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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 ==
 
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First we can reformulate the original problem as
 
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<cmath> \frac{a^2}{2ab^2-b^3+1} = k . </cmath>
 
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Where <math>a</math>, <math>b</math>, <math>k</math> are all positive integers.
 
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We split the solutions in 2 cases:
 
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1) Assume that <math>2ab^2-b^3+1>1</math>. Then we have
 
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<cmath>2ab^2-b^3>0</cmath>
 
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<cmath>b^2(2a-b)>0</cmath>
 
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Since <math>b^2 > 0</math> for all positive integers <math>b</math>, we must also have <math>2a - b > 0</math>, or <math>b < 2a</math> for all positive integers <math>a</math>.
 
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Therefore, the only possible value of <math>b</math> is <math>b = 1</math>.
 
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When <math>b = 1</math>, we have
 
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<cmath> \frac{a^2}{2ab^2-b^3+1} = \frac{a^2}{2a} = \frac{a}{2} = k . </cmath>
 
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Therefore, <math>a = 2k</math> for all positive integers <math>k</math>.
 
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Therefore, for <math>2ab^2-b^3+1>1</math>, the only solution is <math>(a, b) = (2k, 1)</math> for all positive integers <math>k</math>.
 
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2) a) Assume that <math>2ab^2-b^3+1=1</math>. Then we have
 
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<cmath>2ab^2-b^3=0</cmath>
 
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<cmath>b^2(2a-b)=0</cmath>
 
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Therefore we have either <math>b=0</math> or <math>b=2a</math>.
 
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Since we want <math>b</math> to be a positive integer, we must have <math>b=2a</math> for all positive integers <math>a</math>.
 
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When <math>b=2a</math>, we have
 
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<cmath> \frac{a^2}{2ab^2-b^3+1} = \frac{a^2}{8a^3-8a^3+1} = a^2 = k . </cmath>
 
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Therefore, <math>k = a^2</math> for all positive integers <math>a</math>.
 
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Therefore, for <math>2ab^2-b^3+1=1</math> we can have <math>(a, b) = (a, 2a)</math> for all positive integers <math>a</math>.
 
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2) b) Notice that we can reformulate
 
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<cmath> \frac{a^2}{2ab^2-b^3+1} = k . </cmath>
 
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as
 
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<cmath> a^2=k(2ab^2-b^3+1). </cmath>
 
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<cmath> a^2-2kb^2a+k(b^3-1)=0. </cmath>
 
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Using properties of quadratic equations, we have
 
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<cmath>a_1 + a_2 = 2kb^2</cmath>
 
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<cmath>a_1 a_2 = k(b^3-1)</cmath>
 
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From the results in part 2) a), we have <math>a_1 = a</math>, <math>b=2a</math>, and <math>k=a^2</math>, so now we have
 
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<cmath>a + a_2 = 2a^2(2a)^2 = 8a^4</cmath>
 
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<cmath>a a_2 = a^2((2a)^3-1) = 8a^5 - a^2</cmath>
 
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Using either of the equations above, we have <math>a_2 = 8a^4 - a</math>.
 
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Therefore, for <math>2ab^2-b^3+1=1</math> we can have <math>(a, b) = (8a^4 - a, 2a)</math> for all positive integers <math>a</math>.
 
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Therefore, the only pairs of solutions for
 
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<cmath> \frac{a^2}{2ab^2-b^3+1} = k . </cmath>
 
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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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RF
 
  
 
== Resources ==
 
== Resources ==

Latest revision as of 12:56, 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. $\blacksquare$

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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