Solve $x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$ over the integers.

Question

Solve $$x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$$ over the integers.

You can probably guess by now... This problem is adapted from a recent competition.

If there are any other solutions, please post them below. I have provided one if you want to check out.

Answer 1

Not suprisingly,

$$x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$$

$$\iff (x^2y^2 + y^3 + y^2) - (4x^2y + 4y^2 + 4y) + (4x^2 + 4y + 4) = 3$$

$$\iff y^2(x^2 + y + 1) - 4y(x^2 + y + 1) + 4(x^2 + y + 1) = 3$$

$$\iff (y^2 - 4y + 4)(x^2 + y + 1) = 3 \iff (y - 2)^2(x^2 + y + 1) = 3$$

$$\implies (y - 2)^2 \mid 3 \implies (y - 2)^2 \in \{\pm 1, \pm 3\}$$

Having said that, $(y - 2)^2$ is a perfect square. $\implies (y - 2)^2 = 1 \implies \left\{ \begin{align} x^2 + y + 1 = 3\\ y - 2 = \pm 1 \end{align} \right.$.

$$\implies x^2 = -(y - 2) = \mp 1 \implies x^2 = -(y - 2) = 1 \iff \left\{\begin{align} x = \pm 1\\ y = 1\end{align} \right.$$

Answer 2

Write $$x^2(y-2)^2 = -y^3+3y^2-1$$

Since $y-2\mid -y^3+3y^2-1$ and $y\equiv 2\pmod{y-2}$ we have

$$0\equiv -y^3+3y^2-1 \equiv -8+12-1 \equiv 3 \pmod{y-2}$$ So $$y-2\mid 3\implies y-2\in\{1,-1,3,-3\}$$

so $$y\in\{3,1,5,-1\}$$ Checking each of them we are done.