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== | ||
+ | {{solution}} | ||
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+ | ==See Also== | ||
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+ | {{IMO box|year=2017|num-b=5|after=Last Problem} |
Revision as of 01:43, 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}