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Journals / Hacettepe Journal of Mathematics and Statistics / 2019 / Cilt: 48 - Sayı: 4

Generalized Lucas numbers of the form $11x^{2}\mp 1$

Refik Keskin
Ümmügülsüm Öğüt
Pages
1035–1045
Publication Date
2022-10-18
DOI
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AI Summary & Key Takeaways

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⚡ TL;DR(Too Long; Didn't Read)

Moreover, we show that if the equation $V_{n}=11x^{2}-1$ has a solution, then $P\equiv2({mod}8)$ and $P\equiv-1({mod}11).$

📌 3 Key Takeaways

🎯
Objective & Hypothesis:Let $P\geq3$ be an integer and $(V_{n})$ denote generalized Lucas sequence defined by $V_{0}=2,V_{1}=P,$ and $V_{n+1}=PV_{n}-V_{n-1}$ for $n\geq1.$ In this study, we solve the equation $V_{n}=11x^{2}\mp1.$ We show that the equation $V_{n}=11x^{2}+1$ has a solution only when $n=1$ and $P\equiv 1({mod}11)$.
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Key Findings & Data:Let $P\geq3$ be an integer and $(V_{n})$ denote generalized Lucas sequence defined by $V_{0}=2,V_{1}=P,$ and $V_{n+1}=PV_{n}-V_{n-1}$ for $n\geq1.$ In this study, we solve the equation $V_{n}=11x^{2}\mp1.$ We show that the equation $V_{n}=11x^{2}+1$ has a solution only when $n=1$ and $P\equiv 1({mod}11)$.
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Conclusion & Impact:Moreover, we show that if the equation $V_{n}=11x^{2}-1$ has a solution, then $P\equiv2({mod}8)$ and $P\equiv-1({mod}11).$
🔬Design:Research Article
📍Scope:Comprehensive Sample
⚙️Variables:generalized Lucas numbers · congruences · diophantine equation

Abstract

Let $P\geq3$ be an integer and $(V_{n})$ denote generalized Lucas sequence defined by $V_{0}=2,V_{1}=P,$ and $V_{n+1}=PV_{n}-V_{n-1}$ for $n\geq1.$ In this study, we solve the equation $V_{n}=11x^{2}\mp1.$ We show that the equation $V_{n}=11x^{2}+1$ has a solution only when $n=1$ and $P\equiv 1({mod}11)$. Moreover, we show that if the equation $V_{n}=11x^{2}-1$ has a solution, then $P\equiv2({mod}8)$ and $P\equiv-1({mod}11).$

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Cite This Article

Keskin, R., Öğüt, Ü. (2022). Generalized Lucas numbers of the form $11x^{2}\mp 1$. Hacettepe Journal of Mathematics and Statistics, Cilt: 48 - Sayı: 4, 1035–1045. https://ajindex.com/articles/529763

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