One, (20 points) Find all integer solutions to the equation $a^{3} b-a b^{3}+2 a^{2}+2 b^{2}+4=0$.


One, transform the original equation to
$$
2^{2}+\left(a^{2}+b^{2}\right) \times 2+\left(a^{3} b-a b^{3}\right)=0 \text {. }
$$

Let $2=x$, then
$$
x^{2}+\left(a^{2}+b^{2}\right) x+\left(a^{3} b-a b^{3}\right)=0 \text {. }
$$

Considering the above equation as a quadratic equation in $x$, by the quadratic formula we get $x=-b^{2}-a b$ or $x=-a^{2}+a b$.
Thus, $2=-b^{2}-a b$ or $2=-a^{2}+a b$.
From $2=-b^{2}-a b$ and $b \neq 0$, we get
$$
a=-b-\frac{2}{b} \text {. }
$$

According to equation (1), since $a$ and $b$ are integers, $\frac{2}{b}$ must be an integer, hence $b= \pm 1$ or $\pm 2$.
When $b=1$, $a=-3$; when $b=-1$, $a=3$;
When $b=2$, $a=-3$; when $b=-2$, $a=3$.
From $2=-a^{2}+a b$ and $a \neq 0$, we get $b=a+\frac{2}{a}$.
Similarly, we get
When $a=1$, $b=3$; when $a=-1$, $b=-3$;
When $a=2$, $b=3$; when $a=-2$, $b=-3$.
Therefore, the original equation has 8 integer solutions: