Prove that either $m$ divides $n$ or $n$ divides $m$ given that $\operatorname{lcm}(m,n) + \operatorname{gcd}(m,n) = m + n$?
Solution 1:
We may suppose without loss of generality that $m \le n$. If $\text{lcm}(m,n) > n$, then $\text{lcm}(m,n) \ge 2n$, since $\text{lcm}(m,n)$ is a multiple of $n$. But then we have
$\text{lcm}(m,n) < \text{lcm}(m,n)+\gcd(m,n) = m + n \le 2n \le \text{lcm}(m,n)$,
a contradiction. So $\text{lcm}(m,n) = n$.
Solution 2:
Hint $ $ For $\,x = \gcd(m,n),\ {\rm lcm}(m,n) = mn/x,\,$ and your equation is $\,(x-m)(x-n) = 0.$ Thus either $\,x = \gcd(m,n) = m,\,$ so $\,m\mid n,\ $ or $\,x = \gcd(m,n) = n,\,$ so $\,n\mid m.$