<s>[INST] 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.
 [/INST] $A-B=\frac{A}{B}$, so, multiplying by $B$, we have $AB-B^2=A$, so $AB=B^2+A$. 
Since $A$ and $B$ are natural numbers, we have $B \mid B^2+A$, so $B \mid A$. Write $A=CB$, then $CB^2=B^2+CB$, so $CB=B+C$, so $(C-1)(B-1)=1$, so $C=B=2$. 
Hence $A=CB=4$ and since $B=2$, we also have $A+B=6$. 
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