Remarks Regarding Computational Aspects in Algebras Obtained by Cayley–Dickson Process and Some of Their Applications

Abstract

Due to the computational aspects which appear in the study of algebras obtained by the Cayley–Dickson process, it is difficult to obtain nice properties for these algebras. For this reason, finding some identities in such algebras plays an important role in obtaining new properties of these algebras and facilitates computations. In this regard, in the first part of this paper, we present some new identities and properties in algebras obtained by the Cayley–Dickson process. As another remark regarding the computational aspects in these algebras, in the last part of this paper, we solve some quadratic equations in the real division quaternion algebra when their coefficients are some special elements. These special coefficients allowed us to solve interesting quadratic equations, providing solutions directly, without using specialized softs.

1. Introduction

Algebras obtained by the Cayley–Dickson process have at present many applications in various domains, for example, coding theory, computer vision physics, etc. In these algebras it is difficult to obtain nice properties. When their dimension increase, algebras obtained by the Cayley–Dickson process lose commutativity, associativity, and alternativity. For this reason, the study of all kinds of identities on these algebras is very useful for obtaining new properties and relations. Several papers are devoted to the study of these identities ([1,2,3,4] etc.). Therefore, it is very interesting to continue the study of these identities in algebras obtained by the Cayley–Dickson process, since these relations can be helpful to replace the missing commutativity, associativity, and alternativity.

In the following, we consider $\mathcal{A}$, an algebra over the field K, where K is a commutative field with $charK\ne 2$. Algebra $\mathcal{A}$ is called unitary if it contains an identity element with respect to the algebra’s multiplication.

We define the associator of three elements $a,b,c\in$$\mathcal{A}$ as below

$$(a,b,c)=\left(ab\right)c-a\left(bc\right).$$

In an arbitrary algebra over a field K, the following relation is true ([5], (1.2))

$$(ab,c,d)-\left(a,bc,d\right)+\left(a,b,cd\right)=a\left(b,c,d\right)+\left(a,b,c\right)d.$$

(2)

An algebra $\mathcal{A}$ is alternative if the following relations are fulfilled: $a^{2}b=a\left(ab\right)$ and $a{b}^2=\left(ab\right)b,$ for all $a,b\in \mathcal{A}.$ Analgebra $\mathcal{A}$ is flexible if $a\left(ba\right)=\left(ab\right)a=aba,$ for all $a,b\in \mathcal{A}$ and power associative if the subalgebra $<a>$ of $\mathcal{A}$, generated by an arbitrary element $a\in \mathcal{A}$, is associative. We know that an alternative algebra is a flexible and a power associative algebra. If in a unitary algebra $\mathcal{A}\ne K$ we have $x^2+{\alpha }_{x}x+{\beta }_x=0,$ for each $x\in \mathcal{A},$ with ${\alpha }_x,{\beta }_x\in K,$ this algebra is a quadratic algebra. A finite-dimensional algebra $\mathcal{A}$ without zero divisors is called a division algebra (see [6]). The Cayley–Dickson process and the properties of the obtained algebras, will be briefly presented in the following. For other details see [6,7].

Let $\mathcal{A}$ be a finite dimensional unitary algebra over a field $K,$ with a map called a scalar involution

$$\overline{\phantom{x}}:\mathcal{A}\to \mathcal{A},x\to \overline{x}.$$

This map is a linear map and has the following properties

$$\overline{xy}=\overline{y}\overline{x},\overline{\overline{x}}=x,$$

and

$$x+\overline{x},x\overline{x}\in K·1,\mathrm{for}\mathrm{all}x,y\in \mathcal{A}.$$

The element $\overline{x}$ is called the conjugate of the element x. The linear form

$$\mathbf{t}:A\to K,\mathbf{t}\left(x\right)=x+\overline{x}$$

and the quadratic form

$$\mathbf{n}:A\to K, \mathbf{n}\left(x\right)=x\overline{x}$$

are called the trace and the norm of the element x. From here, we have that an algebra $\mathcal{A}$ with a scalar involution is a quadratic algebra.

With $\delta \in K$, a fixed non-zero element, the following algebra multiplication is defined over the vector space

$$\mathcal{A}\oplus \mathcal{A}:\left(a_1,a_2\right)\left(b_1,b_2\right)=\left(a_1{b}_1+\delta {\overline{b}}_2{a}_2,a_2\overline{b_1}+b_2{a}_1\right).$$

(3)

We obtain an algebra structure over $\mathcal{A}\oplus \mathcal{A},$ denoted by $\left(\mathcal{A},\delta \right)$, called the algebra obtained from $\mathcal{A}$ by the Cayley–Dickson process with $dim\left(\mathcal{A},\delta \right)=2dim\mathcal{A}$.

For $x\in \left(\mathcal{A},\delta \right)$, $x=\left(a_1,a_2\right)$, the map

$$\overline{\phantom{x}}:\left(\mathcal{A},\delta \right)\to \left(\mathcal{A},\delta \right),x\to \overline{x}=\left({\overline{a}}_1,-a_2\right),$$

is a scalar involution of the algebra $\left(\mathcal{A},\delta \right)$, which extend the involution $\overline{\phantom{x}}$ of the initial algebra $\mathcal{A}$. We consider the maps

$$\mathbf{t}\left(x\right)=\mathbf{t}\left(a_1\right)$$

and

$$\mathbf{n}\left(x\right)=\mathbf{n}\left(a_1\right)-\delta \mathbf{n}\left(a_2\right)$$

which are the trace and the norm of the element $x\in$$\left(\mathcal{A},\delta \right)$.

Let $\mathcal{A}=K$. If we apply the above process t times, $t\ge 1$, we get an algebra over K, denoted

$$A_t=\left(\frac{{\delta }_1,\cdots ,{\delta }_t}{K}\right).$$

(4)

By induction, the set $\{1,f_2,\cdots ,f_n\},n=2^t$, generates, in this algebra, a basis with the properties:

$$f_i^2={\delta }_{i}1,{}_i\in K,{\delta }_i\ne 0,i=2,\cdots ,n$$

(5)

and

$$f_i{f}_j=-f_j{f}_i={\sigma }_{ij}{f}_k,{\sigma }_{ij}\in K,{\sigma }_{ij}\ne 0,i\ne j,i,j=2,\cdots ,n,$$

(6)

with ${\sigma }_{ij}$ and $f_k$ uniquely determined by $f_i$ and $f_j.$

In each algebra ${\mathcal{A}}_t$ with the basis $\{1,f_2,\cdots ,f_n\}$ satisfying relations $\left(5\right)$ and $\left(6\right),$ we have:

$$f_i\left(f_{i}x\right)={\delta }_{i}x=\left(x{f}_i\right)f_i,$$

(7)

for all $i\in \{1,2,\cdots ,n\}$ and for every $x\in {\mathcal{A}}_t$. ([7], Lemma 4)

From the above, it results that each $x\in$$A_t$ can be written under the form

$$x=x_1+x_2{f}_{2^{t-1}},$$

(8)

where $x_1$ and $x_2\in {\mathcal{A}}_{t-1}$.

