Must x be an integer if 2^x and 3^x are integers?

Show that a real number x for which both 2^x and 3^x are integers must itself be an integer.

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State the problem, from Formal Conjectures

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	1	+# Must x be an integer if 2^x and 3^x are integers?
	2	+
	3	+If $n^x$ is an integer for every positive integer $n$, then $x$ must be an integer; this is a well-known exercise. If only $2^x$ and $3^x$ are known to be integers, the same conclusion is expected, but nobody can prove it: it would follow from the four exponentials conjecture of transcendental number theory. With $2$, $3$ and $5$ in place of $2$ and $3$, it is a theorem.
	4	+
	5	+## Problem
	6	+
	7	+Let $x$ be a real number such that $2^x$ and $3^x$ are both integers. Must $x$ be an integer?
	8	+
	9	+Since $2^x$ is then a positive integer, $x \ge 0$; and if $x = p/q$ is rational, then $2^p = (2^x)^q$ forces $q$ to divide $p$, so $x$ is an integer. The whole difficulty is to rule out irrational $x$.
	10	+
	11	+## Background
	12	+
	13	+Alaoglu and Erdős (1944), studying colossally abundant numbers, remarked that $p^x$ and $q^x$ very likely cannot both be rational for distinct primes $p$, $q$ unless $x$ is an integer, and reported Siegel's result that this holds for three primes. The question was asked on MathOverflow by Alon Amit in 2010; Serge Lang is reported there to have called the $2^x, 3^x$ case a notorious open problem. Kevin Buzzard's answer there relates it to the arithmetic of Grössencharacters, where the analogous statement for all $n$ is a theorem.
	14	+
	15	+## What is known
	16	+
	17	+- The four exponentials conjecture says: if $x_1, x_2$ are linearly independent over $\mathbb{Q}$ and $y_1, y_2$ are linearly independent over $\mathbb{Q}$, then at least one of $e^{x_i y_j}$ ($i, j \in \{1, 2\}$) is transcendental. With $x_1 = 1$, $x_2 = x$, $y_1 = \log 2$, $y_2 = \log 3$ it would show that $x$ is rational, hence an integer.
	18	+- The six exponentials theorem of Lang and Ramachandra (1960s), the same statement with three numbers $y_j$, proves the result when $2^x$, $3^x$ and $5^x$ are all integers. An elementary proof of that case is due to Waldschmidt.
	19	+- If $n^x$ is an integer for every positive integer $n$, then $x$ is an integer; this appeared on the 1971 Putnam competition.
	20	+
	21	+## What would count as progress
	22	+
	23	+- A proof that $2^x, 3^x \in \mathbb{Z}$ forces $x \in \mathbb{Z}$, or the same for any single fixed pair of primes.
	24	+- Any new case of the four exponentials conjecture, for instance for pairs $y_1, y_2$ of logarithms of integers, which is exactly what this question needs.
	25	+- A formal proof of one of the known variants in [Mathoverflow/17560.lean](https://github.com/google-deepmind/formal-conjectures/blob/main/FormalConjectures/Mathoverflow/17560.lean) (the case of all $n$, or of $2$, $3$ and $5$), which would be a step towards formal transcendence theory; the main statement there is `mathoverflow_17560`.
	26	+
	27	+## References
	28	+
	29	+1. L. Alaoglu and P. Erdős, [On highly composite and similar numbers](https://doi.org/10.2307/1990319), Trans. Amer. Math. Soc. 56 (1944), 448–469.
	30	+2. K. Ramachandra, [Contributions to the theory of transcendental numbers I](https://doi.org/10.4064/aa-14-1-65-72), Acta Arith. 14 (1968), 65–72.
	31	+3. A. Amit and others, [If 2^x and 3^x are integers, must x be as well?](https://mathoverflow.net/questions/17560), MathOverflow question 17560 (2010).
	32	+
	33	+## Source
	34	+
	35	+Imported from [Formal Conjectures Mathoverflow/17560](https://github.com/google-deepmind/formal-conjectures/blob/main/FormalConjectures/Mathoverflow/17560.lean), curated by Google DeepMind and contributors. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.