If $2^x $and $3^x$ are integers, must $x$ be as well?

Question

I'm fascinated by this open problem (if it is indeed still that) and every few years I try to check up on its status. Some background: Let $x$ be a positive real number.

1. If $n^x$ is an integer for every $n \in \mathbb{N}$ then $x$ must be an integer. This is a fun little puzzle.
2. If $2^x$, $3^x$ and $5^x$ are integers then $x$ must be an integer. This requires fairly sophisticated tools, and can be derived from the results in e.g. Lang, Algebraic values of meromorphic functions. II., Topology 5, 1966.
3. Finally, if all you know is that $2^x$ and $3^x$ are integers, then as far as I know it is not known if $x$ is forced to be an integer (unbelievable, isn't it?). Although of course one can never be certain, I am quite sure this was still the case as recently as 2003.

So the question is, is that still an open problem, and is there any sort of relevant progress that may provide some hope?

Answer 1

Still open, to the best of my knowledge. The $2^x,3^x,5^x$ result follows from the Six Exponentials Theorem, q.v., and the $2^x,3^x$ would follow from the Four Exponentials Conjecture, q.v.

Answer 2

This question came up recently on the NMBRTHRY mailing list and I can't resist paraphrasing a comment I made there. So, perhaps surprisingly, this question has links to automorphic forms! For if $x$ is a complex number and $||.||^x$ is the associated Grossencharacter of the ideles of $\mathbf{Q}$, that is, the map $\mathbf{A}_\mathbf{Q}^\times/\mathbf{Q}^\times\to\mathbf{C}^\times$ sending an idele to the $x$'th power of its norm, then the assumption that $p^x$ is an integer for all primes $p$ (which is clearly equivalent to the assumption that $n^x$ is an integer for all $n$) implies that the grossencharacter is arithmetic. Now a standard conjecture in the theory of automorphic representations is that an automorphic representation is arithmetic iff it's algebraic, and this conjecture is a theorem for tori, so the theorem in this case says that $||.||^x$ is algebraic which is precisely the statement that $x$ is an integer!

So for tori over general number fields it's a theorem of Waldschmidt that arithmetic implies algebraic for automorphic forms. So in practice we get a vast generalisation of the first question above, where integers can be replaced by algebraic integers and where we can add finite order characters and so on.

As an example, one sees that if $x$ is complex and if there's a number field $E$ in the complexes such that $n^x$ is an integer in $E$, for all $n$, then again the grossencharacter is arithmetic, so algebraic, and hence $x$ must be an integer. I don't know if there's any low-level proof of this (but it follows from standard transcendence theorems). As other examples $n$ can be replaced by the algebraic integers in a number field and so on.

Answer 3

The "fun litle puzzle" was a nightmare to me when I first came across it, which was on the 1971 Putnam exam. I spent a lot of time trying to solve it by doing $2^x,3^x$. When Serge Lang told me after the test that $2^x,3^x$ was a notorious open problem, I didn't know whether to be pleased that my intuition had led me to something widely held to be true, or annoyed that I had spent so much precious exam time on an approach that was doomed.

Anyway, solutions to problem 1 and other Putnam problems are available at various places on the net, in appropriate issues of the American Math Monthly, and in the three volumes of Putnam problems and solutions published by the MAA.

Answer 4

It sure feels like something that should be related to Schanuel's conjecture. Note that this is equivalent to finding integers $m$ and $q$ such that $$x = \frac{\ln m + 2\pi i q}{\ln 3}$$ is not an integer but $2^x$ is. [$3^x$ simplifies to $m$ by construction].

Continuing this, let us actually compute $2^x$. If we split it into real and imaginary parts, a large messy expression ensues. But it naturally splits into 2 reasonable cases, depending on whether $m\gt 0$ or $m \lt 0$. Let's deal with the positive case first. We get $$2^x = m^{\log_3 2}\left( \cos(2q\log_3(2) \pi) + i \sin(2q\log_3(2) \pi)\right) $$ For that to be an integer, it has to at least be real, but unless $2\log_3(2)q$ is an integer, the $\sin$ term will not be $0$. For $q=0$, this is $m^{\log_3 2}$. We can rewrite that as $2^{\log_3 m}$. But we assumed that $m$ was a power of $3$, so $\log_3 m$ (and thus $x$) is an integer.

For the $m \lt 0$, we get the slightly more complicated $$2^x = (-m)^{\log_3 2}\left( \cos(2(2q+1)\log_3(2) \pi ) + i \sin(2(2q+1)\log_3(2) \pi )\right) $$ Since $m\lt 0$, the first term is real, so we need $2 (2q+1)\log_3(2)$ to be an integer for the $\sin$ term to disappear, which cannot happen.

Can someone find a flaw in my reasoning? I somehow expect so, as this did not seem difficult, and I would expect it to be if it's an open problem!