0

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.

Martin Sleziak
  • 56,060
grigori
  • 255

1 Answers1

0

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 ?

user90369
  • 11,696
  • Is there a way to prove that the given substitution exhausts all possible solutions ? –  Nov 15 '16 at 13:17
  • @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 $t=1$). – user90369 Nov 15 '16 at 15:12
  • 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.) –  Nov 15 '16 at 15:15
  • 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. – user90369 Nov 15 '16 at 15:20
  • The question is are you sure to have all solutions. –  Nov 15 '16 at 15:32
  • The parameterizations is the (trivial) proof. Maybe it's somehow strange for you because it's so easy. You can check the function $x^{1/x}$ and compare the value ranges $1<x<e$ and $x>e$ . – user90369 Nov 15 '16 at 15:41
  • Sorry, I don't think it is. Did you read my example ? –  Nov 15 '16 at 15:41
  • Yes. You can use any parameterizations, if they decribe a function identically. The result will always the same. I haven't said, that my parameterization is the only possible (but you can be sure, it's the simpliest here), I only say that it decribes all needed values, which can be seen by the trivial replacement of $x$ and $y$. – user90369 Nov 15 '16 at 15:47
  • There is no need for a downvote because the solution is correct. – user90369 Nov 15 '16 at 15:51
  • Do you agree that my example show a sufficient but not necessary condition ? –  Nov 15 '16 at 15:52
  • I know now what you mean but it's not clear what you want to say with it. The necessary condition here is, that $x$ and $y$ have to be positive integer and this can be decribed by a (any valid) parameterization. – user90369 Nov 15 '16 at 15:56
  • How do you know you are not missing any other branch ? –  Nov 15 '16 at 15:57
  • Because the question concernes the value range $x\geq 1$ for the function $x^{1/x}$ which can be described by $(1+\frac{1}{t})^t$ (=> $1\leq x\leq e$) and $(1+\frac{1}{t})^{t+1}$ (=> $x\geq e$) for $t\in\mathbb{R}_0^+$ including $t\to\infty$. – user90369 Nov 15 '16 at 16:16