Views
2
Downloands
1
Let $P$ and $Q$ be nonzero integers. Generalized Fibonacci and Lucas sequences are defined as follows: $U_{0}(P,Q)=0,U_{1}(P,Q)=1,$ and $ U_{n+1}(P,Q)=PU_{n}(P,Q)+QU_{n-1}(P,Q)$ for $n\geq 1$ and $ V_{0}(P,Q)=2,V_{1}(P,Q)=P,$ and $V_{n+1}(P,Q)=PV_{n}(P,Q)+QV_{n-1}(P,Q)$ for $n\geq 1,$ respectively. In this paper, we assume that $P$ and $Q$ are relatively prime odd positive integers and $P^{2}+4Q>0.$ We determine all indices $n$ such that $U_{n}=(P^{2}+4Q)x^{2}.$ Moreover, we determine all indices $n$ such that $(P^{2}+4Q)U_{n}=x^{2}.$ As a result, we show that the equation $V_{n}^{2}(P,1)+V_{n+1}^{2}(P,1)=x^{2}$ has solution only for $n=2,$ $P=1,$ $x=5$ and $V_{n+1}^{2}(P,-1)=V_{n}^{2}(P,-1)+x^{2}$ has no solutions. Moreover, we solve some Diophantine equations.
Field : Fen Bilimleri ve Matematik
Journal Type : Uluslararası
Article Detail 
Similar Articles
Journal Information
Listen Paper
Cite
Print This Page
Share
Mendeley
