$2^a +1$ is not divisible by $2^b-1$.

Let $a,b>2$ be positive integers. We need to show that $2^a +1$ is not divisible by $2^b-1$.

Could any one give me hint?

Let $b$ be a fixed positive integer. If there is a $k$ such that $2^b-1$ divides $2^k+1$, then there is a smallest such $k$. Call this smallest $k$ by the name $a$.

We first show that $a\lt b$. Suppose to the contrary that $a\ge b$. Note that $$2^a+1=2^{a-b}(2^b-1) +2^{a-b}+1.$$ Thus if $2^b-1$ divides $2^a+1$, then $2^{b}-1$ divides $2^{a-b}+1$, contradicting the minimality of $a$.

It follows that $2^b-1$ divides $2^a+1$ for some $a\lt b$. In particular, $2^b-1\le 2^a+1\le 2^{b-1}+1$.

From $2^b-1\le 2^{b-1}+1$, we conclude that $2^{b}-2^{b-1}=2^{b-1}\le 2$. Thus $b-1\le 1$ and therefore $b\le 2$.

Remark: If $b=1$, then $2^b-1$ divides $2^a+1$ for all $a$. If $b=2$, then $2^b-1$ (that is, $3$) divides $2^a+1$ for all odd values of $a$.