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This question already has an answer here:

$n$ divides $2^{2^n+1}+1$ $\implies n$ divides $2^{2^{2^n+1}+1}+1$?

There are two ways to try to prove this. One is above, the other is its de Morgan counterpart: $n$ doesn't divide $2^{2^{2^n+1}+1}+1 \implies$ $n$ doesn't divide $2^{2^n+1}+1$. Disproving it requires only one example of course.

Tried using $\gcd(2^a+1, 2^b+1) = 2^{\gcd(a, b)}+1$ (where $a$ and $b$ are odd positive integers), stuck on both ends. I figured out that if $n$ divides $2^n+1$ then n divides both $2^{2^n+1}+1$ and $2^{2^{2^n+1}+1}+1$ but this implication doesn't work backwards (e.g. $n=57$).

Would appreciate some help.

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marked as duplicate by davidlowryduda Oct 16 '13 at 0:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You asked a same question a week ago. $\endgroup$ – Next Oct 16 '13 at 0:44
  • $\begingroup$ I did. Is there a way to bump that? If there is, sorry. (The only pointer I got was not enough for me to finish the problem.) $\endgroup$ – mathaway__ Oct 16 '13 at 0:49