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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?

Bill Dubuque
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mathlove
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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 09:51
  • Maybe an inductive proof might help? – dreamer Nov 17 '13 at 09:53
  • @TobiasKildetoft: Well, nothing special. Could you tell me more about your hint? – mathlove Nov 17 '13 at 10:23
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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
  • @Did I am aware of that. But understanding what happens when it holds for a single prime is an important first step. – Tobias Kildetoft Nov 17 '13 at 15:17
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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
  • If so then for any $p,q$ we have $(p/q)^c$ is a rational, i.e. the function $f(x)=x^c$ maps the rationals into the rationals. [Only a restatement, but maybe someone will know something about such power maps, and can conclude the exponent $c$ must be a natural number.] – coffeemath Nov 18 '13 at 18:00
  • Maybe working with Galois extension? I mean, by contradiction if $c=a/b$, with $a,b$ coprime natural numbers and $b\geq 2$, then $[\mathbb{Q}(\sqrt[b]{n^a}):\mathbb{Q}]=1$ for every natural $n$... – PITTALUGA Nov 19 '13 at 15:47
  • Check http://math.stackexchange.com/questions/378130/what-is-the-set-x-in-bbb-r-mid-forall-q-in-bbb-q-qx-in-bbb-q – Macavity Nov 23 '13 at 11:37
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    The problem @TonyK mentions with $2^c$ and $3^c$ is an open question. – Gerry Myerson Nov 24 '19 at 12:28

1 Answers1

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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.)

Community
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Jason DeVito - on hiatus
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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 - on hiatus Nov 20 '13 at 03:23
  • 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 - on hiatus 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
  • @mathlove: I just read http://meta.stackexchange.com/questions/16065/how-does-the-bounty-system-work and it seems like I have a lot of misconceptions. Despite this being CW, I can be given the bounty. I also assumed that if you wrote the answer yourself you could award yourself the bounty. (For something like, only losing half points). I was clearly wrong on both counts! – Jason DeVito - on hiatus Nov 20 '13 at 16:19
  • I've read the following page which answers NO to the question that 'Can I award a bounty to my own answer?'

    http://meta.stackexchange.com/questions/16065/how-does-the-bounty-system-work

     – mathlove Nov 20 '13 at 16:28
  • @mathlove: Yes, I agree with you - I was completely mistaken earlier. – Jason DeVito - on hiatus Nov 20 '13 at 16:32
  • Then, I'll do nothing. Anyway, thanks for your information. It did surprise me. – mathlove Nov 20 '13 at 16:34