\documentclass[12pt]{article} \textwidth = 6.5in \textheight = 9.0in \topmargin = -20pt \evensidemargin = 0pt \oddsidemargin = 0pt \headsep = 25pt \parskip = 10pt \font\smallit=cmti10 \font\smalltt=cmtt10 \font\smallrm=cmr9 \usepackage{amsmath} \usepackage{amssymb} %\usepackage{cite} \raggedbottom \newtheorem{theorem}{Theorem}%[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newenvironment{proof}{\noindent\textit{Proof.} \ignorespaces} {\nopagebreak\null\hfill\hbox{$\Box$}} \DeclareMathOperator{\sgn}{sgn} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \def\thebibliography#1{\section*{\normalsize \refname}%\markright{\refname}% \addcontentsline{toc}{section}{\refname}\list {[\arabic{enumi}]}{\settowidth\labelwidth{[#1]}\leftmargin\labelwidth % {\arabic{enumi}.}{\settowidth\labelwidth{[#1]}\leftmargin\labelwidth \advance\leftmargin\labelsep \usecounter{enumi}} \def\newblock{\hskip .11em plus .33em minus .07em} \sloppy\clubpenalty4000\widowpenalty4000 \sfcode`\.=1000\relax} \let\endthebibliography=\endlist %\date{} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\def\baselinestretch{1.5} \begin{document} \vspace*{-40pt} \centerline{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007), \#A08} \vskip 40pt \begin{center} {\bf CONJUGATE SEQUENCES IN A~FIBONACCI-LUCAS SENSE AND SOME IDENTITIES FOR SUMS OF POWERS OF THEIR ELEMENTS} \vskip 20pt {\bf Roman Witu{\l}a}\\ {\smallit Institute of Mathematics, Silesian University of Technology, Kaszubska 23, 44-100 Gliwice, Poland}\\ {\tt d.slota@polsl.pl} \vskip 10pt {\bf Damian S{\l}ota}\\ {\smallit Institute of Mathematics, Silesian University of Technology, Kaszubska 23, 44-100 Gliwice, Poland}\\ {\tt d.slota@polsl.pl} \end{center} \vskip 30pt \centerline{\smallit Received: 5/30/06, Revised: 11/16/06, Accepted: 2/3/07, Published: 2/9/07} \vskip 30pt \centerline{\bf Abstract} \noindent The aim of this paper is to introduce the notion of a~pair of conjugate sequences in a~Fibonacci-Lucas sense. New identities are generated by means of such sequences (including separate derivations for Fibonacci and Lucas numbers) and by means of Lucas and Vieta-Lucas polynomials. \pagestyle{myheadings} \markright{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007), \#A08\hfill} \thispagestyle{empty} \baselineskip=15pt \vskip 30pt %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% Section %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section*{\normalsize 1. Introduction} \addtocounter{section}{1} The paper, which is a~follow-up of~\cite{wsCEJM}, is still focused on analyzing the notion of a~pair of sequences $\{x_n\}$ and $\{y_n\}$ that are conjugate in a~Fibonacci-Lucas sense (Sections~2 and~4). A~number of identities are derived for such pairs of conjugate sequences (see Lemma~\ref{lem1.7}). In Section~3, by means of Lucas and Vieta-Lucas polynomials, some identities (Le\-mma~\ref{lem2.2}) are presented for sequences $\{z_n\}$ defined by recurrence formula ($z_0,z_1\in \mathbb{C}$): $$ z_{n+2}=a\, z_{n+1} + z_{n},\qquad n\in \mathbb{N}_{0} :=\mathbb{N} \cup \{0\}, $$ and for auxiliary sequences ($\xi_{0,n}=z_n$): $$ \xi_{k,n}=z_{n+k}+(-1)^{k}\, z_{n-k}, \quad k=1,\ldots,n,\quad n\in\mathbb{N}. $$ In Section~5 the decompositions of some special symmetric polynomials of three, four, and five variables are given on the grounds of Vieta-Lucas polynomials. Finally, in Section~6 the decompositions of the polynomials discussed in Section~5 are applied to generate the corresponding identities for the pairs of conjugate sequences in a Fibonacci-Lucas sense. Special cases of such identities for Fibonacci and Lucas numbers are discussed separately~((\ref{41})--(\ref{last})). Several publications were concerned with identities for the sums of the powers of Fibonacci and Lucas numbers (see~\cite{kos,Melham1999a,Melham1999b,Melham2000a,Melham2000b,wsCEJM} and references therein). However, it should be emphasized that the research tasks undertaken in this paper are substantively different from those already discussed elsewhere. \vskip 30pt %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% Section %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section*{\normalsize 2. Conjugate Sequences}%\label{roz1} \addtocounter{section}{1} Let us assume that the elements of sequences $\{ x_n\}$ and $\{ y_n\}$ are determined by means of the same recurrence relation: \begin{equation} x_{n+1}=a\,x_n+b\,x_{n-1},\quad\quad y_{n+1}=a\,y_n+b\,y_{n-1}, \quad n\in\mathbb{N} , \label{47} \end{equation} where $b=\pm 1$, $a\neq 0$, $a^2+4b>0$. First our objective is to find some adjoint-conditions~-- connecting elements of sequences $\{ x_n\}$ and $\{ y_n\}$. These conditions are derived from the following identities for Fibonacci and Lucas numbers: \begin{align*} F_{n-1}+F_{n+1}&=L_n,\\ L_{n-1}+L_{n+1}&=5\,F_n. \end{align*} \begin{lemma} If there exists $A\in\mathbb{R}$ and $n_0\in\mathbb{N}$ such that $x_{n_0}\neq 0$, $$ y_n=x_{n-1}+x_{n+1},\quad\mbox{\rm for}\quad n=n_0-1,n_0+1, $$ and $$ Ax_n=y_{n-1}+y_{n+1},\quad\mbox{\rm for}\quad n=n_0, $$ then, $A=2+2b+a^2$. \end{lemma} \begin{proof} We have \begin{align*} Ax_{n_0}&=y_{n_0-1}+y_{n_0+1}=x_{n_0-2}+2x_{n_0}+x_{n_0+2} \\ &=x_{n_0-2}+abx_{n_0-1}+(2+b+a^2)x_{n_0}=(2+2b+a^2)x_{n_0} \end{align*} so $A=2+2b+a^2$. \end{proof} \begin{lemma} Let $\alpha ,\beta\in\mathbb{C}$, $\alpha\beta\neq 0$, $x_n=u\alpha ^n+v\beta ^n$, $y_n=\zeta\alpha ^n+\xi\beta ^n$, $n\in\mathbb{N} _0$. If $|\alpha |\neq |\beta |$ and identities: \begin{equation} \left\{\begin{array}{l} y_n=x_{n-1}+x_{n+1} \\ (2+2b+a^2)x_n=y_{n-1}+y_{n+1} \end{array}\right. \label{48} \end{equation} hold for every $n\in\mathbb{N}$, then the following identities are satisfied: \begin{equation}\label{pol1} \begin{split} y_n&=(\alpha ^{-1}+\alpha )\,u\,\alpha ^n+(\beta ^{-1}+\beta )\,v\,\beta ^n\\ %\quad\mbox{\rm and}\quad x_n&=\frac{\alpha +\alpha ^{-1}}{2+2b+a^2}\,\zeta\,\alpha ^n + \frac{\beta +\beta ^{-1}}{2+2b+a^2}\,\xi\,\beta ^n, \end{split} \end{equation} for every $n\in\mathbb{N}$. \end{lemma} \begin{proof} We have \begin{multline*} y_n=x_{n-1}+x_{n+1}=u\alpha ^{n-1}+u\alpha ^{n+1}+v\beta ^{n-1}+v\beta ^{n+1}={}\\ {}= u(\alpha ^{-1}+\alpha )\alpha ^n+v(\beta ^{-1}+\beta )\beta ^n. \end{multline*} \end{proof} \begin{corollary} If $b=-1$ then by (\ref{48}) we obtain: $$ ax_n=y_n,\quad\quad n\in\mathbb{N} . $$ \end{corollary} \begin{definition} Sequences $\{ x_n\} _{n=1}^{\infty}$ and $\{ y_n\} _{n=1}^{\infty}$, defined by the same relation $$ x_{n+1}=ax_n+x_{n-1}\quad\mbox{\rm and}\quad y_{n+1}=ay_n+y_{n-1}, $$ are called conjugate sequences in a~Fibonacci-Lucas sense (with parameter~$a$) if the following conditions are satisfied: $$ y_n=x_{n-1}+x_{n+1}, $$ $$ (4+a^2) x_n=y_{n-1}+y_{n+1}, $$ for every $n\in\mathbb{N}$. \end{definition} \begin{remark} If $a=1$, then $\alpha =(1+\sqrt{5})/2$, $\beta =(1-\sqrt{5})/2$. Additionally, if $x_n=u\alpha ^n+v\beta ^n$, $y_n=\zeta\alpha ^n+\xi\beta ^n$, $n\in\mathbb{N} _0$ and identities (\ref{48}) hold, then $$ y_n=\sqrt{5}u\alpha ^n-\sqrt{5}v\beta ^n\quad\mbox{\rm and}\quad x_n= \frac{\sqrt{5}}{5}u\alpha ^n-\frac{\sqrt{5}}{5}v\beta ^n,\quad n\in\mathbb{N} _0. $$ \end{remark} In the language of conjugate sequences, we may recapitulate the discussion to this point as follows: \begin{lemma} If elements of two sequences $\{ x_n\}$ and $\{ y_n\}$ are conjugates in a~Fibo\-na\-cci\-{-Lucas} sense with parameter $a$ and $x_n=u\alpha ^n+v\beta ^n$, $n\in\mathbb{N} _0$, where $\alpha ,\beta$ are roots of the characteristic equation $\mathbb{X} ^2-a\mathbb{X} -1=0$, then $$ \alpha\beta =-1, $$ $$ \alpha +\alpha ^{-1}=-(\beta +\beta ^{-1})=\alpha -\beta =\pm\sqrt{a^2+4} $$ and, consequently, we obtain $$ y_n=\pm\sqrt{a^2+4}(u\alpha ^n-v\beta ^n),\quad n\in\mathbb{N}, $$ or $$ y_n=\pm\frac{1}{\sqrt{a^2+4}}(u\alpha ^n-v\beta ^n),\quad n\in\mathbb{N}. $$ \end{lemma} \begin{proof} We note that $$ (\alpha -\beta )^2+4\alpha\beta =(\alpha +\beta )^2 $$ i.e., $$ \alpha -\beta =\pm\sqrt{a^2+4} $$ and $$ \alpha +\alpha ^{-1}+\beta +\beta ^{-1}= (\alpha +\beta )(\alpha\beta +1)(\alpha\beta )^{-1}=0. $$ \end{proof} \begin{lemma}\label{lem1.7} Let us assume that the elements of sequences $\{ x_n\}$ and $\{ y_n\}$ are conjugates in a~Fibonacci-Lucas sense with parameter $a$, or, more precisely, let us assume that the following conditions are satisfied: \begin{equation} x_{n+1}=ax_n+x_{n-1},\quad\quad y_{n+1}=ay_n+y_{n-1}, \label{adgj1} \end{equation} \begin{equation} y_n=x_{n-1}+x_{n+1} \label{adg1} \end{equation} and \begin{equation} (a^2+4)x_n=y_{n-1}+y_{n+1}. \label{lem1.7-w3} \end{equation} Then, the following identities hold: \begin{equation} z_n+az_{n+3}=(a^2+1)z_{n+2}, \label{aa1} \end{equation} \begin{align} z_{n+4}+cz_{n+2}+z_n&=az_{n+3}+(1+c)z_{n+2}+z_n \nonumber \\ &=(a^2+1+c)z_{n+2}+az_{n+1}+z_n \nonumber \\ &=(a^2+2+c)z_{n+2} \label{aa2} \end{align} for every $c\in\mathbb{C}$ and $z\in\{ x,y\}$, \begin{equation} (a^2+4)x_{n+2}+y_{n-1}=y_{n+3}+y_{n+1}+y_{n-1}=(a^2+3)y_{n+1}, \label{lem1.7-w4} \end{equation} \begin{equation} y_{n+2}+x_{n-1}=x_{n+3}+x_{n+1}+x_{n-1}=(a^2+3)x_{n+1}, \label{lem1.7-w5} \end{equation} \begin{align} x_n+x_{n+4}+ax_{n+5}&=(a^2+1)x_{n+4}+ax_{n+3}+x_n \nonumber \\ &=(a^2+1)x_{n+4}+(a^2+1)x_{n+2} \nonumber \\ &=(a^2+1)y_{n+3}, \label{aa3} \end{align} \begin{equation} x_{n+4}-x_n=x_{n+4}+x_{n+2}-(x_{n+2}+x_n)=y_{n+3}-y_{n+1}=ay_{n+2}, \label{lem1.7-w6} \end{equation} \begin{equation} x_n+ax_{n+5}=a^2y_{n+3}+x_{n+2}, \label{aa4} \end{equation} \begin{align} y_n+y_{n+4}+ay_{n+5}&=(a^2+1)y_{n+4}+ay_{n+3}+y_n \nonumber \\ &=(a^2+1)y_{n+4}+(a^2+1)y_{n+2} \nonumber \\ &=(a^2+1)(a^2+4)x_{n+3}, \label{aa5} \end{align} \begin{align} y_{n+4}-y_n&=y_{n+4}+y_{n+2}-(y_{n+2}+y_n)= \nonumber \\ &=(a^2+4)(x_{n+3}-x_{n+1})=a(a^2+4)x_{n+2} \label{} \end{align} \begin{equation} y_n+ay_{n+5}=a^2(a^2+4)x_{n+3}+y_{n+2}. \label{aa6} \end{equation} However, we have \begin{equation} z_n+z_{n+2}+z_{n+4}+z_{n+6}+a(z_{n+5}+z_{n+7})=(a^2+1)(a^2+4)z_{n+4} \label{pol2} \end{equation} and \begin{equation} z_{n+8}-2z_{n+4}+z_n=a^2(a^2+4)z_{n+4} \label{adgj2} \end{equation} for every $z\in\{ x,y\}$. \end{lemma} In the next section some generalizations of some of the identities from Lemma~\ref{lem1.7} will be presented. \vskip 30pt %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% Section %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section*{\normalsize 3. Lucas and Vieta-Lucas Polynomials} \addtocounter{section}{1} %Modified Chebyshev Polynomials Let us set $$ \Omega _n(x):=2T_n(x/2) $$ and $$ W_n(x):=(-i)^n\Omega _n(ix) $$ for every $n=0,1,2,\ldots$, where $T_n(x)$ means the $n$-th Chebyshev polynomial of the first kind. We note that the polynomials $\Omega_{n}(x)$, called Vieta-Lucas polynomials (see~\cite{Horadam2002a,Robbins1991a}), satisfy the following recurrence relations: $$ \Omega_0(x)=2,\qquad \Omega _1(x)=x, $$ $$ \Omega _{k+2}(x)=x\,\Omega _{k+1}(x)-\Omega _k(x),\qquad k\in\mathbb{N}. $$ Hence, we obtain \begin{align*} \Omega_2(x)&=x^2-2,\qquad \Omega_3(x)=x^3-3x,\qquad \Omega _4(x)=x^4-4x^2+2,\\ \Omega_5(x)&=x^5-5x^3+5x,\qquad \Omega _6(x)=x^6-6x^4+9x^2-2,\\ \Omega_7(x)&=x^7-7x^5+14x^3-7x,\qquad\mbox{\rm etc.} \end{align*} Similarly, we have $$ W_0(x)=2,\qquad W_1(x)=x, $$ $$ W_{n+2}(x)=x\,W_{n+1}(x)+W_n(x), $$ which generate the so-called Lucas polynomials~\cite{Benoumhani2003a,Catalani2004a,Constandache2002a,Filipponi1998a,Horadam2002a}): \begin{align*} W_2(x)&=x^2+2,\qquad W_3(x)=x^3+3x,\qquad W_4(x)=x^4+4x^2+2, \\ W_5(x)&=x^5+5x^3+5x,\qquad W_6(x)=x^6+6x^4+9x^2+2, \\ W_7(x)&=x^7+7x^5+14x^3+7x,\qquad\mbox{\rm etc.} \end{align*} \vspace{0.5cm} \begin{table} \caption{Triangle of the coefficients of polynomials $W_n(x)$} \medskip \noindent \begin{center} \begin{tabular}{l|cccccccccc} & 1 & \hspace{0.07cm} $x$ & \hspace{0.05cm} $x^2$ & \hspace{0.05cm} $x^3$ & \hspace{0.05cm} $x^4$ & \hspace{0.05cm} $x^5$ & \hspace{0.05cm} $x^6$ & \hspace{0.05cm} $x^7$ & \hspace{0.05cm} $x^8$ & \hspace{0.05cm} $x^9$ \\ \hline $W_0(x)$ & 2 & & & & & & & & & \\ $W_1(x)$ & & \hspace{0.07cm} 1 & & & & & & & & \\ $W_2(x)$ & 2 \begin{picture}(-3,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\put(-2,1){\line(2,-1){15}}\put(-2,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 1 & & & & & & & \\ $W_3(x)$ & & \hspace{0.07cm} 3 \begin{picture}(-3,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\put(-2,1){\line(2,-1){15}}\put(-2,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 1 & & & & & & \\ $W_4(x)$ & 2 \begin{picture}(-3,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\put(-2,1){\line(2,-1){15}}\put(-2,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 4 \begin{picture}(-4,0)\put(0,6){\line(2,1){15}}\put(1,12){\line(2,-1){10}}\put(0,1){\line(2,-1){15}}\put(0,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 1 & & & & & \\ $W_5(x)$ & & \hspace{0.07cm} 5 \begin{picture}(-3,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\put(-2,1){\line(2,-1){15}}\put(-2,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 5 \begin{picture}(-4,0)\put(-1,6){\line(2,1){15}}\put(1,12){\line(2,-1){10}}\put(-1,1){\line(2,-1){15}}\put(-1,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 1 & & & & \\ $W_6(x)$ & 2 \begin{picture}(-3,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\put(-2,1){\line(2,-1){15}}\put(-2,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 9 \begin{picture}(-4,0)\put(0,6){\line(2,1){15}}\put(1,12){\line(2,-1){10}}\put(0,1){\line(2,-1){15}}\put(0,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 6 \begin{picture}(-3,0)\put(-1,6){\line(2,1){15}}\put(1,12){\line(2,-1){10}}\put(-1,1){\line(2,-1){15}}\put(-1,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 1 & & & \\ $W_7(x)$ & & \hspace{0.07cm} 7 \begin{picture}(-3,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\put(-2,1){\line(2,-1){15}}\put(-2,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 14 \begin{picture}(-4,0)\put(-3,6){\line(2,1){15}}\put(-1,12){\line(2,-1){10}}\put(-3,1){\line(2,-1){15}}\put(-3,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 7 \begin{picture}(-3,0)\put(-1,6){\line(2,1){15}}\put(1,12){\line(2,-1){10}}\put(-1,1){\line(2,-1){15}}\put(-1,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 