Solve in positive integers the equation $$m^{\frac{1}{n}}+n^{\frac{1}{m}}=2+\frac{2}{mn(m+n)^{\frac{1}{m}+\frac{1}{n}}}.$$
Problem
Source: Bulgarian Autumn Tournament 2023, 12.3
Tags: algebra
VicKmath7
19.11.2023 14:18
There are no solutions. We will show that $m^{\frac{1}{n}}+n^{\frac{1}{m}}>2+\frac{2}{mn}$, which will easily finish as $(m+n)^{\frac{1}{m}+\frac{1}{n}} \geq 1$ (yeah, so troll...). By AM-GM, $m^{\frac{1}{n}}+n^{\frac{1}{m}} \geq 2(m^{m}n^{n})^{\frac{1}{2mn}}$, so we are left to show that $m^m \cdot n^n \geq (1+\frac{1}{mn})^{2mn}$. However, $\lim_{t \rightarrow \infty} (1+\frac{1}{t})^{2t}=e^2$ and $m^m \cdot n^n \geq e^2$ for large $m, n$, so a finite case check suffices.
Marinchoo
21.11.2023 21:42
There are no solutions in positive integers. Without loss of generality, assume that $m\geq n$. If $n\geq 3$, then \[\left(1+\frac{1}{m}\right)^m = \sum\limits_{k=0}^{m} \binom{m}{k} \frac{1}{m^k} < \sum\limits_{k=0}^{m} \frac{1}{k!} < \sum\limits_{k=0}^{\infty} \frac{1}{k!} = e < n\Longrightarrow n^{\frac{1}{m}} > 1+\frac{1}{m}\]Similarly, $m^{\frac{1}{n}} > 1+\frac{1}{n}$, whence $m^{\frac{1}{n}}+n^{\frac{1}{m}} > 2 + \frac{1}{m} + \frac{1}{n} \geq 2+\frac{2}{mn} > 2+\frac{2}{mn(m+n)^{\frac{1}{m}+\frac{1}{n}}}$. The cases $n=1,2$ can be solved directly.