$12 \cdot 36$ Find all positive integer pairs $(a, b)$ that satisfy the equation
$$
2 a^{2}=3 b^{3}
$$

(10th Canadian Mathematics Competition, 1978)
[Solution] Let $(x, y)$ be a pair of positive integers satisfying equation (1), then
$$
2 x^{2}=3 y^{3} \text {. }
$$

Since 2 and 3 are coprime, $x$ must be a multiple of 3.
Let $x=3 x_{1}$, then we have
$$
\begin{aligned}
2 \cdot 3^{2} x_{1}^{2} & =3 y^{3}, \\
2 \cdot 3 x_{1}^{2} & =y^{3} .
\end{aligned}
$$

Thus, $y$ is a multiple of $2 \cdot 3=6$.
Let $y=6 y_{1}$, then we have
$$
\begin{array}{c}
6 x_{1}^{2}=6^{3} y_{1}^{3}, \\
x_{1}^{2}=6^{2} y_{1}^{3} .
\end{array}
$$

Therefore, $x_{1}$ is also a multiple of 6.
Let $x_{1}=6 x_{2}$, then we have

Thus,
$$
\begin{array}{l}
6^{2} x_{2}^{2}=6^{2} y_{1}^{3}, \\
x_{2}^{2}=y_{1}^{3}=c .
\end{array}
$$

Hence, $c$ is both a perfect square and a perfect cube, so $c$ is a sixth power.
Let $c=d^{6}$, then
$$
\begin{array}{c}
x_{2}=d^{3}, \quad y_{1}=d^{2} . \\
x_{1}=6 x_{2}=6 d^{3}, \quad y=6 y_{1}=6 d^{2}, \\
x=3 x_{1}=18 d^{3} .
\end{array}
$$

Therefore, the required pairs are
$$
(a, b)=\left(18 d^{3}, 6 d^{2}\right)
$$

where $d$ is any positive integer.