International Mathematical Olympiad

Solution I Let $(a,b)$ be a pair of positive integers satisfying the condition. Since $k=\frac{a^2}{2a{b}^2-b^3+1}>0$, we have $2a{b}^2-b^3+1>0$ or $a>\frac{b}{2}-\frac{1}{2b^2}$, and hence $a\ge \frac{b}{2}$. Using this, we infer from $k\ge 1$, or $a^2\ge b^2(2a-b)+1$, that $a^2>b^2(2a-b)\ge 0$. Hence $$a>b\text{ or }2a=b.\stackrel{◯}{1}$$ Now consider the two solutions $a_1,a_2$ of the equation $$a^2-2k{b}^{2}a+k(b^3-1)=0\stackrel{◯}{2}$$ for any fixed positive integers $k$ and $b$, and assume that one of them is an integer. Then the other is also an integer because $a_1+a_2=2k{b}^2$. We may assume that $a_1\ge a_2$, and we have $a_1\ge k{b}^2>0$. Furthermore, since $a_1{a}_2=k(b^3-1)$, we get $$0\le a_2=\frac{k(b^3-1)}{a_1}\le \frac{k(b^3-1)}{k{b}^2}<b.$$ Together with (1), we conclude that $a_2=0$ or $a_2=\frac{b}{2}$ (in the latter case $b$ must be even). If $a_2=0$, then $b^3-1=0$, and hence $a_1=2k$ and $b=1$. If $a_2=\frac{b}{2}$, then $k=\frac{b^2}{4}$ and $a_1=\frac{b^4}{2}-\frac{b}{2}$. Therefore the only possibilities are $$(a,b)=(2l,1),(l,2l)\text{ or }(8l^4-l,2l)$$ for some positive integer $l$. All of these pairs satisfy the given condition.

Solution II If $b=1$, it follows from the given condition that $a$ must be even. Let $\frac{a^2}{2a{b}^2-b^3+1}=k$. If $b>1$, then there are two solutions to the equation (②) and one of them is a positive integer. Thus the discriminant $\Delta$ of the equation (②) is a perfect square, that is $\Delta =4k^2{b}^4-4k(b^3-1)$ is a perfect square. Note that, if $b\ge 2$, we have $$(2k{b}^2-b-1{)}^2<\Delta <(2k{b}^2-b+1{)}^2.\stackrel{◯}{3}$$ The proof is given as follows, $$\begin{array}{rl}\Delta -(2k{b}^2-b-1{)}^2& =4k{b}^2-b^2-2b+4k-1\\ & =(4k-1)(b^2+1)-2b\\ & \ge 2(4k-1)b-2b>0,\end{array}$$ $$\begin{array}{rl}(2k{b}^2-b+1{)}^2-\Delta & =4k{b}^2-4k-(b-1{)}^2\\ & =4k(b^2-1)-(b-1{)}^2\\ & >(4k-1)(b^2-1)>0,\end{array}$$ this completes the proof of (3). Since $\Delta$ is a perfect square, it follows from (3) that $$\Delta =4k^2{b}^4-4k(b^3-1)=(2k{b}^2-b{)}^2.$$ Then $4k=b^2$, and hence $b$ must be even. Let $b=2l$. We have $k=l^2$. Together with ②, we have $a=l$ or $8l^4-l$. Therefore the only possibilities are $(a,b)=(2l,1),(l,2l)\text{ or }(8l^4-l,2l)$ for some positive integer $l$. All of these pairs satisfy the given condition.