<s>[INST] How can this equation be solved? I have no idea how to solve this equation.
$$x^2y^2+324y^2+64x^2-36xy^2-16x^2y+144xy = 0 $$
Sorry $x,y \in \mathbb{+Z}$
 [/INST] Consider the equation as a quadratic in $y$, we obtain
$$ y^2 (x^2-36x+324)+y (144x-16x^2) + 64x^2 = 0 $$
First, check that if the coefficient of $y^2$ is 0, then we must have $ x = 18$, which gives $y=8$ (1 solution).
Otherwise, using the quadratic equation, we get that (for $x\neq 18$),
$$y = \frac { 8(x^2-9x \pm 3\sqrt{x^2( 2x - 27)})} {(x-18)^2} $$
Hence, we must have $2x-27 = (2a+1)^2$ for some non-negative $a$. This gives $x = 2a^2 + 2a + 14$. Substituting it into $y$ above, we get
$$ \begin{array} {l l }y & = \frac{ 8 (2a^2 + 2a + 14) ( 2a^2 + 2a + 14 - 9 \pm 3 (2a+1) )  } { (2a^2+2a-4)^2} \\ & = \frac{ 4 (a^2 + a + 7) ( 2a^2 + 2a + 14 - 9 \pm (6a+3) )}{[(a-1)(a+2)]^2} \end{array}$$
Let's first deal with the $+$ case. The numerator actually will cancel out with $(a+2)^2$ in the denominator. Then, using partial fractions, we get
$$ y = 8 + \frac{24}{a-1} + \frac{72}{(a-1)^2}$$
Since $72 = 2^3 \times 3^2$, this means that $(a-1) = -1, 1, 2, 3, 6$.
The $-$ case is similar, and left to you.
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