Prove that either $m$ divides $n$ or $n$ divides $m$ given that $\operatorname{lcm}(m,n) + \operatorname{gcd}(m,n) = m + n$? We are given that $m$ and $n$ are positive integers such that $\operatorname{lcm}(m,n) + \operatorname{gcd}(m,n) = m + n$.
We are looking to prove that one of numbers (either $m$ or $n$) must be divisible by the other.

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 $\text{lcm}(m,n) = n$.
