On the Dynamics of Some Three-Dimensional Systems of Difference Equations

Abstract

This paper looks into the dynamics of nonlinear systems of difference equations, with particular emphasis on fourth-order cases. Analytical solutions are derived for some cases of systems, a tedious task due to the lack of explicit mathematical techniques for their solution. In addition, the qualitative properties of the solutions, such as boundedness and periodicity, are analyzed through theoretical methods and numerical simulations. The results advance our understanding of nonlinear systems, providing important implications for their use in various scientific fields.

1. Introduction

The system of difference equations (SDE) forms an essential part of mathematical modeling and has had a significant impact on scientific fields and applications. Recently, the field has developed significantly as evidenced by the increased interest in scientific activity within this field. This interest confirms the growing value and recognized usefulness of SDEs in addressing complex phenomena and making informed decisions.

In recent research, there has been an increasing focus on formulating solutions for nonlinear SDEs and the extensive analysis of their dynamical behavior. In addition, this field has extended to the application of models in various disciplines which include engineering, physics, biology, and environmental science. Recent studies have discussed various aspects of nonlinear dynamics, such as local and global behavior, boundedness, topological classifications, and chaos analysis [1,2,3]. For example, Khan et al. [4] explored these dynamics in the context of a discrete hepatitis C virus model. Lei and Han [5] investigated the dynamic behavior of a discrete predator–prey model with a fear effect and Allee effect by theoretical analysis and numerical simulation. However, despite the fact that various methods have been found for solving linear difference equations, the aspect surrounding systems of nonlinear difference equations is generally unknown. Despite efforts by several researchers to convert the complicated nonlinear systems into linear expressions, there is a clear discrepancy in the analytical methods for addressing these systems. These efforts have posed a major challenge for researchers who seek to expand their understanding of the behavior of these systems, demonstrate their properties, and obtain solution forms in general.

The increasing interest in nonlinear SDEs focuses on their capacity to capture intricate dynamics and phenomena that linear models might not be able to capture. Nonlinear systems often operate in a rich method, including bifurcations, chaos, and complex attractors, making them useful instruments for comprehending the dynamics of occurrences in the actual world. A particularly active area of research focuses on finding closed-form solutions to nonlinear SDEs, to advance our understanding of dynamical systems and improve predictions with greater accuracy. (see [6,7,8,9,10,11] ). Numerous researchers have made important contributions to the discovery of behavior of difference equations and solved systems, illuminating their dynamics and stability characteristics. For example, Turki and Elsayed [12] provided the form of solutions and the periodic behavior characteristics of some SDEs. The study by Kara et al. [13] highlights the difficulties in solving nonlinear difference equations in closed form. Khaliq et al. [14] discussed the dynamical analysis of a three-dimensional system with two predators in discrete time and one prey. Taskara and Buyuk in [15] investigated the dynamics of some SDEs and proved a relationship between Pell numbers and the solutions of the systems. In this paper, we extend the system in [16] to the following three-dimensional system and delve into studying the behavior of the solution and illustrating explicit solutions for several cases. In addition, boundedness and periodic behavior are explored for specific cases:

$$\begin{array}{cc}\hfill X_{n+1}& ={\displaystyle \frac{Z_{n-1}{X}_{n-3}}{Y_{n-1}(\pm 1\pm Z_{n-1}{X}_{n-3})}},\hfill \\ \hfill Y_{n+1}& ={\displaystyle \frac{X_{n-1}{Y}_{n-3}}{Z_{n-1}(\pm 1\pm X_{n-1}{Y}_{n-3})}},\hfill \\ \hfill Z_{n+1}& ={\displaystyle \frac{Y_{n-1}{Z}_{n-3}}{X_{n-1}(\pm 1\pm Y_{n-1}{Z}_{n-3})}},\hfill \end{array}$$

(1)

where $n=0,1,2,...$ and the initial conditions $X_{n-i}$, $Y_{n-i}$, and $Z_{n-i}$ for $i\in \{-3,-2,-1,0\}$ are arbitrary nonzero real numbers.

The paper is arranged as follows: In Section 2, we define the system and illustrate all the cases derived from system (1). Some theorems and lemmas are shown in this section. Section 3 is designed to present a numerical simulation to validate the effectiveness of our findings. The discussion of the findings is shown in Section 4.

