Suppose $a$ and $b$ are reals such that $a^b=b^a$. If $a$ is algebraic, is $b$ algebraic too?


closed as off-topic by abx, Wojowu, YCor, Felipe Voloch, Ben Linowitz May 19 at 21:20

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    $\begingroup$ No, by Gelfond-Schneider. Take $a=3$ and suitable $b$ for instance. $\endgroup$ – Wojowu May 19 at 16:49
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    $\begingroup$ Picking up on Wojuwo's comment: it's easier for me to contemplate $a^{1/a} = b^{1/b}$. By Gelfond-Schneider, if $b$ is algebraic and irrational, then $b^{1/b}$ will be transcendental. So in the case $a = 3$, we would need a rational $b \neq 3$ to satisfy $b^{1/b} = 3^{1/3}$, and then it's just a matter of the fundamental theorem of arithmetic to rule out this possibility. $\endgroup$ – Todd Trimble May 19 at 18:15
  • $\begingroup$ @ToddTrimble I think it's actually (slightly) easier to keep the problem as stated. If $3^b=b^3$ and $b\neq 3$, it's clear that $b$ is not an integer, and hence that $b$ is irrational (as else $b^3$ is rational and $3^b$ isn't) and hence transcendental (Gelfond-Schneider). (also, it's Wojowu, not Wojuwo :) ) $\endgroup$ – Wojowu May 19 at 19:31
  • $\begingroup$ @Wojowu : thank you for your comment. Maybe you can post it as an answer so that I can accept it. $\endgroup$ – Sylvain JULIEN May 19 at 19:51
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    $\begingroup$ If you say so, Wojowu. I mean, thanks for the additional explanation, but I said easier for me, and that might still be true even after your addition. Chacun a son gout, or however it goes. $\endgroup$ – Todd Trimble May 19 at 20:08

The answer is no. For instance, let $a=3$ and $b\neq 3$ be the real number satisfying $3^b=b^3$. Clearly $b$ is not an integer. It follows that $b$ is irrational -- indeed, if $b$ was a non-integer rational, $3^b$ would be irrational, while $b^3$ would be rational. Finally, $b$ is transcendental, since otherwise $b$ would be algebraic irrational, $b^3$ would be algebraic and $3^b$ would be transcendental by Gelfond-Schneider.

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    $\begingroup$ please do not answer off-topic questions. The question was not yet recognized as off-topic when you posted the answer apparently, so the best course of action may be to delete your answer. $\endgroup$ – user140761 May 20 at 6:07
  • $\begingroup$ oh, actually, you were among people who put it on hold, so my last comment is irrelevant probably. $\endgroup$ – user140761 May 20 at 6:09
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    $\begingroup$ Gelfond-Schneider isn't so common background, so I'd say even if it makes sense to close the question here as non-research level, it's reasonable to have posted an answer (actually this maybe rather should have been migrated to MathSE). $\endgroup$ – YCor May 20 at 22:35

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