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.

Let $a$ be the gcd, and $b$ the lcm. We are told that
$$a+b=m+n.$$
It is a result I hope known to you that the gcd of two positive integers, times their lcm, is equal to the product of the two integers. Thus
$$ab=mn.$$
So $a$ and $b$ are the roots of the same quadratic equation as $m$ and $n$, namely the equation $x^2-(m+n)x+mn=0$. 
Without loss of generality we may assume that $m\le n$. Thus $a=m$ and $b=n$. sinnce $\gcd(m,n)=m$, we conclude that $m$ divides $n$.  
