Last time I got stuck in this problem which I have posted earlier. Today I have come accross to this new situation.
How to solve the diophantine equation $ax^2+hxy+by^2+c=0$ in integers ? Given all of $a,b,c,h\in \mathbb Z$.
The motivation was to solve $30x^2+21y^2-57xy+729=0$ in integers which through MAPLE i got as $(x,y)=(-22,-17), (-10,-11), (10,11), (22,17)$.
On
In general, given an initial solution $x_0,y_0 = m,n$ to,
$$a x^2 + b x y + c y^2 + d = 0\tag1$$
if the discriminant $D= b^2-4ac\,$ is a non-square $D>0$, then an infinite more can be found as,
$$x = mu^2 - 2(b m + 2c n)u v + D m v^2$$
$$y = n u^2 + 2(2a m + b n)u v + D n v^2$$
where $u,v$ solve the Pell equation,
$$u^2 - D v^2 = \pm1\tag2$$
However, if your $D$ is a square, or $D<0$, then $(1)$ will only have a finite number of integer solutions. (In your case, it is a square $D=27^2$.)
Here's the method to solve such an equation ( both theory and equation solver ).
http://www.alpertron.com.ar/QUAD.HTM