Difference between revisions of "2017 IMO Problems/Problem 6"

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==Problem==
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An ordered pair <math>(x, y)</math> of integers is a primitive point if the greatest common divisor of <math>x</math> and <math>y</math> is <math>1</math>. Given a finite set <math>S</math> of primitive points, prove that there exist a positive integer <math>n</math> and integers <math>a_0, a_1, \ldots , a_n</math> such that, for each <math>(x, y)</math> in <math>S</math>, we have:
 
An ordered pair <math>(x, y)</math> of integers is a primitive point if the greatest common divisor of <math>x</math> and <math>y</math> is <math>1</math>. Given a finite set <math>S</math> of primitive points, prove that there exist a positive integer <math>n</math> and integers <math>a_0, a_1, \ldots , a_n</math> such that, for each <math>(x, y)</math> in <math>S</math>, we have:
 
<cmath>a_0x^n + a_1x^{n-1} y + a_2x^{n-2}y^2 + \cdots + a_{n-1}xy^{n-1} + a_ny^n = 1.</cmath>
 
<cmath>a_0x^n + a_1x^{n-1} y + a_2x^{n-2}y^2 + \cdots + a_{n-1}xy^{n-1} + a_ny^n = 1.</cmath>
  
 
==Solution==
 
==Solution==
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{{solution}}
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==See Also==
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{{IMO box|year=2017|num-b=5|after=Last Problem}

Revision as of 00:43, 19 November 2023

Problem

An ordered pair $(x, y)$ of integers is a primitive point if the greatest common divisor of $x$ and $y$ is $1$. Given a finite set $S$ of primitive points, prove that there exist a positive integer $n$ and integers $a_0, a_1, \ldots , a_n$ such that, for each $(x, y)$ in $S$, we have: \[a_0x^n + a_1x^{n-1} y + a_2x^{n-2}y^2 + \cdots + a_{n-1}xy^{n-1} + a_ny^n = 1.\]

Solution

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See Also

{{IMO box|year=2017|num-b=5|after=Last Problem}