All algebras ${\mathcal{A}}_t$ obtained by the Cayley–Dickson process are flexible and power associative, for all $t\ge 1,$ and, in general, are not division algebras, for all $t\ge 1$. For example, if $K=\mathbb{R},$$t=2, {\delta }_1={\delta }_2=-1,$ we obtain the classical division quaternion algebra and for $t=3, {\delta }_1={\delta }_2={\delta }_3=-1,$ we obtain the division octonion algebra.

2. Some Remarks Regarding Algebras Obtained by the Cayley–Dickson Process

In the following, we consider ${\mathcal{A}}_t$ an algebra obtained by the Cayley–Dickson process. In this section we present some new relations, properties and solutions to a quadratic equation in these algebras. In the next section, we will present solutions to a few quadratic equations over quaternion real division algebra, when the coefficients of these equations are some special elements.

Remark 1

([1]). In all algebras ${\mathcal{A}}_t$, obtained by the Cayley–Dickson process, the following identity is satisfied:

$$x^2\left(yx\right)=\left(x^{2}y\right)x.$$

(9)

To obtain identities in these algebras, in [1] the linearization method was presented, a method which we will use in the next sentence.

Proposition 1.

If the field K has an arbitrary characteristic, then, in all algebras ${\mathcal{A}}_t$, obtained by the Cayley–Dickson process, the following identities are satisfied:

$$\left(w^2,v,x\right)+\left(xw,v,w\right)+\left(wx,v,w\right)=0$$

(10)

$$\left(w^2,y,x\right)+\left(x^2,v,w\right)+\left(xw,y,w\right)+\left(wx,y,w\right)+\left(wx,v,x\right)+\left(xw,v,x\right)=0,$$

(11)

where$x,y,v,w\in {\mathcal{A}}_t.$

Proof. 

From the above remark, we have that $x^2\left(yx\right)=\left(x^{2}y\right)x$, for all $x,y\in {\mathcal{A}}_t$. Denoting $T_1=x^2\left(yx\right),T_2=\left(x^{2}y\right)x$, we linearize these relations. For $x+\lambda w,y+\lambda v$, with $v,w\in {\mathcal{A}}_t$, $\lambda \in K$, $\lambda$ nonzero, arbitrary chosen, we obtain

$T_1={(x+\lambda w)}^2\left(\left(y+\lambda v\right)\left(x+\lambda w\right)\right)=$

$=({(x+\lambda w)}^2\left(y+\lambda v\right))\left(x+\lambda w\right)=T_2$. We have

$T_1=\left(x^2+\lambda xw+\lambda wx+{\lambda }^2{w}^2\right)(yx+\lambda yw+\lambda vx+{\lambda }^{2}vw)=$

$={\lambda }^4{w}^2\left(vw\right)+{\lambda }^3\left[w^2\left(yw+vx\right)+\left(xw+wx\right)vw\right]+$

$+{\lambda }^2\left[w^2\left(yx\right)+x^2\left(vw\right)+\left(xw+wx\right)\left(yw+vx\right)\right]+$

$+\lambda \left[\left(xw+wx\right)yx+x^2\left(yw+vx\right)\right]+x^2\left(yx\right).$

$T_2=\left(\left(x^2+\lambda xw+\lambda wx+{\lambda }^2{w}^2\right)\left(y+\lambda v\right)\right)\left(x+\lambda w\right)=$

$=\left[x^{2}y+\lambda \left(x^{2}v+\left(xw+wx\right)y\right)+{\lambda }^2\left(w^{2}y+\left(xw+wx\right)v\right)+{\lambda }^3{w}^{2}v\right]\left(x+\lambda w\right)=$

$={\lambda }^4\left(w^{2}v\right)w+{\lambda }^3\left[\left(w^{2}v\right)x+(w^{2}y+\left(xw+wx\right)v)w\right]+$

$+{\lambda }^2\left[\left(x^{2}v+\left(xw+wx\right)y\right)w+\left(w^{2}y+\left(xw+wx\right)v\right)x\right]+$

$+\lambda \left[\left(x^{2}y\right)w+\left(x^{2}v+\left(xw+wx\right)y\right)x\right]+\left(x^{2}y\right)x.$

Comparing the coefficients of ${\lambda }^3$, we obtain

$\left[w^2\left(yw+vx\right)+\left(xw+wx\right)vw\right]=$

$=\left[\left(w^{2}v\right)x+\left(w^{2}y+\left(xw+wx\right)v\right)w\right]$ and

$w^2\left(yw\right)+w^2\left(vx\right)+\left(xw\right)\left(vw\right)+\left(wx\right)\left(vw\right)=$

$=\left(w^{2}v\right)x+\left(w^{2}y\right)w+\left(\left(xw\right)v\right)w+\left(\left(wx\right)v\right)w$, therefore

$$\left(w^2,v,x\right)+\left(xw,v,w\right)+\left(wx,v,w\right)=0$$

Comparing the coefficients of ${\lambda }^2$, we obtain

$\left[w^2\left(yx\right)+x^2\left(vw\right)+\left(xw+wx\right)\left(yw+vx\right)\right]=$

$=\left[\left(x^{2}v+\left(xw+wx\right)y\right)w+\left(w^{2}y+\left(xw+wx\right)v\right)x\right]$ and $w^2\left(yx\right)+x^2\left(vw\right)+\left(xw\right)\left(yw\right)+\left(xw\right)\left(vx\right)+\left(wx\right)\left(yw\right)+\left(wx\right)\left(vx\right)=$

$=\left(x^{2}v\right)w+\left(\left(xw\right)y\right)w+\left(\left(wx\right)y\right)w+\left(w^{2}y\right)x+\left(\left(xw\right)v\right)x+\left(\left(wx\right)v\right)x$, therefore

$$\left(w^2,y,x\right)+\left(x^2,v,w\right)+\left(xw,y,w\right)+\left(wx,y,w\right)+\left(wx,v,x\right)+\left(xw,v,x\right)=0.$$

Comparing the coefficients of $\lambda$, we obtain

$\left[\left(xw+wx\right)yx+x^2\left(yw+vx\right)\right]=\left[\left(x^{2}y\right)w+\left(x^{2}v+\left(xw+wx\right)y\right)x\right]$ and

$\left(xw\right)\left(yx\right)+\left(wx\right)\left(yx\right)+x^2\left(yw\right)+x^2\left(vx\right)=$

$=\left(x^{2}y\right)w+\left(x^{2}v\right)x+\left(\left(xw\right)y\right)x+\left(\left(wx\right)y\right)x$. It results

$$\left(x^2,y,w\right)+\left(xw,y,x\right)+\left(wx,y,x\right)=0,$$

which is similar to relation $\left(10\right){.}_{}$ □

Remark 2.

The above results are true for a field of an arbitrary characteristic, namely even if char$K\ne 2$. However, if char$K\ne 2$, the proof of relation $\left(10\right)$ is more simple. Indeed, we have $2\left(w^2,v,x\right)+2\left(xw,v,w\right)+2\left(wx,v,w\right)=$

$=\left({\left(w+x\right)}^2,v,\left(x+w\right)\right)+\left({\left(w-x\right)}^2,v,\left(x-w\right)\right)-2\left(x^2,v,x\right)=0$, by using Remark 1.

Proposition 1 and the above remark improved Proposition 2.9 from [8].

Proposition 2.

