Theory and Modern Applications

# Vieta-Pell and Vieta-Pell-Lucas polynomials

## Abstract

In the present paper, we introduce the recurrence relation of Vieta-Pell and Vieta-Pell-Lucas polynomials. We obtain the Binet form and generating functions of Vieta-Pell and Vieta-Pell-Lucas polynomials and define their associated sequences. Moreover, we present some differentiation rules and finite summation formulas.

MSC:11C08, 11B39.

## 1 Introduction

Andre-Jeannin [1] introduced a class of polynomials ${U}_{n}\left(p,q;x\right)$ defined by

${U}_{n}\left(p,q;x\right)=\left(x+p\right){U}_{n-1}\left(p,q;x\right)-q{U}_{n-2}\left(p,q;x\right),\phantom{\rule{1em}{0ex}}n\ge 2,$

with the initial values ${U}_{0}\left(p,q;x\right)=0$ and ${U}_{1}\left(p,q;x\right)=1$.

Vieta-Lucas polynomials were studied as Vieta polynomials by Robbins [2]. Vieta-Fibonacci and Vieta-Lucas polynomials are defined by

$\begin{array}{c}{V}_{n}\left(x\right)=x{V}_{n-1}\left(x\right)-{V}_{n-2}\left(x\right),\phantom{\rule{1em}{0ex}}n\ge 2,\hfill \\ {v}_{n}\left(x\right)=x{v}_{n-1}\left(x\right)-{v}_{n-2}\left(x\right),\phantom{\rule{1em}{0ex}}n\ge 2,\hfill \end{array}$

respectively, where ${V}_{0}\left(x\right)=0$, ${V}_{1}\left(x\right)=1$ and ${v}_{0}\left(x\right)=2$, ${v}_{1}\left(x\right)=x$ [3]. The recursive properties of Vieta-Fibonacci and Vieta-Lucas polynomials were given by Horadam [3].

For $p=0$ and $q=1$, Vieta-Fibonacci polynomials are a special case of the polynomials ${U}_{n}\left(p,q;x\right)$ in [1]. Further, ${U}_{n,m}\left(p,q;x\right)$ in [4] for $p=0$, $q=1$, $m=2$ gives Vieta-Fibonacci polynomials.

Chebyshev polynomials are a sequence of orthogonal polynomials which can be defined recursively. Recall that the n th Chebyshev polynomials of the first kind and second kind are denoted by ${T}_{n}\left(x\right)$ and ${U}_{n}\left(x\right)$, respectively.

It is well known that the Chebyshev polynomials of the first kind and second kind are closely related to Vieta-Fibonacci and Vieta-Lucas polynomials. So, in [5] Vitula and Slota redefined Vieta polynomials as modified Chebyshev polynomials. The related features of Vieta and Chebyshev polynomials are given as

$\begin{array}{c}{V}_{n}\left(x\right)={U}_{n}\left(\frac{1}{2}x\right)\phantom{\rule{1em}{0ex}}\text{[3]},\hfill \\ {v}_{n}\left(x\right)=2{T}_{n}\left(\frac{1}{2}x\right)\phantom{\rule{1em}{0ex}}\left(\text{see [2, 6]}\right).\hfill \end{array}$

For $|x|>1$, we consider ${t}_{n}\left(x\right)$ and ${s}_{n}\left(x\right)$ polynomials by the following recurrence relations:

$\begin{array}{c}{t}_{n}\left(x\right)=2x{t}_{n-1}\left(x\right)-{t}_{n-2}\left(x\right),\phantom{\rule{1em}{0ex}}n\ge 2,\hfill \\ {s}_{n}\left(x\right)=2x{s}_{n-1}\left(x\right)-{s}_{n-2}\left(x\right),\phantom{\rule{1em}{0ex}}n\ge 2,\hfill \end{array}$

where ${t}_{0}\left(x\right)=0$, ${t}_{1}\left(x\right)=1$ and ${s}_{0}\left(x\right)=2$, ${s}_{1}\left(x\right)=2x$. We call ${t}_{n}\left(x\right)$ the n th Vieta-Pell polynomial and ${s}_{n}\left(x\right)$ the n th Vieta-Pell-Lucas polynomial.

