A class of hypersurfaces in $ \mathbb{E}^{n+1}_{s} $ satisfying $ \Delta \vec{H} = \lambda\vec{H} $

Dan Yang; Jinchao Yu; Jingjing Zhang; Xiaoying Zhu; Dan Yang; Jinchao Yu; Jingjing Zhang; Xiaoying Zhu

doi:10.3934/math.2022003

AIMS Mathematics

2022, Volume 7, Issue 1: 39-53. doi: 10.3934/math.2022003

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Research article

A class of hypersurfaces in $\mathbb{E}^{n+1}_{s}$ satisfying $\Delta \vec{H} = \lambda\vec{H}$

1.
School of Mathematics, Liaoning University, Shenyang 110044, China
2.
School of Mechanical Engineering and Automation, Shanghai University, Shanghai 200072, China

Received: 03 July 2021 Accepted: 08 September 2021 Published: 29 September 2021
MSC : 53C40, 53C42

A nondegenerate hypersurface in a pseudo-Euclidean space $\mathbb{E}^{n+1}_{s}$ is called to have proper mean curvature vector if its mean curvature $\vec{H}$ satisfies $\Delta \vec{H} = \lambda \vec{H}$ for a constant $\lambda$ . In 2013, Arvanitoyeorgos and Kaimakamis conjectured ^[1]: any hypersurface satisfying $\Delta \vec{H} = \lambda \vec{H}$ in pseudo-Euclidean space $\mathbb E_{s}^{n+1}$ has constant mean curvature. This paper will give further support evidences to this conjecture by proving that a linear Weingarten hypersurface $M^{n}_{r}$ in $\mathbb E^{n+1}_{s}$ satisfying $\Delta \vec{H} = \lambda \vec{H}$ has constant mean curvature if $M^{n}_{r}$ has diagonalizable shape operator with less than seven distinct principal curvatures.

Keywords:

Citation: Dan Yang, Jinchao Yu, Jingjing Zhang, Xiaoying Zhu. A class of hypersurfaces in $\mathbb{E}^{n+1}_{s}$ satisfying $\Delta \vec{H} = \lambda\vec{H}$ [J]. AIMS Mathematics, 2022, 7(1): 39-53. doi: 10.3934/math.2022003

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Abstract

1. Introduction

Fibonacci polynomials and Lucas polynomials are important in various fields such as number theory, probability theory, numerical analysis, and physics. In addition, many well-known polynomials, such as Pell polynomials, Pell Lucas polynomials, Tribonacci polynomials, etc., are generalizations of Fibonacci polynomials and Lucas polynomials. In this paper, we extend the linear recursive polynomials to nonlinearity, that is, we discuss some basic properties of the bi-periodic Fibonacci and Lucas polynomials.

The bi-periodic Fibonacci $\left \{ f_{n} \left (t\right) \right \}$ and Lucas $\left \{ l_{n} \left (t \right) \right \}$ polynomials are defined recursively by

$\begin{equation*} f_{0}\left (t \right ) = 0, \quad f_{1}\left (t\right ) = 1, \quad {f_{n}\left ( t\right ) = } \begin{cases} ayf_{n-1}\left ( t \right ) +f_{n-2}\left ( t \right ) \quad & n \equiv 0 \pmod{2}, \\ byf_{n-1}\left (t \right ) +f_{n-2}\left (t \right ) \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2, \end{equation*}$

and

$\begin{equation*} l_{0}\left ( t \right ) = 2, \quad l_{1}\left (t \right ) = at, \quad {l_{n}\left ( t \right ) = } \begin{cases} byl_{n-1}\left ( t \right ) +l_{n-2}\left ( t \right ) \quad & n \equiv 0 \pmod{2}, \\ ayl_{n-1}\left ( t \right ) +l_{n-2}\left ( t \right ) \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2, \end{equation*}$

where $a$ and $b$ are nonzero real numbers. For $t = 1$ , the bi-periodic Fibonacci and Lucas polynomials are, respectively, well-known bi-periodic Fibonacci $\left \{f_{n} \right \}$ and Lucas $\left \{l_{n} \right \}$ sequences. We let

