TITLE: Does $2^n-1$ divide $1+2^{2i}$?
QUESTION [6 upvotes]: Let $2^n-1$ be a prime number. If $1<i<n$, I need to prove that $2^n-1$ does not divide $1+2^{{2i}}$. Any comment would be appreciated.

REPLY [2 votes]: You can prove more generally that if $n$ is any integer greater than $2$, then $2^n-1$ does not divide $2^m+1$ for any $m$.  
If $n\gt2$, it's clearly the case that $2^n-1$ does not divide $2^m+1$ for $m\lt n$, since $2^n-1\gt2^m+1$.  
Now consider the smallest $m$ such that $2^n-1$ divides $2^m+1$.  Then $2^n-1$ also divides $2^m+1+2^n-1=2^n(2^{m-n}+1)$, which contradicts the minimality of $m$.  Thus no such $m$ exists.
For $n=2$, it's easy to see that $3\mid2^m+1$ if and only if $m$ is odd.