1 & & \\ $W_8(x)$ & 2 \begin{picture}(-3,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\put(-2,1){\line(2,-1){15}}\put(-2,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 16 \begin{picture}(-4,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\put(-2,1){\line(2,-1){15}}\put(-2,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 20 \begin{picture}(-3,0)\put(-3,6){\line(2,1){15}}\put(-1,12){\line(2,-1){10}}\put(-3,1){\line(2,-1){15}}\put(-3,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 8 \begin{picture}(-3,0)\put(-1,6){\line(2,1){15}}\put(1,12){\line(2,-1){10}}\put(-1,1){\line(2,-1){15}}\put(-1,-1){\line(2,-1){15}}\end{picture} & & \hspace{0.05cm} 1 & \\ $W_9(x)$ & & \hspace{0.07cm} 9 \begin{picture}(-3,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\end{picture} & & \hspace{0.05cm} 30 \begin{picture}(-4,0)\put(-3,6){\line(2,1){15}}\put(-1,12){\line(2,-1){10}}\end{picture} & & \hspace{0.05cm} 27 \begin{picture}(-3,0)\put(-3,6){\line(2,1){15}}\put(-1,12){\line(2,-1){10}}\end{picture} & & \hspace{0.05cm} 9 \begin{picture}(-3,0)\put(-1,6){\line(2,1){15}}\put(1,12){\line(2,-1){10}}\end{picture} & & \hspace{0.05cm} 1 \end{tabular} \hspace{0mm} \begin{tabular}{cc} \multicolumn{2}{c}{Legende:}\\ & b \\ a \begin{picture}(-3,0)\put(-2,6){\line(2,1){15}}\put(0,12){\line(2,-1){10}}\put(-2,1){\line(2,-1){15}}\put(-2,-1){\line(2,-1){15}}\end{picture} & \\ & \hspace{0.2cm} a+b \end{tabular} \end{center} \vspace{0.5cm} \end{table} \begin{lemma} We have $$ \mbox{\rm coeff}\big[ x^2;W_{2k}(x)\big] =k^2,\quad \mbox{\rm coeff}\big[ x^3;W_{2k+1}(x)\big] =\sum\limits_{l=1}^{k}l^2= \frac{1}{6}\, k\, (k+1)\, (2k+1), $$ $$ \mbox{\rm coeff}\big[ x^4;W_{2k}(x)\big] =\sum\limits_{l=1}^{k}(k-l)l^2= \frac{1}{12}\, k^2\, (k^2-1), $$ \begin{equation} W_{n}(x) = \sum_{k=0}^{\lfloor n/2\rfloor} \frac{n}{n-k}\, \binom{n-k}{k}\, x^{n-2k}, \end{equation} \begin{equation} x\sum\limits_{r=1}^{s}W_{2r-1}(x)=W_{2s}(x)-2, \end{equation} and \begin{equation} x\sum\limits_{r=0}^{s}W_{2r}(x)=W_{2s+1}(x). \label{} \end{equation} \end{lemma} The polynomials $W_n(x)$, $n\in\mathbb{N}$, enable the description of many fundamental identities, involving the elements of the sequence $\{ z_n\} _{n=1}^{\infty}$ determined by the following one-parameter recurrence relation: \begin{equation} z_0,z_1\in \mathbb{C}\qquad \mbox{ and } \qquad z_{n+2}=a\,z_{n+1}+z_n,\qquad n\in\mathbb{N}_{0}. \end{equation} Additionally, let us assume that $\xi _{0,n}:=z_n$ and $$ \xi _{k,n}:=z_{n+k}+(-1)^k\,z_{n-k} $$ for every $k=1,2,\ldots ,n-1$. Some of the identities are generalized forms of the identities from Le\-mma~\ref{lem1.7} (especially identities (\ref{adgj1})--(\ref{aa2}) and (\ref{pol2})--(\ref{adgj2})). All the identities presented below seem to be original. \begin{lemma}\label{lem2.2} The following identities hold true \begin{align} \xi _{k+1,n}&=a\xi _{k,n}+\xi _{k-1,n}, \label{asd1} \\ \xi _{k+1,n}&=W_{k-s}(a)\xi _{s+1,n}+W_{k-s-1}(a)\xi _{s,n}, \ \ s=1 \leq k, \label{} \\ \xi _{k,n}&=W_k(a)z_n, \label{} \\ \sum\limits_{r=-s}^{s}z_{n-2r}&=\bigg(\sum\limits_{r=0}^{s}W_{2r}(a)\bigg) z_n=\frac{1}{a}W_{2s+1}(a)z_n, \label{} \\ \sum\limits_{r=1}^{s}\big( z_{n+2r-1}&-z_{n-2r+1}\big) = \bigg(\sum\limits_{r=1}^{s}z_{n+2r-1}\bigg) - \bigg(\sum\limits_{t=1}^{s}z_{n-2r+1}\bigg) = \nonumber \\ &=\bigg(\sum\limits_{r=1}^{s}W_{2r-1}(a)\bigg) z_n= \frac{1}{a}\big( W_{2s}(a)-2\big) . \label{} \end{align} \end{lemma} \begin{proof} First, we note that \begin{align*} z_{n+2k+1}-z_{n-2k-1}&=az_{n+2k}+z_{n+2k-1}-(z_{n-2k+1}-az_{n-2k}) \\ &=a(z_{n+2k}+z_{n-2k})+z_{n+2k-1}-z_{n-2k+1}, \\ z_{n+2k}+z_{n-2k}&=a(z_{n+2k-1}-z_{n-2k+1})+z_{n+2k-2}+z_{n-2k+2}, \end{align*} i.e., that $$ \xi _{k+1,n}=a\xi _{k,n}+\xi _{k-1,n}=W_1(a)\xi _{k,n}+W_0(a)\xi _{k-1,n}. $$ Now, suppose that for some $s\in\mathbb{N} _0$, $s\leqslant k$ we have $$ \xi _{k+1,n}=W_{k-s}(a)\xi _{s+1,n}+W_{k-s-1}(a)\xi _{s,n}. $$ Hence, we get \begin{align*} \xi _{k+1,n}&=W_{k-s}(a)(a\xi _{s,n}+\xi _{s-1,n})+W_{k-s-1}(a)\xi _{s,n} \\ &=W_{k-s-1}(a)\xi _{s,n}+W_{k-s}(a)\xi _{s-1,n}. \end{align*} This leads to \begin{align*} \xi _{k+1,n}&=W_k(a)\xi _{1,n}+W_{k-1}(a)\xi _{0,n} \\ &=(aW_k(a)+W_{k-1}(a))z_n=W_{k+1}(a)z_n. \end{align*} \end{proof} \begin{corollary} We have $$ z_{n+8}-2z_{n+4}+2z_n-2z_{n-4}+z_{n-8}=a^2(a^2+4)W_4(a)z_n, $$ \begin{equation} z_{2k}+(-1)^kz_0=W_k(a)z_k, \label{w2.8} \end{equation} \begin{equation} az_{2k+1}+2(-1)^{k+1}z_0=W_{k+1}(a)z_{k+1}-W_k(a)z_k, \label{w2.9} \end{equation} \begin{equation} a\sum\limits_{k=0}^{2s}z_{2k}=W_{2s+1}(a)z_{2s}, \label{w2.10} \end{equation} \begin{equation} a\sum\limits_{k=0}^{s}z_{2k+1}=W_{s+1}(a)z_{s+1}+((-1)^s-1)z_0, \label{w2.11} \end{equation} \begin{equation} a\bigg(\sum\limits_{k=s}^{2s-1}z_{2k}-\sum\limits_{l=0}^{s-1}z_{2l}\bigg) = (W_{2s}(a)-2)z_{2s-1}, \label{w2.12} \end{equation} \begin{equation} a\sum\limits_{k=0}^{4s}z_k=W_{2s+1}(a)z_{2s}+W_{2s-1}(a)z_{2s-1}, \label{w2.13} \end{equation} \begin{align} a\sum\limits_{k=1}^{s}W_k(a)z_k&=a\sum\limits_{k=1}^{s}z_{2k}+ \frac{a}{2}((-1)^s-1)z_0 \nonumber \\ &=\left\{\begin{array}{ll} W_{s+1}(a)z_s-az_0 & \mbox{\rm for $s\in 2\mathbb{N}$}, \\ W_s(a)z_{s-1}-az_s-2az_0 & \mbox{\rm for $s\in 2\mathbb{N} -1.$} \end{array}\right. \label{w2.14} \end{align} \end{corollary} \vskip 30pt %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% Section %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section*{\normalsize 4. Some Generalizations of Conjugate Sequences Concept} \addtocounter{section}{1} Let us assume that the sequences $\{ x_n\}$ and $\{ y_n\}$ are determined by the same recurrence equations~(\ref{47}). The following result is a~response to an attempt to substitute conditions~(\ref{48}) by some other conditions connecting the elements of the sequences $\{ x_n\}$ and~ $\{ y_n\}$. \begin{lemma} Let $A,B,C,D,\alpha ,\beta\in\mathbb{C}$, $|\alpha |\neq |\beta |$. Let us assume that there exist $u,v\in\mathbb{C}$ so that for every $n\in\mathbb{N}$ the following identities are satisfied: \begin{equation} x_n=A\alpha ^n+B\beta ^n \label{1} \end{equation} \begin{equation} y_n=C\alpha ^n+D\beta ^n \label{2} \end{equation} \begin{equation} x_n+x_{n+2}+x_{n+4}+x_{n+6}=u\cdot y_{n+3} \label{3} \end{equation} and \begin{equation} y_n+y_{n+2}+y_{n+4}+y_{n+6}=v\cdot x_{n+3}. \label{4} \end{equation} If $ABCD\neq 0$ then $-b=\alpha\beta =\pm 1$. Moreover, if $AC\neq 0$ (resp. $BD\neq 0$) then $$ u=\frac{A}{C}(\alpha +\alpha ^{-1})[(\alpha +\alpha ^{-1})^2-2] \quad\quad\mbox{\rm and}\quad\quad v=\frac{C}{A}(\alpha +\alpha ^{-1}) [(\alpha +\alpha ^{-1})^2-2] $$ (resp., $ u=\frac{B}{D}(\beta +\beta ^{-1})[(\beta +\beta ^{-1})^2-2] \quad\mbox{\rm and}\quad v=\frac{D}{B}(\beta +\beta ^{-1}) [(\beta +\beta ^{-1})^2-2] $). \end{lemma} \begin{proof} The following four identities from (\ref{3}) and (\ref{4}) can be generated: \begin{equation} \left\{\begin{array}{l} A\frac{\alpha ^8-1}{\alpha ^3(\alpha ^2-1)}= A\frac{\alpha ^4-\alpha ^{-4}}{\alpha -\alpha ^{-1}}=uC, \\[1ex] B\frac{\beta ^8-1}{\beta ^3(\beta ^2-1)}= B\frac{\beta ^4-\beta ^{-4}}{\beta -\beta ^{-1}}=uD, \\[1ex] C\frac{\alpha ^8-1}{\alpha ^3(\alpha ^2-1)}= C\frac{\alpha ^4-\alpha ^{-4}}{\alpha -\alpha ^{-1}}=vA, \\[1ex] D\frac{\beta ^8-1}{\beta ^3(\beta ^2-1)}= D\frac{\beta ^4-\beta ^{-4}}{\beta -\beta ^{-1}}=uB. \end{array}\right. \label{5} \end{equation} Hence $$ A\bigg(\frac{\alpha ^4-\alpha ^{-4}}{\alpha -\alpha ^{-1}}\bigg) ^2= uvA\quad\quad\mbox{\rm and}\quad\quad A\bigg(\frac{\beta ^4-\beta ^{-4}}{\beta -\beta ^{-1}}\bigg) ^2=uvB. $$ If $ABCD\neq 0$ then $$ \bigg(\frac{\alpha ^4-\alpha ^{-4}}{\alpha -\alpha ^{-1}}\bigg) ^2= \bigg(\frac{\beta ^4-\beta ^{-4}}{\beta -\beta ^{-1}}\bigg) ^2, $$ i.e., $$ \alpha ^3+\alpha +\alpha ^{-1}+\alpha ^{-3}= \pm (\beta ^3+\beta +\beta ^{-1}+\beta ^{-3}) $$ or \begin{multline*} (\alpha +\alpha ^{-1})^3-2(\alpha +\alpha ^{-1})\pm \big((\beta +\beta ^{-1})^3-2(\beta +\beta ^{-1})\big)= {}\\ {}=\big((\alpha +\alpha ^{-1})\pm (\beta +\beta ^{-1})\big)%\times{}\\ %{}\times \big((\alpha +\alpha ^{-1})^2\mp (\beta +\beta ^{-1})(\alpha +\alpha ^{-1})+ (\beta +\beta ^{-1})^2-2\big), \end{multline*} while $$ \triangle _{\alpha +\alpha ^{-1}}=(\beta +\beta ^{-1})^2- 4(\beta +\beta ^{-1})^2+8=-3\beta ^{-2}\bigg[\beta ^4-\frac{2}{3}\beta ^2+1\bigg] <0. $$ Therefore $$ \alpha +\alpha ^{-1}=\pm (\beta +\beta ^{-1}), $$ i.e., $$ \alpha ^2\pm (\beta +\beta ^{-1})\alpha +1=0, $$ which implies for plus sign $$ (\alpha +\beta )(\alpha +\beta ^{-1})= 0,\quad\mbox{\rm i.e.,}\quad \alpha\beta =-1 $$ and for minus sign $$ (\alpha -\beta )(\alpha -\beta ^{-1})= 0,\quad\mbox{\rm i.e.,}\quad \alpha\beta =1 $$ (since $|\alpha |\neq |\beta |$). \end{proof} \begin{corollary} If $ABCD\neq 0$ and $\alpha\beta =1$ then $$ \alpha +\alpha ^{-1}=\alpha +\beta =\beta ^{-1}+\beta $$ and $$ AD=BC. $$ So, sequences $\{ x_n\}$ and $\{ y_n\}$ are linearly dependent; or, more precisely, we have $x_n=\frac{A}{D}y_n$, $n\in\mathbb{N}$. \end{corollary} \begin{corollary} If $ABCD\neq 0$ and $\alpha\beta =-1$ then $$ \alpha +\alpha ^{-1}=\alpha -\beta =-(\beta ^{-1}+\beta ) $$ and $$ AD+BC=0. $$ Moreover, we have $$ x_n+x_{n+2}=\frac{B}{D}(\beta -\alpha )y_{n+1} \quad\quad\mbox{\rm and}\quad\quad y_n+y_{n+2}=\frac{D}{B}(\beta -\alpha )x_{n+1}; $$ or $$ x_{n}^{\ast}+x_{n+2}^{\ast}=(\beta -\alpha )^2y_{n+1} \quad\quad\mbox{\rm and}\quad\quad y_n+y_{n+2}=x_{n+1}^{\ast} $$ where $x_n^{\ast}:=\frac{D}{B}(\beta -\alpha )x_n$, $n\in\mathbb{N}$. In the sequel if $B(\beta -\alpha )=D$ or $D(\beta -\alpha )=B$ then sequences $\{ x_n\}$ and $\{ y_n\}$ are conjugates in a~Fibonacci-Lucas sense with parameter $a=\alpha +\beta$. \end{corollary} \begin{proof} We have \begin{align*} x_n+x_{n+2}&=A(1+\alpha ^2)\alpha ^n+B(1+\beta ^2)\beta ^n=\\ &=-\frac{BC}{D}\alpha (\alpha +\alpha ^{-1})\alpha ^n+ B\beta (\beta +\beta ^{-1})\beta ^n= \\ &=\frac{B}{D}(\beta -\alpha )[C\alpha ^{n+1}+D\beta ^{n+1}]= \frac{B}{D}(\beta -\alpha )y_{n+1}. \end{align*} \end{proof} \vskip 30pt %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% Section %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section*{\normalsize 5. Decompositions of Some Special Polynomials of Many Variables} \addtocounter{section}{1} In this section an attempt is made at decomposing some symmetric polynomials of~$3$, $4$, and~$5$ variables into factors (such decompositions may easily be generalized to polynomials of~$6$ or even more variables). The decompositions are derived by means of modified Chebyshev polynomials $\Omega _n(x)$ (Vieta-Lucas polynomials). \begin{lemma} The following identities hold \begin{align} &p_n^{(3)}(x,y,z)\!:\!=(x\!+y\!+z)^n\!-(x\!+y)^n\!-(x\!+z)^n\!- (y\!+z)^n\!+x^n\!+y^n\!+z^n= \nonumber \\ &=nxyz\bigg[\frac{(x\!+y\!+z)^{n-1}}{xy\!+xz\!+yz}\!- (xy\!+xz\!+yz)^{(n-3)/2}\Omega _{n-1} \bigg(\frac{x\!+y\!+z}{(xy\!+xz\!+yz)^{1/2}}\bigg) \nonumber \\ &\hspace{0.5cm} +F_{3,n}\big( xyz,x\!+y\!+z,xy,x\!+ y,xz,x\!+z,yz,y\!+z\big)\bigg] \label{2.1} %\\ \end{align} %&\mbox{\rm (in the following formulas the compact form %of the respective identities is given)} \nonumber \\ (in the following formulas, the compact form of the respective identities is given) \begin{equation*} %& =\left\{\begin{array}{lll} 0, & \mbox{\rm for} & n\leqslant 2, \\ 6xyz, & \mbox{\rm for} & n=3, \\ 12xyz(x\!+y\!+z), & \mbox{\rm for} & n=4, \\ 10xyz\big[ 2(x\!+y\!+z)^2\!-xy\!-xz\!-yz\big] , & \mbox{\rm for} & n=5, \\ 30xyz(x\!+y\!+z)\big[ (x\!+y\!+z)^2\!-xy\!-xz\!-yz\big] , & \mbox{\rm for} & n=6, \\ 7xyz\big[ 6(x\!+y\!+z)^4\!-9(xy\!+xz\!+yz)(x\!+y\!+z)^2 && \\ \hspace{0.5cm} +2(xy\!+xz\!+yz)^2\!-xyz(x\!+y\!+z)\big] , & \mbox{\rm for} & n=7, \\ 56xyz(x\!+y\!+z)\big[ \big( (x\!+y\!+z)^2\!-xy\!-xz\!-yz\big) ^2\! && \\ \hspace{0.5cm}- \frac{1}{2}xyz(x\!+y\!+z)\big] , & \mbox{\rm for} & n=8, \\ 18xyz\big[ 4(x\!+y\!+z)^6\!-10(xy\!+xz\!+yz)(x\!+y\!+z)^4 && \\ \hspace{0.5cm} +8(xy\!+xz\!+yz)^2(x\!+y\!+z)^2\!-(xy\!+xz\!+yz)^3 &&\\ \hspace{0.5cm}- 4xyz(x\!+y\!+z)^3 +xyz\big( xy(x\!+y)\!+xz(x\!+z) &&\\ \hspace{0.5cm} +yz(y\!+z)\big) \!+ \frac{10}{3}(xyz)^2\big] , & \mbox{\rm for} & n=9. \end{array}\right. %\nonumber \end{equation*} Moreover, we have \begin{align*} &F_{3,n}\big( xyz,x\!+y\!+z,xy,x\!+y,xz,x\!+z,yz,y\!+z\big) = \\ &=\left\{\begin{array}{lll} 0, & \mbox{\rm for} & n\leqslant 6, \\ -xyz(x\!+y\!+z), & \mbox{\rm for} & n=7, \\ -\frac{7}{2}xyz(x\!+y\!+z)^2, & \mbox{\rm for} & n=8, \\ -8xyz(x\!+y\!+z)^3\!+2xyz\big( xy(x\!+y) &&\\ \hspace*{5mm}+xz(x\!+z)\!+yz(y\!+z)\big) \!+ \frac{20}{3}(xyz)^2, & \mbox{\rm for} & n=9, \\ -15xyz(x\!+y\!+z)^4\!+9xyz\big( xy(x\!+y)^2 && \\ \hspace*{5mm} +xz(x\!+z)^2\!+yz(y\!+z)^2\big)+48x^2y^2z^2(x\!+y\!+z), & \mbox{\rm for} & n=10, \\ -25xyz(x\!+y\!+z)^5\!+25xyz\big( xy(x\!+y)^3 && \\ \hspace*{5mm}+xz(x\!+z)^3\!+yz(y\!+z)^3\big) -3xyz\big( x^3(y\!+z)^2 &&\\ \hspace*{5mm}+y^3(x\!+z)^2\!+z^3(x\!+y)^2\big) \!+92(xyz)^2\times && \\ \hspace{5mm} \times\big( (x\!+y)^2\!+(x\!+z)^2\!+(y\!+z)^2\big) &&\\ \hspace*{5mm}+ 117(xyz)^3(x\!+y\!+z), & \mbox{\rm for} & n=11, \end{array}\right. \end{align*} and \begin{multline*} F_{3,12}=-\frac{77}{2}x^2y^2z^2(x+y+z)^6+{}\\ {}+ 55x^2y^2z^2\big( xy(x+y)^4+xz(x+z)^4+yz(y+z)^4\big)-{} \\ -\frac{33}{2}x^2y^2z^2\big( x^2y^2(x+y)^2+ x^2z^2(x+z)^2+y^2z^2(y+z)^2\big) + {}\\ {}+484x^3y^3z^3(x+y+z)^3 -319x^3y^3z^3(xy+xz+yz)(x+y+z)+%{}\\ %{}+ \frac{6699}{2}x^4y^4z^4. \end{multline*} \end{lemma} \begin{remark} We note that $xyz$ divides $p_n^{(3)}(x,y,z)$ for every $n\in\mathbb{N}$. \end{remark} \begin{lemma} We have \begin{align} &p_n^{(4)}(x,y,z,u) :=(x +y +z +u)^n - (x +y +z)^n -(x +y +u)^n - \nonumber \\ &-(x +z +u)^n -(y +z +u)^n+(x +y)^n +(x +z)^n +(x +u)^n + \nonumber\\ &(y +z)^n ++(y +u)^n +(z +u)^n -x^n -y^n -z^n -u^n= \nonumber \\ &=n(n -1)xyzu\bigg[ \frac{(x +y +z +u)^{n -2}}{xy +xz +xu +yz +yu +zu}-\nonumber \\ &-(xy +xz +xu +yz +yu +zu)^{(n -4)/2}%\times \nonumber \\ %&\times \Omega _{n-2}\bigg(\frac{x +y +z +u}{(xy +xz +xu + yz +yu +zu)^{1/2}}\bigg) + \nonumber \\ &+F_{4,n}(xyzu,x +y +z +u,xyz,x +y +z,xyu,x +y +u,xzu,\nonumber \\ &\hspace*{5mm}x +z +u,yzu,y +z +u)\bigg] = \label{aa7} \end{align} \begin{equation*} =\left\{\begin{array}{lll} 0, & \mbox{\rm for} & n\leqslant 3, \\ 24xyzu, & \mbox{\rm for} & n=4, \\ 60xyzu(x\!