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The problem is finding all possible solutions to $$x^y=y^x$$ where $x, y \in \mathbb{N}$ and $y>x$.

If it's transformed $x^y$ into $x^{y-x}x^x$ then you can say that $y^x=xy^x$. Then, if I was able to prove that $y=kx$ I got the solution (in fact, there`s a unique solution $x=2, y=4$). But I'm not sure of the implication $$y^x=qx^x \rightarrow y=kx$$ with $x, y, k, q \in \mathbb{N}$

If x is proportional to y, then, from the first equation we obtain $$x={k}^{\frac {1} {k-1}}$$ If we analize this function we see that there is a natural pair $(2,2)$ and for bigger $x$ the function tends asintotically to 1, then that´s the unique solution. But the problem is to see if k is natural or simply rational, and if there is a rational pair for k.

What can be the way to solve this?

Mario Ibáñez
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$x^y=y^x\iff\sqrt[x]x=\sqrt[y]y\iff\dfrac{\ln x}x=\dfrac{\ln y}y$ . Now, plot the graphic of the function $f(t)=\sqrt[t]t$ or $g(t)=\dfrac{\ln t}t$ , and tell me what you notice. :-)

Lucian
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  • That tends asintotically to 1 and that there is no integer pair? – Mario Ibáñez Feb 14 '14 at 14:04
  • @MarioIbáñez: Notice how the graphic resembles the form of a hill, whose the top is at $t=e<2.72$. So the only possible solutions are for $x\leqslant2.72$. Obviously, $x=0$ and $x=1$ are easily excludable, since $0^y=y^0\iff 0=1$, which is absurd, and $1^y=y^1\iff y=1\iff x=y=1$, which contradicts $y>x$. Finally, $2^4=4^2$ is the only natural solution to our equation. – Lucian Feb 14 '14 at 14:15