In all algebras ${\mathcal{A}}_t$, obtained by the Cayley–Dickson process, the following identity is satisfied:

$$\left(x^m,y,x^n\right)=0,m,n\in \mathbb{N};$$

Proof. 

Algebras ${\mathcal{A}}_t$ are power-associative and for each $x\in {\mathcal{A}}_t,$ we have $x^2-t\left(x\right)x+n\left(x\right)=0$. From here, we have that $x^m={\alpha }_m\left(x\right)x+{\beta }_m\left(x\right),{\alpha }_m\left(x\right),{\beta }_m\left(x\right)\in K$. It results that

$\left(x^m,y,x^n\right)=$$\left({\alpha }_m\left(x\right)x+{\beta }_m\left(x\right),y,{\alpha }_n\left(x\right)x+{\beta }_n\left(x\right)\right)=0{.}_{}$ □

Proposition 3.

Let ${\mathcal{A}}_t$, be an algebra obtained by the Cayley–Dickson process. We consider the elements $x,y\in {\mathcal{A}}_t,$ such that the following equality holds:

$${(x\overline{y}+y\overline{x})}^2={\gamma }^2\mathbf{n}\left(x\right)\mathbf{n}\left(y\right),\gamma \in K-\left\{0\right\}.$$

(12)

Then the algebra $<x,y>,$ generated by x and $y,$ has dimension $1.$

Proof. 

We prove that ${[x\overline{y}+y\overline{x}]}^2={\gamma }^2\mathbf{n}\left(x\right)\mathbf{n}\left(y\right)={\gamma }^2\left(x\overline{x}\right)\left(y\overline{y}\right),$ for all $x,y\in A,$ if and only if $x=sy,$$s\in K$. Since $x\overline{y}+y\overline{x}=x\overline{y}+\overline{x\overline{y}}\in K$, therefore, if $x=sy,$ we have relation $\left(12\right)$. Now, supposing that relation $\left(12\right)$ is true and if there is not an element $s\in K$ such that $x=sy,$ then, for $a,b\in K,$$a\ne 0,b\ne 0,$ we have $ax+by\ne 0.$ If we have $ax+by=0,$ it results $x=-\frac{b}{a}y,$ false. It results that

$$\left(ax+by\right)\overline{\left(ax+by\right)}\ne 0.$$

(13)

from relation $\left(13\right),$ we have

$$a^2\left(x\overline{x}\right)+abx\overline{y}+bay\overline{x}+b^{2}y\overline{y}\ne 0.$$

(14)

In relation $\left(14\right),$ we take $a=y\overline{y}$ and we obtain

$$\left(y\overline{y}\right)\left(x\overline{x}\right)+bx\overline{y}+by\overline{x}+b^2\ne 0.$$

(15)

If we take $b=-{\gamma }^{-1}\left(x\overline{y}+y\overline{x}\right)\in K,$ with $b\ne 0$ and we replace this value in relation $\left(15\right),$ we get ${\gamma }^2\left(x\overline{x}\right)\left(y\overline{y}\right)-{(x\overline{y}+y\overline{x})}^2\ne 0,$ which it is not true. It results that there is an element $v\in K$ such that $x=vy$. From here, we have that the dimension of the algebra $<x,y>$ is 1. □

The above proposition generalized Proposition 2.8 from [9].

Proposition 4.

Let ${\mathcal{A}}_t$ be an algebra obtained by the Cayley–Dickson process over a field with characteristic not 2 and $a\in$ ${\mathcal{A}}_t$.

(1) If $a\in$${\mathcal{A}}_t-K,$ then the solution in ${\mathcal{A}}_t$ of the equation $x^2=a$ are the solutions in K of the following system:

$$\left\{\begin{array}{c}2x_0^2-\mathbf{n}\left(x\right)=a_0\\ 2x_0{x}_i=a_i,i\in \{1,2,\cdots ,2^t-1\}\end{array}\right.,$$

where $a={\displaystyle \sum _{i=0}^{2^t-1}}{a}_i{f}_i,x={\displaystyle \sum _{i=0}^{2^t-1}}{x}_i{f}_i,a_i,x_i\in K$.

(2) If $a\in$$K,$ then the solution in ${\mathcal{A}}_t$ of the equation $x^2=a$ are the solutions in K of the following system:

$$\left\{\begin{array}{c}2x_0^2-\mathbf{n}\left(x\right)=a\\ 2x_0{x}_i=0,i\in \{1,2,\cdots ,2^t-1\}\end{array}\right..$$

Proof. 

(1) Let $a\in$${\mathcal{A}}_t-K$, $a={\displaystyle \sum _{i=0}^{2^t-1}}{a}_i{f}_i$. Since ${\mathcal{A}}_t$ is a quadratic algebra, we have that $x^2=2x_{0}x-\mathbf{n}\left(x\right)=a$, therefore $2x_0^2-\mathbf{n}\left(x\right)=a_0$ and $2x_0{x}_i=a_i,i\in \{1,2,\cdots ,2^t-1\}$.

(2) Let $a\in$ K. It results $x^2=2x_{0}x-\mathbf{n}\left(x\right)=a$. We get $2x_0^2-\mathbf{n}\left(x\right)=a$ and $2x_0{x}_i=0,i\in \{1,2,\cdots ,2^t-1\}$. If $x_0=0,$ then $\mathbf{n}\left(x\right)=-a$. If $x_0\ne 0,$ then $x_i=0,$ for all $i\in \{1,2,\cdots ,2^t-1\},$ and $x_0$ is the solution in K of the equation $x_0^2=a$. □

Remark 3.

Let $a\in$${\mathcal{A}}_t$. We denote $a=a_0+\overrightarrow{a},$$a_0\in K$. Therefore all elements $b\in$${\mathcal{A}}_t$ of the form $b=\tau +\theta \overrightarrow{a},\tau ,\theta \in K,$ commute with a. (see [8], Proposition 2.11 and Remark 2.12).

3. A Quadratic Equation over Real Quaternions

Solving equations with coefficients in algebras ${\mathcal{A}}_t,$ obtained by the Cayley–Dickson, is not an easy problem, even if t is small, $t\in \{2,3\},$ that means we have coefficients in quaternion or in octonion algebras. There are many papers devoted to the study of the roots of polynomial equations with coefficients in ${\mathcal{A}}_t$. In [10], the author studied zeros of polynomials over Cayley–Dickson algebras, without providing a clear algorithm to find them. In the papers [11,12,13,14,15,16,17,18,19] the authors studied roots of quaternionic and octonionic polynomials and provided laboriously methods to find these roots. To find these roots “by hand” is not so easy and, in most cases, are used specialized softs. Even so, the above-mentioned methods are done only for quaternions and octonions, since in the proof are used Moufang identities, which are true only in alternative algebras. To generalize these algorithms to all algebras ${\mathcal{A}}_t$ can be a real challenge. In this section, we consider quadratic equations over real division quaternion algebra, when the coefficients are some special elements. These special coefficients allowed us to solve interesting quadratic equations, providing solutions directly, without using specialized softs. For example, in [13], the authors find roots of a polynomial of degree n, with quaternionic coefficients, by using its companion polynomial, which is polynomial of degree $2n$ with real coefficients. For finding roots of the companion polynomial we used some special softs.

