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 by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions
  • <url>Forum/viewtopic.php?p=262#262 Discussion on AoPS/MathLinks</url>
Invalid username
Login to AoPS