Find all pairs of natural numbers, $(x,y)$, satisfying $xy=x+y+(x,y)$, with $x\le y$. Where $(x,y)$ is the gcd of $x$ and $y$.
My attempt:- Let $(x,y)=d$, then $x=dt$ and $y=dk$ for some integers $t,k$ and $(t,k)=1$. If $t = k = 1$ then $x=y=d$ and hence we get a pair $(3,3)$ which satisfies the above equation. So from here we can assume that $t\neq k$ and also $t\lt k$ as $x\leq y$. Now the equation $$xy=x+y+(x,y)$$ can be written as $$d^{2}tk=d(t+k+1)$$ which reduces to $$dtk=t+k+1$$ And this implies $d\mid t+k+1$ , $t\mid k+1$ and $k\mid t+1$. The latter two conclusions give $t=1$ and $k=2$ or $t=2$ and $k=3$ .This means $t+k+1= 4$ or $6$, this implies $d= 1$ or $2$ or $3$ or $4$ or $6$ .Possibilities ,$d=3,4$ and $6$ can be easily ruled out. So $d=1$or $2$ and this gives us the pairs $(2,3$ and $(2,4)$. So the only pairs satisfyng the above equation are $(2,3)$, $(2,4)$ and $(3,3)$. Does this look correct?I think, I have complicated this too much. Can this problem be solved without going too much into divisibility? Thank you.
$xy \leq y+y+y=3y$ so $ x\leq 3$. That should make the solution simpler.