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Let $x$ is a real number such that $n^x\in\Bbb Z,$ for every positive integer $n.$ Prove that $x$ is an integer.

I got that problem here and it looks difficult, I tried writing $x$ as $\lfloor x\rfloor+\{x\}$ (where $\lfloor x\rfloor$ and $\{x\}$ are the floor and the fractional part of $x,$ respectively), then $$n^x=n^{\lfloor x\rfloor}n^{\{x\}}.$$ Since $n^{\lfloor x\rfloor}>0,$ then $$n^{x-\lfloor x\rfloor}=n^{\{x\}},$$ so that $n^{\{x\}}$ must be a rational number for every positive integer $n,$ but I don't know how to prove that $\{x\}$ must necessarily be $0$ ($0\leq\{x\}<1$).

Any help is really appreciated!

CIJ
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  • Any irrational power will not resolve as a simple root, though I don't know if you would need to prove that for your class. It is pretty obvious, though. That leaves you with rational powers. If $n^{\frac1k}\in\mathbb{Z}$, what do you know of $(n+1)^{\frac1k}$? Do you know anything at all about it? – Terra Hyde Jul 17 '15 at 02:54
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    The Gelfond–Schneider theorem states "If $a$ and $b$ are algebraic numbers with $a \ne 0,1$ and $b$ irrational, then any value of $a^b$ is a transcendental number.$ – Steven Alexis Gregory Jul 17 '15 at 13:57
  • @TerraHyde How do you show it if the power is irrational? For example $2^x = 3$ if $x = \log_2 3$, and $x$ is definitely irrational. – Stephen Montgomery-Smith Jul 17 '15 at 16:28
  • Now that I am thinking about it, I am not sure. The same rule applies to the irrationals as to the rationals about $n+1$, but I am not sure how to show it. Hmm. – Terra Hyde Jul 17 '15 at 20:40
  • For the case when $x = p/q$ is rational, you can just use $n=2$. Because if $2^{p/q} = m \in \mathbb N$, then $2^p = m^q$. Therefore $m$ must be a power of $2$ (because each integer has a unique prime factorization), and you quickly conclude $q | p$. – Stephen Montgomery-Smith Jul 17 '15 at 22:17
  • Of course, when $x$ is algebraic, the proof is "easy" (btw thank you for the link @StevenGregory), but what happens when $x$ is trascendental?, can we give an integer $m>0$ such that $m^x$ is not an integer? – CIJ Jul 20 '15 at 05:14
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    @CarlosIsraelJrl. I don't know. I tried to think of an argument by contrapositive or contradiction but failed. I figured I could at least point out the reference and see if it helped someone smarter than I am. – Steven Alexis Gregory Jul 25 '15 at 14:21

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