Difference between revisions of "2017 IMO Problems/Problem 6"
Tigerzhang (talk | contribs) m (→solution) |
|||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
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== | ||
+ | {{solution}} | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2017|num-b=5|after=Last Problem} |
Revision as of 00:43, 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:
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See Also
{{IMO box|year=2017|num-b=5|after=Last Problem}