I have wondered for a while if there are any interesting rational solutions to $a^b = b^a$. I have tried but cannot find any solutions other than $a=2$ and $b=4$, or vice versa.
Thank you in advance.
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$\begingroup$ Don't close this problem as it received many votes and some good answers. $\endgroup$– GH from MOCommented May 25, 2019 at 19:20
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2 Answers
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There is an infinite number of rational solutions $$a=\left(\frac{n+1}{n}\right)^n,\;\;b=\left(\frac{n+1}{n}\right)^{n+1},\;\;n\in\mathbb{Z},\;\;0\neq n\neq -1.$$
For a proof that these are all the rational solutions of $a^b=b^a$ with $a\neq b$, see Marta Sved's article (1990). As she describes, this question has a long history, it was first answered by Euler in 1748 and has been generalized in various ways.
I show a screen shot from Euler's proof that there is an infinite number of rational solutions (Euler uses the word "innumerabilia" -- uncountable, obviously not in the technical sense of the word).
answered May 23, 2019 at 16:13
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$\begingroup$ There's only a beginning of proof in your link, not a full proof. $\endgroup$ Commented May 23, 2019 at 16:20
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3$\begingroup$ Thank you very much. I had been looking at this problem for so long, it feels stupid how simple the solution is. $\endgroup$ Commented May 23, 2019 at 17:16
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$\begingroup$ @NajibIdrissi --- changed the link to a complete proof $\endgroup$ Commented May 23, 2019 at 18:49
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1$\begingroup$ Innumerabilia in Latin means countless and that sounds like a word that Euler would have used to mean infinite. I don't know what Latin word is used by modern users of Latin for the mathematical notion of uncountable. $\endgroup$ Commented May 23, 2019 at 20:30
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There are infinitely many solutions.
For example, take any positive integer $n$ and set $c=(n+1)/n, a=c^n, b=c^{n+1}$. Then $$ a^b=c^{nb}=c^{nc^{n+1}} $$ and $$ b^a=c^{(n+1)a}=c^{(n+1)c^n}. $$
These two expressions are equal because
$$ nc^{n+1}=(nc)c^n=(n+1)c^n. $$
answered May 23, 2019 at 16:13