Find the integer solutions of the equation

$$
x^{2}=y^{2}\left(x+y^{4}+2 y^{2}\right)
$$
If $x=0$, then $y=0$ and conversely, if $y=0$, then $x=0$. It follows that $(x, y)=(0,0)$ is a solution of the problem. Assume $x \neq 0$ and $y \neq 0$ satisfy the equation. The equation can be transformed in the form $x^{2}-x y^{2}=y^{6}+2 y^{4}$. Then $4 x^{2}-4 x y^{2}+y^{4}=4 y^{6}+9 y^{4}$ and consequently $\left(\frac{2 x}{y^{2}}-1\right)^{2}=4 y^{2}+9 \quad$ (1). Obviously $\frac{2 x}{y^{2}}-1$ is integer. From (1), we get that the numbers $\frac{2 x}{y^{2}}-1,2 y$ and 3 are Pythagorean triplets. It follows that $\frac{2 x}{y^{2}}-1= \pm 5$ and $2 y= \pm 4$. Therefore, $x=3 y^{2}$ or $x=-2 y^{2}$ and $y= \pm 2$. Hence $(x, y)=(12,-2),(x, y)=(12,2),(x, y)=(-8,-2)$ and $(x, y)=(-8,2)$ are the possible solutions. By substituting them in the initial equation we verify that all the 4 pairs are solution. Thus, together with the couple $(x, y)=(0,0)$ the problem has 5 solutions.