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arXiv 0901.2076math.NT
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$y^2=x^n+k$上等差数列的有理点

Rational points in arithmetic progression on $y^2=x^n+k$

Maciej Ulas

摘要:

设$C$是由方程$y^2=f(x)$给出的超椭圆曲线,其中$f\in\Z[x]$和$f$没有重根。如果$i=1,2,..., n$的数$x_{i}$是等差数列,我们说$i=1,2,..., n$的点$P_{i}=(x_{i}, y_{i})\in C(\Q)$是等差数列。本文证明了存在一个多项式$k\in\Z[t]$,它具有这样的性质:在椭圆曲线$\cal{E}: y^2=x^3+k(t)$(定义在域$\Q(t)$上)上,我们可以找到等差数列中的四个点,它们在曲线$\cal{E}$上的所有$\Q(t)$ -有理点的群中是独立的.特别地,这个结果推广了Lee和Vélez从\cite{LeeVel}得到的一些早期结果。我们还表明,如果$n\in\N$是奇数,那么有无穷多个$k$的这样一个属性,曲线$y^2=x^n+k$有四个有理点的算术级数。在$n$是偶数的情况下,我们可以找到无穷多个$k$,使得在曲线$y^2=x^n+k$上有六个等差数列中的有理点。

英文摘要:

Let $C$ be a hyperelliptic curve given by the equation $y^2=f(x)$, where $f\in\Z[x]$ and $f$ hasn't multiple roots. We say that points $P_{i}=(x_{i}, y_{i})\in C(\Q)$ for $i=1,2,..., n$ are in arithmetic progression if the numbers $x_{i}$ for $i=1,2,..., n$ are in arithmetic progression. In this paper we show that there exists a polynomial $k\in\Z[t]$ with such a property that on the elliptic curve $\cal{E}: y^2=x^3+k(t)$ (defined over the field $\Q(t)$) we can find four points in arithmetic progression which are independent in the group of all $\Q(t)$-rational points on the curve $\cal{E}$. In particular this result generalizes some earlier results of Lee and Vélez from \cite{LeeVel}. We also show that if $n\in\N$ is odd then there are infinitely many $k$'s with such a property that on the curves $y^2=x^n+k$ there are four rational points in arithmetic progressions. In the case when $n$ is even we can find infinitely many $k$'s such that on the curves $y^2=x^n+k$ there are six rational points in arithmetic progression.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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