Difference between revisions of "2017 IMO Problems/Problem 6"
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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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| + | ==See Also== | ||
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| + | {{IMO box|year=2017|num-b=5|after=Last Problem}} | ||
Latest revision as of 02:09, 19 November 2023
Problem
An ordered pair 
of integers is a primitive point if the greatest common divisor of 
and 
is 
. Given a finite set 
of primitive points, prove that there exist a positive integer 
and integers 
such that, for each in , we have:
![\[a_0x^n + a_1x^{n-1} y + a_2x^{n-2}y^2 + \cdots + a_{n-1}xy^{n-1} + a_ny^n = 1.\]](http://latex.artofproblemsolving.com/c/3/3/c33a8093712f7fef24e6545476a95c7877da7250.png)
Solution
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See Also
| 2017 IMO (Problems) • Resources | ||
| Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Problem |
| All IMO Problems and Solutions | ||
