$y^2+5xy+6x^2-9x-4y=0$ where $x$ and $y$ are integers Need some help solving this for integers $x$ and $y$:
$$
y^2+5xy+6x^2-9x-4y=0
$$
I managed to make something like this:
$$
(y+3x-4)(y+2x)=x\\
(y+3x)(y+2x-3)=y
$$
Find integers for $x$ and $y$ that satisfy the equations above.
But, what do I do next, or is this a bad approach?
 A: The equation defines a conic section, and as a first step you could put it into a more standard form. Let $u,v$ be variables to replace $x,y$, where $x=u+a$ and $y=v+b$. Here, $a,b$ are fixed (but not yet known) integers. This turns the equation into:
$$(v+b)^2+5(u+a)(v+b)+6(u+a)^2-9(u+a)-4(v+b)=0$$
$$v^2+5uv+6u^2+(2b+5a-4)v+(5b+12a-9)u+(b^2+5ab+6a^2-9a-4b)=0$$
We would like to choose $a,b$ so that the $v$ and $u$ terms vanish. A little linear algebra gives that this happens when $a=2$ and $b=-3$. So you have:
$$v^2+5uv+6u^2-3=0$$
$$v^2+5uv+6u^2=3$$
Now, much easier to factor:
$$(v+2u)(v+3u)=3$$
And you could use the various integer factorizations of $3$ to solve for possible pairs $(u,v)$, and then convert back to pairs $(x,y)$.
A: $$  (y+3x -3)(y+2x - 1)= y^2 + 5 yx + 6 x^2 -9x-4y +3  $$
$$  (y+3x -3)(y+2x - 1)= 3  $$
A: Hint:  written as a quadratic in $\,y\,$, the discriminant of $\;y^2+(5x-4)y+3x(2x-3)\,$ must be a perfect square: $\;\Delta_y=(5x-4)^2-12x(2x-3)=x^2 - 4 x + 16=(x-2)^2+12\,$.
A: We have $$(y+3x)(y+2x)-(9x+4y)=0$$
so if we put $a=y+3x$ and $b=y+2x$ we get $$ab-a-3b=0\implies b={a\over a-3}$$
so $a-3\mid a$ and thus $a-3\mid 3$ so $a-3\in\{-3,-1,1,3\}$ so $a\in\{0,2,4,6\}$ and corresponding $b\in\{0,-2,4,2 \}$ Now for each pair $(a,b)$ solve for $x$ and $y$...