2. The Main Results

In this section, we substantiate some theorems and lemmas related to the behavior of some cases derived from the system (1) such as boundedness, periodicity, and solutions forms. A nonlinear difference equation is any equation of the form

$$Y_{n+1}=f(Y_n,Y_{n-1},...),n=0,1,2,...$$

(2)

where $y_n$ is the value of y in generation n and where the recursion function f depends on nonlinear combinations of its arguments (f may involve quadratics, exponentials, reciprocals, powers of the $x_n$’s, etc.) [17].

Definition 1 

([18]). Assume $I_X, I_Y$ and $I_Z$ are any intervals of real numbers and $S:I_X^3\times I_Y^3\times I_Z^3\to I_X,F:I_X^3\times I_Y^3\times I_Z^3\to I_Y, T:I_X^3\times I_Y^3\times I_Z^3\to I_Z$ are continuously differentiable functions. Then, for each initial condition ($X_i,Y_i,Z_i$) $\in I_X\times I_Y\times I_Z$ for $i\in \{-3,-2,-1,0\}$, the system of difference Equations (1)

$$\begin{array}{cc}\hfill X_{n+1}& =S(X_n,X_{n-1},X_{n-2},X_{n-3},Y_n,Y_{n-1},Y_{n-2},Y_{n-3},Z_n,Z_{n-1},Z_{n-2},Z_{n-3}),\hfill \\ \hfill Y_{n+1}& =F(X_n,X_{n-1},X_{n-2},X_{n-3},Y_n,Y_{n-1},Y_{n-2},Y_{n-3},Z_n,Z_{n-1},Z_{n-2},Z_{n-3}),\hfill \\ \hfill Z_{n+1}& =T(X_n,X_{n-1},X_{n-2},X_{n-3},Y_n,Y_{n-1},Y_{n-2},Y_{n-3},Z_n,Z_{n-1},Z_{n-2},Z_{n-3}),\hfill \end{array}$$

(3)

has a unique solution ${\{X_n,Y_n,Z_n\}}_{n=-3}^{\infty }$.

Definition 2 

([18]). A sequence $\left\{Y_n\right\}$ is said to be periodic with period p if $Y_{n+p}=Y_n$ for $n=0,1,2,\dots$.

2.1. The First Case

This section is designed to discuss the form of solutions of the following system:

$$\begin{array}{cc}\hfill X_{n+1}& ={\displaystyle \frac{Z_{n-1}{X}_{n-3}}{Y_{n-1}(1+Z_{n-1}{X}_{n-3})}},\hfill \\ \hfill Y_{n+1}& ={\displaystyle \frac{X_{n-1}{Y}_{n-3}}{Z_{n-1}(1+X_{n-1}{Y}_{n-3})}},\hfill \\ \hfill Z_{n+1}& ={\displaystyle \frac{Y_{n-1}{Z}_{n-3}}{X_{n-1}(1+Y_{n-1}{Z}_{n-3})}},\hfill \end{array}$$

(4)

where $n=0,1,2,...$ and the initial conditions are arbitrary nonzero real numbers.

Theorem 1. 

Suppose that $\{X_n,Y_n,Z_n\}$ are solutions of the system (4). Then, for $n\ge 0$, the solutions of system (4) can take the following form:

$$\begin{array}{cc}\hfill X_{12n-3}& ={\displaystyle \frac{a_3{\prod }_{i=0}^{n-1}(1+(6i+2)b_1{c}_3)(1+(6i+4)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(6i+6)c_1{a}_3)}{{\prod }_{i=0}^{n-1}(1+(6i+1)a_1{b}_3)(1+(6i+3)c_1{a}_3)(1+(6i+5)b_1{c}_3)}},\hfill \\ \hfill Y_{12n-3}& ={\displaystyle \frac{b_3{\prod }_{i=0}^{n-1}(1+(6i+2)c_1{a}_3)(1+(6i+4)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(6i+6)a_1{b}_3)}{{\prod }_{i=0}^{n-1}(1+(6i+1)b_1{c}_3)(1+(6i+3)a_1{b}_3)(1+(6i+5)c_1{a}_3)}},\hfill \\ \hfill Z_{12n-3}& ={\displaystyle \frac{c_3{\prod }_{i=0}^{n-1}(1+(6i+2)a_1{b}_3)(1+(6i+4)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(6i+6)b_1{c}_3)}{{\prod }_{i=0}^{n-1}(1+(6i+1)c_1{a}_3)(1+(6i+3)b_1{c}_3)(1+(6i+5)a_1{b}_3)}},\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill X_{12n-2}& ={\displaystyle \frac{a_2{\prod }_{i=0}^{n-1}(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(6i+6)c_0{a}_2)}{{\prod }_{i=0}^{n-1}(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}},\hfill \\ \hfill Y_{12n-2}& ={\displaystyle \frac{b_2{\prod }_{i=0}^{n-1}(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(6i+6)a_0{b}_2)}{{\prod }_{i=0}^{n-1}(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}},\hfill \\ \hfill Z_{12n-2}& ={\displaystyle \frac{c_2{\prod }_{i=0}^{n-1}(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(6i+6)b_0{c}_2)}{{\prod }_{i=0}^{n-1}(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}},\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill X_{12n-1}& =a_1\prod _{i=0}^{n-1}{\displaystyle \frac{(1+(6i+1)b_1{c}_3)(1+(6i+3)a_1{b}_3)(1+(6i+5)c_1{a}_3)}{(1+(6i+2)c_1{a}_3)(1+(6i+4)b_1{c}_3)(1+(6i+6)a_1{b}_3)}},\hfill \\ \hfill Y_{12n-1}& =b_1\prod _{i=0}^{n-1}{\displaystyle \frac{(1+(6i+1)c_1{a}_3)(1+(6i+3)b_1{c}_3)(1+(6i+5)a_1{b}_3)}{(1+(6i+2)a_1{b}_3)(1+(6i+4)c_1{a}_3)(1+(6i+6)b_1{c}_3)}},\hfill \\ \hfill Z_{12n-1}& =c_1\prod _{i=0}^{n-1}{\displaystyle \frac{(1+(6i+1)a_1{b}_3)(1+(6i+3)c_1{a}_3)(1+(6i+5)b_1{c}_3)}{(1+(6i+2)b_1{c}_3)(1+(6i+4)a_1{b}_3)(1+(6i+6)c_1{a}_3)}},\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill X_{12n}& =a_0\prod _{i=0}^{n-1}{\displaystyle \frac{(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}{(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2)(1+(6i+6)a_0{b}_2)}},\hfill \\ \hfill Y_{12n}& =b_0\prod _{i=0}^{n-1}{\displaystyle \frac{(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}{(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2)(1+(6i+6)b_0{c}_2)}},\hfill \\ \hfill Z_{12n}& =c_0\prod _{i=0}^{n-1}{\displaystyle \frac{(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}{(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2)(1+(6i+6)c_0{a}_2)}},\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill X_{12n+1}& ={\displaystyle \frac{c_1{a}_3{\prod }_{i=0}^{n-1}(1+(6i+2)a_1{b}_3)(1+(6i+4)c_1{a}_3)(1+(6i+6)b_1{c}_3)}{b_1{\prod }_{i=0}^n(1+(6i+1)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(6i+3)b_1{c}_3)(1+(6i+5)a_1{b}_3)}},\hfill \\ \hfill Y_{12n+1}& ={\displaystyle \frac{a_1{b}_3{\prod }_{i=0}^{n-1}(1+(6i+2)b_1{c}_3)(1+(6i+4)a_1{b}_3)(1+(6i+6)c_1{a}_3)}{c_1{\prod }_{i=0}^n(1+(6i+1)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(6i+3)c_1{a}_3)(1+(6i+5)b_1{c}_3)}},\hfill \\ \hfill Z_{12n+1}& ={\displaystyle \frac{b_1{c}_3{\prod }_{i=0}^{n-1}(1+(6i+2)c_1{a}_3)(1+(6i+4)b_1{c}_3)(1+(6i+6)a_1{b}_3)}{a_1{\prod }_{i=0}^n(1+(6i+1)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(6i+3)a_1{b}_3)(1+(6i+5)c_1{a}_3)}},\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill X_{12n+2}& ={\displaystyle \frac{c_0{a}_2{\prod }_{i=0}^{n-1}(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2)(1+(6i+6)b_0{c}_2)}{b_0{\prod }_{i=0}^n(1+(6i+1)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}},\hfill \\ \hfill Y_{12n+2}& ={\displaystyle \frac{a_0{b}_2{\prod }_{i=0}^{n-1}(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2)(1+(6i+6)c_0{a}_2)}{c_0{\prod }_{i=0}^n(1+(6i+1)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}},\hfill \\ \hfill Z_{12n+2}& ={\displaystyle \frac{b_0{c}_2{\prod }_{i=0}^{n-1}(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2)(1+(6i+6)a_0{b}_2)}{a_0{\prod }_{i=0}^n(1+(6i+1)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}},\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill X_{12n+3}& ={\displaystyle \frac{b_1{c}_3{c}_1{\prod }_{i=0}^n(1+(6i+1)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(6i+3)c_1{a}_3)(1+(6i+5)b_1{c}_3)}{a_1{b}_3{\prod }_{i=0}^n(1+(6i+2)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(6i+4)a_1{b}_3)(1+(6i+6)c_1{a}_3)}},\hfill \\ \hfill Y_{12n+3}& ={\displaystyle \frac{c_1{a}_3{a}_1{\prod }_{i=0}^n(1+(6i+1)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(6i+3)a_1{b}_3)(1+(6i+5)c_1{a}_3)}{b_1{c}_3{\prod }_{i=0}^n(1+(6i+2)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(6i+4)b_1{c}_3)(1+(6i+6)a_1{b}_3)}},\hfill \\ \hfill Z_{12n+3}& ={\displaystyle \frac{a_1{b}_3{b}_1{\prod }_{i=0}^n(1+(6i+1)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(6i+3)b_1{c}_3)(1+(6i+5)a_1{b}_3)}{c_1{a}_3{\prod }_{i=0}^n(1+(6i+2)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(6i+4)c_1{a}_3)(1+(6i+6)b_1{c}_3)}},\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill X_{12n+4}& ={\displaystyle \frac{b_0{c}_2{c}_0{\prod }_{i=0}^n(1+(6i+1)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}{a_0{b}_2{\prod }_{i=0}^n(1+(6i+2)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(6i+4)a_0{b}_2)(1+(6i+6)c_0{a}_2)}},\hfill \\ \hfill Y_{12n+4}& ={\displaystyle \frac{c_0{a}_2{a}_0{\prod }_{i=0}^n(1+(6i+1)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}{b_0{c}_2{\prod }_{i=0}^n(1+(6i+2)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(6i+4)b_0{c}_2)(1+(6i+6)a_0{b}_2)}},\hfill \\ \hfill Z_{12n+4}& ={\displaystyle \frac{a_0{b}_2{b}_0{\prod }_{i=0}^n(1+(6i+1)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}{c_0{a}_2{\prod }_{i=0}^n(1+(6i+2)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(6i+4)c_0{a}_2)(1+(6i+6)b_0{c}_2)}},\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill X_{12n+5}& ={\displaystyle \frac{b_3{b}_1{c}_3{\prod }_{i=0}^n(1+(6i+2)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(6i+4)b_1{c}_3)(1+(6i+6)a_1{b}_3)}{c_1{a}_3{\prod }_{i=0}^n(1+(6i+1)b_1{c}_3)(1+(6i+3)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(6i+5)c_1{a}_3)}},\hfill \\ \hfill Y_{12n+5}& ={\displaystyle \frac{c_3{c}_1{a}_3{\prod }_{i=0}^n(1+(6i+2)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(6i+4)c_1{a}_3)(1+(6i+6)b_1{c}_3)}{a_1{b}_3{\prod }_{i=0}^n(1+(6i+1)c_1{a}_3)(1+(6i+3)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(6i+5)a_1{b}_3)}},\hfill \\ \hfill Z_{12n+5}& ={\displaystyle \frac{a_3{a}_1{b}_3{\prod }_{i=0}^n(1+(6i+2)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(6i+4)a_1{b}_3)(1+(6i+6)c_1{a}_3)}{b_1{c}_3{\prod }_{i=0}^n(1+(6i+1)a_1{b}_3)(1+(6i+3)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(6i+5)b_1{c}_3)}},\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill X_{12n+6}& ={\displaystyle \frac{b_2{b}_0{c}_2{\prod }_{i=0}^n(1+(6i+2)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(6i+4)b_0{c}_2)(1+(6i+6)a_0{b}_2)}{c_0{a}_2{\prod }_{i=0}^n(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(6i+5)c_0{a}_2)}},\hfill \\ \hfill Y_{12n+6}& ={\displaystyle \frac{c_2{c}_0{a}_2{\prod }_{i=0}^n(1+(6i+2)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(6i+4)c_0{a}_2)(1+(6i+6)b_0{c}_2)}{a_0{b}_2{\prod }_{i=0}^n(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(6i+5)a_0{b}_2)}},\hfill \\ \hfill Z_{12n+6}& ={\displaystyle \frac{a_2{a}_0{b}_2{\prod }_{i=0}^n(1+(6i+2)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(6i+4)a_0{b}_2)(1+(6i+6)c_0{a}_2)}{b_0{c}_2{\prod }_{i=0}^n(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(6i+5)b_0{c}_2)}},\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill X_{12n+7}& ={\displaystyle \frac{a_1{b}_3{\prod }_{i=0}^n(1+(6i+1)c_1{a}_3)(1+(6i+3)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(6i+5)a_1{b}_3)}{c_3{\prod }_{i=0}^n(1+(6i+2)a_1{b}_3)(1+(6i+4)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(6i+6)b_1{c}_3)}},\hfill \\ \hfill Y_{12n+7}& ={\displaystyle \frac{b_1{c}_3{\prod }_{i=0}^n(1+(6i+1)a_1{b}_3)(1+(6i+3)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(6i+5)b_1{c}_3)}{a_3{\prod }_{i=0}^n(1+(6i+2)b_1{c}_3)(1+(6i+4)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(6i+6)c_1{a}_3)}},\hfill \\ \hfill Z_{12n+7}& ={\displaystyle \frac{c_1{a}_3{\prod }_{i=0}^n(1+(6i+1)b_1{c}_3)(1+(6i+3)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(6i+5)c_1{a}_3)}{b_3{\prod }_{i=0}^n(1+(6i+2)c_1{a}_3)(1+(6i+4)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(6i+6)a_1{b}_3)}},\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill X_{12n+8}& ={\displaystyle \frac{a_0{b}_2{\prod }_{i=0}^n(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(6i+5)a_0{b}_2)}{c_2{\prod }_{i=0}^n(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(6i+6)b_0{c}_2)}},\hfill \\ \hfill Y_{12n+8}& ={\displaystyle \frac{b_0{c}_2{\prod }_{i=0}^n(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(6i+5)b_0{c}_2)}{a_2{\prod }_{i=0}^n(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(6i+6)c_0{a}_2)}},\hfill \\ \hfill Z_{12n+8}& ={\displaystyle \frac{c_0{a}_2{\prod }_{i=0}^n(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(6i+5)c_0{a}_2)}{b_2{\prod }_{i=0}^n(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(6i+6)a_0{b}_2)}},\hfill \end{array}$$