The relations below are obvious

$\begin{array}{c}{s}_{n}\left(x\right)=2{T}_{n}\left(x\right),\hfill \\ {t}_{n+1}\left(x\right)={U}_{n}\left(x\right).\hfill \end{array}$

The first few terms of ${t}_{n}\left(x\right)$ and ${s}_{n}\left(x\right)$ are as follows:

$\begin{array}{ll}{t}_{2}\left(x\right)=2x,& {s}_{2}\left(x\right)=4{x}^{2}-2,\\ {t}_{3}\left(x\right)=4{x}^{2}-1,& {s}_{3}\left(x\right)=8{x}^{3}-6x,\\ {t}_{4}\left(x\right)=8{x}^{3}-4x,& {s}_{4}\left(x\right)=16{x}^{4}-16{x}^{2}+2,\\ {t}_{5}\left(x\right)=16{x}^{4}-12{x}^{2}+1,& {s}_{5}\left(x\right)=32{x}^{5}-40{x}^{3}+10x,\\ {t}_{6}\left(x\right)=32{x}^{5}-32{x}^{3}+6x,& {s}_{6}\left(x\right)=64{x}^{6}-96{x}^{4}+36{x}^{2}-2,\\ {t}_{7}\left(x\right)=64{x}^{6}-80{x}^{4}+24{x}^{2}-1,& {s}_{7}\left(x\right)=128{x}^{7}-224{x}^{5}+112{x}^{3}-14x.\end{array}$

The aim of this paper is to determine the recursive key features of Vieta-Pell and Vieta-Pell-Lucas polynomials. In conjunction with these properties, we examine their interrelations and define their associated sequences. Furthermore, we present some differentiation rules and summation formulas.

## 2 Main results

Some fundamental recursive properties of Vieta-Pell and Vieta Pell-Lucas polynomials are given in this section.

### Characteristic equation

Vieta-Pell and Vieta-Pell-Lucas polynomials have the following characteristic equation:

${\lambda }^{2}-2x\lambda +1=0$

with the roots α and β

$\begin{array}{c}\alpha =x+\sqrt{{x}^{2}-1},\hfill \\ \beta =x-\sqrt{{x}^{2}-1.}\hfill \end{array}$

Also, α and β satisfy the following equations:

$\begin{array}{c}\alpha +\beta =2x,\hfill \\ \alpha \beta =1,\hfill \\ \alpha -\beta =\Delta =2\sqrt{{x}^{2}-1}.\hfill \end{array}$
(1)

### Binet form

By appropriate procedure, we can easily find the Binet forms as

${t}_{n}\left(x\right)=\frac{{\alpha }^{n}-{\beta }^{n}}{\alpha -\beta },$
(2)
${s}_{n}\left(x\right)={\alpha }^{n}+{\beta }^{n}.$
(3)

### Generating function

Vieta-Pell and Vieta-Pell-Lucas polynomials can be defined by the following generating functions:

$\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}{t}_{n}\left(x\right){y}^{n}=y{\left(1-2xy+{y}^{2}\right)}^{-1},\hfill \\ \sum _{n=0}^{\mathrm{\infty }}{s}_{n}\left(x\right){y}^{n}=\left(2-2xy\right){\left(1-2xy+{y}^{2}\right)}^{-1}.\hfill \end{array}$

### Negative subscript

We can also extend the definition of ${t}_{n}\left(x\right)$ and ${s}_{n}\left(x\right)$ to the negative index

$\begin{array}{c}{t}_{-n}\left(x\right)=-{t}_{n}\left(x\right),\hfill \\ {s}_{-n}\left(x\right)={s}_{n}\left(x\right).\hfill \end{array}$

### Simson formulas

$\begin{array}{c}{t}_{n+1}\left(x\right){t}_{n-1}\left(x\right)-{t}_{n}^{2}\left(x\right)=-1,\hfill \\ {s}_{n+1}\left(x\right){s}_{n-1}\left(x\right)-{s}_{n}^{2}\left(x\right)=4\left({x}^{2}-1\right).\hfill \end{array}$

We arrange the first ten coefficients of ${t}_{n}\left(x\right)$ in Table 1. Let $T\left(n,j\right)$ denote the element in row n and column j, where $j⩾0$, $n⩾1$. As seen from the Table 1, it is obvious that

$T\left(n,0\right)=2T\left(n-2,0\right)+T\left(n-1,0\right)$

can be written like the coefficients of Pell polynomials in [7]. Moreover,

$\sum _{j=0}^{⌊\frac{n-1}{2}⌋}T\left(n,j\right)=n.$

For example, for $n=8$ we can find

$\sum _{j=0}^{3}T\left(8,j\right)=T\left(8,0\right)+T\left(8,1\right)+T\left(8,2\right)+T\left(8,3\right)=8.$

