Diophantine equations with products of consecutive terms in Lucas sequences

被引:22
|
作者
Luca, F
Shorey, TN
机构
[1] Univ Nacl Autonoma Mexico, Inst Matemat, Morelia 58089, Michoacan, Mexico
[2] Tata Inst Fundamental Res, Sch Math, Bombay 400005, Maharashtra, India
关键词
Lucas sequences; primitive divisors; arithmetic progressions;
D O I
10.1016/j.jnt.2004.08.007
中图分类号
O1 [数学];
学科分类号
0701 ; 070101 ;
摘要
In this paper, we show that if (u(n))(n) >= 1 is a Lucas sequence, then the Diophantine equation u(n) (.) u(n+1) (.....) u(n+k) = y(m) in integers n >= 1, k >=, 4, m >= 2 and y with vertical bar y vertical bar > 1 has only finitely many solutions. We also determine all such solutions when (u(n))(n >= 1) is the sequence of Fibonacci numbers and when u(n) = (x(n) - 1)/(x - 1) for all n >= 1 with some integer x > 1. (c) 2004 Elsevier Inc. All rights reserved.
引用
收藏
页码:298 / 311
页数:14
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