$p^q=q^p$ for $p, q\in\mathbb Q$

Question

Just saw a neat trick (the fact that $2^4=4^2$ 'saved the bacon' in the proof here: How do I compare $\sqrt{2}$ and $\pi^{1/ \pi}$?). That led me to ask:

What is known about pairs of positive rational numbers $p,q\in\mathbb Q$, $p, q>0$ such that $p^q=q^p$?

For instance, is $p=2, q=4$ the only nontrivial ($p\ne q$) solution?

No, there are in fact infinitely many solutions with $p\neq q$ and both of them rational. — Tobias Kildetoft, Dec 05 '17 at 10:11
See this thread and the ones linked to it. Ross Millikan's answer is relevant to your question. If $a/(a-1)$ is an integer then we have $x^y=y^x$ for $y=a^{a/(a-1)}, x=y/a$. Observe that here both $x$ and $y$ are then rational. The case $a/(a-1)=2$ gives $y=4, x=2$. When $a/(a-1)=4$ we get $a=4/3$, $y=256/81$ and $x=64/27$. — Jyrki Lahtonen, Dec 05 '17 at 10:19
@JyrkiLahtonen Do you happen to know if those are the only ones? — , Dec 05 '17 at 10:26
@JyrkiLahtonen all solutions, proved, in Bennett Reznick http://www.math.ubc.ca/~bennett/monthly013-021.pdf formula (6) — Will Jagy, Dec 05 '17 at 19:10

score 2 · Answer 1 · edited Dec 25 '17 at 14:34

2

Just to document the answer that @WillJagy gave in the comments:

"all solutions, proved, in Bennett Reznick http://www.math.ubc.ca/~bennett/monthly013-021.pdf, formula (6)".

which is:

$$x_n = \left(1+\frac{1}{n}\right)^n, \qquad y_n=\left(1+\frac{1}{n}\right)^{n+1}$$

edited Dec 25 '17 at 14:34

Sangchul Lee

answered Dec 25 '17 at 14:25

1 Answers1