Find all positive rational solutions to $x^y = y^x$

Asked May 08 '14 at 02:34

Active May 09 '14 at 04:15

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Note: This is essentially the same as What are the positive rational solutions of $x^{(x+y)} = (x+y)^y$?, but that did not have any good answers.

In my answer to Understanding the graph for $x^y = y^x$, I showed that a parameterization of $x^y = y^x$ is $x = r^{1/(r-1)}$ and $y = r^{r/(r-1)}$. Setting $r = 1+1/n$, where $n$ is a positive integer, gives $x = (1+1/n)^n$ and $y = (1+1/n)^{n+1}$ as rational solutions.

For $n=1$, $x=2$ and $y=4$, which is well known.

For $n=2$, $x=9/4$ and $y=27/8$, which is not well known.

To check the $n=2$ case, $(9/4)^{27/8} =((3/2)^2)^{27/8} =(3/2)^{54/8} =(3/2)^{27/4} $ and $(27/8)^{9/4} =((3/2)^3)^{9/4} =(3/2)^{27/4} $.

I believe that these are all the positive rational solutions, but I do not have a proof.

edited Apr 13 '17 at 12:21

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asked May 08 '14 at 02:34

marty cohen

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you also have $x = y = $anything. I don't think these are included in the ones you gave? – oxeimon May 08 '14 at 02:36
If any rational number can be written in the form $1+1/n$, then yes. This is not true, and hence, the conclusion is false. – May 08 '14 at 02:42
The only rational solutions are $(\pm 2, \pm 4)$. See this and this. – glebovg May 08 '14 at 02:44
Not true. I added the example $x = 9/4$ and $y = 27/8$. – marty cohen May 09 '14 at 04:15
Aryabhata's answer here gives links to a proof of the fact that for $(x,y)$ to be rational it is, indeed, sufficient and necessary that $r=1+1/n$, $n$ integer. The integer $n$ does not have to be positive. For example $n=-2$ gives $x=4$, $y=2$ (you get the swapped solutions that way). – Jyrki Lahtonen May 09 '14 at 04:57

Find all positive rational solutions to $x^y = y^x$

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