Let $\mathbb{H}\left(-1,-1\right)$ be the real division quaternion algebra, the algebra of the elements of the form $a=a_0·1+a_1{e}_1+a_2{e}_2+a_3{e}_3,$ where $a_i\in \mathbb{R},i\in \{0,1,2,3\}$ and the elements of basis $\{1,e_1,e_2,e_3\}$ given by the following multiplication table:

·	1	$e_1$	$e_2$	$e_3$
1	1	$e_1$	$e_2$	$e_3$
$e_1$	$e_1$	$-1$	$e_3$	$-e_2$
$e_2$	$e_2$	$-e_3$	$-1$	$e_1$
$e_3$	$e_3$	$e_2$	$-e_1$	$-1$

The norm of a real quaternion a is $\mathbf{n}\left(a\right)=a_1^2+a_2^2+a_3^2+a_4^2.$ For the real quaternion algebra, in [20], the Fibonacci quaternions were defined:

$$F_n=f_n+f_{n+1}{e}_1+f_{n+2}{e}_2+f_{n+3}{e}_3$$

and the formula of the norm was given:

$$\mathbf{n}\left(F_n\right)=F_n{\overline{F}}_n=3f_{2n+3}.$$

Here ${\overline{F}}_n=f_n-f_{n+1}{e}_1-f_{n+2}{e}_2-f_{n+3}{e}_3$ is the conjugate of the element $F_n$ in the algebra $\mathbb{H}\left(-1,-1\right)$.

In [19], the authors presented specific formulae for solving the following equation

$$x^2+bx+c=0,$$

(16)

with $b,c\in \mathbb{H}\left(-1,-1\right)$.

Proposition 5

([19] Theorem 2.3). Solutions to the quadratic equation

$$x^2+bx+c=0,b,c\in \mathbb{H}\left(1,1\right)$$

are:

Case 1. If $b,c\in \mathbf{R}$ and $b^2<4c$ then:

$$x=\frac{1}{2}(-b+e{e}_1+f{e}_2+g{e}_3)$$

(17)

with $e^2+f^2+g^2=4c-b^2$ where $e,f,g\in \mathbf{R}$.

Case 2. If $b,c\in \mathbf{R}$ and $b^2\ge 4c$ then:

$$x={\displaystyle \frac{-b\pm \sqrt{b^2-4c}}{2}}$$

(18)

Case 3. If $b\in \mathbf{R},c\notin \mathbf{R}$ then:

$$x=\frac{-b}{2}\pm \frac{\theta }{2}\mp \frac{c_1}{\theta }{e}_1\mp \frac{c_2}{\theta }{e}_2\mp \frac{c_3}{\theta }{e}_3,$$

(19)

where $c=c_0+c_1{e}_1+c_2{e}_2+c_3{e}_3$ and

$$\theta =\sqrt{\frac{b^2-4c_0+\sqrt{{\left(b^2-4c_0\right)}^2+16\left(c_1^2+c_2^2+c_3^2\right)}}{2}}$$

Case 4. If $b\notin \mathbf{R}$ then:

$$x={\displaystyle \frac{(-Re(b\left)\right)}{2}}-{(b^{\prime }+T)}^{-1}(c^{\prime }-N)$$

(20)

with $b^{\prime }=b-Re\left(b\right)=Im\left(b\right)$, $c^{\prime }=c-{\displaystyle \frac{\left(Re\right(b\left)\right)}{2}}(b-{\displaystyle \frac{\left(Re\right(b\left)\right)}{2}})$, where $(T,N)$ are chosen in the following way:

(1) $T=0,N=(\mathrm{\Gamma }\pm \sqrt{{\mathrm{\Gamma }}^2-4\mathrm{\Psi }})/2,$ if $\mathrm{\Delta }=0,{\mathrm{\Gamma }}^2\ge 4\mathrm{\Psi };$

(2) $T=\pm \sqrt{2\sqrt{E}-\mathrm{\Gamma }},N=\sqrt{\mathrm{\Psi }}$ if $\mathrm{\Delta }=0,{\mathrm{\Gamma }}^2<4\mathrm{\Psi };$

(3) $T=\pm \sqrt{z},N=(T^3+\mathrm{\Gamma }T+\mathrm{\Delta })/2T$ if $\mathrm{\Delta }\ne 0$ and z is the unique positive solution to the equation:

$$z^3+2\mathrm{\Gamma }{z}^2+({\mathrm{\Gamma }}^2-4\mathrm{\Psi })z-{\mathrm{\Delta }}^2=0,$$

(21)

where $\mathrm{\Gamma }=|b^{\prime }{|}^2+2Re\left(c^{\prime }\right),$$\mathrm{\Psi }=|c^{\prime }{|}^2$ and $\mathrm{\Delta }=2Re\left(\overline{b^{\prime }}{c}^{\prime }\right)$.

By using the above results, we will give solutions to the monic quadratic equation $x^2+bx+c=0$, where $b,c\in \mathbb{H}\left(-1,-1\right)$ are two Fibonacci quaternion elements. We will show that, by using such elements, we can obtain a much simpler form of the solutions. Due to the chosen coefficients, we are interested only in cases 3 or 4 of Proposition 5. To be able to compute solutions, we must compute first the following elements:

$$b^{\prime }=Im\left(b\right),$$

$$c^{\prime }=c-{\displaystyle \frac{\left(Re\right(b\left)\right)}{2}}(b-{\displaystyle \frac{\left(Re\right(b\left)\right)}{2}}),$$

$$\mathrm{\Gamma }=\mathbf{n}\left(b^{\prime }\right)+2Re\left(c^{\prime }\right),$$

$$\mathrm{\Delta }=2Re\left(\overline{b^{\prime }}{c}^{\prime }\right),$$

$$\mathrm{\Psi }=\mathbf{n}\left(c^{\prime }\right),$$

The above elements help us to compute parameters $T,N$ and then to obtain the solutions. The expression of the solutions depends on the parameters $T,N,$$\mathrm{\Delta }$. The first step is to see when $\mathrm{\Delta }$ is 0 or not. To compute $\mathrm{\Delta }$, we will compute first $b^{\prime }$ and $c^{\prime }$, where

$$b=F_n=f_n+f_{n+1}{e}_1+f_{n+2}{e}_2+f_{n+3}{e}_3$$

$$c=F_m=f_m+f_{m+1}{e}_1+f_{m+2}{e}_2+f_{m+3}{e}_3$$

$$b^{\prime }=b-Re\left(b\right)=Im\left(b\right)$$

$$b^{\prime }=f_{n+1}{e}_1+f_{n+2}{e}_2+f_{n+3}{e}_3$$

We obtain the following formulae, whose demonstrations can be found in Appendix A.

$$c^{\prime }=[(4f_m-f_n^2)+(4f_{m+1}-2f_n{f}_{n+1})e_1+(4f_{m+2}-2f_n{f}_{n+2})e_2+(4f_{m+3}-2f_n{f}_{n+3})e_3]/4$$

(22)

$$\mathrm{\Delta }=2(f_{n+1}{f}_{m+1}+f_{n+2}{f}_{m+2}+f_{n+3}{f}_{m+3})-f_n(f_{n+1}^2+f_{n+2}^2+f_{n+3}^2).$$

(23)

Remark 4.

