<s>[INST] Prove $2^b-1$ does not divide $2^a + 1$ for $a,b>2$ I'm trying to prove $2^b-1$ does not divide $2^a + 1$ for $a,b>2$. Can someone give a hint in the right direction for this?
 [/INST] Assume otherwise and let $a$ be the nonnegative integer such that $2^b-1\mid 2^a+1$ for some $b>2$. 
If $2^a+1=k(2^b-1)$ with $k\ge 1$, then $$2^{a-b}+1 = 2^a+1-2^{a-b}(2^b-1)=(k-2^{a-b})(2^b-1).$$
If $a\ge b$, this shows that $2^b-1\mid 2^{a-b}+1$, contradicting minimality of $a$.
Therefore $a\le b-1$. But then  we have $$2^b-1=2^{b-1}+2^{b-1}-1\ge 2^{b-1}+3>2^a+1$$ contradicting $2^b-1\mid 2^a+1$.

Note that $b=2$ allows us to find $2^2-1=3\mid 9=2^3+1$ or $33=2^5+1$ and so on.
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