Diophantine Equations of degree 3 and 4

Question

I have these two Diophantine equations: $$0 = x^2 y - 36 x^2 - 12 x y + 36 x + 36 y - 9$$ and $$0 = x^2y^2 - 36 x^2 - 12 x y^2 + 36 x + 36 y^2 - 9$$ Each of them is slightly different from each other, but the first one is of degree 3 while the one is degree 4. Can any one of them be solved without factoring and if so, how? I graphed it but nothing really stood out.

Answer 1

I know that you asked for solutions without factorizing, but factorizing is the best way - if not the only one - to solve diophantine equations.

From $(1)$ we obtain $$y= \bigg(\frac{6x-3}{x-6}\bigg)^2$$

Since $y\in\mathbb Z$, so does $$\frac{6x-3}{x-6}=\frac{6\cdot (x-6)+33}{x-6}=6+\frac{33}{x-6}$$ Hence, $x-6\mid 11\cdot 3$, which implies that

$$x-6=\left\{ \begin{array}{ll} \pm 33\\ \pm11\\ \pm3\\ \pm1 \end{array} \right.$$

It's easy to end it now (solve each equation separately). The second case is very similar.

Answer 2

Your first equation can be written as $$y(x-6)^2-(x-3)^2=0$$ and the second as $$y^2(x-6)^2-(x-3)^2=0$$