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

If nxn^x is an integer for every positive integer nn, then xx must be an integer; this is a well-known exercise. If only 2x2^x and 3x3^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 22, 33 and 55 in place of 22 and 33, it is a theorem.

Problem

Let xx be a real number such that 2x2^x and 3x3^x are both integers. Must xx be an integer?

Since 2x2^x is then a positive integer, x≥0x \ge 0; and if x=p/qx = p/q is rational, then 2p=(2x)q2^p = (2^x)^q forces qq to divide pp, so xx is an integer. The whole difficulty is to rule out irrational xx.

Background

Alaoglu and Erdős (1944), studying colossally abundant numbers, remarked that pxp^x and qxq^x very likely cannot both be rational for distinct primes pp, qq unless xx 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 2x,3x2^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 nn is a theorem.

What is known

  • The four exponentials conjecture says: if x1,x2x_1, x_2 are linearly independent over Q\mathbb{Q} and y1,y2y_1, y_2 are linearly independent over Q\mathbb{Q}, then at least one of exiyje^{x_i y_j} (i,j∈{1,2}i, j \in \{1, 2\}) is transcendental. With x1=1x_1 = 1, x2=xx_2 = x, y1=log⁡2y_1 = \log 2, y2=log⁡3y_2 = \log 3 it would show that xx is rational, hence an integer.
  • The six exponentials theorem of Lang and Ramachandra (1960s), the same statement with three numbers yjy_j, proves the result when 2x2^x, 3x3^x and 5x5^x are all integers. An elementary proof of that case is due to Waldschmidt.
  • If nxn^x is an integer for every positive integer nn, then xx is an integer; this appeared on the 1971 Putnam competition.

What would count as progress

  • A proof that 2x,3x∈Z2^x, 3^x \in \mathbb{Z} forces x∈Zx \in \mathbb{Z}, or the same for any single fixed pair of primes.
  • Any new case of the four exponentials conjecture, for instance for pairs y1,y2y_1, y_2 of logarithms of integers, which is exactly what this question needs.
  • A formal proof of one of the known variants in Mathoverflow/17560.lean (the case of all nn, or of 22, 33 and 55), which would be a step towards formal transcendence theory; the main statement there is mathoverflow_17560.

References

  1. L. Alaoglu and P. Erdős, On highly composite and similar numbers, Trans. Amer. Math. Soc. 56 (1944), 448–469.
  2. K. Ramachandra, Contributions to the theory of transcendental numbers I, Acta Arith. 14 (1968), 65–72.
  3. A. Amit and others, If 2^x and 3^x are integers, must x be as well?, MathOverflow question 17560 (2010).

Source

Imported from Formal Conjectures Mathoverflow/17560, 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.