(1) The solutions to Equation $\left(16\right),$ in case 4, depends when Δ is zero or not. The question is when $\mathrm{\Delta }=0$? From its expression it is clear that, in the most of cases, we have $\mathrm{\Delta }\ne 0$.

(2) When $n=m$, in relation $\left(23\right)$, we have $(f_{n+1}{f}_{m+1}+f_{n+2}{f}_{m+2}+f_{n+3}{f}_{m+3})=(f_{n+1}^2+f_{n+2}^2+f_{n+3}^2)$. From here, relation $\mathrm{\Delta }=0$ implies $\left(2-f_n\right)\left(f_{n+1}^2+f_{n+2}^2+f_{n+3}^2\right)=0$, therefore, $f_n=2$. It results that $n=m=3$. In this particular case, we obtain particular solutions to Equation $\left(16\right)$, solutions which will be presented at the end of this paper. Due to the monotonicity of the Fibonacci sequence, in the rest of cases, we have $\mathrm{\Delta }\ne$ 0.

(3) From above, we remark that the solutions to Equation $\left(16\right)$, when are involved two Fibonacci quaternions as coefficients, can be obtained by using subcases 3 and 4 from Proposition 5. To obtain these solutions, root of the polynomial of the third degree $\left(21.\right)$ must be obtained.

For Γ and Ψ, we have

$$\mathrm{\Gamma }=3f_{2n+3}-\frac{3}{2}{f}_n^2+2f_m$$

(24)

and

$$\mathrm{\Psi }=3f_{2m+3}+\frac{f_n^2}{16}(f_n^2+4f_{n+1}^2+4f_{n+2}^2+4f_{n+3}^2)-$$

$$-\frac{f_n}{2}(f_m{f}_n+2f_{m+1}{f}_{n+1}+2f_{m+2}{f}_{n+2}+2f_{m+3}{f}_{n+3})$$

(25)

In the known references, we can find some examples of roots of equations with coefficients in $\mathbb{H}\left(-1,-1\right)$. Most of them are “served”, with the roots already known. By using the above results, we will find the roots for the following equations, equations which are not in the above-described situation:

$$x^2+F_{3}x+F_3=0,$$

(26)

where $F_3=f_3+f_4{e}_1+f_5{e}_2+f_6{e}_3$ is a Fibonacci quaternion;

$$x^2+F_{0}x+F_0=0,$$

(27)

where $F_0=e_1+e_2+2e_3$ is a Fibonacci quaternion

$$x^2+F_{0}x+F_1=0,$$

(28)

where $F_0=e_1+e_2+2e_3$ and $F_1=1+e_1+2e_2+3e_3$ are Fibonacci quaternions.

Proposition 6.

Equation $\left(26\right)$ has the following solutions:

$$x_1=-1+\frac{\alpha }{98}\left(F_3-f_3\right),$$

$$x_2=-1+\frac{\beta }{98}\left(F_3-f_3\right),$$

where $\alpha =49+\sqrt{2499}$ and $\beta =49-\sqrt{2499}$.

Proof. 

We have $n=m=3$, therefore $b=c,$ then $\mathrm{\Delta }=0$, from Remark 4. Replacing in formula $\left(25\right),$ for $\mathrm{\Psi }$, we obtain:

$\mathrm{\Psi }=(f_3{}^2+f_4^2+f_5^2+f_6^2)+\frac{f_3^2}{16}(f_3^2+4f_4^2+4f_5^2+4f_6^2)-$

-$\frac{f_n}{2}(f_3^2+2f_4^2+2f_5^2+2f_6^2)$

$\mathrm{\Psi }=(2^2+3^2+5^2+8^2)+\frac{1}{4}(2^2+4·3^2+4·5^2+4·8^2)-$

$-\frac{2}{2}(2^2+2·3^2+2·5^2+2·8^2)$

$\mathrm{\Psi }=102+\frac{396}{4}-200=1.$

Replacing in formula $\left(24\right),$ we obtain:

$$\mathrm{\Gamma }=3f_{2n+3}-\frac{3}{2}{f}_n^2+2f_m$$

$$\mathrm{\Gamma }=3f_9-\frac{3}{2}{f}_3^2+2f_3$$

$$\mathrm{\Gamma }=102-6+4=100$$

We are in the first subcase of case 4 from Proposition 5. Therefore $T=0$, $N=(\mathrm{\Gamma }\pm \sqrt{{\mathrm{\Gamma }}^2-4\mathrm{\Psi }})/2$. $N=(100\pm \sqrt{(10000-4})/2$, $N=50\pm \left(\sqrt{9996}\right)/2$, $N=50\pm \sqrt{2499}$. It results the following solutions:

$x=(-Re\left(b\right))/2-{(b\prime +T)}^{-1}(c\prime -N)$

$x=-1-{(3e_1+5e_2+8e_3)}^{-1}\left(\right[(4·2-2^2)+(4·3-2·2·3)e_1+$

$+(4·5-2·2·5)e_2+(4·8-2·2·8)·e_3\left)\right]/4-(50\pm \sqrt{2499}))$

$x=-1-{(3e_1+5e_2+8e_3)}^{-1}(1-(50\pm \sqrt{2499}))$

$x=-1-{(3e_1+5e_2+8e_3)}^{-1}(49\mp \sqrt{2499}))$

We have ${(3e_1+5e_2+8e_3)}^{-1}=-\frac{1}{98}\left(3e_1+5e_2+8e_3\right)$. Denoting with $\alpha =49+\sqrt{2499}$ and $\beta =49-\sqrt{2499}$, we obtain the following solutions:

$x_1=-1+\frac{\alpha }{98}\left(3e_1+5e_2+8e_3\right)=-1+\frac{\alpha }{98}\left(F_3-f_3\right),$

$x_2=-1+\frac{\beta }{98}\left(3e_1+5e_2+8e_3\right)=-1+\frac{\beta }{98}\left(F_3-f_3\right){.}_{}$ □

Proposition 7.

Equation $\left(27\right)$ has the following solutions:

$$x_1=\sqrt{3}-\frac{\sqrt{3}+3}{6}{F}_0,$$

$$x_2=-\sqrt{3}-\frac{3-\sqrt{3}}{6}{F}_0.$$

Proof. 

