由斐波那契三元组诱导的椭圆曲线上的秩与整数点
Ranks and integer points on elliptic curves induced by Fibonacci triples
AI总结：
该研究由斐波那契三元组构造椭圆曲线，证明奇数k≥3时曲线秩≥2，奇数k≥3且秩为2时整数点对应霍加特-伯格姆问题解，还得出奇偶k对应椭圆族一般秩，提出启发式秩分布。
AI中文摘要：
设$F_n$和$L_n$分别表示斐波那契数和卢卡斯数，考虑椭圆曲线$E_k:y^2=(F_{2k}x+1)(F_{2k+2}x+1)(F_{2k+4}x+1)$，这些曲线自然源于正则丢番图三元组$\{F_{2k},F_{2k+2},F_{2k+4}\}$。对于奇数$k$，我们给出有理点$Q_k=\bigg( -\frac{F_{k-1}}{L_kF_{k+1}F_{k+2}}, \frac{F_{2k+1}}{L_kF_{k+1}F_{k+2}} \bigg)$；对所有奇数$k\boldsymbol{\text{≥}}3$，该点与标准点$P_k=(0,1)$线性无关，特别地，$\text{rank}E_k(\boldsymbol{\text{Q}})\boldsymbol{\text{≥}}2$。此外，若奇数$k\boldsymbol{\text{≥}}3$且$\text{rank}E_k(\boldsymbol{\text{Q}})=2$，则$E_k$上的所有整数点恰好是霍加特-伯格姆扩展问题两个已知解对应的点。通过参数化两条圆锥曲线$L^2-5F^2=\boldsymbol{\text{±}}4$并应用单射特化准则，我们还证明，对应的单参数椭圆族在奇数$k$情形下的一般秩为2，偶数$k$情形下为1。最后，我们讨论计算数据并提出启发式秩分布：秩1、2、3的概率分别为1/4、1/2、1/4，秩至少为4的密度为0。
英文摘要：
Let $F_n$ and $L_n$ denote the Fibonacci and Lucas numbers, respectively, and consider \[ E_k:\qquad y^2=(F_{2k}x+1)(F_{2k+2}x+1)(F_{2k+4}x+1). \] These elliptic curves arise naturally from the regular Diophantine triples \[ \{F_{2k},F_{2k+2},F_{2k+4}\}. \] For odd $k$, we exhibit the rational point \[ Q_k=\left( -\frac{F_{k-1}}{L_kF_{k+1}F_{k+2}}, \frac{F_{2k+1}}{L_kF_{k+1}F_{k+2}} \right). \] For every odd $k\geq 3$, this point is independent of the standard point $P_k=(0,1)$; in particular, $\operatorname{rank}E_k(\mathbb{Q})\geq 2$. Moreover, if $k\geq 3$ is odd and $\operatorname{rank}E_k(\mathbb{Q})=2$, then all integer points on $E_k$ are exactly the points arising from the two known solutions of the Hoggatt-Bergum extension problem. By parametrizing the two conics $L^2-5F^2=\pm4$ and applying an injective specialization criterion, we also show that the corresponding one-parameter elliptic families have generic ranks $2$ in the odd case and $1$ in the even case. Finally, we discuss computational data and propose the heuristic rank distribution $1/4,1/2,1/4$ for ranks $1,2,3$, respectively, with density zero for rank at least $4$.
