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. 
 A: $a$ divides $b \iff \exists k\in\mathbb{Z}$ such that $b=ka$.
Now, $ab$ divides $a+b\iff \exists K\in\mathbb{Z}$ s.t $a+b=Kab$.
If $a=b\Rightarrow a+a=Kaa\Rightarrow 2a=Ka^2$, remember that $a\neq0$. Then
$2=Ka$ where you have two ways. 
What integer values should $K$ take to satisfy the equation? 
A: We have 
$$ab \mid a+b \implies ab \mid b( a+b) \implies  ab \mid b( a+b) - ab \implies  ab \mid b^2 \implies  a \mid b$$
In a similar way we can show 
$$b \mid a$$ and so 
$$a=b$$
If we substitute  $b$ by $a$ in $ab \mid a+b $ we get
$$a^2 \mid 2a \implies a \mid 2 \implies a \in \{1,2\}.$$ Both $(a,b)\in\{(1,1), (2,2)\}$ are solutons.
A: $abx=(a+b)$, 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 $a=b=1$