$\begin{equation*} {\varsigma \left ( n \right ) = } \begin{cases} 0 \quad & n \equiv 0 \pmod{2}, \\ 1 \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2. \end{equation*}$

In ^[1], the scholars give the Binet formulas of the bi-periodic Fibonacci and Lucas polynomials as follows:

$\begin{equation} f_{n}\left (t \right ) = \frac{a^{\varsigma \left (n+1 \right) }}{ \left( ab \right )^{\left \lfloor \frac{n}{2} \right \rfloor } }\left (\frac{\sigma^{n} \left ( t\right ) -\tau ^{n}\left ( t \right ) }{\sigma\left (t \right ) -\tau \left ( t \right ) }\right), \end{equation}$

(1.1)

and

$\begin{equation} l_{n}\left (t \right ) = \frac{a^{\varsigma \left (n \right) }}{ \left( ab \right )^{\left \lfloor \frac{n+1}{2} \right \rfloor } }\left (\sigma ^{n}\left (t\right ) +\tau ^{n}\left ( t\right ) \right), \end{equation}$

(1.2)

where $n\ge 0$ , $\sigma\left (t \right)$ , and $\tau \left (t \right)$ are zeros of $\lambda ^{2}-abt\lambda -ab$ . This is $\sigma \left (t \right) = \frac{abt +\sqrt{a^{2} b^{2} t^{2}+4ab} }{2}$ and $\tau \left (t \right) = \frac{abt -\sqrt{a^{2} b^{2} t^{2}+4ab} }{2}$ . We note the following algebraic properties of $\sigma\left (t \right)$ and $\tau \left (t \right)$ :

$\begin{equation*} \sigma \left ( t \right ) +\tau \left (t \right ) = abt, \quad \sigma \left (t \right ) -\tau \left ( t \right ) = \sqrt{a^{2} b^{2} t^{2} +4ab}, \quad \sigma\left ( t \right ) \tau \left ( t\right ) = -ab. \end{equation*}$

Many scholars studied the properties of bi-periodic Fibonacci and Lucas polynomials; see ^[2,3,4,5,6]. In addition, many scholars studied the power sums problem of second-order linear recurrences and its divisible properties; see ^[7,8,9,10].

Taking $a = b = 1$ and $t = 1$ , we obtain the Fibonacci $\left \{ F_{n} \right \}$ or Lucas $\left \{ L_{n} \right \}$ sequence. Melham ^[11] proposed the following conjectures:

Conjecture 1. Let $m\ge 1$ be an integer, then the sum

$\begin{equation*} L_{1} L_{3} L_{5}\cdots L_{2m+1}\sum\limits_{k = 1}^{n} F_{2k}^{2m+1} \end{equation*}$

can be represented as $\left (F_{2n+1}-1 \right)^{2}R_{2m-1}\left (F_{2n+1} \right)$ , including $R_{2m-1}\left (t \right)$ as a polynomial with integer coefficients of degree $2m-1$ .

Conjecture 2. Let $m\ge 1$ be an integer, then the sum

$\begin{equation*} L_{1} L_{3} L_{5}\cdots L_{2m+1}\sum\limits_{k = 1}^{n} L_{2k}^{2m+1} \end{equation*}$

can be represented as $\left (L_{2n+1}-1 \right)Q_{2m}\left (L_{2n+1} \right)$ , where $Q_{2m}\left (t \right)$ is a polynomial with integer coefficients of degree $2m$ .

In ^[12], the authors completely solved the Conjecture $2$ and discussed the Conjecture $1$ . Using the definition and properties of bi-periodic Fibonacci and Lucas polynomials, the power sums problem and their divisible properties are studied in this paper. The results are as follows:

Theorem 1. We get the identities

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k}^{2m+1} \left (t\right ) = \frac{a^{2m+1} }{b\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j}\left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1\right ) }\left ( t \right ) - f_{2j+1}\left ( t \right ) }{l_{2j+1} \left ( t \right ) } \right ), \end{align}$

(1.3)

