If $A$ and $B$ are natural numbers such that $A-B = \dfrac{A}{B}$, what is the value of $A+B$? If $A$ and $B$ are natural numbers such that $A-B = \dfrac{A}{B}$,  what is the value of $A+B$?
The answer is $6$. I tried all the basic maths functions but was not able to get it. I would like to know if its a trial and error based question or it can be derived.

We have: 
$A = B + \dfrac{A}{B}\implies AB - B^2 - A = 0$
$\implies A = \dfrac{B^2}{B-1} = B+1 + \dfrac{1}{B-1}$
$\implies B-1 = 1 \implies B = 2$
$\implies A = 4$ 
$\therefore A+B = 6$