Let ${P}_{n}$ denote the n th Pell number, so we have

$\sum _{j=1}^{⌊\frac{n-1}{2}⌋}|T\left(n,j\right)|={P}_{n}.$

### 2.1 Interrelations of ${t}_{n}\left(x\right)$ and ${s}_{n}\left(x\right)$

Most of the equations below can be obtained by using the Binet form and convenient routine operations

${t}_{n+1}\left(x\right)-{t}_{n-1}\left(x\right)={s}_{n}\left(x\right)=2x{t}_{n}\left(x\right)-2{t}_{n-1}\left(x\right),$
(4)
$\begin{array}{c}{s}_{n+1}\left(x\right)-{s}_{n-1}\left(x\right)=4\left({x}^{2}-1\right){t}_{n}\left(x\right),\hfill \\ {t}_{n}\left(x\right){s}_{n}\left(x\right)={t}_{2n}\left(x\right),\hfill \\ {s}_{n}^{2}\left(x\right)+4\left({x}^{2}-1\right){t}_{n}^{2}\left(x\right)=2{s}_{2n}\left(x\right),\hfill \\ {s}_{n}^{2}\left(x\right)-4\left({x}^{2}-1\right){t}_{n}^{2}\left(x\right)=4,\hfill \\ {t}_{n+1}^{2}\left(x\right)-{t}_{n}^{2}\left(x\right)={t}_{2n+1}\left(x\right),\hfill \\ {s}_{n+1}^{2}\left(x\right)-{s}_{n}^{2}\left(x\right)=4\left({x}^{2}-1\right){t}_{2n+1}\left(x\right),\hfill \\ {s}_{n+1}^{2}\left(x\right)+{s}_{n}^{2}\left(x\right)=2x{s}_{2n+1}\left(x\right)+4,\hfill \\ {t}_{n+1}^{2}\left(x\right)-{t}_{n-1}^{2}\left(x\right)=2x{t}_{2n}\left(x\right),\hfill \\ {s}_{n}\left(x\right){s}_{n+1}\left(x\right)-4\left({x}^{2}-1\right){t}_{n}\left(x\right){t}_{n+1}\left(x\right)=4x,\hfill \\ {s}_{n}\left(x\right){s}_{n+1}\left(x\right)+4\left({x}^{2}-1\right){t}_{n}\left(x\right){t}_{n+1}\left(x\right)=2{s}_{2n+1}\left(x\right),\hfill \\ {s}_{4n}\left(x\right)-2=4\left({x}^{2}-1\right){t}_{2n}^{2}\left(x\right),\hfill \\ {t}_{n+1}\left(x\right)-x{t}_{n}\left(x\right)=\frac{1}{2}{s}_{n}\left(x\right),\hfill \\ 2{s}_{n+1}\left(x\right)-2x{s}_{n}\left(x\right)=4\left({x}^{2}-1\right){t}_{n}\left(x\right).\hfill \end{array}$
(5)

Proposition 1 ${s}_{n}\left(2{x}^{2}-1\right)-{s}_{n}^{2}\left(x\right)=-2$.

Proof Consider the expression ${s}_{n}\left(2{x}^{2}-1\right)$. Then α, β, Δ are replaced by ${\alpha }^{\ast }$, ${\beta }^{\ast }$, ${\Delta }^{\ast }$, respectively. So, ${\alpha }^{\ast }={\alpha }^{2}$, ${\beta }^{\ast }={\beta }^{2}$, ${\Delta }^{\ast }=2x\Delta$ and by using the Binet form, the proof is completed. □

### 2.2 Associated sequences

Definition 1 The k th associated sequences $\left\{{t}_{n}^{\left(k\right)}\left(x\right)\right\}$ and $\left\{{s}_{n}^{\left(k\right)}\left(x\right)\right\}$ of $\left\{{t}_{n}\left(x\right)\right\}$ and $\left\{{s}_{n}\left(x\right)\right\}$ are defined by, respectively ($k\ge 1$)

${t}_{n}^{\left(k\right)}\left(x\right)={t}_{n+1}^{\left(k-1\right)}\left(x\right)-{t}_{n-1}^{\left(k-1\right)}\left(x\right),$
(6)
${s}_{n}^{\left(k\right)}\left(x\right)={s}_{n+1}^{\left(k-1\right)}\left(x\right)-{s}_{n-1}^{\left(k-1\right)}\left(x\right),$
(7)

where ${t}_{n}^{\left(0\right)}\left(x\right)={t}_{n}\left(x\right)$ and ${s}_{n}^{\left(0\right)}\left(x\right)={s}_{n}\left(x\right)$.

