$\forall n\in\mathbb N:n^x\in\mathbb Q$ implies $x\in\mathbb Z$ - elementary proof?

Question

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?

Answer 1

Not exactly elementary, but it does follow from Wilkie’s conjecture, which was recently proven in https://arxiv.org/abs/2202.05305.

Indeed, suppose $x$ is irrational, and WLOG suppose that $0\leq x\leq 1$. Look at the graph of the function $f(t)=t^{x}$. It is definable in $\mathbb{R}_{exp}$ and is transcendental (i.e does not contain semialgebraic curves). It contains the points $(1,1^x),\dots,(H,H^x)$ which are all of height $\leq H$ due to the assumption that $x\leq 1$. Thus the graph of $f$ contains polynomialy many (in $H$) points of height $\leq H$ which contradicts Wilkie’s conjecture, which states that the growth should be polylogarithmic in $H$.

Answer 2

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.

Answer 3

The problem states:

(1) $x$ is fixed, possibly from $\mathbb{R}$

(2) $n$ ranges over all $\mathbb{N}$. In other words, $n$ can be any natural number.

(3) Assume $n^x\in\mathbb{Q}$. By (2), the condition $n^x\in\mathbb{Q}$ must hold for all natural $n$.

(4) Then necessarily one must conclude $x\in\mathbb{Z}$

An elementary proof that uses only the Fundamental Theorem Of Arithmetics is as follows:

Let $a,b\in\mathbb{N}$ be some naturals at our disposal. Then the assumption (3) states:

$n^x=\frac{a}{b}$

Denote prime factors of $a$ by $P_a$ with multiplicities $p_a$. This way one can write $a$ in the form

$a=\prod P_a^{p_a}$

Similarly, define primes $P_b$ along with multiplicities $p_b$ so that

$b=\prod P_b^{p_b}$

This is just the Fundamental Theorem Of Arithmetics. Do notice that multiplicities are natural numbers.

Number $n$ can be any natural number. The result must hold for all $n$. So assume $n$ is prime itself. Say, $n=2$.

The assumption (3) now reads

$\prod P_b^{p_b} 2^{x}=\prod P_a^{p_a}$

If $x$ was real, one could write it as a sum of its integer part $\left[x\right]$ and its fractional part $\left\{x\right\}$ as $x=\left[x\right]+\left\{x\right\}$. Do notice that the fractional part $\left\{x\right\}$ is less than $1$ and positive.

The assumption (3) now becomes

$\prod P_b^{p_b} 2^{\left[x\right]}2^{\left\{x\right\}}=\prod P_a^{p_a}$

There is nothing wrong with the factor $2^{\left[x\right]}$ -- it is of the form "prime to some integer power". However, the factor $2^{\left\{x\right\}}$ makes no sense -- it would break the Fundamental Theorem Of Arithmetics unless $2^{\left\{x\right\}}$ was itself of the form "prime to some integer power". Do notice that the right hand side is a natural number -- hence so is the product on the left hand side. However, $2^{\left\{x\right\}}$ is less than $2$ and it cannot be written in the form "prime to some integer power". This contradicts the Fundamental Theorem Of Arithmetics.

Hence one concludes that $x$ must be an integer -- with no fractional part.