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k+1}^{2m+1} \left ( t \right ) = \frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{f_{\left ( 2n+2\right )\left (2j+1 \right ) }\left (t\right ) -f_{2\left ( 2j+1 \right ) }\left (t\right ) }{l_{2j+1}\left ( t \right ) } \right ), \end{align}$

(1.4)

$\begin{align} & \sum\limits_{k = 1}^{n}l_{2k}^{2m+1} \left (t \right ) = \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{l_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left (t \right ) -l_{2j+1}\left ( t\right ) }{l_{2j+1}\left ( t\right ) } \right ), \end{align}$

(1.5)

$\begin{align} &\sum\limits_{k = 1}^{n}l_{2k+1}^{2m+1} \left (t\right ) = \frac{ a ^{m+1} }{b^{m+1} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j} \left ( \frac{l_{\left ( 2n+2\right )\left ( 2j+1\right ) }\left ( t \right ) -l_{2\left ( 2j+1 \right ) }\left ( t \right ) }{l_{2j+1}\left ( t \right ) } \right ), \end{align}$

(1.6)

where $n$ and $m$ are positive integers.

Theorem 2. We get the identities

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k}^{2m} \left ( t \right ) = \frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j} \frac{f_{2j\left ( 2n+1 \right ) }\left (t \right ) }{f_{2j}\left (t\right ) } -\frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} }\binom{2m}{m}\left ( -1 \right )^{m}\left ( n+ \frac{1}{2 } \right ) , \end{align}$

(1.7)

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k+1}^{2m} \left ( t \right ) = \frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m} \binom{2m}{m-j} \left ( \frac{f_{2j\left ( 2n+2\right ) }\left ( t \right ) -f_{4j }\left ( t \right ) }{f_{2j}\left ( t \right ) } \right )-\frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} }\binom{2m}{m}n , \end{align}$

(1.8)

$\begin{align} & \sum\limits_{k = 1}^{n}l_{2k}^{2m} \left ( t \right ) = \sum\limits_{j = 0}^{m} \binom{2m}{m-j} \frac{f_{2j\left ( 2n+1 \right ) }\left ( t \right ) }{l_{2j+1}\left (t \right ) }-2^{2m-1}-\binom{2m}{m}\left ( n+\frac{1}{2} \right ) , \end{align}$

(1.9)

$\begin{align} &\sum\limits_{k = 1}^{n}l_{2k+1}^{2m} \left ( t \right ) = \frac{ a ^{m} }{b^{m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j} \left ( \frac{f_{2j\left ( 2n+2\right ) }\left ( t \right ) -f_{4j }\left (t \right ) }{f_{2j}\left ( t \right ) } \right ) -\frac{a^{m} }{b^{m} } \binom{2m}{m} \left ( -1 \right )^{m}n , \end{align}$

(1.10)

where $n$ and $m$ are positive integers.

As for application of Theorem 1, we get the following:

Corollary 1. We get the congruences:

$\begin{equation} bl_{1} \left ( t \right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t \right ) \equiv 0\pmod{f_{2n+1}\left ( t \right ) -1}, \end{equation}$

(1.11)

and

$\begin{equation} al_{1} \left ( t\right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}l_{2k}^{2m+1}\left ( t \right ) \equiv 0\pmod{l_{2n+1}\left (t \right ) -at}, \end{equation}$

(1.12)

where $n$ and $m$ are positive integers.

Taking $t = 1$ in Corollary 1, we have the following conclusions for bi-periodic Fibonacci $\left \{ f_{n} \right \}$ and Lucas $\left \{ l_{n} \right \}$ sequences.

Corollary 2. We get the congruences:

$\begin{equation} bl_{1} l_{3} \cdots l_{2m+1} \sum\limits _{k = 1}^{n}f_{2k}^{2m+1} \equiv 0\pmod{f_{2n+1} -1}, \end{equation}$

(1.13)

and

$\begin{equation} al_{1} l_{3} \cdots l_{2m+1} \sum\limits _{k = 1}^{n}l_{2k}^{2m+1} \equiv 0\pmod{l_{2n+1} -a }, \end{equation}$

(1.14)

where $n$ and $m$ are nonzero real numbers.