Presently,

$\begin{array}{c}{t}_{n}^{\left(1\right)}\left(x\right)={s}_{n}\left(x\right)\phantom{\rule{1em}{0ex}}\left(\text{by (4)}\right),\hfill \\ {s}_{n}^{\left(1\right)}\left(x\right)={\Delta }^{2}{t}_{n}\left(x\right)\phantom{\rule{1em}{0ex}}\left(\text{by (5)}\right)\hfill \end{array}$

are the members of the first associated sequences $\left\{{t}_{n}^{\left(1\right)}\left(x\right)\right\}$ and $\left\{{s}_{n}^{\left(1\right)}\left(x\right)\right\}$. If (6) and (7) are applied repeatedly, the results emerge

$\begin{array}{c}{t}_{n}^{2j}\left(x\right)={s}_{n}^{\left(2j-1\right)}\left(x\right)={\Delta }^{2j}{t}_{n}\left(x\right),\hfill \\ {t}_{n}^{\left(2j-1\right)}\left(x\right)={s}_{n}^{\left(2j-2\right)}\left(x\right)={\Delta }^{2j-2}{s}_{n}\left(x\right).\hfill \end{array}$

### Some special values of ${t}_{n}\left(x\right)$ and ${s}_{n}\left(x\right)$

$\begin{array}{c}\left\{\begin{array}{l}{t}_{n}\left(-x\right)={\left(-1\right)}^{n+1}{t}_{n}\left(x\right),\\ {t}_{n}\left(1\right)=n,\\ {t}_{n}\left(-1\right)={\left(-1\right)}^{n+1}n,\\ {t}_{2n}\left(0\right)=0,\\ {t}_{2n-1}\left(0\right)={\left(-1\right)}^{n+1},\end{array}\hfill \\ \left\{\begin{array}{l}{s}_{n}\left(1\right)=2,\\ {s}_{n}\left(-1\right)=2{\left(-1\right)}^{n},\\ {s}_{2n}\left(0\right)=2{\left(-1\right)}^{n},\\ {s}_{2n-1}\left(0\right)=0.\end{array}\hfill \end{array}$

### 2.3 Differentiation formulas

$\begin{array}{c}\frac{d{s}_{n}\left(x\right)}{dx}=2n{t}_{n}\left(x\right),\hfill \\ \frac{d{t}_{n}\left(x\right)}{dx}=\frac{n{s}_{n}\left(x\right)-2x{t}_{n}\left(x\right)}{2\left({x}^{2}-1\right)},\hfill \\ \frac{{d}^{2}{s}_{n}\left(x\right)}{d{x}^{2}}=n\left(\frac{n{s}_{n}\left(x\right)-2x{t}_{n}\left(x\right)}{{x}^{2}-1}\right).\hfill \end{array}$

Since the derivation function of ${s}_{n}\left(x\right)$ is a polynomial, all of the derivatives must exist for all real numbers. Thus, we can give the following formulas.

Proposition 2

$\begin{array}{c}\frac{{d}^{2}{s}_{n}\left(x\right)}{d{x}^{2}}{|}_{{}_{x=1}}=\frac{2}{3}\left({n}^{4}-{n}^{2}\right),\hfill \\ \frac{{d}^{2}{s}_{n}\left(x\right)}{d{x}^{2}}{|}_{{}_{x=-1}}=\frac{2}{3}{\left(-1\right)}^{n-1}\left({n}^{2}-{n}^{4}\right).\hfill \end{array}$

Proof If we take the limit on ${s}_{n}^{″}\left(x\right)=n\left(\frac{n{s}_{n}\left(x\right)-2x{t}_{n}\left(x\right)}{{x}^{2}-1}\right)$, we have the numerical value of ${s}_{n}^{″}$ at $x=1$ and $x=-1$.

