$a,b$ be two positive integers , where $b>2 $ , then is it possible that $2^b-1 \mid 2^a+1$?

Question

If $a,b$ be two positive integers , where $b>2 $ , then is it possible that $2^b-1\mid2^a+1$ ? I have figured out that if $2^b-1\mid 2^a+1$, then $2^b-1\mid 2^{2a}-1$ , so $b\mid2a$ and also $a >b$ ; but nothing else. Please help. Thanks in advance

Answer 1

Assume that it is possible.Then obviously $a>b$ so write $a=bx+r$ with $r \leq b-1$ .Then : $$2^b-1 \mid 2^{bx}-1$$

Multiply by $2^r$ to get : $$2^b-1 \mid 2^{bx+r}-2^r=2^a-2^r$$

But we known that $2^b-1 \mid 2^a+1$ so subtracting them we get :

$$2^b-1 \mid 2^r+1$$

This means that : $$2^b-1 \leq 2^r+1 \leq 2^{b-1}+1$$ $$2^{b-1} \leq 2$$ so $$b \leq 2$$ a contradiction .

Answer 2

If $b$ is odd, then $b\mid 2a$ implies $b\mid a$, then $2^b-1\mid 2^a-1$ and hence $2^b-1\nmid 2^a+1$ (as $2^a+1$ is between $2^a-1$ and $2^a-1+(2^b-1)$).

If $b$ is even, $b=2c$ say, then $c\mid a$, hence $2^c-1\mid 2^a-1$ and $2^c-1\mid 2^b-1$. As $c>1$ this shows $d:=\gcd(2^a-1,2^b-1)>1$ (and of course odd) and so $\gcd(2^a+1,2^b-1)<\frac{2^b-1}{d}<2^b-1$.