Taking $a = b = 1$ and $t = 1$ in Corollary 1, we have the following conclusions for bi-periodic Fibonacci $\left \{ F_{n} \right \}$ and Lucas $\left \{ L_{n} \right \}$ sequences.

Corollary 3. We get the congruences:

$\begin{equation} L_{1} L_{3} \cdots L_{2m+1} \sum\limits _{k = 1}^{n}F_{2k}^{2m+1} \equiv 0\pmod{F_{2n+1} -1}, \end{equation}$

(1.15)

and

$\begin{equation} L_{1} L_{3} \cdots L_{2m+1} \sum\limits _{k = 1}^{n}L_{2k}^{2m+1} \equiv 0\pmod{L_{2n+1} -1 }, \end{equation}$

(1.16)

where $n$ and $m$ are nonzero real numbers.

2. Proofs of theorems

To begin, we will give several lemmas that are necessary in proving theorems.

Lemma 1. We get the congruence

$\begin{equation*} f_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -f_{2j+1 } \left ( t \right ) \equiv 0 \pmod{f_{2n+1 } \left ( t \right )-1}, \end{equation*}$

where $n$ and $m$ are nonzero real numbers.

Proof. We prove it by complete induction for $j\ge 0$ . This clearly holds when $j = 0$ . If $j = 1$ , we note that $abf_{3\left (2n+1 \right) } \left (t \right) = \left (a^{2} b^{2} t^{2} +4ab \right) f_{2n+1}^{3}\left (t \right)-3abf_{2n+1} \left (t\right)$ and we obtain

$\begin{equation*} \begin{split} & f_{3\left ( 2n+1 \right ) }\left ( t \right )-f_{3 }\left ( t \right ) = \left ( abt^{2}+4 \right ) f_{2n+1}^{3}\left ( t\right )-3f_{2n+1} \left (t \right )-\left ( abt^{2}+4 \right )f_{1}^{3} \left (t \right )+3f_{1} \left (t\right ) \\ & = \left ( abt^{2}+4 \right )\left ( f_{2n+1}\left ( t \right )-f_{1}\left ( t \right )\right )\left ( f_{2n+1}^{2}\left (t \right )+ f_{2n+1}\left ( t \right )f_{1}\left ( t \right )+ f_{1}^{2}\left ( t \right ) \right )-3\left ( f_{2n+1} \left ( t\right ) -f_{1}\left ( t \right ) \right ) \\ & = \left ( abt^{2}+4 \right )\left ( f_{2n+1}\left ( t\right )-1\right )\left ( f_{2n+1}^{2}\left (t \right )+ f_{2n+1}\left ( t \right )f_{1}\left ( t \right )+ f_{1}^{2}\left ( t \right ) \right )-3\left ( f_{2n+1} \left ( t \right ) -1 \right ) \\ & \equiv 0 \pmod{f_{2n+1}\left (t \right ) -1 } . \end{split} \end{equation*}$

This is obviously true when $j = 1$ . Assuming that Lemma 1 holds if $j = 1, 2, \dots, k$ , that is,

$\begin{equation*} f_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -f_{2j+1 } \left ( t \right ) \equiv 0 \pmod{f_{2n+1 } \left ( t \right )-1}. \end{equation*}$

If $j = k+1 \ge 2$ , we have

$\begin{equation*} l_{2\left ( 2n+1 \right ) } \left ( t \right ) f_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t \right ) = f_{\left ( 2n+1 \right )\left ( 2j+3 \right ) }\left ( t \right ) +abf_{\left ( 2n+1 \right )\left ( 2j-1 \right ) }\left ( t \right ), \end{equation*}$

and

$\begin{equation*} \begin{split} abl_{2\left ( 2n+1 \right ) } \left ( t \right ) = \left ( a^{2} b^{2} t^{2} +4ab \right )f_{2n+1}^{2}\left ( t \right )-2ab \equiv \left ( a^{2} b^{2} t^{2} +4ab \right )f_{1}^{2}\left ( t \right )-2ab \pmod{f_{2n+1}\left ( t \right ) -1 }. \end{split} \end{equation*}$

