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.

$\newcommand{\l}{\operatorname{lcm}}$
There are already good answer to this question, but I like to demonstrate an important technique. First note that the equation is homogeneous since $$\gcd(k m, k n) +\l(k m, k n) = k(\gcd(m,n)+\l(m,n))$$ and $$km+kn=k(m+n)$$ for $k>0$. Also the conclusion is homogeneous since $$k m \mid k n \quad \text{ if and only if }\quad m \mid n$$
Then it's enough to deal with the case where $\gcd(m,n)=1$. The equation becomes $mn + 1 = m + n$ which factors as $(m-1)(n-1)=0$, so $m=1$ or $n=1$.