Now, for $n=0,m=0$, we have $b=F_0=e_1+e_2+2e_3$ and $c=F_0=e_1+e_2+2e_3$. It results $\mathrm{\Delta }=2(f_{n+1}{f}_{m+1}+f_{n+2}{f}_{m+2}+f_{n+3}{f}_{m+3})-f_n(f_{n+1}^2+f_{n+2}^2+f_{n+3}^2)=$

$=2(1+1+4)=12$. For $\mathrm{\Psi }$, we obtain $\mathrm{\Psi }$=$3f_{2m+3}$+$\frac{f_n^2}{16}(f_n^2$+$4f_{n+1}^2$+$4f_{n+2}^2$+$4f_{n+3}^2)-$$\frac{f_n}{2}(f_m{f}_n$+$2f_{m+1}{f}_{n+1}$+$2f_{m+2}{f}_{n+2}$+$2f_{m+3}{f}_{n+3})=3·2=6.\mathrm{\Gamma }=3f_{2n+3}-\frac{3}{2}{f}_n^2+2f_m=6$$z^3+2\mathrm{\Gamma }{z}^2+({\mathrm{\Gamma }}^2-4\mathrm{\Psi })z-{\mathrm{\Delta }}^2=0\Rightarrow z^3+12z^2+\left(36-48\right)z-144=0\Rightarrow$

$z^3+12z^2-12z-144=0\to z=12\to T_{1,2}=\pm 2\sqrt{3}$

We have $b=b^{\prime }=c=c^{\prime }$, therefore, for $T_1=2\sqrt{3}$, we get

$N_1=(T_1^3+\mathrm{\Gamma }{T}_1+\mathrm{\Delta })/2T_1=\frac{24\sqrt{3}+12\sqrt{3}+12}{4\sqrt{3}}=\frac{9\sqrt{3}+3}{\sqrt{3}}=9+\sqrt{3},$ therefore $x_1=-{(b^{\prime }+2\sqrt{3})}^{-1}(c^{\prime }-9-\sqrt{3})=$

$=-{\left(2\sqrt{3}+e_1+e_2+2e_3\right)}^{-1}\left(-9-\sqrt{3}+e_1+e_2+2e_3\right)=$

$=-\frac{1}{18}\left(2\sqrt{3}-e_1-e_2-2e_3\right)\left(-9-\sqrt{3}+e_1+e_2+2e_3\right)=$

$=-\frac{1}{18}[-18\sqrt{3}+\left(3\sqrt{3}+9\right)F_0]=\sqrt{3}-\frac{\sqrt{3}+3}{6}{F}_0$

For $T_1=-2\sqrt{3},$ we have $N_2=\frac{-24\sqrt{3}-12\sqrt{3}+12}{-4\sqrt{3}}=\frac{36\sqrt{3}-12}{4\sqrt{3}}=9-\sqrt{3},$ therefore

$x_2=-{\left(-2\sqrt{3}+e_1+e_2+2e_3\right)}^{-1}\left(-9+\sqrt{3}+e_1+e_2+2e_3\right)=$

$=-\frac{1}{18}\left(-2\sqrt{3}-e_1-e_2-2e_3\right)\left(-9+\sqrt{3}+e_1+e_2+2e_3\right)=$

$=\frac{1}{18}\left(2\sqrt{3}+e_1+e_2+2e_3\right)\left(-9+\sqrt{3}+e_1+e_2+2e_3\right)=$

$=\frac{1}{18}[-18\sqrt{3}+6+\left(-9+3\sqrt{3}\right)F_0-6]=$

$=-\sqrt{3}-\frac{3-\sqrt{3}}{6}{F}_0{.}_{}$ □

Proposition 8.

Equation $\left(28\right)$ has the following solutions:

$$x_{1,2}=-\frac{{\overline{F}}_0+T_{1,2}}{\mathbf{n}\left(F_0+T_{1,2}\right)}(F_1-N_{1,2}),$$

where δ is the real positive solution of the equation $z^3+16z^2+4z-324=0$ and $T_{1,2}=\pm \sqrt{\delta },N_{1,2}=(T_{1,2}^3+8T_{1,2}+18)/2T_{1,2}$.

Proof. 

Now, we take $n=0,m=1$. We have $F_0=e_1+e_2+2e_3$ and $F_1=1+e_1+2e_2+3e_3$. We have $b=b^{\prime }=F_0$ and $c=c^{\prime }=F_1$. We have $\mathrm{\Delta }=2(f_{n+1}{f}_{m+1}+f_{n+2}{f}_{m+2}+f_{n+3}{f}_{m+3})-f_n(f_{n+1}^2+f_{n+2}^2+f_{n+3}^2)=2(1+2+6)=18$. For $\mathrm{\Psi }$, we obtain $\mathrm{\Psi }$=$3f_{2m+3}$+$\frac{f_n^2}{16}(f_n^2$+$4f_{n+1}^2$+$4f_{n+2}^2$+$4f_{n+3}^2)$-$\frac{f_n}{2}(f_m{f}_n$+$2f_{m+1}{f}_{n+1}$+$2f_{m+2}{f}_{n+2}$+$2f_{m+3}{f}_{n+3})=3·5=15.\mathrm{\Gamma }=3f_{2n+3}-\frac{3}{2}{f}_n^2+2f_m=3·2-\frac{3}{2}·0+2=8z^3+2\mathrm{\Gamma }{z}^2+({\mathrm{\Gamma }}^2-4\mathrm{\Psi })z-{\mathrm{\Delta }}^2=0\Rightarrow z^3+16z^2+4z-324=0$. This equation has a real positive solution $\delta \in \left(3,4\right).$ We have $T_{1,2}=\pm \sqrt{\delta },N_{1,2}=(T_{1,2}^3+8T_{1,2}+18)/2T_{1,2}$. Therefore, we get $x_{1,2}=-{(F_0+T_{1,2})}^{-1}(F_1-N_{1,2})=-\frac{{\overline{F}}_0+T_{1,2}}{\mathbf{n}\left(F_0+T_{1,2}\right)}$ ${(F_1-N_{1,2})}_{}.$

□

We consider now the equation

$$x^2+xb+c=0,$$

(29)

with $b,c\in \mathbb{H}\left(-1,-1\right)$, which are not in the cases described in Proposition 5, but can be reduced to these situations. Let $p\left(x\right)=x^2+xb+c$. We remark that $\overline{p\left(x\right)}=\overline{x^2+xb+c}={\overline{x}}^2+\overline{b}\overline{x}+\overline{c}$. Therefore, if $x_0$ is a solution to Equation $\left(29\right),$ then its conjugate ${\overline{x}}_0$ is a solution to the equation

$${\overline{x}}^2+\overline{b}\overline{x}+\overline{c}=0,$$

(30)

called the conjugate of Equation $\left(16\right)$. Conversely, if $x_0$ is a solution to Equation $\left(30\right)$, then ${\overline{x}}_0$ is a solution to $\left(29\right)$. This equation can be solved by using Proposition 5. We consider the particular case when $b,c$ are Fibonacci quaternion elements. For this purpose, we must compute the elements

$${\overline{b}}^{\prime }=Im\left(\overline{b}\right)$$

$${\overline{c}}^{\prime }=\overline{c}-(Re\left(\overline{b}\right)/2)(\overline{b}-\left(Re\left(\overline{b}\right)\right)/2),$$

$$\overline{\mathrm{\Gamma }}=\mathbf{n}\left({\overline{b}}^{\prime }\right)+2Re\left({\overline{c}}^{\prime }\right),$$

$$\overline{\mathrm{\Delta }}=2Re\left(b^{\prime }{\overline{c}}^{\prime }\right),$$

$$\overline{\mathrm{\Psi }}=\mathbf{n}\left({\overline{\mathbf{c}}}^{\prime }\right).$$