We have

$\begin{equation*} \begin{split} &f_{\left ( 2n+1 \right )\left ( 2k+3 \right ) } \left (t\right ) -f_{ 2k+3 } \left ( t\right )\\ & = l_{2\left ( 2n+1 \right ) }\left ( t \right ) f_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left ( t \right )-abf_{\left (2n+1\right )\left ( 2k-1 \right ) } \left ( t \right ) -l_{2} \left ( t \right )f_{2k+1}\left ( t \right )+abf_{2k-1} \left ( t \right ) \\ & \equiv \left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left (t \right ) -2 \right )f_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left (t \right )-abf_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left (t \right ) \\ &\quad -\left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left ( t \right ) -2 \right ) f_{2k+1} \left ( t \right )+abf_{2k-1}\left ( t \right ) \\ & \equiv \left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left ( t \right ) -2 \right )\left ( f_{\left ( 2n+1 \right ) \left ( 2k+1 \right ) }\left ( t \right )-f_{2k+1}\left ( t\right ) \right ) -ab\left ( f_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left (t \right ) -f_{2k-1}\left ( t \right ) \right ) \\ & \equiv 0 \pmod{f_{2n+1}\left (t \right ) -1 } . \end{split} \end{equation*}$

This completely proves Lemma 1. □

Lemma 2. We get the congruence

$\begin{equation*} al_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -al_{2j+1 } \left ( t \right ) \equiv 0 \pmod{l_{2n+1 } \left ( t \right )-at}, \end{equation*}$

where $n$ and $m$ are nonzero real numbers.

Proof. We prove it by complete induction for $j\ge 0$ . This clearly holds when $j = 0$ . If $j = 1$ , we note that $al_{3\left (2n+1 \right) } \left (t \right) = b l_{2n+1}^{3}\left (t \right)+3al_{2n+1} \left (t \right)$ and we obtain

$\begin{equation*} \begin{split} & a l_{3\left ( 2n+1 \right ) }\left (t \right )-al_{3 }\left ( t \right ) = b l_{2n+1}^{3}\left ( t \right )+3al_{2n+1} \left ( t \right )-bl_{1}^{3} \left (t \right )-3al_{1} \left ( t \right ) \\ & = \left ( l_{2n+1}\left ( t \right )-l_{1}\left ( t \right )\right )\left ( bl_{2n+1}^{2}\left ( t \right )+ bl_{2n+1}\left ( t \right )l_{1}\left ( t \right )+ bl_{1}^{2}\left (t \right ) \right )-3a\left ( l_{2n+1} \left ( t \right ) -l_{1}\left ( t\right ) \right ) \\ & = \left ( l_{2n+1}\left ( t \right )-at\right )\left ( bl_{2n+1}^{2}\left ( t \right )+ bayl_{2n+1}\left ( t \right )+ ba^{2}t^{2} \right )-3a\left ( l_{2n+1} \left ( t \right ) -at \right )\\ & \equiv 0 \pmod{l_{2n+1}\left ( t \right ) -at } . \end{split} \end{equation*}$

This is obviously true when $j = 1$ . Assuming that Lemma 2 holds if $j = 1, 2, \dots, k$ , that is,

$\begin{equation*} al_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left (t \right ) -al_{2j+1 } \left ( t \right ) \equiv 0 \pmod{l_{2n+1 } \left ( t \right )-at}. \end{equation*}$

If $j = k+1 \ge 2$ , we have

$\begin{equation*} l_{2\left ( 2n+1 \right ) } \left (t \right ) l_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t\right ) = l_{\left ( 2n+1 \right )\left ( 2j+3 \right ) }\left ( t\right ) +l_{\left ( 2n+1 \right )\left ( 2j-1 \right ) }\left ( t \right ), \end{equation*}$

and

$\begin{equation*} \begin{split} & al_{2\left ( 2n+1 \right ) } \left ( t \right ) = bl_{2n+1}^{2}\left ( t \right )+2a \equiv bl_{1}^{2}\left ( t \right )+2a \pmod{l_{2n+1}\left ( t \right ) -at }. \end{split} \end{equation*}$

