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Consider the following two problems:

  1. Show that if for some $x\in\mathbb R$ and for each $n\in\mathbb N$ we have $n^x\in\mathbb N$, then $x\in\mathbb N$.
  2. Show that if for some $x\in\mathbb R$ and for each $n\in\mathbb N$ we have $n^x\in\mathbb Q$, then $x\in\mathbb Z$.

The first of those is a somewhat infamous Putnam problem (A6 from 1971) and there is an elementary proof of this using calculus of differences and mean value results, which you can read here.

As mentioned in an answer here, the second problem follows from the six exponentials theorem, even if we only require $2^x,3^x,5^x$ to be rational. However, this solution is very non-elementary, and I suspect that using all values of $n$ we might be able to give an easier proof, just like we can for the first problem (though I'm aware the linked proof doesn't generalize).

Is it possible to solve the second problem with elementary means?

Wojowu
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    There is a recent writeup that's tangential to this question: https://link.springer.com/article/10.1007/s12045-018-0676-1. It gives a relatively self-contained proof of the specific claim that $2^x, 3^x, 5^x \in \mathbb Q \implies x \in \mathbb Z$. The proof is decidedly non-elementary in that it relies on complex analysis, but it appears to be at least accessible at an undergrad level. The paper does use the Putnam problem as motivation, so it hints that at least those authors were not aware of an elementary approach to your question. – Erick Wong Sep 20 '18 at 14:23

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Not an answer, but some ideas that could help. The first thing to notice is that if $n^x\in\mathbb Q$ for all $n\in\mathbb N$, then this is also true for $n\in\mathbb Q_+$. Now, let us use $S$ to denote the set of all $x$ with this property: $$ S=\{ x\in\mathbb R \mid \forall r\in\mathbb Q : r^x\in\mathbb Q \} $$ With this notation, what we need is to prove that $S \subseteq \mathbb Z$.

Notice that we have $\mathbb Z \subseteq S$. Moreover, given $x, y \in S$, we have the following closure properties:

  • $x + y \in S$, because $r^{x+y} = r^x \cdot r^y$
  • $-x \in S$, because $r^{-x} = \frac1{r^x}$
  • $x \cdot y \in S$ because $r^{x\cdot y} = (r^x)^y$

So my idea is to proceed by contradiction and assume that there exists $x \in S \setminus \mathbb Z$, then use the above properties to construct another value $y \in S$ which exhibits a contradiction. Notice for this case that we can not have $x \in \mathbb Q \setminus \mathbb Z$. This can be proved by taking $r$ to be a prime number and proving that $r^x$ is irrational. Thus, $x$ must be irrational, which means that based on the above closure properties we can show that $S$ is dense in $\mathbb R$. And this is where I'm stuck for now.