# If $n^c\in\mathbb N$ for every $n\in\mathbb N$, then $c$ is a non-negative integer?

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?

-
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
@901301: In my opinion, it does not seem to help. – mathlove Nov 17 '13 at 10:24
@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
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
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

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).