We have

$\begin{equation*} \begin{split} &al_{\left ( 2n+1 \right )\left ( 2k+3 \right ) } \left ( t\right ) -al_{\left ( 2k+3 \right ) } \left ( t \right )\\ & = a\left ( l_{2\left ( 2n+1 \right ) }\left ( t \right ) l_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left (t \right )-l_{\left (2n+1\right )\left ( 2k-1 \right ) } \left ( t \right ) \right )-a\left ( l_{2} \left (t \right )l_{2k+1}\left ( t \right )-l_{2k-1} \left (t \right ) \right ) \\ & \equiv \left ( bl_{ 1}^{2} \left ( t \right ) +2a \right )l_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left ( t \right )-al_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left ( t \right ) -\left ( bl_{ 1}^{2} \left ( t \right ) +2a \right ) l_{2k+1} \left ( t \right )+al_{2k-1}\left ( t \right ) \\ & \equiv \left ( abt^{2} +2 \right )\left ( al_{\left ( 2n+1 \right ) \left ( 2k+1 \right ) }\left ( t\right )-al_{2k+1}\left ( t \right ) \right ) - \left ( al_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left ( t\right ) -al_{2k-1}\left (t \right ) \right ) \\ & \equiv 0 \pmod{l_{2n+1}\left ( t \right ) -at } . \end{split} \end{equation*}$

This completely proves Lemma 2. □

Proof of Theorem 1. We only prove (1.3), and the proofs for other identities are similar.

$\begin{equation*} \begin{split} \sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t \right )& = \sum\limits_{k = 1}^{n}\left (\frac{a^{\varsigma \left ( 2k+1 \right ) } }{\left (ab \right )^{ \left \lfloor \frac{2k}{2} \right \rfloor} }\cdot \left ( \frac{\sigma ^{2k}\left ( t \right ) - \tau ^{2k} \left (t \right ) }{\sigma\left ( t \right ) - \tau \left (t \right ) } \right ) \right ) ^{2m+1} \\ & = \frac{a^{2m+1} }{\left ( \sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{k = 1}^{n}\frac{\left ( \sigma ^{2k}\left (t \right ) - \tau ^{2k} \left ( t \right ) \right )^{2m+1} }{\left ( ab \right )^{\left ( 2m+1 \right )k } } \\ & = \frac{a^{2m+1} }{\left ( \sigma\left (t\right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{k = 1}^{n}\sum\limits _{j = 0}^{2m+1}\left ( -1 \right )^{j} \binom{2m+1}{j} \frac{ \sigma^{2k\left ( 2m+1-j \right ) } \left ( t \right ) \tau ^{2kj} \left ( t \right ) }{\left ( ab \right )^{\left ( 2m+1 \right )k } } \\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{2m+1}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m+1-2j \right ) }\left (t \right ) }{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } }{\frac{\left ( ab \right )^{2m+1-2j} }{\sigma ^{2\left ( 2m+1-2j \right ) }\left (t \right ) } -1} \right )\\ & = \frac{a^{2m+1} }{\left (\sigma\left (t \right ) -\tau \left (t\right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } }{\frac{\left ( ab \right )^{2m+1-2j} }{\sigma ^{2\left ( 2m+1-2j \right ) }\left (t \right ) } -1} - \frac{1-\frac{\sigma ^{2n\left ( 2j-1-2m \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2j-1-2m\right )n } } }{\frac{\left ( ab \right )^{2j-1-2m} }{\sigma ^{2\left ( 2j-1-2m \right ) }\left ( t \right ) } -1}\right )\\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{\frac{\sigma^{2 \left ( 2m+1-2j \right ) }\left ( t\right ) }{\left ( ab \right )^{ 2m+1-2j } }- \frac{\sigma ^{\left ( 2n+2 \right ) \left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( n+1 \right ) \left ( 2m+1-2j \right ) } } +1- \frac{\sigma ^{2n \left ( 2j-1-2m \right ) }\left (t \right ) }{\left ( ab \right )^{ \left ( 2j-1-2m \right ) n }} }{1- \frac{\sigma^{2 \left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ \left ( 2m+1-2j \right ) }}} \right )\\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j}\\ & \quad \times \left ( \frac{\sigma^{2m+1-2j}\left ( t \right ) - \tau ^{2m+1-2j}\left ( t \right ) -\frac{\sigma^{\left ( 2n+1 \right ) \left ( 2m+1-2j \right ) }\left ( t \right )}{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } +\frac{\tau ^{\left ( 2n+1 \right ) \left ( 2m+1-2j \right ) }\left (t \right )}{\left ( ab \right )^{ \left ( 2m+1-2j \right )n } }}{-\sigma ^{2m+1-2j}\left ( t \right ) -\tau ^{2m+1-2j}\left ( t \right ) } \right ) \\ & = \frac{a^{2m+1} }{b\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j}\left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1\right ) }\left (t \right ) - f_{2j+1}\left (t \right ) }{l_{2j+1} \left ( t \right ) } \right ) . \end{split} \end{equation*}$

