Number of natural numbers $x$ and $y$ with $x\leq y$ satisfying $xy = x + y +\operatorname{GCD}(x,y)$

Question

PRMO 2014 Problem 11 :

For natural numbers $x$ and $y$ let $(x,y)$ denote the greatest common divisor of $x$ and $y$. How many pairs of natural numbers $x$ and $y$ with $x\leq y$ satisfy the equation $xy = x + y + (x,y)$?

My solution :

Assume that $x = aq$ and $y = bq$ where $(a,b) = 1$

Then -

$abq = a + b + 1 $

$\implies (aq-1)(bq-1) = q $

I am unable to solve the question further.

This diophantine is different from the one's in which the value of q is constant and given. Here it's variable so, I think that some sort of range would have to be figured out and then we'll proceed with hit and trial.

I tried transforming it into a quadratic in $q $ and equating the discriminant as a perfect square but none of it is actually helping.

So, how shall I be proceeding further ?

Answer 1

You are on the right track, but seem to have made a mistake in the bookkeeping. With $d:=(x,y)$ and $x=ad$ and $y=bd$, so that $a,b\geq1$ and $(a,b)=1$, you get $$xy=x+y+(x,y)\qquad\Rightarrow\qquad abd^2=ad+bd+d,$$ and hence $$(ad-1)(bd-1)=d+1.$$ If $d=1$ then $(a-1)(b-1)=2$ shows that $\{x,y\}=\{2,3\}$.

If $d\geq2$ then $ad-1,bd-1\geq1$ and so $$d+1=(ad-1)(bd-1)\geq ad-1\geq(a-1)d+1,$$ and so by symmetry $a,b\leq2$. For $a=b=1$ we find $d=3$ and so $x=y=3$. For $\{a,b\}=\{1,2\}$ we find $d=2$ and so $\{x,y\}=\{2,4\}$.