# $a^b=b^a$ and algebraicity [closed]

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

• No, by Gelfond-Schneider. Take $a=3$ and suitable $b$ for instance. May 19, 2019 at 16:49
• 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. May 19, 2019 at 18:15
• @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 :) ) May 19, 2019 at 19:31
• @Wojowu : thank you for your comment. Maybe you can post it as an answer so that I can accept it. May 19, 2019 at 19:51
• 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. May 19, 2019 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.