Pythagorean triples containing generalized Lucas numbers
Author(s) -
Zafer Şi̇ar,
Refi̇k Keski̇n
Publication year - 2018
Publication title -
turkish journal of mathematics
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.454
H-Index - 27
eISSN - 1303-6149
pISSN - 1300-0098
DOI - 10.3906/mat-1702-102
Subject(s) - mathematics , fibonacci number , combinatorics , diophantine equation , lucas number , pythagorean triple , lucas sequence , prime (order theory) , coprime integers , prime factor , prime number , fibonacci polynomials , pythagorean theorem , geometry , orthogonal polynomials , difference polynomials
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.
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