□

Proof of Theorem 2. We only prove (1.7), and the proofs for other identities are similar.

$\begin{equation*} \begin{split} \sum\limits _{k = 1}^{n}f_{2k}^{2m }\left ( t \right )& = \sum\limits_{k = 1}^{n}\left (\frac{a^{\varsigma \left ( 2k +1\right ) } }{\left (ab \right )^{ \left \lfloor \frac{2k}{2} \right \rfloor} }\cdot \left ( \frac{\sigma ^{2k}\left ( t\right ) - \tau ^{2k} \left ( t \right ) }{\sigma\left ( t\right ) - \tau \left (t \right ) } \right ) \right ) ^{2m} \\ & = \frac{a^{2m} }{\left ( \sigma\left (t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{k = 1}^{n}\frac{\left ( \sigma^{2k}\left ( t\right ) - \tau ^{2k} \left ( t \right ) \right )^{2m} }{\left ( ab \right )^{ 2mk } } \\ & = \frac{a^{2m} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{k = 1}^{n}\sum\limits _{j = 0}^{2m}\left ( -1 \right )^{j} \binom{2m}{j} \frac{ \sigma^{2k\left ( 2m-j \right ) } \left (t \right ) \tau ^{2kj} \left ( t \right ) }{\left ( ab \right )^{2mk } }\\ & = \frac{a^{2m} }{\left (\sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m} }\sum\limits _{j = 0}^{2m}\left ( -1 \right )^{j} \binom{2m }{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m-2j \right )n } } }{\frac{\left ( ab \right )^{2m-2j} }{\sigma ^{2\left ( 2m-2j \right ) }\left ( t \right ) } -1} \right ) \end{split} \end{equation*}$

$\begin{equation*} \begin{split} & = \frac{a^{2m} }{\left (\sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m-2j \right )n } } }{\frac{\left ( ab \right )^{2m-2j} }{\sigma ^{2\left ( 2m-2j \right ) }\left ( t \right ) } -1} +\frac{1-\frac{\sigma ^{2n\left ( 2j-2m \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2j-2m\right )n } } }{\frac{\left ( ab \right )^{2j-2m} }{\sigma ^{2\left ( 2j-2m \right ) }\left (t\right ) } -1}\right )\\ & \quad +\frac{a^{2m} }{\left ( \sigma \left ( t \right ) -\tau \left (t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n\\ & = \frac{a^{2m} }{\left (\sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m}{j} \left ( \frac{\frac{\sigma ^{2 \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ 2m -2j } }- \frac{\sigma ^{\left ( 2n+2 \right ) \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( n+1 \right ) \left ( 2m -2j \right ) } } -1+\frac{\sigma^{2n \left ( 2j -2m \right ) }\left ( t \right ) }{\left ( ab \right )^{ \left ( 2j -2m \right ) n }} }{1- \frac{\sigma^{2 \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ 2m-2j }}} \right )\\ & \quad +\frac{a^{2m} }{\left (\sigma \left ( t \right ) -\tau \left ( t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n \\ & = \frac{a^{2m } }{\left ( \sigma\left ( t \right ) -\tau \left ( t \right ) \right )^{2m } }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m }{j} \left ( \frac{\sigma^{2m -2j}\left ( t \right ) - \tau ^{2m -2j}\left (t\right ) -\frac{\sigma^{\left ( 2n+1 \right ) \left ( 2m -2j \right ) }\left (t \right )}{\left ( ab \right )^{n\left ( 2m -2j \right ) } } +\frac{\tau ^{\left ( 2n+1 \right ) \left ( 2m-2j \right ) }\left (t \right )}{\left ( ab \right )^{n\left ( 2m-2j \right ) } }}{ \tau ^{2m -2j}\left ( t \right )-\sigma ^{2m -2j}\left ( t \right ) } \right ) \\ & \quad +\frac{a^{2m} }{\left ( \sigma \left (t \right ) -\tau \left ( t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n\\ & = \frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j}\left ( \frac{f_{2j\left ( 2n+1 \right ) }\left ( t \right ) - f_{ 2j}\left (t \right ) }{f_{ 2j} \left (t \right ) } \right ) +\frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n . \end{split} \end{equation*}$

