<s>[INST] $2^a +1$ 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?
 [/INST] 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 $a$. 
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