2003 IMO Problems/Problem 2

Problem

(Aleksander Ivanov, Bulgaria) Determine all pairs of positive integers $(a,b)$ such that $\frac{a^2}{2ab^2-b^3+1}$ is a positive integer.

Solution

The only solutions are of the form $(a,b) = (2n,1)$ , $(a,b) = (n,2n)$ , and $(8n^4-n,2n)$ for any positive integer $n$ .

First, we note that when $b=1$ , the given expression is equivalent to $a/2$ , which is an integer if and only if $a$ is even.

Now, suppose that $(a,b)$ is a solution not of the form $(2n,1)$ . We have already given all solutions for $b=1$ ; then for this new solution, we must have $b>1$ . Let us denote $\frac{a^2}{2ab^2-b^3+1} = k .$ Denote $P(t) = t^2 - 2kb^2 t + k(b^3-1) .$ Since $k(b^3-1) >0$ , and $a$ is a positive integer root of $P$ , there must be some other root $a'$ of $P$ .

Without loss of generality, let $a' \ge a$ . Then $a^2 \le aa' = k(b^3-1)$ , so $k = \frac{a^2}{2ab^2-b^3+1} \le k \frac{b^3-1}{2ab^2-b^3+1},$ or $2ab^2 - (b^3-1) \le b^3-1,$ which reduces to $a \le b - \frac{1}{b^2} < b .$ It follows that $0 < 2ab^2 -b^3 + 1 \le a^2 < b^2 ,$ or $0 < (2a-b)b^2 + 1 < b^2 .$ Since $a$ and $b$ are integers, this can only happen when $2a-b=0$ , so $(a,b)$ can be written as $(n,2n)$ , and $k = n^2$ . It follows that $a' = \frac{k(b^3-1)}{a} = 8n^4-n.$ Since $a'$ is the other root of $P$ , it follows that $(a',b)$ also satisfies the problem's condition. Therefore the solutions are exactly the ones given at the solution's start. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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2003 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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2003 IMO Problems/Problem 2

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