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Dergiler / 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
Sayfa
1035–1045
Yayın Tarihi
2022-10-18
DOI
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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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Amaç & Hipotez: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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Temel Bulgular & Veri: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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Sonuç & Katkı:Moreover, we show that if the equation $V_{n}=11x^{2}-1$ has a solution, then $P\equiv2({mod}8)$ and $P\equiv-1({mod}11).$
🔬Tür:Araştırma Makalesi
📍Kapsam:Genişletilmiş Akademik Örneklem
⚙️Değişkenler: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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Bu Makaleye Atıf Yap

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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Kaynakça (20)

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