# Difference between revisions of "2014 USAMO Problems/Problem 3"

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==Solution== | ==Solution== | ||

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+ | Consider an elliptic curve with a generator <math>g</math>, such that <math>g</math> is not a root of <math>0</math>. By repeatedly adding <math>g</math> to itself under the standard group operation, with can build <math>g, 2g, 3g, \ldots</math> as well as <math>-g, -2g, -3g, \ldots</math>. If we let <cmath>P_k = (3k-2014)g</cmath> then we can observe that collinearity between <math>P_a</math>, <math>P_b</math>, and <math>P_c</math> occurs only if <math>P_a + P_b + P_c = 0</math> (by definition of the group operation), which is equivalent to <math>(3a-2014)g + (3b-2014)g + (3c-2014)g = (3a+3b+3c-3*2014)g = 0</math>, or <math>3a + 3b + 3c = 3*2014</math>, or <math>a + b + c = 2014</math>. We know that all these points <math>P_k</math> exist because <math>3k-2014</math> is never 0 for integer <math>k</math>, so that none of these points need to be point at infinity (the identity element of the group). |

## Revision as of 19:44, 29 April 2014

## Problem

Prove that there exists an infinite set of points in the plane with the following property: For any three distinct integers and , points , , and are collinear if and only if .

## Solution

Consider an elliptic curve with a generator , such that is not a root of . By repeatedly adding to itself under the standard group operation, with can build as well as . If we let then we can observe that collinearity between , , and occurs only if (by definition of the group operation), which is equivalent to , or , or . We know that all these points exist because is never 0 for integer , so that none of these points need to be point at infinity (the identity element of the group).