Pythagorean triples containing generalized lucas numbers
dc.contributor.author | Siar, Z. and Keskin, R. | |
dc.date.accessioned | 2021-04-08T12:07:39Z | |
dc.date.available | 2021-04-08T12:07:39Z | |
dc.date.issued | 2018 | |
dc.identifier | 10.3906/mat-1702-102 | |
dc.identifier.issn | 13000098 | |
dc.identifier.uri | https://www.scopus.com/inward/record.uri?eid=2-s2.0-85050719669&doi=10.3906%2fmat-1702-102&partnerID=40&md5=b85825eb3db66fecd3bfa4d829fb4b4a | |
dc.identifier.uri | http://acikerisim.bingol.edu.tr/handle/20.500.12898/4402 | |
dc.description.abstract | Let P and Q be nonzero integers. Generalized Fibonacci and Lucas sequences are defined as follows: U0(P,Q) = 0;U1(P,Q) = 1, and Un+1(P,Q) = PUn(P,Q)+QUn-1(P,Q) for n ≥ 1 and V0(P,Q) = 2, V1(P,Q) = P, and Vn+1(P,Q) = PVn(P,Q) + QVn-1(P;Q) for n ≥ 1; respectively. In this paper, we assume that P and Q are relatively prime odd positive integers and P2 + 4Q > 0: We determine all indices n such that Un = (P2 + 4Q)x2: Moreover, we determine all indices n such that (P2 +4Q)UNn = x2: As a result, we show that the equation V2 n (P; 1)+V2 n+1(P; 1) = x2 has solution only for n = 2; P = 1; x = 5 and V 2 n+1(P;-1) = V 2 n (P;-1) + x2 has no solutions. Moreover, we solve some Diophantine equations. ©TÜBITAK. | |
dc.language.iso | English | |
dc.source | Turkish Journal of Mathematics | |
dc.title | Pythagorean triples containing generalized lucas numbers |
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