If $n>1$, can $2^n-1$ be a power of $n$? I have tried some modular arithmetic but there is probably an easy solution that I am missing.
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7If you mean if $2^n-1=n^a$ for some integer $a>0$, the answer is no, as special case of Catalan's conjecture-Mihăilescu's theorem. – Nulhomologous May 22 '20 at 15:15
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I do, yes! Thank you – AnalysisStudent0414 May 22 '20 at 15:17
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2This does not discard there is an elementary argument for this special case. For example, it is easy to show that $a$ must be even... – Nulhomologous May 22 '20 at 15:27
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1Yep, I am leaving the question open. But first and foremost it's nice to know that it's true! – AnalysisStudent0414 May 22 '20 at 15:31
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@AnalysisStudent0414 It;s best to delete dupes of FAQs like this. – Bill Dubuque May 22 '20 at 15:53
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@Gone: It's not an actual duplicate since it's a weaker claim, which, in fact, allows for a more elementary proof. – quasi May 22 '20 at 15:58
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Well, it doesn't let me delete it since others have put time and effort. I'll just leave it as it is! – AnalysisStudent0414 May 22 '20 at 16:00
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I was ready to post a more elementary proof but it was already closed. – quasi May 22 '20 at 16:02
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@quasi Please see Coping with abstract duplicate questions. Likely you can understand the concerns from the analogous issues in your sci.math days (eternal September, etc). – Bill Dubuque May 22 '20 at 16:04
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@Gone: My point is that a proof can be had for this statement that doesn't even need Fermat's little Theorem, so requires less of the reader. – quasi May 22 '20 at 16:06
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@quasi If you have something novel to add beyond the answers in the linked dupe(s) then please post another answer there. There are currently 17 listed dupes of that, but likely many more, so I'd be surprised if we don't already have most known proofs. – Bill Dubuque May 22 '20 at 16:10
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@quasi What novelty do you have in mind? – Bill Dubuque May 22 '20 at 16:14
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@Gone: Requring $2^n-1$ to be a power of $n$ allows for an easier contradiction than just assuming $2^n-1$ is a multiple of $n$. If you open the question, I'll post it (and you can always close it again if you choose to). – quasi May 22 '20 at 16:19
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@quasi No, we can only close once. Please elaborate so I can decide on reopening. – Bill Dubuque May 22 '20 at 16:23
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@Bill Dubuque: My proof is more elementary than the proofs in the linked reference, and moreover, I prove that if $n > 1$, then $2^n-1$ can't be of the form $a^b$, where $a,b$ are positive integers with $b > 1$ – quasi May 22 '20 at 16:28
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@quasi Please give the details since that too may be a dupe. – Bill Dubuque May 22 '20 at 16:28
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@Bill Dubuque: If that result is a dupe, please give a link. – quasi May 22 '20 at 16:29
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@quasi Ah, misread it as a divisor. I don't know what proof you have in mind so I can't assess if it is a dupe or not. – Bill Dubuque May 22 '20 at 18:13
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@Bill Dubuque: If the question of whether $2^n-1$ can be a nontrivial perfect power is not a dupe, then why would you need to see my proof first (which by the way, is really simple)? What's the big deal? – quasi May 22 '20 at 18:22
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@quasi Because that too is a FAQ (e.g. here and here, etc), so your proof is likely to be a dupe. If you think you have a novel proof of that then please add it to one of those more specific questions. – Bill Dubuque May 22 '20 at 18:29
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@Bill Dubuque: The key issue was whether the claim is a dupe, and as you've now shown, in fact it is. And yes, for the answers in the links you've now provided, my proof is essentially the same as at least one of them, so I'll now forget about this question. Thanks for taking the time to address my concerns. – quasi May 22 '20 at 18:36
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@quasi It was time well spent, since now readers have links (see sidebar) to other approaches, thanks to you. This is one of the main values the site provides - to efficiently link together related results to aid in learning (and to possibly spark interest in topics that one might not have realized were related). – Bill Dubuque May 22 '20 at 18:38
