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?

  • 7
    $\begingroup$ can I get a hint for the first one? $\endgroup$ – faridrb Mar 9 '10 at 4:46
  • 2
    $\begingroup$ I've hit this with the wiki-hammer, per the consensus of tea.mathoverflow.net/discussion/8/… $\endgroup$ – Scott Morrison Mar 9 '10 at 6:41
  • 4
    $\begingroup$ Is there a generalization of this to arbitrary collections of primes? $\endgroup$ – Harry Gindi Mar 9 '10 at 8:20
  • 4
    $\begingroup$ @fpqc: go and find out what the six exponentials theorem is and then all will be much clearer. If you look at the wikipedia page, the hint is that x_i=log(p_i). $\endgroup$ – Kevin Buzzard Mar 9 '10 at 8:32
  • 4
    $\begingroup$ @jef: the best hint I can think of is "calculus of differences". $\endgroup$ – Alon Amit Mar 9 '10 at 18:41

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.

  • 2
    $\begingroup$ Do I recall correctly that this follows from Schanuel's conjecture? $\endgroup$ – Marty Mar 9 '10 at 2:45
  • $\begingroup$ @Marty: yes, according to Wikipedia. $\endgroup$ – Qiaochu Yuan Mar 9 '10 at 4:48
  • 2
    $\begingroup$ This answer was given by Gerry. I just added the link to Wikipedia. $\endgroup$ – Alon Amit Mar 9 '10 at 18:47

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.

  • $\begingroup$ Very interesting! Just to make sure I understand - does this generalize the "n^x for all n" version only, or can it be applied to "2,3,5" and "2,3" as well? $\endgroup$ – Alon Amit Mar 9 '10 at 18:42
  • $\begingroup$ Oh, "arithmetic" means "n^x for all n", but in the proof that arithmetic implies algebraic he only uses finitely many n. $\endgroup$ – Kevin Buzzard Mar 9 '10 at 19:00
  • 1
    $\begingroup$ I'll confess that I was the one that brought it up on the NMBRTHRY list. After it was pointed out that it was the six exponentials theorem, I made a big "duh", since Lang was my advisor when he proved the above referenced result, and I knew about then. $\endgroup$ – Victor Miller May 2 '10 at 1:50

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.

  • $\begingroup$ Here's a JSTOR link to the relevant page in the Monthly: jstor.org/stable/2318375?seq=6 $\endgroup$ – PersonX Jun 14 '10 at 11:28
  • 8
    $\begingroup$ Just a remark: The "fun little puzzle" appears as a problem in Robert Young's "Excursions in Calculus" (Dolciani Mathematical Expositions #13) with the subtitle "A Putnam competition problem that no contestant solved!" $\endgroup$ – Todd Eisworth Jan 9 '12 at 16:39

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!

  • $\begingroup$ Er... you assumed that m was a power of 3 to prove that m is a power of 3. $\endgroup$ – Qiaochu Yuan Mar 10 '10 at 3:46
  • $\begingroup$ Right - fixing that. $\endgroup$ – Jacques Carette Mar 10 '10 at 3:47
  • $\begingroup$ You are still assuming that m is a power of 3. This is not necessary if a counterexample to the conjecture exists. $\endgroup$ – Qiaochu Yuan Mar 10 '10 at 4:58
  • 1
    $\begingroup$ Indeed - I'll have to think about it some more. I think my argument at least rules out some cases (so reducing the problem to positive m and q=0), but then get 'stuck'. I was originally thinking that the hard cases would come from $q\neq 0$ or $m\lt 0$! $\endgroup$ – Jacques Carette Mar 10 '10 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.