USA IMO 2003

First Solution. (Based on work by Anders Kaseorg) Rewrite equation $(\ast )$ as $a^2-2a{b}^{2}k=-b^{3}k+k$. Adding $b^4{k}^2$ to both sides completes the square on the left-hand side and gives $$(k{b}^2-a{)}^2=b^4{k}^2-b^{3}k+k,$$ or $$(2k{b}^2-2a{)}^2=(2b^{2}k{)}^2-2b(2b^{2}k)+4k.$$ Completing the square on the right-hand side gives $$(2k{b}^2-2a{)}^2=(2b^{2}k-b{)}^2+4k-b^2.$$ or, $$y^2-x^2=4k-b^2,$$ where $x=2k{b}^2-b$ and $y=2k{b}^2-2a$. If $4k=b^2$, then either $x=y$ or $x=-y$. In the former case, $b=2a$; in the latter case, $4k{b}^2-b=2a$, that is, $b^4-b=2a$. Because $k=b^2/4$ is an integer if and only if $b$ is even, we get the solutions $$\left(\frac{b}{2},b\right)\text{and}\left(\frac{b^4-b}{2},b\right)$$ for any even $b$; that is, $(a,b)=(t,2t)$ and $(a,b)=(8t^4-t,2t)$ for all positive integers $t$. If $4k<b^2$, then $y^2<x^2$, so $y^2\le (x-1{)}^2$ (since $x$ is clearly positive). Thus, $$4k-b^2\le (x-1{)}^2-x^2=-2x+1=-4k{b}^2+2b+1,$$ or, $4k(b^2+1)\le b^2+2b+1<3b^2+1$. Because $4k>3$, this is a contradiction. Similarly, if $4k>b^2$, then $y^2>x^2$, so $y^2\ge (x+1{)}^2$. Thus, $$4k-b^2\ge (x+1{)}^2-x^2=2x+1=4k{b}^2-2b+1,$$ or $4k(b^2-1)+(b-1{)}^2\le 0$. We must have $b=1$ and $$k=\frac{a^2}{2a-1+1}=\frac{a}{2}.$$ This is an integer whenever $a$ is even, so we get the solutions $(a,b)=(2t,1)$ for all positive integers $t$.

Second Solution. (Based on work by Po-Ru Loh) Assume that $b=1$. Then $$\frac{a^2}{2a{b}^2-b^3+1}=\frac{a^2}{2a}=\frac{a}{2}$$ is a positive integer if and only if $a$ is even. Thus, $(a,b)=(2t,1)$ are solutions of the problem for all positive integers $t$. Now we assume that $b>1$. Viewing equation $(\ast )$ as a quadratic in $a$, replace $a$ by $x$ to consider the equation $$x^2-2b^{2}kx+(b^3-1)k=0(\ast {)}^{\prime }$$ for fixed positive integers $b$ and $k$. Its roots are $$x=\frac{2b^{2}k\pm \sqrt{4b^4{k}^2-4b^{3}k+4k}}{2}=b^{2}k\pm \sqrt{b^4{k}^2-b^{3}k+k}.$$ Assume that $x_1=a$ is an integer root of equation $({\ast }^{\prime })$. Then $b^4{k}^2-b^{3}k+k$ must be a perfect square. We claim that $${\left(b^{2}k-\frac{b}{2}-\frac{1}{2}\right)}^2<b^4{k}^2-b^{3}k+k<{\left(b^{2}k-\frac{b}{2}+\frac{1}{2}\right)}^2.$$ Note first that $$\begin{array}{rl}{\left(b^{2}k-\frac{b}{2}-\frac{1}{2}\right)}^2& =b^4{k}^2-2b^{2}k{\left(\frac{b}{2}+\frac{1}{2}\right)}^2+\frac{1}{4}(b+1{)}^2\\ & =b^4{k}^2-b^{3}k-b^{2}k+\frac{1}{4}(b+1{)}^2.\end{array}$$ To establish the first inequality in our claim, it suffices to show that $$-b^{2}k+\frac{1}{4}(b+1{)}^2<k,$$ or, $(b+1{)}^2<4(b^2+1)k$, which is evident as $(b+1{)}^2<2(b^2+1)$ and $k\ge 1$. Note also that $$\begin{array}{rl}{\left(b^{2}k-\frac{b}{2}+1\right)}^2& =b^4{k}^2+2b^{2}k\left(\frac{1}{2}-\frac{b}{2}\right)+\frac{1}{4}(1-b{)}^2\\ & >b^4{k}^2-b^{3}k+b^{2}k\\ & >b^4{k}^2-b^{3}k+k,\text{as }b>1,\end{array}$$ which establishes the second inequality in our claim. Because all of $b,k$, and $\sqrt{b^4{k}^2-b^{3}k+k}$ are positive integers, we conclude from our claim that $b^{2}k-\frac{b}{2}$ is an integer and that $$b^4{k}^2-b^{3}k+k={\left(b^{2}k-\frac{b}{2}\right)}^2=b^{4}k-b^{3}k+\frac{b^2}{4},$$ and so $k=b^2/4$. Thus, $b=2t$ for some positive integer $t$. The two solutions of the equation $(\ast )$ becomes $$x=b^{2}k\pm \left(b^{2}k-\frac{b}{2}\right),$$ that is. $x=t$ or $x=8t^4-t$. Hence, $(a,b)=(t,2t)$ and $(a,b)=(8t^4-t,2t)$ are the possible solutions of the problem, in addition to the solutions $(2t,1)$.

Third Solution. Because both $k$ and $a^2$ are positive, $2a{b}^2-b^3+1>0$, or, $$2a>b-\frac{1}{b^2}.$$ Because $a$ and $b$ are positive integers, we have $2a\ge b$. Because $k$ is a positive integer, $a^2\ge 2a{b}^2-b^3+1$, or, $a^2\ge b^2(2a-b)+1$. Because $$a^2>b^2(2a-b)\ge 0,$$ we have $$a>b\text{or}2a=b.(†)$$ We consider again the quadratic equation $(\ast )$ for fixed positive integers $b$ and $k$, and assume that $x_1=a$ is an integer root of equation $(\ast )$. Then the other root $x_2$ is also an integer because $x_1+x_2=2b^{2}k$. Without loss of generality, we assume that $x_1\ge x_2$. Then $x_1\ge b^{2}k>0$. Furthermore, because $x_1{x}_2=(b^3-1)k$, we obtain $$0\le x_2=\frac{(b^3-1)k}{x_1}\le \frac{(b^3-1)k}{b^{2}k}<b.$$ If $x_2=0$, then $b^3-1=0$, and so $x_1=2k$ and $(a,b)$ can be written in the form of $(2t,1)$ for some integers $t$. If $x_2>0$, then $(a,b)=(x_2,b)$ is a pair of positive integers satisfying the equations $(†)$ and $(\ast )$. We conclude that $2x_2=b$, and so $$k=\frac{x_2^2}{2x_2{b}^2-b^3+1}=x_2^2=\frac{b^2}{4},$$ and $x_1=b^4/2-b/2$. Thus, $(a,b)$ can be written in the form of either $(t,2t)$ or $(8t^3-t,2t)$ for some positive integers $t$.