Most odd degree hyperelliptic curves have no finite rational points

You need to know: Polynomial, degree of a polynomial.

Background: Let {\cal P} be the set of odd-degree polynomials P(x)=\sum\limits_{i=0}^{2g+1} a_i x^{2g+1-i} with rational coefficients a_i and leading coefficient a_0=1. With change of variables x'=u^2x, y'=u^{2g+1}y, we can have new coefficients a'_i=u^{2i}a_i, i=1,2,\dots,2g+1, and, by selecting u to be the common denominator of a_1, a_2, \dots, a_{2g+1}, we can make all coefficients integers. After this, define height of P\in{\cal P} by H(P) = \max\{|a_1|, |a_2|^{1/2}, \dots, |a_{2g+1}|^{1/(2g+1)}\}. For fixed integer g>0 and real X>0, let \mu(X,g) be a fraction of the polynomials P\in {\cal P} of degree 2g+1 and height less than X for which the equation y^2=P(x) has no rational solutions.

The Theorem: On 1st February 2013, Bjorn Poonen and Michael Stoll submitted to arxiv a paper in which they proved that \lim\limits_{g\to\infty}(\lim\limits_{X\to\infty}\mu(X,g)) = 1.

Short context: Set of real solutions to y^2=P(x) for P \in {\cal P} is known as odd degree hyperelliptic curve, and rational solutions are called finite rational points on this curve. In this terminology, the Theorem states that most odd degree hyperelliptic curves have no finite rational points. Moreover, Poonen and Stoll also proved for “almost all” polynomials P\in{\cal P} in the same sense as in the Theorem, there is an explicit algorithm, with polynomial P as an input, and output certifying that there are indeed no rational solutions to y^2=P(x). In other words, there exists a universal method able to solve almost all equations in the form y^2=P(x), \, P \in {\cal P} at once!

Links: Free arxiv version of the original paper is here, journal version is here.

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