12. (44th IMO Problem) Find all pairs of positive integers, such that $\frac{a^{2}}{2 a b^{2}-b^{3}+1}$ is a positive integer.

 untranslated part:
(第 44 届 IMO 试题) 求所有的正整数对, 使得 $\frac{a^{2}}{2 a b^{2}-b^{3}+1}$ 为正整数. 

Corrected translation:
12. (44th IMO Problem) Find all pairs of positive integers such that $\frac{a^{2}}{2 a b^{2}-b^{3}+1}$ is a positive integer.
12. Let $(a, b)$ be a solution that satisfies the conditions.

Since $k=\frac{a^{2}}{2 a b^{2}-b^{3}+1}>0$, we have $2 a b^{2}-b^{3}+1>0$, which means $a>\frac{b}{2}-\frac{1}{2 b^{2}}$.
Therefore, $a \geqslant \frac{b}{2}$.
Given $k \geqslant 1$, i.e., $a^{2} \geqslant b^{2}(2 a-b)+1$, then $a^{2}>b^{2}(2 a-b) \geqslant 0$.
Therefore, $a>b$, or $2 a=b$.
Assume $a_{1}$ and $a_{2}$ are the two solutions of $a^{2}-2 k b^{2} a+k\left(b^{3}-1\right)=0$ for fixed positive integers $k$ and $b$, and assume one of them is an integer. Since $a_{1}+a_{2}=2 k b^{2}$, the other is also an integer.
Without loss of generality, assume $a_{1} \geqslant a_{2}$, then $a_{1} \geqslant k b^{2}>0$.
Also, since $a_{1} a_{2}=k\left(b^{3}-1\right)$, we have $0 \leqslant a_{2}=\frac{k\left(b^{3}-1\right)}{a_{1}} \leqslant \frac{k\left(b^{3}-1\right)}{k b^{2}}<b$.
Combining with (1), we get $a_{2}=0$ or $a_{2}=\frac{b}{2}$ (where $b$ must be even).
If $a_{2}=0$, then $b^{3}-1=0$. Therefore, $a_{1}=2 k, b=1$.
If $a_{2}=\frac{b}{2}$, then $k=\frac{b^{2}}{4} \cdot a_{1}=\frac{b^{3}}{2}-\frac{b}{2}$.
Thus, the possible solutions are
$(a, b)=(2 l, 1)$ or $(l, 2 l)$ or $\left(8 l^{4}-l, 2 l\right)$, where $l$ is some positive integer.
Verification shows that all these pairs satisfy the given conditions.