Find the pairs of integers $(a, b)$ such that $2^{a}+1$ is divisible by $2^{b}-1$.
If $b=1$, then $2^{b}-1=1$ divides everything, so $(a, 1)$ works. Next, if $b=2$, then $2^{b}-1=3$ divides $2^{1}+1=3$ and for any odd $a$, $2^{a}+1$ is divisible by 3, so $(2 k+1,2)$ works.

Now, let's take $b \geq 3$. First, if $a<b$, then $2^{a}+1<2^{b}-1$, and it is impossible to have $2^{b}-1 \mid 2^{a}+1$. If $a \geq b$, I perform the Euclidean division of $a$ by $b$: $a=b q+r$. Then I see that:

$$
2^{a}+1=\left(2^{b}-1\right)\left(2^{a-b}+2^{a-2 b}+\cdots+2^{a-q b}\right)+2^{r}+1
$$

![](https://cdn.mathpix.com/cropped/2024_05_10_8c44cd8b5c585c15e58bg-294.jpg?height=63&width=780&top_left_y=1967&top_left_x=267)

The only solutions are therefore $(a, 1)$ and $(2 k+1,2)$.