# Number of all pairs of natural numbers

The question is to find out the number of all pairs of natural numbers $(x,y)$ which satisfy the equation $2x(xy-2y-3)=(x+y)(3x+y)$

It is clear that there exists no solution for $x=y$.I tried for small numbers but couldnot find one which satisfies the condition.So i deceided to go reverse and prove that l.h.s is greater than or less than r.h.s. for all natural numbers.But i couldnot go ahead.i tried to make it as a quadratic equation of x and solve for it.But that too didnot helped me.Any way to go on with this problem?Thanks.

The equation can be expressed as $y^2+y(-2x^2+8x)+(3x^2+6x)=0$. Using the quadratic, we find that $y=x^2-4x\pm\sqrt{x^4-8x^3+13x^2-6x}$. Therefore, rational solutions will exist when $x^4-8x^3+13x^2-6x=x(x-6)(x-1)^2$ is square, which itself will occur when $x(x-6)$ is square. We find solutions at $(6,12), (-2,0)$ and $(-2,24)$ as well as $(0,0)$ and $(1,-3)$ (as they make the previous equation equal to $0$). So the number of solutions will be proportionate to the number of solutions of $x(x-6)=k^2$ for some natural $k$ (I will leave this for you to figure out...)