${s}_{n}^{″}\left(1\right)=\frac{n}{2}\underset{x\to 1}{lim}\frac{n{s}_{n}\left(x\right)-2x{t}_{n}\left(x\right)}{\left(x-1\right)}.$

Since ${s}_{n}\left(1\right)=2$, ${t}_{n}\left(1\right)=n$, apply L’Hôpital’s rule:

$\begin{array}{rcl}{s}_{n}^{″}\left(1\right)& =& \frac{n}{2}\underset{x\to 1}{lim}\frac{\frac{d}{dx}\left(n{s}_{n}\left(x\right)-2x{t}_{n}\left(x\right)\right)}{\frac{d}{dx}\left(x-1\right)}\\ =& \frac{n}{2}\underset{x\to 1}{lim}\frac{d}{dx}\left(n{s}_{n}\left(x\right)-2x{t}_{n}\left(x\right)\right)\\ =& \frac{n}{2}\underset{x\to 1}{lim}\left(2{n}^{2}{t}_{n}\left(x\right)-2{t}_{n}\left(x\right)-2x\frac{d}{dx}{t}_{n}\left(x\right)\right)\\ =& \frac{n}{2}\left(2{n}^{2}{t}_{n}\left(1\right)-2{t}_{n}\left(1\right)-2\underset{x\to 1}{lim}\frac{d}{dx}{t}_{n}\left(x\right)\right)\\ =& \frac{n}{2}\left(2{n}^{3}-2n\right)-\frac{1}{2}\underset{x\to 1}{lim}\frac{d}{dx}\left(2n{t}_{n}\left(x\right)\right)\\ =& {n}^{4}-{n}^{2}-\frac{1}{2}{s}_{n}^{″}\left(1\right).\end{array}$

So, the proof for $x=-1$ is similar. □

### 2.4 Some summation formulas

In this section we deal with the matrix

$\mathbf{V}=\left[\begin{array}{cc}2x& -1\\ 1& 0\end{array}\right].$

By induction, we have

${\mathbf{V}}^{m}=\left[\begin{array}{cc}{t}_{m+1}\left(x\right)& -{t}_{m}\left(x\right)\\ {t}_{m}\left(x\right)& -{t}_{m-1}\left(x\right)\end{array}\right].$
(8)

So, the matrix V generates Vieta-Pell and Vieta-Pell-Lucas polynomials. Hence,

$\left[\begin{array}{c}{t}_{m+1}\left(x\right)\\ {t}_{m}\left(x\right)\end{array}\right]={\mathbf{V}}^{m}\left[\begin{array}{c}1\\ 0\end{array}\right]$
(9)

and from (1.10) in [8], we get

${t}_{m}\left(x\right)=\left[\begin{array}{cc}1& 0\end{array}\right]{\mathbf{V}}^{m-1}\left[\begin{array}{c}1\\ 0\end{array}\right].$
(10)

It is known that

${\mathbf{V}}^{m+n}={\mathbf{V}}^{m}{\mathbf{V}}^{n}.$
(11)

From (8) and (11), the elementary formulas for ${t}_{n}\left(x\right)$ are obvious

$\begin{array}{c}{t}_{m+n+1}\left(x\right)={t}_{m+1}\left(x\right){t}_{n+1}\left(x\right)-{t}_{m}\left(x\right){t}_{n}\left(x\right),\hfill \\ {t}_{m+n}\left(x\right)={t}_{m+1}\left(x\right){t}_{n}\left(x\right)-{t}_{m}\left(x\right){t}_{n-1}\left(x\right),\hfill \\ {t}_{m+n-1}\left(x\right)={t}_{m}\left(x\right){t}_{n}\left(x\right)-{t}_{m-1}\left(x\right){t}_{n-1}\left(x\right).\hfill \end{array}$

If we use the matrix technique for summation in [8], we get the first finite summation as follows.

Proposition 3

1. (i)

${\sum }_{n=1}^{m}{t}_{n}\left(x\right)=\frac{{t}_{m+1}\left(x\right)-{t}_{m}\left(x\right)-1}{2\left(x-1\right)}$,

2. (ii)

${\sum }_{n=1}^{m}{s}_{n}\left(x\right)=\frac{{s}_{m+1}\left(x\right)-{s}_{m}\left(x\right)+2-2x}{2\left(x-1\right)}$.

