a + 1}{2^b -1} = \mbox{non integer} + \mbox{some integer}$$
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