Example 19 (IMO-44 Problem) Find all pairs of positive integers $(a, b)$ such that $\frac{a^{2}}{2 a b^{2}-b^{3}+1}$ is a positive integer.
Solution 1 Let $(a, b)$ be a pair of positive integers satisfying the condition. Since
$$
k=\frac{a^{2}}{2 a b^{2}-b^{3}+1}>0,
$$

it follows that $2 a b^{2}-b^{3}+1>0, a>\frac{b}{2}-\frac{1}{2 b^{2}}$, thus $a \geqslant \frac{b}{2}$. Combining this with $k \geqslant 1$, we have
$$
a^{2} \geqslant b^{2}(2 a-b)+1,
$$

hence $a^{2}>b^{2}(2 a-b) \geqslant 0$. Therefore, $a>b$ or $2 a=b$.
Now let $a_{1}, a_{2}$ be the two solutions of the quadratic equation in $a$ (with $k, b$ fixed)
$$
a^{2}-2 k b^{2} a+k\left(b^{3}-1\right)=0
$$

and assume one of them is an integer, then by $a_{1}+a_{2}=2 k b^{2}$, the other solution is also an integer. Without loss of generality, let $a_{1} \geqslant a_{2}$, then $a_{1} \geqslant k b^{2}>0$, further, by $a_{1} a_{2}=k\left(b^{3}-1\right)$, we get
$$
0 \leqslant a_{2}=\frac{k\left(b^{3}-1\right)}{a_{1}} \leqslant \frac{k\left(b^{3}-1\right)}{k b^{2}}1$ when, we denote $\frac{a^{2}}{2 a b^{2}-b^{3}+1}=k$, then the quadratic equation in $a$ (2) has a positive integer solution, thus the discriminant of the equation in $a$ is a perfect square, i.e., $\Delta=4 k^{2} b^{4}-4 k\left(b^{3}-1\right)$ is a perfect square.
Notice that, when $b \geqslant 2$, we have
$$
\left(2 k b^{2}-b-1\right)^{2}0 ; \\
\left(2 k b^{2}-b+1\right)^{2}-\Delta & =4 k b^{2}-4 k-(b-1)^{2}=4 k\left(b^{2}-1\right)-(b-1)^{2}
\end{aligned}
$$
In
real
mathematics
we
have
$$
>(4 k-1)\left(b^{2}-1\right)>0 .
$$

Therefore, (3) holds.
Using the fact that $\Delta$ is a perfect square and (3), we know
$$
\Delta=4 k^{2} b^{4}-4 k\left(b^{3}-1\right)=\left(2 k b^{2}-b\right)^{2},
$$

thus, $4 k=b^{2}$, and hence $b$ is even. Let $b=2 l$, then $k=l^{2}$, using (2) we can find $a=l$ or $8 l^{4}-l$.
In summary, the pairs $(a, b)=(2 l, 1),(l, 2 l)$ or $\left(8 l^{4}-l, 2 l\right)$, where $l$ is a positive integer, satisfy the condition. Direct verification shows that they meet the requirements.