N53 (44-2, Bulgaria) 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 that satisfy 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 $k \geqslant 1$, i.e., $a^{2} \geqslant b^{2}(2 a-b)+1$, we know that $a^{2}>b^{2}(2 a-b) \geqslant 0$. Therefore,
$a>b$ or $2 a=b$.
Now let $a_{1}, a_{2}$ be the two solutions of the equation in $a$ (with $k, b$ fixed)
$$
a^{2}-2 k b^{2} a+k\left(b^{3}-1\right)=0
$$

where one of the solutions is an integer. Then, by $a_{1}+a_{2}=2 k b^{2}$, the other solution is also an integer. Without loss of generality, assume $a_{1} \geqslant a_{2}$, then $a_{1} \geqslant k b^{2}>0$, further, by $a_{1} a_{2}=k\left(b^{3}-1\right)$, we get
$$
0 \leqslant a_{2}=\frac{k\left(b^{3}-1\right)}{a_{1}} \leqslant \frac{k\left(b^{3}-1\right)}{k b^{2}}0$, then from the previous result, $a_{2}=\frac{b}{2}$, then $k=\frac{b^{2}}{4}, a_{1}=\frac{b^{4}}{2}-\frac{b}{2}$. At this point, $b$ is even.
In summary, we can only have
$$
(a, b)=(2 l, 1),(l, 2 l) \text { or }\left(8 l^{4}-l, 2 l\right),
$$

where $l$ is a positive integer. Direct verification shows that the above forms of $(a, b)$ all meet the requirements.
Solution 2 When $b=1$, by the condition, $a$ is even.
When $b>1$, 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. That is,
$$
\Delta=4 k^{2} b^{4}-4 k\left(b^{3}-1\right)
$$

is a perfect square.
Notice that, when $b \geqslant 2$, we have
$$
\left(2 k b^{2}-b-1\right)^{2}0 ; \\
\left(2 k b^{2}-b+1\right)^{2}-\Delta & =4 k b^{2}-4 k+(b-1)^{2} \\
& >0
\end{aligned}
$$

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

Thus, $4 k=b^{2}$, hence $b$ is even, set $b=2 l$, then $k=l^{2}$. Using (2), we can solve for $a=l$ $8 l^{1}-l$.

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.