# Find all $x,y\in \mathbb{N}$, such that $x^{y}=y^{y-x}$

Find all $$x,y\in \mathbb{N}$$, such that $$x^{y}=y^{y-x}$$.

I have found the solutions $$(1,1)$$ and $$(2,4)$$. I've been trying to prove that there aren't any more. I tried splitting the cases by parity, but nothing interesting seemed to come out.

I also tried rearranging it into:

$$x=y^{\frac{y-x}{y}}$$.

Is there any way to proceed with this problem?

Thank you very much

• Does this answer your question? Find all Pairs of Integers $(x,y)$ , $x \gt y \ge 2$ such that $x^y=y^{x-y}$ Dec 7, 2021 at 8:51
• @Saad: That’s not the same equation though, is it (although similar methods may apply)? Dec 7, 2021 at 9:51
• @Mindlack Indeed, I'm not sure it can be called a duplicate, but the idea of setting $y=tx$ comes from the comment there, after which one actually reaches $y = t^{-\frac 1t}$ so the question becomes the possibly easier : for which rationals $t$ is it true that $t^{-\frac 1t}$ is an integer? Dec 7, 2021 at 10:38
• Hint: Using Maple (still running) the first few solutions are x=1, y=1 x=2, y=4 x=9, y=27 x=64, y=256 x=625, y=3125 Interesting. All powers of the same prime. Dec 7, 2021 at 10:44
• @Teresa Lisbon: how do you get this? By the same method, I can find infinitely many solutions parametrized by $t=y/x$ (an integer). Dec 7, 2021 at 10:45

Actually, the function can be written as $$\left(\frac{y}{x}\right)^{y} = y^{x}$$ If you let $$t = \frac{y}{x}$$ then we can get $$t^{t} = y$$ So if you apply $$y = t^{t}$$ and $$x = t^{t - 1}$$ with $$t$$ an integer you get infinite pairs of solutions. Furthermore, if $$t$$ is not an integer then $$t = q/p$$ which is irreducible. Then $$y = (q/p)^{q/p}$$ is an integer so $$y^{p} = (q/p)^q$$ is an integer, which is not possible. So $$(x, y)$$ is a solution to the equation iff $$(x, y) = (t^{t-1}, t^t)$$ with $$t$$ an integer, and there are infinitely many pairs.