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Prove that there exists an infinite set of points $$ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots $$ in the plane with the following property: For any three distinct integers $a,b,$ and $c$, points $P_a$, $P_b$, and $P_c$ are collinear if and only if $a+b+c=2014$

it is said this

This problem is USAMO 2014 problem,and some discuss in mathlinks:

By Now,I can't see solve this problem.Thank you

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If you read the AOPS link, they told you that people came up with $P(a) = (a - 2014/3, (a - 2014/3)^3) $. – Calvin Lin May 11 '14 at 13:22

Let $C$ be an elliptic curve. The (abelian) group law on $C$ is given by $a+ b+ c=0$ with $a,b,c$ distinct iff $a,b,c$ are collinear. Let $Q$ be a point on $C$ such that $\langle Q\rangle\cong\mathbb Z$. Let $P_n=(3n-2014)Q$. Then $P_a,P_b,P_c$ with $a,b,c$ distinct are collinear iff $((3a-2014)+(3b-2014)+(3c-2014))Q=0$, i.e. iff $a+b+c=2014$.

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