We obtain

$${\overline{b}}^{\prime }=Im\left(\overline{b}\right)=-f_{n+1}{e}_1-f_{n+2}{e}_2-f_{n+3}{e}_3,$$

$${\overline{c}}^{\prime }=[(4f_m-f_n^2)-(4f_{m+1}-2f_n{f}_{n+1})e_1-(4f_{m+2}-2f_n{f}_{n+2})e_2-(4f_{m+3}-2f_n{f}_{n+3})e_3]/4,$$

$$\overline{\mathrm{\Gamma }}=3f_{2n+3}-\frac{3}{2}{f}_n^2+2f_m,$$

$$\overline{\mathrm{\Delta }}=2(f_{n+1}{f}_{m+1}+f_{n+2}{f}_{m+2}+f_{n+3}{f}_{m+3})-f_n(f_{n+1}^2+f_{n+2}^2+f_{n+3}^2),$$

$$\begin{array}{ccc}\hfill \overline{\mathrm{\Psi }}& =& 3f_{2m+3}+\frac{f_n^2}{16}(f_n^2+4f_{n+1}^2+4f_{n+2}^2+4f_{n+3}^2)-\hfill \\ & & -\frac{f_n}{2}(f_m{f}_n+2f_{m+1}{f}_{n+1}+2f_{m+2}{f}_{n+2}+2f_{m+3}{f}_{n+3}).\hfill \end{array}$$

From the above remarks, we will solve the following equations:

$$x^2+x{F}_3+F_3=0,$$

(31)

where $F_3=f_3+f_4{e}_1+f_5{e}_2+f_6{e}_3$ is a Fibonacci quaternion;

$$x^2+x{F}_0+F_0=0,$$

(32)

where $F_0=e_1+e_2+2e_3$ is a Fibonacci quaternion

$$x^2+x{F}_0+F_1=0,$$

(33)

where $F_0=e_1+e_2+2e_3$ and $F_1=1+e_1+2e_2+3e_3$ are Fibonacci quaternions.

Proposition 9.

Equation $\left(31\right)$ has the following solutions:

$$z_1=-1-\frac{\alpha }{98}\left({\overline{F}}_3-f_3\right)$$

and

$$z_2=-1-\frac{\beta }{98}\left({\overline{F}}_3-f_3\right),$$

where $\alpha =49+\sqrt{2499}$ and $\beta =49-\sqrt{2499}$.

Proof. 

For Equation $\left(31\right)$, we consider its conjugate ${\overline{x}}^2+{\overline{F}}_3\overline{x}+{\overline{F}}_3=0$. From the above formulae and Proposition 6, we have, $\overline{\mathrm{\Delta }}=0,$$\overline{\mathrm{\Psi }}=1,\overline{\mathrm{\Gamma }}=100,T=0$, $N=(\overline{\mathrm{\Gamma }}\pm \sqrt{{\overline{\mathrm{\Gamma }}}^2-4\overline{\mathrm{\Psi }}})/2$ therefore, $N=(100\pm \sqrt{(10000-4})/2$, $N=50\pm \left(\sqrt{9996}\right)/2$, $N=50\pm \sqrt{2499}$, ${\overline{b}}^{\prime }=-3e_1-5e_2-8e_3$ and $\overline{c}\prime =1$. It results the following solutions:

$x=(-Re\left(\overline{b}\right))/2-{(\overline{b}\prime +T)}^{-1}(\overline{c}\prime -N).$

$x=-1+{(3e_1+5e_2+8e_3)}^{-1}(1-(50\pm \sqrt{2499}))$

$x=-1+{(3e_1+5e_2+8e_3)}^{-1}(49\mp \sqrt{2499}))$

We have ${(3e_1+5e_2+8e_3)}^{-1}=-\frac{1}{98}\left(3e_1+5e_2+8e_3\right)$. Denoting with $\alpha =49+\sqrt{2499}$ and $\beta =49-\sqrt{2499}$, we obtain the following solutions:

$x_1=-1-\frac{\alpha }{98}\left(3e_1+5e_2+8e_3\right)=-1-\frac{\alpha }{98}\left(F_3-f_3\right),$

$x_2=-1-\frac{\beta }{98}\left(3e_1+5e_2+8e_3\right)=-1-\frac{\beta }{98}\left(F_3-f_3\right).$ From here, solutions of Equation (31) are $z_1={\overline{x}}_1=-1-\frac{\alpha }{98}\left({\overline{F}}_3-f_3\right)$ and $z_2={\overline{x}}_2=-1-\frac{\beta }{98}\left({\overline{F}}_3-f_3\right){.}_{}$ □

Proposition 10.

Equation $\left(32\right)$ has the following solutions:

$$z_1=\sqrt{3}-\frac{\sqrt{3}+3}{6}{F}_0$$

and

$$z_2=-\sqrt{3}+\frac{9+\sqrt{3}}{18}{F}_0.$$

Proof. 

For Equation $\left(32\right)$, we consider its conjugate ${\overline{x}}^2+{\overline{F}}_0\overline{x}+{\overline{F}}_0=0$. From the above formulae and Proposition 7, we have, $\overline{\mathrm{\Delta }}=12,$$\overline{\mathrm{\Psi }}=6,\overline{\mathrm{\Gamma }}=6$. Equation $z^3+2\mathrm{\Gamma }{z}^2+({\mathrm{\Gamma }}^2-4\mathrm{\Psi })z-{\mathrm{\Delta }}^2=0$ is the same $z^3+12z^2-12z-144=0$, with solution $z=12$, therefore, $T_{1,2}=\pm 2\sqrt{3}$. Equation $\left(32\right)$ has solution $x=(-Re\left(\overline{b}\right))/2-{(\overline{b}\prime +T)}^{-1}(\overline{c}\prime -N)$. We have $\overline{b}={\overline{b}}^{\prime }=\overline{c}={\overline{c}}^{\prime }=-e_1-e_2-2e_3$, therefore, for $T_1=2\sqrt{3}$, we have

$N_1=(T_1^3+\mathrm{\Gamma }{T}_1+\mathrm{\Delta })/2T_1=\frac{24\sqrt{3}+12\sqrt{3}+12}{4\sqrt{3}}=\frac{9\sqrt{3}+3}{\sqrt{3}}=9+\sqrt{3},$ therefore $x_1=-{(\overline{b}\prime +2\sqrt{3})}^{-1}(\overline{c}\prime -9-\sqrt{3})=$

$=-{\left(2\sqrt{3}-e_1-e_2-2e_3\right)}^{-1}\left(-9-\sqrt{3}-e_1-e_2-2e_3\right)=$

$=\frac{1}{18}\left(2\sqrt{3}+e_1+e_2+2e_3\right)\left(9+\sqrt{3}+e_1+e_2+2e_3\right)=$

$=\frac{1}{18}[18\sqrt{3}+\left(3\sqrt{3}+9\right)F_0]=\sqrt{3}+\frac{\sqrt{3}+3}{6}{F}_0$

For $T_1=-2\sqrt{3},$ we have $N_2=\frac{-24\sqrt{3}-12\sqrt{3}+12}{-4\sqrt{3}}=\frac{36\sqrt{3}-12}{4\sqrt{3}}=9-\sqrt{3},$ therefore,

$x_2=-{\left(-2\sqrt{3}-e_1-e_2-2e_3\right)}^{-1}\left(-9+\sqrt{3}-e_1-e_2-2e_3\right)=$

$=-\frac{1}{18}\left(-2\sqrt{3}+e_1+e_2+2e_3\right)\left(-9+\sqrt{3}-e_1-e_2-2e_3\right)=$

$=\frac{1}{18}[-18\sqrt{3}+\left(-9-\sqrt{3}\right)F_0]=$

$=-\sqrt{3}-\frac{9+\sqrt{3}}{18}{F}_0$. From here, solutions of Equation $\left(32\right)$ are $z_1={\overline{x}}_1=\sqrt{3}+\frac{\sqrt{3}+3}{6}{\overline{F}}_0=\sqrt{3}-\frac{\sqrt{3}+3}{6}{F}_0$ and $z_2={\overline{x}}_2=-\sqrt{3}-\frac{9+\sqrt{3}}{18}{\overline{F}}_0=-\sqrt{3}+\frac{9+\sqrt{3}}{18}{F}_0{.}_{}$ □

Proposition 11.

