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Find all positive integers such that x^y=y^x. Given the graph of x^a and a^x intecepts once at a, it will intercept again and large x as a^x dominates for large x(how could I prove this). Also playing around I have found the solution for x^a=a^a, is a and r^(r/r-1), but this doesn't really answer the question does it as I have defined x=ar, and for a fixed a I can't find x except the obvious solution.

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marked as duplicate by barak manos, Dietrich Burde, Martin Sleziak, user299912, Daniel Fischer Nov 15 '16 at 13:39

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Use $\displaystyle x:=(1+\frac{1}{t})^t$ and $\displaystyle y:=(1+\frac{1}{t})^{t+1}$ which solves the equation $x^y=y^x$ for all $x,y>1$ .

When are $x,y$ positive integers ?

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  • $\begingroup$ Is there a way to prove that the given substitution exhausts all possible solutions ? $\endgroup$ – Yves Daoust Nov 15 '16 at 13:17
  • $\begingroup$ @Yves Daoust: Replace $x$ and $y$ by their parameterizations, that's the proof. Every solution with $1<x<e$ and $y>e$ is included and therefore also positive integers, which are in this value range (exist only one: for $t=1$). $\endgroup$ – user90369 Nov 15 '16 at 15:12
  • $\begingroup$ Substitution only shows that the condition is sufficient, not that it is necessary. (For example, $xy=1$ is solved by $x=e^t,y=e^{-t}$, but this does not exhaust the solutions.) $\endgroup$ – Yves Daoust Nov 15 '16 at 15:15
  • $\begingroup$ Which condition do you mean and why should the condition (which you mean) has to be necessary instead of sufficient ? --- The question is about the integers. With the parameterizations one has all solutions and therefore also for the wished (special) values. $\endgroup$ – user90369 Nov 15 '16 at 15:20
  • $\begingroup$ The question is are you sure to have all solutions. $\endgroup$ – Yves Daoust Nov 15 '16 at 15:32

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