Suppose that $a$ and $b$ are positive integers such that $ab$ divides $a + b$. Suppose that $a$ and $b$ are positive integers such that $ab$ divides $a + b$.
Prove that $a = b$, then prove that $a$ is either $1$ or $2$.
I was thinking using the theorem:

If $n|a$ and $n|b$,then $n|(ax+by)$ for any $x,y\in \mathbb Z$.

Then I can derive that $ab$ divides $a$ and $b$. I could not go further. However, this doesn't seem like can help me solve the question I had. 

$abx=(a+b)$, where $x$ is greater than or equal to $1$ ($x$ cannot be $0$) 
$$x = \frac{1}{a} + \frac{1}{b}$$
This implies that $x$ must be $1$ or $2$. 
($2$ is the greatest value possible for sum of two reciprocals of natural numbers)
so if $x=1$, then $\frac{1}{a}$ and $\frac{1}{b}$ must both be $\frac{1}{2}$ and $a=b=2$.
so if $x=2$ then $\frac{1}{a}$ and $\frac{1}{b}$ must both be $1$ and $a=b=1$