Equation $\left(33\right)$ has the following solutions:

$$z_{1,2}=-\frac{{\overline{F}}_0+T_{1,2}}{\mathbf{n}\left({\overline{F}}_0+T_{1,2}\right)}(F_1-N_{1,2}),$$

where δ is the real positive solution of the equation $z^3+16z^2+4z-324=0$ and $T_{1,2}=\pm \sqrt{\delta },N_{1,2}=(T_{1,2}^3+8T_{1,2}+18)/2T_{1,2}$.

Proof. 

For Equation $\left(33\right)$, we consider its conjugate ${\overline{x}}^2+{\overline{F}}_0\overline{x}+{\overline{F}}_1=0$. From the above formulae and Proposition 8, we have $\overline{\mathrm{\Delta }}=18,$$\overline{\mathrm{\Psi }}=15,\overline{\mathrm{\Gamma }}=8.$ We have $F_0=e_1+e_2+2e_3$, $F_1=1+e_1+2e_2+3e_3$, $\overline{b}={\overline{b}}^{\prime }=-e_1-e_2-2e_3$, ${\overline{c}}^{\prime }=\overline{c}=1-e_1-2e_2-3e_3.$ It results in the following equation $z^3+2\mathrm{\Gamma }{z}^2+({\mathrm{\Gamma }}^2-4\mathrm{\Psi })z-{\mathrm{\Delta }}^2=0\Rightarrow z^3+16z^2+4z-324=0$. This equation has a real positive solution $\delta \in \left(3,4\right).$ We have $T_{1,2}=\pm \sqrt{\delta },$

$N_{1,2}=(T_{1,2}^3+8T_{1,2}+18)/2T_{1,2}$. Therefore, we get $x_{1,2}=-{({\overline{F}}_0+T_{1,2})}^{-1}({\overline{F}}_1-N_{1,2})=-\frac{F_0+T_{1,2}}{\mathbf{n}\left({\overline{F}}_0+T_{1,2}\right)}({\overline{F}}_1-N_{1,2})$. From here, we obtain solutions for Equation $\left(33\right)$, the same as for $\left(28\right)$, namely $z_{1,2}=-\frac{{\overline{F}}_0+T_{1,2}}{\mathbf{n}\left({\overline{F}}_0+T_{1,2}\right)}(F_1-N_{1,2}){.}_{}$

Now, we consider the equation

$$x^2+Ax+xB+C=0,$$

(34)

with $A,B,C\in \mathbb{H}\left(-1,-1\right)$, which are not in the cases described in Proposition 5, but can be reduced to these situations. We have that $x^2+Ax+xB+C={\left(x+B\right)}^2+\left(A-B\right)\left(x+B\right)-AB+C$. Denoting $y=x+B,\sigma =A-B$ and $\eta =-AB+C,$ it results the following equation

$$y^2+\sigma y+\eta =0.$$

(35)

which is a $\left(16\right)$-type equation. Therefore, if z is a solution to Equation $\left(34\right),$ then $z+B$ is a solution to Equation $\left(35\right)$ and if w is a solution to Equation $\left(35\right),$ therefore $w-B$ is a solution to Equation $\left(34\right)$. □

Proposition 12.

Solutions to the equation

$$x^2+F_{0}x+x{F}_0+F_m=0,$$

(36)

where $F_0=e_1+e_2+2e_3$ and $F_m=f_m+f_{m+1}{e}_1+f_{m+2}{e}_2+f_{m+3}{e}_3$ are Fibonacci quaternions,

$$x_{1,2}=\pm \frac{\theta }{2}-F_0\mp \frac{c_1}{\theta }{e}_1\mp \frac{c_2}{\theta }{e}_2\mp \frac{c_3}{\theta }{e}_3,$$

where θ=$\sqrt{2}\sqrt{\sqrt{3f_{2m+3}+6f_m+36}-f_m-6},c_1=f_{m+1},c_2=f_{m+2},c_3=f_{m+3}$

Proof. 

From the above, Equation $\left(36\right)$ is equivalent to $y^2+\eta =0$, with $\eta =-F_0^2+F_m=6+F_m$. Now, we are in case 3 of Proposition 5. We have $b=0,c_0=6+f_m,c_1=f_{m+1},c_2=f_{m+2},c_3=f_{m+3}$. Therefore,

$$\theta =\sqrt{\frac{-4\left(6+f_m\right)+4\sqrt{3f_{2m+3}+6f_m+36}}{2}}=\sqrt{2}\sqrt{\sqrt{3f_{2m+3}+6f_m+36}-f_m-6}.$$

The solutions to the equation $y^2+\eta =0$ are

$$x_{1,2}=\pm \frac{\theta }{2}\mp \frac{c_1}{\theta }{e}_1\mp \frac{c_2}{\theta }{e}_2\mp \frac{c_3}{\theta }{e}_3.$$

Solutions to Equation $\left(36\right)$ are $x_{1,2}-F_0{.}_{}$ □

Proposition 13.

With the above notations, solutions to the equation

$$x^2+F_{0}x+x{F}_0+F_2=0,$$

(37)

are

$$x_1=1-2e_1-\frac{5}{2}{e}_2-\frac{9}{2}{e}_3$$

and

$$x_2=-1+\frac{1}{2}{e}_2+\frac{1}{2}{e}_3.$$

Proof. 

We take $m=2$ and we get $\theta =\sqrt{2}\sqrt{\sqrt{3f_7+6f_2+36}-f_2-6}=$

$=\sqrt{2}\sqrt{\sqrt{3·13+6+36}-1-6}$=$\sqrt{2}\sqrt{\sqrt{81}-1-6}$=$\sqrt{2}·\sqrt{2}=2.$ We obtain the following solutions: $x_{1,2}=\pm 1-F_0\mp e_1\mp \frac{3}{2}{e}_2\mp \frac{5}{2}{e}_3.$ It results $x_{1,2}=\pm 1-e_1-e_2-2e_3\mp e_1\mp \frac{3}{2}{e}_2\mp \frac{5}{2}{e}_3$. We have $x_1=1-2e_1-\frac{5}{2}{e}_2-\frac{9}{2}{e}_3$ and $x_2=-1+\frac{1}{2}{e}_2+\frac{1}{2}{e}_3$. □

4. Conclusions

In this paper, we gave some new identities and properties in algebras obtained by the Cayley–Dickson process and we provide more examples of quadratic equations over real division quaternion algebras with coefficients Fibonacci quaternion elements. These special coefficients allowed us to solve interesting quadratic equations, provided solutions directly, without using specialized softs. As further research, we intend to study some types of the equation over algebras obtained by the Cayley–Dickson in higher dimensions and to find methods to solve these equations.