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Supposing that a real number $c$ is given, is the following true?

"If $n^c$ is a natural number for every natural number $n$, then $c$ is a non-negative integer."

Though this seems true, I can't prove that. Can anyone help?

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Hint: What does it tell you if this holds for some specified natural number, for example a prime? –  Tobias Kildetoft Nov 17 '13 at 9:51
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@901301: In my opinion, it does not seem to help. –  mathlove Nov 17 '13 at 10:24
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@TobiasKildetoft Any proof should use that $n^c$ is an integer, for several integers n, since otherwise the result fails. –  Did Nov 17 '13 at 11:27
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This is an excellent question! It feels like it should have a simple proof, but I can't see any. –  TonyK Nov 17 '13 at 12:54
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Perhaps even the following is true: If $2^c$ and $3^c$ are both integers, then $c$ is an integer. –  TonyK Nov 17 '13 at 19:39

1 Answer 1

up vote 9 down vote accepted
+50

A variant of this question was asked on Mathoverflow here by Alon Amit. As Gerry Myerson answers, in particular, it's apparently sufficient to know that only $2^c$ and $3^c$ and $5^c$ are all integers. It's apparently unknown whether it's sufficient to know that $2^c$ and $3^c$ are integers.

He also mentions that the original question (using $n$ instead of $2,3,5$) was actually a 1971 Putnam problem and Chris Phan provides a link to the solution. (It's problem A6).

(Community wiki because I've done nothing.)

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Thank you very much for great information! –  mathlove Nov 19 '13 at 17:04
    
@mathlove: You're welcome! Incidentally, since this is CW, I don't think you can award me the bounty (or maybe you can??) In any case, I don't need it. So if there is something you or I can do to minimize how much rep this costs you, let me know! (For example, I have no qualms with deleting this answer and letting you write your own answer.) –  Jason DeVito Nov 20 '13 at 3:23
    
Well, ok, so if I'm going to do nothing, then nothing happens, right? Is it OK to you or not? –  mathlove Nov 20 '13 at 15:01
    
@mathlove: Well, I was just trying to look out for your bounty. I am ok with anything you decide to do (or not do). –  Jason DeVito Nov 20 '13 at 16:11
    
Thanks. What I said depends on my understanding that a bounty never comes back. –  mathlove Nov 20 '13 at 16:16

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