Difference between revisions of "2003 IMO Problems/Problem 2"

Revision as of 11:40, 18 August 2025

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 1

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. $\blacksquare$

Solution 2

First we can reformulate the original problem as $\frac{a^2}{2ab^2-b^3+1} = k .$ Where $a$ , $b$ , $k$ are all positive integers.

We split the solutions in 2 cases:

1) Assume that $2ab^2-b^3+1>1$ . Then we have $2ab^2-b^3>0$ $b^2(2a-b)>0$ Since $b^2 > 0$ for all positive integers $b$ , we must also have $2a - b > 0$ , or $b < 2a$ for all positive integers $a$ . Therefore, the only possible value of $b$ is $b = 1$ .

When $b = 1$ , we have $\frac{a^2}{2ab^2-b^3+1} = \frac{a^2}{2a} = \frac{a}{2} = k .$ Therefore, $a = 2k$ for all positive integers $k$ .

Therefore, for $2ab^2-b^3+1>1$ , the only solution is $(a, b) = (2k, 1)$ for all positive integers $k$ .

2) a) Assume that $2ab^2-b^3+1=1$ . Then we have $2ab^2-b^3=0$ $b^2(2a-b)=0$ Therefore we have either $b=0$ or $b=2a$ .

Since we want $b$ to be a positive integer, we must have $b=2a$ for all positive integers $a$ .

When $b=2a$ , we have $\frac{a^2}{2ab^2-b^3+1} = \frac{a^2}{8a^3-8a^3+1} = a^2 = k .$ Therefore, $k = a^2$ for all positive integers $a$ .

Therefore, for $2ab^2-b^3+1=1$ we can have $(a, b) = (a, 2a)$ for all positive integers $a$ .

2) b) Notice that we can reformulate $\frac{a^2}{2ab^2-b^3+1} = k .$ as $a^2=k(2ab^2-b^3+1).$ $a^2-2kb^2a+k(b^3-1)=0.$

Using properties of quadratic equations, we have $a_1 + a_2 = 2kb^2$ $a_1 a_2 = k(b^3-1)$ From the results in part 2) a), we have $a_1 = a$ , $b=2a$ , and $k=a^2$ , so now we have $a + a_2 = 2a^2(2a)^2 = 8a^4$ $a a_2 = a^2((2a)^3-1) = 8a^5 - a^2$ Using either of the equations above, we have $a_2 = 8a^4 - a$ .

Therefore, for $2ab^2-b^3+1=1$ we can have $(a, b) = (8a^4 - a, 2a)$ for all positive integers $a$ .

Therefore, the only pairs of solutions for $\frac{a^2}{2ab^2-b^3+1} = k .$ are $(a, b) = {(2k, 1), (a, 2a), (8a^4 - a, 2a)}$ for all positive integers $a$ and $k$ . $\blacksquare$

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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>

@@ Line 85: / Line 85: @@
 <cmath> \frac{a^2}{2ab^2-b^3+1} = k . </cmath>
 are <math>(a, b) = {(2k, 1), (a, 2a), (8a^4 - a, 2a)}</math> for all positive integers <math>a</math> and <math>k</math>. <math>\blacksquare</math>
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