+y\!+z\!+u), & \mbox{\rm for} & n=5, \\ 60xyzu\big[ 2(x\!+y\!+z\!+u)^2-{}&&\\ \hspace{0.5cm}-(xy\!+xz\!+xu\!+yz\!+yu\!+zu)\big] , & \mbox{\rm for} & n=6, \\ 210xyzu(x\!+y\!+z\!+u)\big[ (x\!+y\!+z\!+u)^2-{} & & \\ \hspace{0.5cm}-(xy\!+xz\!+xu\!+yz\!+yu\!+zu)\big] , & \mbox{\rm for} & n=7, \\ 56xyzu\big[ 6(x\!+y\!+z\!+u)^4\!-9(xy\!+xz\!+xu+ & & \\ \hspace{0.5cm} +yz\!+yu\!+zu)(x\!+y\!+z\!+u)^2+2(xy\!+xz+&&\\ \hspace{0.5cm} +xu\!+yz\!+yu\!+zu)^2\!-xyz(x\!+y\!+z)- && \\ \hspace{0.5cm} -xyu(x\!+y\!+u)\!-xuz(x\!+u\!+z)-&&\\ \hspace{0.5cm} -yuz(y\!+u\!+z)\!-3xyzu\big] = && \\ \hspace{0.5cm} =28xyzu\big[ 5(x\!+y\!+z\!+u)^4\!+ 5(x^2\!+y^2\!+z^2\!+u^2)\times&&\\ \hspace{0.5cm} \times(x\!+y\!+z\!+u)^2+2(x^2\!+y^2\!+z^2\!+u^2)^2-&&\\ \hspace{0.5cm} -30xyzu\!-10xyz(x\!+y\!+z)-10xyu(x\!+y\!+u)- && \\ \hspace{0.5cm} -10xuz(x\!+u\!+z)\!-10uyz(u\!+y\!+z)- && \\ \hspace{0.5cm} -4(x^2y^2\!+x^2u^2\!+x^2z^2\!+ y^2u^2\!+y^2z^2\!+u^2z^2)\big] , & \mbox{\rm for} & n=8, \\ \end{array}\right. %\nonumber \end{equation*} and for $n=9$: \begin{multline*} =252xyzu(x +y +z +u)\big[ 2(x +y +z +u)^4 -4(xy +xz +xu +yz +yu +zu)(x +y +z +u)^2+ {} \\ {}+2(xy +xz +xu +yz +yu +zu)^2 - xyz(x +y +z)^2 -xuz(x +u +z)^2-xuy(x +u +y)^2 - {}\\ -uyz(u +y +z)^2 -42xyuz(x +y +u +z)\big] , \end{multline*} where \begin{multline*} F_{4,n}(xyzu,x\!+y\!+z\!+u,xyz,x\!+y\!+z,xyu, x\!+y\!+u,xzu,x\!+z\!+u,yzu,y\!+z\!+u)= \\ =\left\{ \begin{array}{lll} 0 & \mbox{\rm for} & n\leqslant 7, \\ -xyz(x +y +z) -xyu(x +y +u) - &&\\ \hspace{5mm} -xuz(x +u +z) -yuz(y +u +z) -3xyzu, & \mbox{\rm for} & n=8, \\ -\frac{7}{2}xyz(x +y +z)^2 -\frac{7}{2}xyu(x +y +u)^2 &&\\ \hspace{5mm}-\frac{7}{2}xuz(x +u +z)^2 -\frac{7}{2}yuz(y +u +z)^2- && \\ \hspace{5mm} -147xyuz(x +y +u +z) & \mbox{\rm for} & n=9. \end{array} \right. \end{multline*} \end{lemma} \begin{remark} We note that $xyzu$ divides $p_n^{(4)}(x,y,z,u)$ for every $n\in\mathbb{N}$. \end{remark} \begin{lemma} We have the following decompositions: \begin{multline*} p_n^{(5)}(x,y,z,u,v) : =(x +y +z +u +v)^n - (x +y +z +u)^n -(x +y +z +v)^n - \\ -(x +y +u +v)^n-(x +z +u +v)^n -(z +y +u +v)^n +(x +y +z)^n + (x +y +u)^n + \\ +(x +y +v)^n +(x +z +u)^n+(x +z +v)^n +(x +u +v)^n +(y +z +u)^n + (y +z +v)^n + \\ +(y +u +v)^n +(z +u +v)^n-(x +y)^n -(x +z)^n -(x +u)^n -(x +v)^n - (y +z)^n - \\ -(y +u)^n -(y +v)^n -(z +u)^n-(z +v)^n -(u +v)^n +x^n +y^n +z^n +u^n +v^n= \\ =\left\{\begin{array}{ll} 0 & \mbox{\rm for $n\leqslant 4$}, \\[1ex] 5!xyzuv & \mbox{\rm for $n=5$}, \\[1ex] \frac{6!}{2!}xyzuv(x +y +z +u +v) & \mbox{\rm for $n=6$}, \\[1ex] \frac{7!}{2!3!}xyzuv\big[ 3(x +y +z +u +v)^2 - &\\ \hspace*{5mm}-x^2 -y^2 -z^2 -u^2 -v^2\big] = & \\[1ex] \hspace*{2mm}=\frac{7!}{3!}xyzuv[(x +y +z +u +v)^2 +xy +xz +&\\ \hspace*{5mm}+xu +xv +yz +yu +yv +zu +zv +uv] & \mbox{\rm for $n=7$}. \end{array}\right. \end{multline*} \end{lemma} \vskip 30pt %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% Section %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section*{\normalsize 6. Applications} \subsection*{\normalsize 6.1 Identities for the Powers of Elements of Conjugate Sequences} \addtocounter{section}{1} We now present some identities obtained by applying identity~(\ref{2.1}) to the elements of sequences $\{ x_n\}$ and $\{ y_n\}$ satisfying (\ref{adgj1})--(\ref{lem1.7-w3}). \begin{itemize} \item[a)] For $x=x_{n+4}$, $y=-x_n$, $z=-ay_{n+2}$ and odd powers only: \begin{equation*} x_{n+4}^k-x_n^k-a^ky_{n+2}^k=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1$}, \\[1ex] 3ax_ny_{n+2}x_{n+4} & \mbox{\rm for $k=3$}, \\[1ex] 5ax_ny_{n+2}x_{n+4}(x_nx_{n+4}+a^2y_{n+2}^2) & \mbox{\rm for $k=5$}, \\[1ex] 7ax_ny_{n+2}x_{n+4}(x_nx_{n+4}+a^2y_{n+2}^2)^2 & \mbox{\rm for $k=7$}, \\[1ex] 9ax_ny_{n+2}x_{n+4}\big[\frac{1}{3}(ax_ny_{n+2}x_{n+4})^2- &\\ \hspace*{5mm}-(x_nx_{n+4}+a^2y_{n+2}^2)^3\big] & \mbox{\rm for $k=9$}; \end{array}\right. \end{equation*} \item[b)] for $x=x_{n+4}$, $y=-x_n$, $z=ay_{n+2}$: \begin{align*} (2a)^k&y_{n+2}^k+x_{n+4}^k+(-1)^kx_n^k-(ay_{n+2}-x_n)^k-(2ay_{n+2}+x_n)^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 2$}, \\[1ex] -6ax_ny_{n+2}x_{n+4} & \mbox{\rm for $k=3$}, \\[1ex] -24a^2x_ny_{n+2}^2x_{n+4} & \mbox{\rm for $k=4$}, \\[1ex] -10ax_ny_{n+2}x_{n+4}[7a^2y_{n+2}^2+x_nx_{n+4}] & \mbox{\rm for $k=5$}, \\[1ex] -60a^2x_ny_{n+2}^2x_{n+4}[3a^2y_{n+2}^2+x_nx_{n+4}] & \mbox{\rm for $k=6$}, \\[1ex] -112a^2x_ny_{n+2}^2x_{n+4}[(3a^2y_{n+2}^2+x_nx_{n+4})^2+&\\ \hspace*{5mm}+a^2x_ny_{n+2}^2x_{n+4}] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[c)] for $x=x_{n+4}$, $y=x_n$, $z=-ay_{n+2}$: \begin{align*} x_{n+4}^k&-(a^2+2)^kx_{n+2}^k+2^kx_n^k+(-a)^ky_{n+2}^k-(x_n-ay_{n+2})^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 2$}, \\[1ex] -6ax_ny_{n+2}x_{n+4} & \mbox{\rm for $k=3$}, \\[1ex] -24ax_n^2y_{n+2}x_{n+4} & \mbox{\rm for $k=4$}, \\[1ex] -10ax_ny_{n+2}x_{n+4}[9x_n^2-ay_{n+2}x_{n+4}] & \mbox{\rm for $k=5$}, \\[1ex] -60ax_n^2y_{n+2}x_{n+4}[5x_n^2-ay_{n+2}x_{n+4}] & \mbox{\rm for $k=6$}, \\[1ex] -112ax_n^2y_{n+2}x_{n+4}[(5x_n^2-ay_{n+2}x_{n+4})^2+&\\ \hspace*{5mm}+ax_n^2y_{n+2}x_{n+4}] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[d)] for $x=x_{n+4}$, $y=x_n$, $z=ay_{n+2}$: \begin{align*} (2^k-1)&(ay_{n+2}+x_n)^k-(2x_n+ay_{n+2})^k-(x_n+2ay_{n+2})^k+{}\\ &+x_{n+4}^k+x_n^k+a^ky_{n+2}^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 2$}, \\[1ex] 6ax_ny_{n+2}x_{n+4} & \mbox{\rm for $k=3$}, \\[1ex] 24ax_ny_{n+2}x_{n+4}^2 & \mbox{\rm for $k=4$}, \\[1ex] 10ax_ny_{n+2}x_{n+4}[7x_{n+4}^2-ax_ny_{n+2}] & \mbox{\rm for $k=5$}, \\[1ex] 60ax_ny_{n+2}x_{n+4}^2[3x_{n+4}^2-ax_ny_{n+2}] & \mbox{\rm for $k=6$}, \\[1ex] 112ax_ny_{n+2}x_{n+4}^2[(3x_{n+4}^2-ax_ny_{n+2})^2-&\\ \hspace*{5mm}-ax_ny_{n+2}x_{n+4}^2] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[e)] for $x=y_{n+4}$, $y=-y_n$, $z=-a(a^2+4)x_{n+2}$ and odd powers only: \begin{align*} &y_{n+4}^k-y_n^k-(a(a^2+4))^kx_{n+2}^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1$}, \\[1ex] 3a(a^2+4)y_nx_{n+2}y_{n+4} & \mbox{\rm for $k=3$}, \\[1ex] 5a(a^2+4)y_nx_{n+2}y_{n+4}(a^2(a^2+4)^2x_{n+2}^2+y_ny_{n+4}) & \mbox{\rm for $k=5$}, \\[1ex] 7a(a^2+4)y_nx_{n+2}y_{n+4}(a^2(a^2+4)^2x_{n+2}^2+y_ny_{n+4})^2 & \mbox{\rm for $k=7$}, \\[1ex] 9a(a^2+4)y_nx_{n+2}y_{n+4}[(a^2(a^2+4)^2x_{n+2}^2+y_ny_{n+4})^3+ & \\ \hspace{1cm} +\frac{1}{3}(a(a^2+4)y_nx_{n+2}y_{n+4})^2] & \mbox{\rm for $k=9$}; \end{array}\right. \end{align*} \item[f)] for $x=y_{n+4}$, $y=-y_n$, $z=a(a^2+4)x_{n+2}$: \begin{align*} (2a&(a^2+4))^kx_{n+2}^k+y_{n+4}^k+(-1)^ky_n^k-(a(a^2+4)x_{n+2}-y_n)^k- \\ &-(2a(a^2+4)x_{n+2}+y_n)^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 2$}, \\[1ex] -6a(a^2+4)y_nx_{n+2}y_{n+4} & \mbox{\rm for $k=3$}, \\[1ex] -24a^2(a^2+4)^2y_nx_{n+2}^2y_{n+4} & \mbox{\rm for $k=4$}, \\[1ex] -10a(a^2+4)y_nx_{n+2}y_{n+4}\times&\\ \hspace*{5mm}\times[7a^2(a^2+4)^2x_{n+2}^2+y_ny_{n+4}] & \mbox{\rm for $k=5$}, \\[1ex] -60a^2(a^2+4)^2y_nx_{n+2}^2y_{n+4}\times&\\ \hspace*{5mm}\times[3a^2(a^2+4)^2x_{n+2}^2+y_ny_{n+4}] & \mbox{\rm for $k=6$}, \\[1ex] -112a^2(a^2+4)^2y_nx_{n+2}^2y_{n+4}\times&\\ \hspace*{5mm}\times[(3a^2(a^2+4)^2x_{n+2}^2+y_ny_{n+4})^2+ & \\ \hspace{5mm} +a^2(a^2+4)^2y_nx_{n+2}^2y_{n+4}] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[g)] for $x=y_{n+4}$, $y=y_n$, $z=-a(a^2+4)x_{n+2}$: \begin{align*} y_{n+4}^k&-(a^2+2)^ky_{n+2}^k+2^ky_n^k+{}\\ &+(-a(a^2+4))^kx_{n+2}^k- (y_n-a(a^2+4)x_{n+2})^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 2$}, \\[1ex] -6a(a^2+4)y_nx_{n+2}y_{n+4} & \mbox{\rm for $k=3$}, \\[1ex] -24a(a^2+4)y_n^2x_{n+2}y_{n+4} & \mbox{\rm for $k=4$}, \\[1ex] -10a(a^2+4)y_nx_{n+2}y_{n+4}\times&\\ \hspace*{5mm}\times[7y_n^2+a(a^2+4)x_{n+2}y_{n+4}] & \mbox{\rm for $k=5$}, \\[1ex] -60a(a^2+4)y_n^2x_{n+2}y_{n+4}\times&\\ \hspace*{5mm}\times[3y_n^2+a(a^2+4)x_{n+2}y_{n+4}] & \mbox{\rm for $k=6$}, \\[1ex] -112a(a^2+4)y_n^2x_{n+2}y_{n+4}\times&\\ \hspace*{5mm}\times[(3y_n^2+a(a^2+4)x_{n+2}y_{n+4})^2+ & \\ \hspace{1cm} +a(a^2+4)y_n^2x_{n+2}y_{n+4}] & \mbox{\rm for $k=7$}; \end{array}\right. \end{align*} \item[h)] for $x=x_n$, $y=ax_{n+1}$, $z=ax_{n+3}$: \begin{align*} x_{n+4}^k&+a^kx_{n+3}^k-[1+(a^2+1)^k]x_{n+2}^k-a^ky_{n+2}^k+a^kx_{n+1}^k+x_n^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 2$}, \\[1ex] 6a^2x_nx_{n+1}x_{n+3} & \mbox{\rm for $k=3$}, \\[1ex] 12a^2x_nx_{n+1}x_{n+3}x_{n+4} & \mbox{\rm for $k=4$}, \\[1ex] 10a^2x_nx_{n+1}x_{n+3}\times&\\ \hspace*{5mm}\times[2x_{n+4}^2-ax_nx_{n+1}-ax_{n+2}x_{n+3}] & \mbox{\rm for $k=5$}, \\[1ex] 30a^2x_nx_{n+1}x_{n+3}x_{n+4}\times&\\ \hspace*{5mm}\times[a^2x_{n+3}^2+x_{n+2}x_{n+4}-ax_nx_{n+1}] & \mbox{\rm for $k=6$}; \end{array}\right. \end{align*} \item[i)] for $x=y_n$, $y=ay_{n+1}$, $z=ay_{n+3}$: \begin{align*} y_{n+4}^k&+a^ky_{n+3}^k-[(a^2+1)^k+1]y_{n+2}^k+\\ &+a^ky_{n+1}^k+y_n^k- a^k(a^2+4)^kx_{n+2}^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 2$}, \\[1ex] 6a^2y_ny_{n+1}y_{n+3} & \mbox{\rm for $k=3$}, \\[1ex] 12a^2y_ny_{n+1}y_{n+3}y_{n+4}, & \mbox{\rm for $k=4$}, \\[1ex] 10a^2y_ny_{n+1}y_{n+3}\times&\\ \hspace*{5mm}\times[2y_{n+4}^2-a(y_ny_{n+1}+y_{n+2}y_{n+3})], & \mbox{\rm for $k=5$}, \\[1ex] 30a^2y_ny_{n+1}y_{n+3}y_{n+4}\times&\\ \hspace*{5mm}\times[y_{n+4}^2-a(y_ny_{n+1}+y_{n+2}y_{n+3})], & \mbox{\rm for $k=6$}, \\[1ex] 56a^2y_ny_{n+1}y_{n+3}y_{n+4}\times&\\ \hspace*{5mm}\times[(y_{n+4}^2-a(y_ny_{n+1}+y_{n+2}y_{n+3}))^2- & \\ \hspace{1cm} -\frac{1}{2}a^2y_ny_{n+1}y_{n+3}y_{n+4}] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[j)] for $x=x_{n-1}$, $y=x_{n+1}$, $z=y_{n+2}$: \begin{align*} y_{n+2}^k&+[(4+a^2)^k-(3+a^2)^k+1]x_{n+1}^k+x_{n-1}^k-y_n^k-(x_{n+1}+y_{n+2})^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 2$}, \\[1ex] 6x_{n-1}x_{n+1}y_{n+2} & \mbox{\rm for $k=3$}, \\[1ex] 12(2+2b+a^2)x_{n-1}x_{n+1}^2y_{n+2} & \mbox{\rm for $k=4$}, \\[1ex] 10x_{n-1}x_{n+1}y_{n+2}\times&\\ \hspace*{5mm}\times[(2(4+a^2)^2-(3+a^2)^2)x_{n+1}^2-x_{n-1}y_{n+2}] & \mbox{\rm for $k=5$}, \\[1ex] 30(4+a^2)x_{n-1}x_{n+1}^2y_{n+2}\times&\\ \hspace*{5mm}\times[(2a^2+7)x_{n+1}^2-x_{n-1}y_{n+2}] & \mbox{\rm for $k=6$}, \\[1ex] 56(a^2+4)x_{n-1}x_{n+1}^2y_{n+2}\times&\\ \hspace*{5mm}\times[((2a^2+7)x_{n+1}^2-x_{n-1}y_{n+2})^2- & \\ \hspace{1cm} -\frac{1}{2}(4+a^2)x_{n-1}x_{n+1}^2y_{n+2}] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[k)] for $x=bx_n$, $y=y_{n-1}$, $z=y_{n+1}$, $b\in\mathbb{C}$: (then $ bx_n+y_{n+1}=(b+1)x_n+x_{n+2} \ \mbox{\rm and}\ bx_n+y_{n-1}= (b+1)x_n+x_{n-2} $) we obtain \begin{align*} [(4&+b+a^2)^k-(4+a^2)^k+b^k]x_n^k+y_{n-1}^k+y_{n+1}^k-{}\\ &-(bx_n+y_{n+1})^k- (bx_n+y_{n-1})^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 2$}, \\[1ex] 6bx_ny_{n-1}y_{n+1} & \mbox{\rm for $k=3$}, \\[1ex] 12b(4+b+a^2)x_n^2y_{n-1}y_{n+1} & \mbox{\rm for $k=4$}, \\[1ex] 10bx_ny_{n-1}y_{n+1}[(2(4+b+a^2)^2-&\\ \hspace*{5mm}-b(4+a^2))x_n^2-y_{n-1}y_{n+1}] & \mbox{\rm for $k=5$}, \\[1ex] 30b(4+b+a^2)x_n^2y_{n-1}y_{n+1}\times & \\ \hspace{5mm} \times [((4+b+a^2)^2-b(4+a^2))x_n^2-y_{n-1}y_{n+1}] & \mbox{\rm for $k=6$}, \\[1ex] 56b(4+b+a^2)x_n^2y_{n-1}y_{n+1}\times & \\ \hspace{5mm} \times [(\{ (4+b+a^2)^2-b(4+a^2)\} x_n^2- & \\ \hspace{5mm} -y_{n-1}y_{n+1})^2-\frac{1}{2}b(4+b+a^2)x_n^2y_{n-1}y_{n+1}] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[l)] for $x=w_n$, $y=bw_{n+2}$, $z=w_{n+4}$, where $b\in\mathbb{C}$ and $w\in\{ x,y\}$: \begin{align*} &w_{n+4}^k+[(a^2+2+b)^k-(a^2+2)^k+b^k]w_{n+2}^k+w_n^k- \\ &\hspace*{5mm}-(bw_{n+2}+w_n)^k-((a^2+2+b)w_{n+2}-w_n)^k= \\ &=\left\{\begin{array}{ll} 0, & \mbox{\rm for $k\leqslant 2$}, \\[1ex] 6bw_nw_{n+2}w_{n+4}, & \mbox{\rm for $k=3$}, \\[1ex] 12(a^2+2+b)bw_nw_{n+2}^2w_{n+4}, & \mbox{\rm for $k=4$}, \\[1ex] 10bw_nw_{n+2}w_{n+4}[(2(a^2+2+b)^2-&\\ \hspace*{2mm}-b(a^2+2))w_{n+2}^2-w_nw_{n+4}], & \mbox{\rm for $k=5$}, \\[1ex] 30b(a^2+2+b)w_nw_{n+2}^2w_{n+4}\times & \\ \hspace{2mm} \times [((a^2+2+b)^2-b(a^2+2))w_{n+2}^2-w_nw_{n+4}], & \mbox{\rm for $k=6$}, \\[1ex] 56b(a^2+2+b)w_nw_{n+2}^2w_{n+4}\times & \\ \hspace{2mm} \times [(((a^2+2+b)^2-b(a^2+2))w_{n+2}^2-w_nw_{n+4})^2- & \\ \hspace{2mm} -\frac{1}{2}b(a^2+2+b)w_nw_{n+2}^2w_{n+4}] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[m)] for $x=(a^2+3)x_{n+5}$, $y=-y_{n+2}$, $z=-x_{n-1}$: \begin{align*} (a^2&+3)^k[x_{n+5}^k+a^ky_{n+3}^k-(-1)^kx_{n+1}^k]+(-y_{n+2})^k+(-x_{n-1})^k- \\ &-((a^2+3)x_{n+5}-y_{n+2})^k-((a^2+3)x_{n+5}-x_{n-1})^k= \\ &=\left\{\begin{array}{ll} 0, & \mbox{\rm for $k\leqslant 2$}, \\[1ex] 6(a^2+3)x_{n-1}y_{n+2}x_{n+5}, & \mbox{\rm for $k=3$}, \\[1ex] 12a(a^2+3)^2x_{n-1}y_{n+2}y_{n+3}x_{n+5}, & \mbox{\rm for $k=4$}, \\[1ex] 10(a^2+3)x_{n-1}y_{n+2}x_{n+5}\times & \\ \hspace{1cm} \times [2a^2(a^2+3)^2y_{n+3}^2-x_{n-1}y_{n+2}+ & \\ \hspace{1.5cm} +(a^2+3)^2x_{n+1}x_{n+5}], & \mbox{\rm for $k=5$}, \\[1ex] 30a(a^2+3)^2x_{n-1}y_{n+2}y_{n+3}x_{n+5}\times & \\ \hspace{1cm} \times [a^2(a^2+3)^2y_{n+3}^2-x_{n-1}y_{n+2}+ & \\ \hspace{1.5cm} +(a^2+3)^2x_{n+1}x_{n+5}], & \mbox{\rm for $k=6$}, \\[1ex] 56a(a^2+3)^2x_{n-1}y_{n+2}y_{n+3}x_{n+5}\times & \\ \hspace{1cm} \times [(a^2(a^2+3)^2y_{n+3}^2-x_{n-1}y_{n+2}+ & \\ \hspace{1.5cm} +(a^2+3)^2x_{n+1}x_{n+5})^2- & \\ \hspace{1.5cm} -\frac{1}{2}a(a^2+3)^2x_{n-1}y_{n+2}y_{n+3}x_{n+5}] & \mbox{\rm for $k=8$}. \end{array}\right. \end{align*} \end{itemize} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\subsection*{\normalsize 6.2} Below, the identities generated by (\ref{aa7}) and the elements of sequences $\{ x_n\}$ and $\{ y_n\}$ satisfying relations (\ref{adgj1})--(\ref{lem1.7-w3}) will be given: \begin{itemize} \item[a)] for $x=x_n,\quad y=x_{n+2},\quad z=x_{n+2},\quad u=x_{n+4}$ \begin{align*} \big[ (a^2&+2b+2)^k-2(a^2+2b+1)^k+(a^2+2b)^k+2^k\big] x_{n+2}^k+ \\ &+2y_{n+1}^k+2y_{n+3}^k-(y_{n+1}+x_{n+2})^k-(y_{n+3}+x_{n+2})^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 3$}, \\[1ex] 24x_nx_{n+2}^2x_{n+4} & \mbox{\rm for $k=4$}, \\[1ex] 60(a^2+2b+2)x_nx_{n+2}^3x_{n+4} & \mbox{\rm for $k=5$}, \\[1ex] 60x_nx_{n+2}^2x_{n+4}\big[ 2\big( (a^2+2b+2)^2-& \\ \hspace{5mm} -(a^2+2b+\frac{1}{2})\big) x_{n+2}^2-x_nx_{n+4}\big] & \mbox{\rm for $k=6$}, \\[1ex] 210(a^2+2b+2)x_nx_{n+2}^3x_{n+4}\big[\big( (a^2+2b+2)^2- & \\ \hspace{5mm} -2(a^2+2b+\frac{1}{2})\big) x_{n+2}^2-x_nx_{n+4}\big] & \mbox{\rm for $k=7$}; \end{array}\right. \end{align*} \item[b)] for $x=w_n,\quad y=w_{n+2},\quad