# 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. In this problem we can do it by an alternative method a^2/2ab^2-b^3+1>=1 a^2>=2ab^2-b^3+1 a^2-2ab+b^2>=1/b (a-b)^2>=1/b The solutions are a>=2 and b>=1 are all the solutions

## Resources

 2003 IMO (Problems) • Resources Preceded byProblem 1 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 3 All IMO Problems and Solutions