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I propose a conjecture variant of Cayley-Bacharach's theorem.

I'm an electrical engineer, I am not a mathematician. I don't know how to prove this result. Could you give a solution or let me know some more information for the conjecture:

Conjecture: Assume that two curves $C_1$ and $C_2$ in the projective plane meet in $\frac{d^2+3d}{2}$ (different) points, as they do in general over an algebraically closed field. Then every curve of degree $d$ that passes through any $\frac{d^2+3d}{2}-1$ of the points also passes through the $\frac{d^2+3d}{2}$ th point.

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    $\begingroup$ What is $d$? Is it somehow related to the degrees of $C_1$ and $C_2$? $\endgroup$ – abx Sep 21 '15 at 17:38
  • $\begingroup$ Degrees of $C_1$ and $C_2$ and $d$ are independent. Example a curve $C_1$ degree 7 and a curve $C_2$ degree 2 meet at 14 points = $\frac{4^2+3.4}{2}$ => every curve of degree 4 passes through any 13 points of the points also passes through 14th point. $\endgroup$ – Oai Thanh Đào Sep 21 '15 at 18:29
  • $\begingroup$ Dear Dr. @abx , thank to You very much. I edited the question to clear. $\endgroup$ – Oai Thanh Đào Sep 21 '15 at 19:09
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I do not understand how your $d$ is related to the degrees of $C_1$ and $C_2$. Anyway, many variants and generalizations of the classical Cayley-Bacharach for cubics are known. One of them is the statement below, whose proof can be found in the beautiful paper by Eisenbud, Green and Harris

Cayley-Bacharach Theorems and conjectures, Bull. Amer. Math. Soc. 33 (1996).

Theorem (Generalized Cayley-Bacharach). Let $C_1$ and $C_2$ be two curves in $\mathbb{P}^2$ of degrees $d$ and $e$, respectively. Assume that they intersect in a collection of $d \cdot e$ distinct points $\Gamma = \{p_1, \ldots, p_{de}\}$. If $C \subset \mathbb{P}^2$ is a curve of degree $d+e-3$ containing all but one point of $\Gamma$, then $C$ contains all of $\Gamma$.

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  • $\begingroup$ Dear Dr. @FrancescoPolizzi , I minor change the question. I think the conjecture and the theorem are different. I am sorry if the conjecture is wrong. $\endgroup$ – Oai Thanh Đào Sep 21 '15 at 19:07

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