□

Proof of Corollary 1. First, from the definition of $f_{n}\left (t \right)$ and binomial expansion, we easily prove $\left (f_{2n+1}\left (t \right)-1, a^{2} b^{2} t^{2} +4ab \right) = 1$ . Therefore, $\left (f_{2n+1}\left (t \right)-1, \left (a^{2} b^{2} t^{2} +4ab \right)^{m} \right) = 1$ . Now, we prove (1.11) by Lemma 1 and (1.3):

$\begin{equation*} \begin{split} &b l_{1} \left ( t\right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t\right ) \\ & = l_{1} \left ( t\right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right ) \left ( \frac{a^{2m+1} }{\left ( \sigma\left (t \right ) -\tau \left (t \right ) \right )^{2m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j} \left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t \right ) -f_{2j+1}\left ( t \right ) }{l_{2j+1}\left (t\right ) } \right ) \right )\\ & \equiv 0\pmod{f_{2n+1}\left ( t \right ) -1}. \end{split} \end{equation*}$

Now, we use Lemma 2 and (1.5) to prove (1.12):

$\begin{equation*} \begin{split} &al_{1} \left ( t\right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t\right )\sum\limits _{k = 1}^{n}l_{2k}^{2m+1}\left ( t \right ) \\ & = l_{1} \left ( t \right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right ) \left ( \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{al_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t\right ) -al_{2j+1}\left (t \right ) }{l_{2j+1}\left ( t \right ) } \right ) \right )\\ & \equiv 0\pmod{l_{2n+1}\left ( t \right ) -at}. \end{split} \end{equation*}$

□

3. Conclusions

In this paper, we discuss the power sums of bi-periodic Fibonacci and Lucas polynomials by Binet formulas. As corollaries of the theorems, we extend the divisible properties of the sum of power of linear Fibonacci and Lucas sequences to nonlinear Fibonacci and Lucas polynomials. An open problem is whether we extend the Melham conjecture to nonlinear Fibonacci and Lucas polynomials.

Use of AI tools declaration

The authors declare that they did not use Artificial Intelligence (AI) tools in the creation of this paper.

Acknowledgments

The authors would like to thank the editor and referees for their helpful suggestions and comments, which greatly improved the presentation of this work. All authors contributed equally to the work, and they have read and approved this final manuscript. This work is supported by Natural Science Foundation of China (12126357).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

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AIMS Mathematics

A class of hypersurfaces in $\mathbb{E}^{n+1}_{s}$ satisfying $\Delta \vec{H} = \lambda\vec{H}$

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1. Introduction

2. Proofs of theorems

3. Conclusions

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A class of hypersurfaces in En+1s \mathbb{E}^{n+1}_{s} satisfying Δ→H=λ→H \Delta \vec{H} = \lambda\vec{H}

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1. Introduction

2. Proofs of theorems

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A class of hypersurfaces in $\mathbb{E}^{n+1}_{s}$ satisfying $\Delta \vec{H} = \lambda\vec{H}$