Find the sum of the areas of all rectangles whose area is tripled when three units are added to the height and two units are added to the length

Question

A rectangle has all sides of integer length. When three units are added to the height and two units to the length, the area of the rectangle is tripled. What is the sum of all the original areas of such rectangles.

I thought I'd start by calling the sides of the rectangle $x$ and $y$. Then I'd continue by :

$(x+3)(y+2) = 3xy$. But how do I proceed?

Answer 1

We have $$2xy-2x-3y=6$$ so $$(2x-3)(y-1)=9\ .$$ Since $x$ and $y$ are integers, $2x-3$ and $y-1$ must be factors of $9$. Can you take it from here?

Answer 2

$2xy-2x-3y-6=0\iff x=\dfrac{3y+6}{2(y-1)}$

Clearly $y$ must be even $=2z$ say,

$x=\dfrac{3z+3}{2z-1}$

As $2z-1$ is odd,

$\implies2z-1$ must divide $2(3z+3)=3(2z-1)+9$

Can you take it from here?