Some Diophantine Problems Related to k-Fibonacci Numbers

被引:1
|
作者
Trojovsky, Pavel [1 ]
Hubalovsky, Stepan [2 ]
机构
[1] Univ Hradec Kralove, Fac Sci, Dept Math, Hradec Kralove 50003, Czech Republic
[2] Univ Hradec Kralove, Fac Sci, Dept Appl Cybernet, Hradec Kralove 50003, Czech Republic
关键词
k-Fibonacci number; k-Lucas number; Galois theory; Diophantine equation; POWERS;
D O I
10.3390/math8071047
中图分类号
O1 [数学];
学科分类号
0701 ; 070101 ;
摘要
Let k >= 1 be an integer and denote (F-k,F-n) n as the k-Fibonacci sequence whose terms satisfy the recurrence relation F-k,F-n=kF(k,n-1)+F-k,F-n-2, with initial conditions F-k,F-0=0 and F-k,F-1=1. In the same way, the k-Lucas sequence (L-k,L-n)(n) is defined by satisfying the same recursive relation with initial values L-k,L-0=2 and L-k,L-1=k. The sequences(F-k,F-n)(n >= 0) and (L-k,L-n)(n >= 0) were introduced by Falcon and Plaza, who derived many of their properties. In particular, they proved that F-k,n(2)+F-k,n+1(2)=F-k,F-2n+1 and F-k,n+1(2)-F-k,n-1(2)=kF(k,2n), for all k >= 1 and n >= 0. In this paper, we shall prove that if k>1 and F-k,n(s)+F-k,n+1(s) is an element of(F-k,F-m)(m >= 1) for infinitely many positive integers n, then s=2. Similarly, that if F-k,n+1(s)-F-k,n-1(s) is an element of(kF(k,m))(m >= 1) holds for infinitely many positive integers n, then s=1 or s=2. This generalizes a Marques and Togbe result related to the case k=1. Furthermore, we shall solve the Diophantine equations F-k,F-n=L-k,L-m, F-k,F-n=F-n,F-k and L-k,L-n=L-n,L-k.
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页数:10
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