Proof (i) Let the matrix A,

$\mathbf{A}=\mathbf{I}+\mathbf{V}+{\mathbf{V}}^{2}+\cdots +{\mathbf{V}}^{n-2}+{\mathbf{V}}^{n-1}$

be the series of matrices. Then we have

$\mathbf{VA}=\mathbf{V}+{\mathbf{V}}^{2}+\cdots +{\mathbf{V}}^{n-1}+{\mathbf{V}}^{n}.$

Hence,

$\begin{array}{r}\left(\mathbf{V}-\mathbf{I}\right)\mathbf{A}={\mathbf{V}}^{n}-\mathbf{I},\\ \mathbf{A}={\left(\mathbf{V}-\mathbf{I}\right)}^{-1}\left({\mathbf{V}}^{n}-\mathbf{I}\right)\\ \phantom{\mathbf{A}}=\frac{1}{2\left(x-1\right)}\left[\begin{array}{cc}{t}_{n+1}\left(x\right)-{t}_{n}\left(x\right)-1& {t}_{n-1}\left(x\right)-{t}_{n}\left(x\right)+1\\ {t}_{n}\left(x\right)-{t}_{n-1}\left(x\right)-1& {t}_{n-2}\left(x\right)-{t}_{n-1}\left(x\right)+2x-1\end{array}\right],\\ \sum _{n=1}^{m}{t}_{n}\left(x\right)=\left[\begin{array}{cc}1& 0\end{array}\right]\mathbf{A}\left[\begin{array}{c}1\\ 0\end{array}\right]\phantom{\rule{1em}{0ex}}\left(\text{by (10)}\right)\\ \phantom{\sum _{n=1}^{m}{t}_{n}\left(x\right)}=\frac{1}{2\left(x-1\right)}\left[\begin{array}{cc}{t}_{m+1}\left(x\right)-{t}_{m}\left(x\right)-1& {t}_{m-1}\left(x\right)-{t}_{m}\left(x\right)+1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \phantom{\sum _{n=1}^{m}{t}_{n}\left(x\right)}=\frac{1}{2\left(x-1\right)}\left[{t}_{m+1}\left(x\right)-{t}_{m}\left(x\right)-1\right].\end{array}$

Thus, the proof is completed.

1. (ii)
$\begin{array}{rcl}\sum _{n=1}^{m}{s}_{n}\left(x\right)& =& \sum _{n=1}^{m}\left({\alpha }^{n}+{\beta }^{n}\right)\phantom{\rule{1em}{0ex}}\left(\text{by (3)}\right)\\ =& \alpha \left(\frac{1-{\alpha }^{m}}{1-\alpha }\right)+\beta \left(\frac{1-{\beta }^{m}}{1-\beta }\right)\\ =& \frac{\left(\alpha +\beta \right)-2\alpha \beta -\left({\alpha }^{m+1}+{\beta }^{m+1}\right)+\alpha \beta \left({\alpha }^{m}+{\beta }^{m}\right)}{2-2x}\\ =& \frac{2x-2-\left({\alpha }^{m+1}+{\beta }^{m+1}\right)+{\alpha }^{m}+{\beta }^{m}}{2\left(1-x\right)}\phantom{\rule{1em}{0ex}}\left(\text{by (1)}\right)\\ =& \frac{{s}_{m+1}\left(x\right)-{s}_{m}\left(x\right)+2-2x}{2\left(x-1\right)}\phantom{\rule{1em}{0ex}}\left(\text{by (3)}\right).\end{array}$

This completes the proof. □

Theorem 1 Let V be a square matrix such that ${\mathbf{V}}^{2}=2x\mathbf{V}-\mathbf{I}$. Then, for all $n\in {\mathbb{Z}}^{+}$,

${\mathbf{V}}^{n}={t}_{n}\left(x\right)\mathbf{V}-{t}_{n-1}\left(x\right)\mathbf{I},$

where ${t}_{n}\left(x\right)$ is the nth Vieta-Pell polynomial and I is a unit matrix.

Proof The proof is obvious from induction. □

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## Acknowledgements

The authors thank the referees for their valuable suggestions, which improved the standard of the paper.

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### Corresponding author

Correspondence to Feyza Yalcin.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The authors declare that the research was realized in collaboration with the same responsibility and contributions. Both authors read and approved the final manuscript.

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Tasci, D., Yalcin, F. Vieta-Pell and Vieta-Pell-Lucas polynomials. Adv Differ Equ 2013, 224 (2013). https://doi.org/10.1186/1687-1847-2013-224