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This question already has an answer here:

I'm stuck with the following number theory problem:

Show that $(2^{2^m} + 1 , 2^{2^n}+1) = 1$ for any distinct pair of positive integers $n,m$

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marked as duplicate by lulu, MooS, Dietrich Burde, mrp, Especially Lime May 4 '17 at 10:29

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.

  • $\begingroup$ what have you tried? do you know any statements about gcd that might be helpful here? $\endgroup$ – supinf May 4 '17 at 9:35
  • $\begingroup$ Though it is interesting that the duplicate provides three answers, but none has the same approach as mine. $\endgroup$ – MooS May 4 '17 at 9:39
  • $\begingroup$ @MooS Yes, this is nice. But there are more duplicates out there, probably even with your approach. $\endgroup$ – Dietrich Burde May 4 '17 at 9:40
  • $\begingroup$ Yes, this seems highly likely to me :) $\endgroup$ – MooS May 4 '17 at 9:41
  • $\begingroup$ @DietrichBurde, MooS Don't forget to upvote those new users.. $\endgroup$ – reuns May 4 '17 at 9:53
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If $p$ is a (obviously odd) prime divisor of $2^{2^n}+1$, wee see that $2^{2^n} = -1 \mod p$ and thus $2^{2^{n+1}} = 1 \mod p$. This shows $2^{n+1}=ord_p(2)$, in particular $n$ is uniquely determined by $p$. Thus if $p$ divides $2^{2^n}+1$ and $2^{2^m}+1$ we get $n=m$.

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WLOG let $m>n$ and $m=n+c, c>0$ and if $a^{2^n}+1=r$

Now $$a^{2^m}+1=(a^{2^n})^{2^c}+1=(r-1)^{2^c}+1\equiv2\pmod r$$

$$\implies\left(a^{2^m}+1,a^{2^n}+1\right)=\left(2,a^{2^n}+1\right)$$

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