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I just answered another question on $2^4 = 4^2.$

Does anyone know how to find the question where someone had a decent conjecture on all rational solutions to $x^y = y^x?$

Turns out all rational solutions are, with an integer $n,$

$$ x = \left( 1 + \frac{1}{n} \right)^n, \; \; y = \left( 1 + \frac{1}{n} \right)^{n+1}, $$ where $$ x^y = y^x = x^x \left( 1 + \frac{1}{n} \right)^x $$ need not be rational. I would say $n \neq -1$ in any case. Not quite sure whether $n \leq -2$ give repeats of the solutions with $n \geq 1.$ Alright, given $n \geq 1,$ it is easy enough to see that we get a repeat from $-1-n.$ So, we might as well restrict to $n \geq 1.$

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  • $\begingroup$ I'll add that there is also this question, which asks about the same equation for integers, but the answers provide also some information about rational solutions. And there are lots of linked questions to that one. (In fact, it is linked from the post mentioned by Daniel Fischer.) $\endgroup$ – Martin Sleziak Jan 18 '15 at 5:55
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I guess you mean this one, found per google search:

rational solutions x^y = y^x site:math.stackexchange.com

rational solutions x^y = y^x site:math.stackexchange.com

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    $\begingroup$ You are my new god. Do you have any religious principles that would make that a problem? $\endgroup$ – Will Jagy Jan 18 '15 at 1:37
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    $\begingroup$ It would force me to seriously doubt my existence. But that's not a problem. $\endgroup$ – Daniel Fischer Jan 18 '15 at 1:39
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    $\begingroup$ I think you did the right thing studying mathematics. $\endgroup$ – Will Jagy Jan 18 '15 at 1:40
  • $\begingroup$ I tried your string in Giggle, it actually worked, exponents and everything. $\endgroup$ – Will Jagy Jan 18 '15 at 4:56
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    $\begingroup$ Maybe also using quotes to get the exact phrase might be useful for similar searches: rational solutions "x^y=y^x" site:math.stackexchange.com. $\endgroup$ – Martin Sleziak Jan 18 '15 at 7:34

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