z=w_{n+4},\quad u=w_{n+6}$, $w\in \{x,y\}$ respectively: \begin{align*} (&a^2+4)^kx_{n+5}^k+(a^2+4)^k((a^2+2)^k+1)x_{n+3}^k+(a^2+4)^kx_{n+1}^k- y_{n+6}^k-\\ &-y_n^k +((a^2+2)^k-(a^2+3)^k-1)(y_{n+4}^k+y_{n+2}^k)+(y_n+y_{n+6})^k- \\ &-[(a^2+2)(a^2+4)x_{n+3}-y_{n+2}]^k-[(a^2+2)(a^2+4)x_{n+3}-y_{n+4}]^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 3$}, \\[1ex] 24y_ny_{n+2}y_{n+4}y_{n+6} & \mbox{\rm for $k=4$}, \\[1ex] 60(a^2+2)(a^2+4)y_ny_{n+2}x_{n+3}y_{n+4}y_{n+6} & \mbox{\rm for $k=5$}, \\[1ex] 60y_ny_{n+2}y_{n+4}y_{n+6}[2(a^2+2)^2(a^2+4)^2x_{n+3}^2- & \\ \hspace{5mm} -(a^2+3)y_{n+2}y_{n+6}-(a^2+4)x_{n+1}y_{n+4}-y_ny_{n+2}] & \mbox{\rm for $k=6$}, \\[1ex] 210(a^2+2)(a^2+4)y_ny_{n+2}x_{n+3}y_{n+4}y_{n+6}\times & \\ \hspace{5mm} \times[(a^2+2)^2(a^2+4)^2x_{n+3}^3-(a^2+3)y_{n+2}y_{n+6} & \\ \hspace{5mm} -(a^2+4)x_{n+1}y_{n+4}-y_ny_{n+2}] & \mbox{\rm for $k=7$}, \end{array}\right. \end{align*} \begin{align*} y&_{n+5}^k+((a^2+2)^k+1)y_{n+3}^k+y_{n+1}^k-x_{n+6}^k-x_n^k+ \\ &+((a^2+2)^k-(a^2+3)^k-1)(x_{n+4}^k+x_{n+2}^k)+(x_n+x_{n+6})^k- \\ &-((a^2+2)y_{n+3}-x_{n+4})^k-((a^2+2)y_{n+3}-x_{n+2})^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k\leqslant 3$}, \\[1ex] 24x_nx_{n+2}x_{n+4}x_{n+6} & \mbox{\rm for $k=4$}, \\[1ex] 60(a^2+2)x_nx_{n+2}x_{n+4}x_{n+6}y_{n+3} & \mbox{\rm for $k=5$}, \\[1ex] 60x_nx_{n+2}x_{n+4}x_{n+6}[2(a^2+2)^2y_{n+3}^2 & \\ \hspace{5mm} -(a^2+3)x_{n+2}x_{n+6}-y_{n+1}x_{n+4}-x_nx_{n+2}] & \mbox{\rm for $k=6$}, \\[1ex] 210(a^2+2)x_nx_{n+2}x_{n+4}x_{n+6}y_{n+3}\times & \\ \hspace{5mm} \times [(a^2+2)^2y_{n+3}^2-(a^2+3)x_{n+2}x_{n+6} & \\ \hspace{5mm} -y_{n+1}x_{n+4}-x_nx_{n+2}] & \mbox{\rm for $k=7$}. \end{array}\right. \end{align*} \end{itemize} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection*{\normalsize 6.2 Identities for Powers of Fibonacci and Lucas Numbers} Applications of identity (\ref{2.1}) to generate some identities for Fibonacci and Lucas numbers will now be presented. \begin{itemize} \item[a)] For $x=F_n$, $y=F_{n+1}$, $z=F_{n+2}$: \begin{align*} 2^k&F_{n+2}^k-L_{n+1}^k-F_{n+3}^k+F_n^k+F_{n+1}^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2$}, \\[1ex] 6F_nF_{n+1}F_{n+2}, & \mbox{\rm for $k=3$}, \\[1ex] 24F_nF_{n+1}F_{n+2}^2, &\mbox{\rm for $k=4$}, \\[1ex] 10F_nF_{n+1}F_{n+2}[7F_{n+2}^2-F_nF_{n+1}], & \mbox{\rm for $k=5$}, \\[1ex] 60F_nF_{n+1}F_{n+2}^2[3F_{n+2}^2-F_nF_{n+1}], & \mbox{\rm for $k=6$}, \\[1ex] 14F_nF_{n+1}F_{n+2}[(6F_{n+2}^2-F_nF_{n+1})^2- & \\ \hspace*{5mm}-5F_{n+2}^2(F_{n+2}^2+F_nF_{n+1})], & \mbox{\rm for $k=7$}, \\[1ex] 112F_nF_{n+1}F_{n+2}^2[(3F_{n+2}^2-F_nF_{n+1})^2-&\\ \hspace*{5mm}-F_nF_{n+1}F_{n+2}^2], & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[b)] for $x=F_n$, $y=F_{n+1}$, $z=F_{n+3}$: \begin{align*} F_{n+4}^k&-(2^k+1)F_{n+2}^k-L_{n+2}^k+F_n^k+F_{n+1}^k+F_{n+3}^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2$}, \\[1ex] 6F_nF_{n+1}F_{n+3}, & \mbox{\rm for $k=3$}, \\[1ex] 12F_nF_{n+1}F_{n+3}F_{n+4}, & \mbox{\rm for $k=4$}, \\[1ex] 10F_nF_{n+1}F_{n+3}\times&\\ \hspace*{5mm}\times[2F_{n+4}^2-F_{n+2}F_{n+3}-F_nF_{n+1}], & \mbox{\rm for $k=5$}, \\[1ex] 30F_nF_{n+1}F_{n+3}F_{n+4}\times&\\ \hspace*{5mm}\times[F_{n+4}^2-F_{n+2}F_{n+3}-F_nF_{n+1}], & \mbox{\rm for $k=6$}, \\[1ex] 56F_nF_{n+1}F_{n+3}F_{n+4}\times&\\ \hspace*{5mm}\times[(F_{n+4}^2-F_nF_{n+1}-F_{n+2}F_{n+3})^2- & \\ \hspace{5mm} -\frac{1}{2}F_nF_{n+1}F_{n+3}F_{n+4}], & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[c)] for $x=F_n$, $y=F_{n+1}$, $z=F_{n+4}$: \begin{align*} &L_{n+3}^k+F_{n+4}^k-2^kF_{n+3}^k-(1+3^k)F_{n+2}^k+F_{n+1}^k+F_n^k= \\ &=\left\{\!\!\begin{array}{ll} 0 & \mbox{\rm for $k=1,2$}, \\[1ex] 6F_nF_{n+1}F_{n+4}, & \mbox{\rm for $k=3$}, \\[1ex] 12F_nF_{n+1}L_{n+3}F_{n+4}, & \mbox{\rm for $k=4$}, \\[1ex] 10F_nF_{n+1}F_{n+4}[2L_{n+3}^2-F_nF_{n+1}-F_{n+2}F_{n+4}] & \\ \hspace{2mm} =2F_nF_{n+1}F_{n+4}[9L_{2n+6}-L_{2n+1}-16(-1)^n], & \mbox{\rm for $k=5$}, \\[1ex] 30F_nF_{n+1}L_{n+3}F_{n+4}[L_{n+3}^2-F_nF_{n+1}-F_{n+2}F_{n+4}] & \\ \hspace{2mm} =6F_nF_{n+1}L_{n+3}F_{n+4}[4L_{2n+6}-L_{2n+1}-6(-1)^n], & \mbox{\rm for $k=6$}, \\[1ex] 56F_nF_{n+1}L_{n+3}F_{n+4}[(L_{n+3}^2-F_nF_{n+1}-F_{n+2}F_{n+4})^2 & \\ \hspace{5mm} -\frac{1}{2}F_nF_{n+1}L_{n+3}F_{n+4}] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[d)] for $x=F_n$, $y=F_{n+2}$, $z=F_{n+4}$: \begin{align*} F_{n+4}^k&-L_{n+3}^k+(4^k-3^k+1)F_{n+2}^k-L_{n+1}^k+F_n^k =\\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2$}, \\[1ex] 6F_nF_{n+2}F_{n+4}, & \mbox{\rm for $k=3$}, \\[1ex] 48F_nF_{n+2}^2F_{n+4}, & \mbox{\rm for $k=4$}, \\[1ex] 10F_nF_{n+2}F_{n+4}[29F_{n+2}^2-F_nF_{n+4}] & \\ \hspace{2mm} =2F_nF_{n+2}F_{n+4}[28L_{2n+4}-51(-1)^n] & \\ \hspace{2mm} =10F_nF_{n+2}F_{n+4}[28F_{n+2}^2+(-1)^n], & \mbox{\rm for $k=5$}, \\[1ex] 120F_nF_{n+2}^2F_{n+4}[13F_{n+2}^2-F_nF_{n+4}] & \\ \hspace{2mm} =24F_nF_{n+2}^2F_{n+4}[12L_{2n+4}-19(-1)^n] & \\ \hspace{2mm} =120F_nF_{n+2}^2F_{n+4}[12F_{n+2}^2+(-1)^n], & \mbox{\rm for $k=6$}, \\[1ex] 224F_nF_{n+2}^2F_{n+4}[(13F_{n+2}^2-F_nF_{n+4})^2-&\\ \hspace*{5mm}-2F_nF_{n+2}^2F_{n+4}] & \mbox{\rm for $k=8$}; \end{array}\right. \end{align*} \item[e)] for $x=2F_n$, $y=F_{n+2}$, $z=F_{n+5}$: \begin{align*} &F_{n+5}^k-2^kF_{n+4}^k+(6^k-5^k+1)F_{n+2}^k-(F_{n+1}+3F_n)^k+2^kF_n^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2$}, \\[1ex] 12F_nF_{n+2}F_{n+5}, & \mbox{\rm for $k=3$}, \\[1ex] 144F_nF_{n+2}^2F_{n+5}, & \mbox{\rm for $k=4$}, \\[1ex] 20F_nF_{n+2}F_{n+5}[72F_{n+2}^2-4F_nF_{n+4}-F_{n+2}F_{n+5}] & \\ \hspace{2mm} =4F_nF_{n+2}F_{n+5}[71L_{2n+4}-10F_{2n+5}-112(-1)^n], & \mbox{\rm for $k=5$}, \\[1ex] 360F_nF_{n+2}^2F_{n+5}[36F_{n+2}^2-4F_nF_{n+4}-F_{n+2}F_{n+5}] & \\ \hspace{2mm} =360F_nF_{n+2}^2F_{n+5}[7L_{2n+4}-2F_{2n+5}-8(-1)^n], & \mbox{\rm for $k=6$}; \end{array}\right. \end{align*} \item[f)] for $x=F_n$, $y=F_{n+3}$, $z=F_{n+4}$: \begin{align*} (&F_n+F_{n+5})^k-F_{n+5}^k+F_{n+4}^k+F_{n+3}^k-(2^k+3^k)F_{n+2}^k+F_n^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2$}, \\[1ex] 6F_nF_{n+3}F_{n+4}, & \mbox{\rm for $k=3$}, \\[1ex] 12F_nF_{n+3}F_{n+4}(F_n+F_{n+5}), & \mbox{\rm for $k=4$}, \\[1ex] 10F_nF_{n+3}F_{n+4}[2(F_n+F_{n+5})^2-F_{2n+6}+2(-1)^n], & \mbox{\rm for $k=5$}, \\[1ex] 30F_nF_{n+3}F_{n+4}(F_n+F_{n+5})\times&\\ \hspace*{5mm}\times[(F_n+F_{n+5})^2-F_{2n+6}+2(-1)^n], & \mbox{\rm for $k=6$}; \end{array}\right. \end{align*} \item[g)] for $x=2F_n$, $y=F_{n+3}$, $z=F_{n+4}$: \begin{align*} F_{n+4}^k&+F_{n+3}^k+(5F_{n+2})^k+(2F_n)^k-F_{n+5}^k-{}\\ &-(F_n+2F_{n+2})^k- (F_n+3F_{n+2})^k =\\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2$}, \\[1ex] 12F_nF_{n+3}F_{n+4}, & \mbox{\rm for $k=3$}, \\[1ex] 120F_nF_{n+2}F_{n+3}F_{n+4}, & \mbox{\rm for $k=4$}, \\[1ex] 20F_nF_{n+3}F_{n+4}\times&\\ \hspace*{5mm}\times[F_{n+2}^{2}-2F_{n}F_{n+5}-F_{n+3}F_{n+4}]\\ \hspace{2mm} =4F_nF_{n+3}F_{n+4}\times&\\ \hspace*{5mm}\times[45L_{2n+4}-4L_{2n+3}-79(-1)^{n}], & \mbox{\rm for $k=5$}, \\[1ex] 300F_nF_{n+2}F_{n+3}F_{n+4}\times&\\ \hspace*{5mm}\times[25F_{n+2}^{2}-2F_{n}F_{n+5}-F_{n+3}F_{n+4}]\\ \hspace*{2mm}=60F_nF_{n+2}F_{n+3}F_{n+4}\times&\\ \hspace*{5mm}\times[20L_{2n+4}-4L_{2n+3}-29(-1)^{n}], & \mbox{\rm for $k=6$}; \end{array}\right. \end{align*} \item[h)] for $x=L_n$, $y=L_{n+1}$, $z=L_{n+2}$: \begin{align*} 2^kL_{n+2}^k&-5^k\cdot F_{n+1}^k-L_{n+3}^k+L_n^k+L_{n+1}^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2$}, \\[1ex] 6L_nL_{n+1}L_{n+2}, & \mbox{\rm for $k=3$}, \\[1ex] 24L_nL_{n+1}L_{n+2}^2, & \mbox{\rm for $k=4$}, \\[1ex] 10L_nL_{n+1}L_{n+2}[7L_{n+2}^2-L_nL_{n+1}], & \mbox{\rm for $k=5$}, \\[1ex] 60L_nL_{n+1}L_{n+2}^2[3L_{n+2}^2-L_nL_{n+1}], & \mbox{\rm for $k=6$}; \end{array}\right. \end{align*} \item[i)] for $x=L_n$, $y=L_{n+1}$, $z=L_{n+3}$: \begin{align*} L_{n+4}^k&-5^k\cdot F_{n+2}^k-(2^k+1)L_{n+2}^k+L_n^k+L_{n+1}^k+L_{n+3}^k= \\ &=\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2$}, \\[1ex] 6L_nL_{n+1}L_{n+3}, & \mbox{\rm for $k=3$}, \\[1ex] 12L_nL_{n+1}L_{n+3}L_{n+4}, & \mbox{\rm for $k=4$}, \\[1ex] 10L_nL_{n+1}L_{n+3}\times&\\ \hspace*{5mm}\times[2L_{n+4}^2-L_{n+2}L_{n+3}-L_nL_{n+1}], & \mbox{\rm for $k=5$}, \\[1ex] 30L_nL_{n+1}L_{n+3}L_{n+4}\times&\\ \hspace*{5mm}\times[L_{n+4}^2-L_{n+2}L_{n+3}-L_nL_{n+1}], & \mbox{\rm for $k=6$}. \end{array}\right. \end{align*} \end{itemize} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\subsection*{} In the next step, identity~(\ref{aa7}) shall be applied to generate some selected identities for Fibonacci and Lucas numbers. \begin{itemize} \item[a)] For $x=F_n$, $y=F_{n+1}$, $z=F_{n+3}$, $u=F_{n+5}$ we obtain the following identities: \begin{multline} F_{n+6}^k-F_{n+5}^k-(2^k+1)F_{n+4}^k-(4^k-3^k+1)F_{n+3}^k+(2^k+1)F_{n+2}^k- \\ -F_{n+1}^k-F_n^k+L_{n+2}^k+L_{n+4}^k+(F_n+F_{n+5})^k-(F_n+L_{n+4})^k= \\ =\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2,3$}, \\[1ex] 24F_nF_{n+1}F_{n+3}F_{n+5}, & \mbox{\rm for $k=4$,} \\[1ex] 60F_nF_{n+1}F_{n+3}F_{n+5}F_{n+6}, & \mbox{\rm for $k=5$,} \\[1ex] 60F_nF_{n+1}F_{n+3}F_{n+5}(2F_{n+6}^2-F_{n+5}F_{n+4}- & \\ \hspace{5mm} -F_{n+3}F_{n+2}-F_{n+1}F_n), & \mbox{\rm for $k=6$,} \\[1ex] 210F_nF_{n+1}F_{n+3}F_{n+5}F_{n+6}(F_{n+6}^2- & \\ \hspace{5mm} -F_{n+5}F_{n+4}-F_{n+3}F_{n+2}-F_{n+1}F_n), & \mbox{\rm for $k=7$,} \end{array}\right. \label{41} \end{multline} where the following identity was utilized: $F_{n+1}+F_{n+5}=3F_{n+3}$ and $L_{n+1}+L_{n+5}=3L_{n+3}$; \item[b)] for $x=L_n$, $y=L_{n+1}$, $z=L_{n+3}$, $u=L_{n+5}$: \begin{multline} L_{n+6}^k-L_{n+5}^k-(2^k+1)L_{n+4}^k-(4^k-3^k+1)L_{n+3}^k+ (2^k+1)L_{n+2}^k- \\ -L_{n+1}^k-L_n^k+ 5^kF_{n+2}^k+5^kF_{n+4}^k+(L_n+L_{n+5})^k-(L_n+5F_{n+4})^k= \\ =\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2,3$}, \\[1ex] 24L_nL_{n+1}L_{n+3}L_{n+5} & \mbox{\rm for $k=4$,} \\[1ex] 60L_nL_{n+1}L_{n+3}L_{n+5}L_{n+6} & \mbox{\rm for $k=5$,} \\[1ex] 60L_nL_{n+1}L_{n+3}L_{n+5}[2L_{n+6}^2-L_{n+5}L_{n+4}- & \\ \hspace{5mm} -L_{n+3}L_{n+2}-L_{n+1}L_n], & \mbox{\rm for $k=6$;} \\ \end{array}\right.\label{42} \end{multline} \item[c)] for $x=F_n$, $y=F_{n+2}$, $z=F_{n+4}$, $u=F_{n+6}$: \begin{multline} L_{n+5}^k+(3^k+1)L_{n+3}^k+L_{n+1}^k-F_{n+6}^k-F_n^k+ (3^k-4^k-1)(F_{n+4}^k+F_{n+2}^k)+ \\ +(F_n+F_{n+6})^k-(2F_{n+4}+3F_{n+2})^k-(3F_{n+4}+2F_{n+2})^k= \\ =\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2,3$}, \\[1ex] 24F_nF_{n+2}F_{n+4}F_{n+6} & \mbox{\rm for $k=4$}, \\[1ex] 180F_nF_{n+2}F_{n+4}F_{n+6}L_{n+3} & \mbox{\rm for $k=5$}, \\[1ex] 60F_nF_{n+2}F_{n+4}F_{n+6}[18L_{n+3}^2- & \\ \hspace{5mm} -4F_{n+2}F_{n+6}-L_{n+1}F_{n+4}-F_nF_{n+2}] & \mbox{\rm for $k=6$}, \\[1ex] 630F_nF_{n+2}F_{n+4}F_{n+6}L_{n+3}[9L_{n+3}^2- & \\ \hspace{5mm} -4F_{n+2}F_{n+6}-L_{n+1}F_{n+4}-F_nF_{n+2}] & \mbox{\rm for $k=7$}; \end{array}\right. \end{multline} \item[d)] for $x=L_n$, $y=L_{n+2}$, $z=L_{n+4}$, $u=L_{n+6}$: \begin{multline}\label{last} (5F_{n+5})^k+(15^k+5^k)F_{n+3}^k+(5F_{n+1})^k-L_{n+6}^k -(4^k-3^k+1)L_{n+4}^k- \\ -(4^k-3^k+1)L_{n+2}^k-L_{n}^{k}+(6L_{n+2}+2L_{n+1})^k- \\ -(4L_{n+2}+L_{n+5})^k-(4L_{n+4}-L_{n+1})^k= \\ =\left\{\begin{array}{ll} 0 & \mbox{\rm for $k=1,2,3$}, \\[1ex] 24L_nL_{n+2}L_{n+4}L_{n+6} & \mbox{\rm for $k=4$}, \\[1ex] 900L_nL_{n+2}L_{n+4}L_{n+6}F_{n+3} & \mbox{\rm for $k=5$}, \\[1ex] 60F_nF_{n+2}F_{n+4}F_{n+6}[18L_{n+3}^2- & \\ \hspace{5mm} -4F_{n+2}F_{n+6}-L_{n+1}F_{n+4}-F_nF_{n+2}] & \mbox{\rm for $k=6$}, \\[1ex] 630F_nF_{n+2}F_{n+4}F_{n+6}L_{n+3}[9L_{n+3}^2- & \\ \hspace{5mm} -4F_{n+2}F_{n+6}-L_{n+1}F_{n+4}-F_nF_{n+2}] & \mbox{\rm for $k=7$}. \end{array}\right. \end{multline} \end{itemize} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{thebibliography}{99} \footnotesize \bibitem{Benoumhani2003a} M.~Benoumhani, \textit{A~Sequence of Binomial Coefficients Related To Lucas and Fibonacci Numbers}, J.~Int. Sequences \textbf{6} (2003), 1--10. \bibitem{Catalani2004a} M.~Catalani, \textit{Identities for Fibonacci and Lucas Polynomials Derived from a~Book of Gould}, arXiv:math.CO/0407105 (2004), 1--7. \bibitem{Constandache2002a} A.~Constandache, A.~Das, F.~Toppan, \textit{Lucas Polynomials and a~Standard Lax Representation for he Polytropic Gas Dynamics}, Letters in Math. Phys. \textbf{60} (2002), 197--209. \bibitem{Filipponi1998a} P.~Filipponi, A.~F.~Horadam, \textit{First Derivative Sequences of Extended Fibonacci and Lucas Polynomials}, in: G.~E.~Bergum et al. (eds.), Applications of Fibonacci Numbers \textbf{7} (1998), 115--128. \bibitem{grz} R.~Grzymkowski, R.~Witu{\l}a, \textit{Calculus Methods in Algebra, Part One}, WPKJS, Gliwice 2000 (in Polish). \bibitem{Horadam2002a} A.~F.~Horadam, \textit{Vieta Polynomials}, Fibonacci Quarterly \textbf{40} (2002), 223--232. \bibitem{kos} T.~Koshy, \textit{Fibonacci and Lucas Numbers with Applications}, Wiley, New York 2001. \bibitem{Melham1999a} R.~S.~Melham, \textit{Sums of Certain Products of Fibonacci and Lucas Numbers~-- Part~I}, Fibonacci Quaeterly \textbf{37} (1999), 248--251. \bibitem{Melham1999b} R.~S.~Melham, \textit{Families of Identities Involving Sums of Powers of the Fibonacci and Lucas Numbers}, Fibonacci Quarterly \textbf{37} (1999), 315--319. \bibitem{Melham2000a} R.~S.~Melham, \textit{Sums of Certain Products of Fibonacci and Lucas Numbers~-- Part~II}, Fibonacci Quarterly \textbf{38} (2000), 3--7. \bibitem{Melham2000b} R.~S.~Melham, \textit{Alternating Sums of Fourth Powers of Fibonacci and Lucas Numbers}, Fibonacci Quarterly \textbf{38} (2000), 254--259. \bibitem{mod} P.~S.~Modenov, \textit{Special Problems of Elementary Mathematics}, Sovietskaja Nauka, Moscow 1957 (in Russian). \bibitem{pas} S.~Paszkowski, \textit{Numerical Applications of Chebyshev Polynomials and Series}, PWN, Warsaw 1975 (in Polish). \bibitem{riv} T.~Rivlin, \textit{Chebyshef Polynomials from Approximation Theory to Algebra and Number Theory}, Wiley, New York 1990. \bibitem{Robbins1991a} N.~Robbins, \textit{Vieta's Triangular Array and a~Related Family of Polynomials}, Int. J.~Math.~\& Math. Sci. \textbf{14} (1991), 239--244. \bibitem{wsJMAA} R.~Witu{\l}a, D.~S{\l}ota, \textit{On Modified Chebyshev Polynomials}, J.~Math. Anal. Appl. \textbf{324} (2006), 321--343. \bibitem{wsCEJM} R.~Witu{\l}a, D.~S{\l}ota, \textit{Cauchy, Ferrers-Jackson and Chebyshev Polynomials and identities for the powers of elements of some conjugate recurrence sequences}, Central Eur. J.~Math. \textbf{4} (2006), 531--546. \end{thebibliography} \end{document}