3. Find all pairs of positive integers $(a, b)$, satisfying $a\left|b^{2}, b\right| a^{2},(a+1) \mid\left(b^{2}+1\right)$.
(2012, Serbian Mathematical Olympiad)
Given $b^{2}=c a$.
Then the original problem is transformed into
$$
\begin{array}{l}
b^{2}=c a \mid a^{4} \text { and }(a+1) \mid(c a+1) \\
\Leftrightarrow c \mid a^{3} \text { and }(a+1) \mid(c-1) .
\end{array}
$$

Let $c=d(a+1)+1\left(d \in \mathbf{Z}_{+}\right)$.
Since $a^{3} \equiv-1 \equiv-c(\bmod a+1)$,
$c \equiv 1(\bmod a+1)$,
we know $\frac{a^{3}}{c} \equiv-1(\bmod a+1)$,
which means there exists $e \in \mathbf{Z}_{+}$, such that $\frac{a^{3}}{c}=e(a+1)-1$.
$$
\begin{array}{l}
\text { Hence } a^{3}=c(e(a+1)-1) \\
=(d(a+1)+1)(e(a+1)-1) .
\end{array}
$$

Expanding and simplifying, we get
$$
a^{2}-a+1=d e(a+1)+(e-d),
$$

which implies $e-d \equiv a^{2}-a+1 \equiv 3(\bmod a+1)$,
so $e-d=k(a+1)+3$, and
$$
d e=a-2-k(k \in \mathbf{Z}) \text {. }
$$

Considering four cases: $k \in \mathbf{Z}_{+}$, $k=0$, $k=-1$, and $k \leqslant-2$, we find
$$
\begin{aligned}
(a, b)= & \left(t^{2}, t\right),\left((t+1)^{2}\left(t^{2}+2 t\right),\right. \\
& \left.(t+1)\left(t^{2}+2 t\right)^{2}\right),\left(t^{2}, t^{3}\right)(t \in \mathbf{N}) .
\end{aligned}
$$