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Question:

$\alpha \in \mathbb{R}$, if $n^{\alpha}$ is an integer for $\forall n \in \mathbb{N}$, then show that $\alpha$ is a nonnegative integer.

I totally don't know how the procedure should go. Besides, I got this problem in the calculus textbook, but I cannot understand why the problem is included in the textbook.

Can you give me some simple key points to this kind of problem? Thanks for your advice.

[EDIT] I modified some expressions in order not to have various interpretations.

user1551
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ToBY
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    I'm not sure why this is in a calculus textbook either, but suppose $\alpha$ is negative. Then $n^\alpha$ is between $0$ and $1$, so not an integer. – pancini Feb 09 '20 at 07:43
  • @ElliotG I got that part. But I'm still wondering why $\alpha$ should be nothing but an integer. – ToBY Feb 09 '20 at 07:46
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    That's clearly false, since we can have $n = 4$, $\alpha = \frac{1}{2}$, and we have $4^\frac{1}{2} = 2 \in \mathbb{Z}$. – Clement Yung Feb 09 '20 at 07:51
  • Oh yeah that makes no sense. I mean $8^{1/3}=2$, but $1/3$ is not an integer. – pancini Feb 09 '20 at 07:51
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    Think about what a reasonable interpretation should be (not one with easy counterexamples): for $\alpha \in \mathbf R$, if $n^\alpha$ is an integer for every positive integer $n$ then $\alpha$ is a nonnegative integer. You can’t disprove that by making a specific choice of $n$. – KCd Feb 09 '20 at 07:58
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    The page https://mathoverflow.net/questions/17560/if-2x-and-3x-are-integers-must-x-be-as-well shows how this solved problem is closely related to harder and unsolved problems. – KCd Feb 09 '20 at 08:07

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