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

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is a positive integer.
 
is a positive integer.
  
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== Solution 1 ==
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== Solution ==
 
The only solutions are of the form <math>(a,b) = (2n,1)</math>, <math>(a,b) = (n,2n)</math>, and <math>(8n^4-n,2n)</math> for any positive integer <math>n</math>.
 
The only solutions are of the form <math>(a,b) = (2n,1)</math>, <math>(a,b) = (n,2n)</math>, and <math>(8n^4-n,2n)</math> for any positive integer <math>n</math>.
  
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<cmath> a' = \frac{k(b^3-1)}{a} = 8n^4-n. </cmath>
 
<cmath> a' = \frac{k(b^3-1)}{a} = 8n^4-n. </cmath>
 
Since <math>a'</math> is the other root of <math>P</math>, it follows that <math>(a',b)</math> also satisfies the problem's condition.  Therefore the solutions are exactly the ones given at the solution's start.  <math>\blacksquare</math>
 
Since <math>a'</math> is the other root of <math>P</math>, it follows that <math>(a',b)</math> also satisfies the problem's condition.  Therefore the solutions are exactly the ones given at the solution's start.  <math>\blacksquare</math>
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== Solution 2 ==
 
== Solution 2 ==

Revision as of 12:54, 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

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 $k$. $\blacksquare$

RF

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>
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