<s>[INST] Number theory fun problem Say $a,b > 2 $ are integers. Then we have that $2^a + 1$ is not divisible by $2^b - 1$.
Any thoughts on how to tackle this problem???
 [/INST] Let $a = bq + r$ where $0 \leq r < b$. Hence, $$2^a+1 = 2^{bq} \cdot 2^r +1 = (2^{bq}-1)\cdot 2^r + 2^r+1$$
But $2^b-1$ divides $2^{bq}-1$ since $(x-1)$ divides $x^m-1$ for all $m \in \mathbb{N}$.
Hence, if $2^b-1$ divides $2^a+1$, then $2^b-1$ has to divide $2^r+1$. But $0 \leq r < b$. Hence, we need $$2^b-1 \leq 2^r + 1 \leq 2^{b-1} + 1$$
i.e. $$2^b - 2^{b-1} \leq 2$$
This is possible only if $b=1$ or $2$. But we are given that $b > 2$.
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