One, (50 points) Find all non-zero integers $a, b$, such that $\frac{(a, b)\left(a b^{2}+1\right)}{a^{2} b+2}$ is an integer. Here, $(a, b)$ denotes the greatest common divisor of $a$ and $b$.
(1) If $a=2 b$, then $(a, b)=|b|$,
$$
\frac{d\left(a b^{2}+1\right)}{a^{2} b+2} \in \mathbf{Z} \Leftrightarrow \frac{|b|\left(2 b^{3}+1\right)}{4 b^{3}+2} \in \mathbf{Z}
$$
$$
\Leftrightarrow \frac{|b|}{2} \in \mathbf{Z}.
$$
Let $b=2 t(t \in \mathbf{Z}, t \neq 0)$. Then $a=4 t$. Thus,
$$
\left\{\begin{array}{l}a=4 t, \\ b=2 t\end{array}(t= \pm 1, \pm 2, \cdots)\right.
$$
is a solution to this problem.
(2) If $a>2 b$, then from equation (1) we get
$$
\left|a^{2} b+2\right| \leqslant a-2 b.
$$
From equation (2), we can get $b<0$, (if $b>0$, then from equation (2) we get $a^{2}+2 \leqslant a^{2} b+2=\left|a^{2} b+2\right| \leqslant a-2 b \leqslant a-2$, which contradicts $b>0$ and $a>2 b$).
(i) If $a=1$, then
$$
\frac{2 b-1}{b+2} \in \mathbf{Z} \Leftrightarrow \frac{2(b+2)-5}{b+2} \in \mathbf{Z} \Leftrightarrow \frac{5}{b+2} \in \mathbf{Z}
$$
$$
\Leftrightarrow b+2= \pm 1, \pm 5 \Leftrightarrow b=-3,-1,3,-7.
$$
Thus, $(a, b)=(1,-7),(1,-3),(1,-1)$, $(1,3)$ are also solutions to this problem.
(ii) If $a=-1$, then
$$
\frac{2 b+1}{b+2} \in \mathbf{Z} \Leftrightarrow \frac{2(b+2)-3}{b+2} \in \mathbf{Z} \Leftrightarrow \frac{3}{b+2} \in \mathbf{Z}
$$
$$
\Leftrightarrow b+2= \pm 1, \pm 3 \Leftrightarrow b=-3,-1,-5,1.
$$
Thus, $(a, b)=(-1,-5),(-1,-3)$, $(-1, -1)$, $(-1,1)$ are also solutions to this problem.
(iii) If $|a| \geqslant 2$, then from equation (1) we get
$$
-\left(a^{2} b+2\right) \leqslant a-2 b,
$$

i.e., $a^{2} b+2+a-2 b \geqslant 0$,
$$
\begin{array}{l}
0 \leqslant\left(a^{2}-2\right) b+a+2 \leqslant-\left(a^{2}-2\right)+a+2, \\
a^{2}-a-4 \leqslant 0, \\
(a-3)(a+2) \leqslant-2 .
\end{array}
$$

Therefore, $a=2$.
Substituting into equation (1) we get
$$
\begin{array}{l}
\frac{2-2 b}{4 b+2} \in \mathbf{Z} \Leftrightarrow \frac{1-b}{2 b+1} \in \mathbf{Z} \\
\Leftrightarrow \frac{2-2 b}{2 b+1} \in \mathbf{Z}(2 \text { and } 2 b+1 \text { are coprime }) \\
\Leftrightarrow \frac{3-(2 b+1)}{2 b+1} \in \mathbf{Z} \Leftrightarrow \frac{3}{2 b+1} \in \mathbf{Z} \\
\Leftrightarrow 2 b+1= \pm 1, \pm 3 .
\end{array}
$$

If $b<0$, and $|a| \neq 1$, then from equation (3) we get
$$
a^{2} b+2 \leqslant 2 b-a,
$$

i.e., $\left(a^{2}-2\right) b+a+2 \leqslant 0$.
Thus, $a^{2}-2+a+2 \leqslant\left(a^{2}-2\right) b+a+2 \leqslant 0$,

i.e., $a(a+1) \leqslant 0$.
This contradicts $a \neq 0$ and $|a| \neq 1$.
(iii) If $b>0$, and $|a|=1$, then this has already been discussed in (2) (i) and (ii).