(16)

where $X_0=a_0$, $X_{-1}=a_1$, $X_{-2}=a_2$, $X_{-3}=a_3$, $Y_0=b_0$, $Y_{-1}=b_1$, $Y_{-2}=b_2$, $Y_{-3}=b_3$, $Z_0=c_0$, $Z_{-1}=c_1$, $Z_{-2}=c_2$ and $Z_{-3}=c_3$.

Proof. 

The results for $n=0$ are obviously true. Now for $n>0$, assume the results hold for $n-1$ and they are provided as follows:

$$\begin{array}{cc}\hfill X_{12n-15}& ={\displaystyle \frac{a_3{\prod }_{i=0}^{n-2}(1+(6i+2)b_1{c}_3)(1+(6i+4)a_1{b}_3){\prod }_{i=0}^{n-3}(1+(6i+6)c_1{a}_3)}{{\prod }_{i=0}^{n-2}(1+(6i+1)a_1{b}_3)(1+(6i+3)c_1{a}_3)(1+(6i+5)b_1{c}_3)}},\hfill \\ \hfill Y_{12n-15}& ={\displaystyle \frac{b_3{\prod }_{i=0}^{n-2}(1+(6i+2)c_1{a}_3)(1+(6i+4)b_1{c}_3){\prod }_{i=0}^{n-3}(1+(6i+6)a_1{b}_3)}{{\prod }_{i=0}^{n-2}(1+(6i+1)b_1{c}_3)(1+(6i+3)a_1{b}_3)(1+(6i+5)c_1{a}_3)}},\hfill \\ \hfill Z_{12n-15}& ={\displaystyle \frac{c_3{\prod }_{i=0}^{n-2}(1+(6i+2)a_1{b}_3)(1+(6i+4)c_1{a}_3){\prod }_{i=0}^{n-3}(1+(6i+6)b_1{c}_3)}{{\prod }_{i=0}^{n-2}(1+(6i+1)c_1{a}_3)(1+(6i+3)b_1{c}_3)(1+(6i+5)a_1{b}_3)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-14}& ={\displaystyle \frac{a_2{\prod }_{i=0}^{n-2}(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2){\prod }_{i=0}^{n-3}(1+(6i+6)c_0{a}_2)}{{\prod }_{i=0}^{n-2}(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}},\hfill \\ \hfill Y_{12n-14}& ={\displaystyle \frac{b_2{\prod }_{i=0}^{n-2}(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2){\prod }_{i=0}^{n-3}(1+(6i+6)a_0{b}_2)}{{\prod }_{i=0}^{n-2}(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}},\hfill \\ \hfill Z_{12n-14}& ={\displaystyle \frac{c_2{\prod }_{i=0}^{n-2}(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2){\prod }_{i=0}^{n-3}(1+(6i+6)b_0{c}_2)}{{\prod }_{i=0}^{n-2}(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-13}& =a_1\prod _{i=0}^{n-2}{\displaystyle \frac{(1+(6i+1)b_1{c}_3)(1+(6i+3)a_1{b}_3)(1+(6i+5)c_1{a}_3)}{(1+(6i+2)c_1{a}_3)(1+(6i+4)b_1{c}_3)(1+(6i+6)a_1{b}_3)}},\hfill \\ \hfill Y_{12n-13}& =b_1\prod _{i=0}^{n-2}{\displaystyle \frac{(1+(6i+1)c_1{a}_3)(1+(6i+3)b_1{c}_3)(1+(6i+5)a_1{b}_3)}{(1+(6i+2)a_1{b}_3)(1+(6i+4)c_1{a}_3)(1+(6i+6)b_1{c}_3)}},\hfill \\ \hfill Z_{12n-13}& =c_1\prod _{i=0}^{n-2}{\displaystyle \frac{(1+(6i+1)a_1{b}_3)(1+(6i+3)c_1{a}_3)(1+(6i+5)b_1{c}_3)}{(1+(6i+2)b_1{c}_3)(1+(6i+4)a_1{b}_3)(1+(6i+6)c_1{a}_3)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-12}& =a_0\prod _{i=0}^{n-2}{\displaystyle \frac{(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}{(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2)(1+(6i+6)a_0{b}_2)}},\hfill \\ \hfill Y_{12n-12}& =b_0\prod _{i=0}^{n-2}{\displaystyle \frac{(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}{(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2)(1+(6i+6)b_0{c}_2)}},\hfill \\ \hfill Z_{12n-12}& =c_0\prod _{i=0}^{n-2}{\displaystyle \frac{(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}{(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2)(1+(6i+6)c_0{a}_2)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-11}& ={\displaystyle \frac{c_1{a}_3{\prod }_{i=0}^{n-2}(1+(6i+2)a_1{b}_3)(1+(6i+4)c_1{a}_3)(1+(6i+6)b_1{c}_3)}{b_1{\prod }_{i=0}^{n-1}(1+(6i+1)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(6i+3)b_1{c}_3)(1+(6i+5)a_1{b}_3)}},\hfill \\ \hfill Y_{12n-11}& ={\displaystyle \frac{a_1{b}_3{\prod }_{i=0}^{n-2}(1+(6i+2)b_1{c}_3)(1+(6i+4)a_1{b}_3)(1+(6i+6)c_1{a}_3)}{c_1{\prod }_{i=0}^{n-1}(1+(6i+1)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(6i+3)c_1{a}_3)(1+(6i+5)b_1{c}_3)}},\hfill \\ \hfill Z_{12n-11}& ={\displaystyle \frac{b_1{c}_3{\prod }_{i=0}^{n-2}(1+(6i+2)c_1{a}_3)(1+(6i+4)b_1{c}_3)(1+(6i+6)a_1{b}_3)}{a_1{\prod }_{i=0}^{n-1}(1+(6i+1)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(6i+3)a_1{b}_3)(1+(6i+5)c_1{a}_3)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-10}& ={\displaystyle \frac{c_0{a}_2{\prod }_{i=0}^{n-2}(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2)(1+(6i+6)b_0{c}_2)}{b_0{\prod }_{i=0}^{n-1}(1+(6i+1)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}},\hfill \\ \hfill Y_{12n-10}& ={\displaystyle \frac{a_0{b}_2{\prod }_{i=0}^{n-2}(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2)(1+(6i+6)c_0{a}_2)}{c_0{\prod }_{i=0}^{n-1}(1+(6i+1)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}},\hfill \\ \hfill Z_{12n-10}& ={\displaystyle \frac{b_0{c}_2{\prod }_{i=0}^{n-2}(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2)(1+(6i+6)a_0{b}_2)}{a_0{\prod }_{i=0}^{n-1}(1+(6i+1)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-9}& ={\displaystyle \frac{b_1{c}_3{c}_1{\prod }_{i=0}^{n-1}(1+(6i+1)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(6i+3)c_1{a}_3)(1+(6i+5)b_1{c}_3)}{a_1{b}_3{\prod }_{i=0}^{n-1}(1+(6i+2)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(6i+4)a_1{b}_3)(1+(6i+6)c_1{a}_3)}},\hfill \\ \hfill Y_{12n-9}& ={\displaystyle \frac{c_1{a}_3{a}_1{\prod }_{i=0}^{n-1}(1+(6i+1)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(6i+3)a_1{b}_3)(1+(6i+5)c_1{a}_3)}{b_1{c}_3{\prod }_{i=0}^{n-1}(1+(6i+2)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(6i+4)b_1{c}_3)(1+(6i+6)a_1{b}_3)}},\hfill \\ \hfill Z_{12n-9}& ={\displaystyle \frac{a_1{b}_3{b}_1{\prod }_{i=0}^{n-1}(1+(6i+1)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(6i+3)b_1{c}_3)(1+(6i+5)a_1{b}_3)}{c_1{a}_3{\prod }_{i=0}^{n-1}(1+(6i+2)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(6i+4)c_1{a}_3)(1+(6i+6)b_1{c}_3)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-8}& ={\displaystyle \frac{b_0{c}_2{c}_0{\prod }_{i=0}^{n-1}(1+(6i+1)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}{a_0{b}_2{\prod }_{i=0}^{n-1}(1+(6i+2)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(6i+4)a_0{b}_2)(1+(6i+6)c_0{a}_2)}},\hfill \\ \hfill Y_{12n-8}& ={\displaystyle \frac{c_0{a}_2{a}_0{\prod }_{i=0}^{n-1}(1+(6i+1)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}{b_0{c}_2{\prod }_{i=0}^{n-1}(1+(6i+2)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(6i+4)b_0{c}_2)(1+(6i+6)a_0{b}_2)}},\hfill \\ \hfill Z_{12n-8}& ={\displaystyle \frac{a_0{b}_2{b}_0{\prod }_{i=0}^{n-1}(1+(6i+1)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}{c_0{a}_2{\prod }_{i=0}^{n-1}(1+(6i+2)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(6i+4)c_0{a}_2)(1+(6i+6)b_0{c}_2)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-7}& ={\displaystyle \frac{b_3{b}_1{c}_3{\prod }_{i=0}^{n-1}(1+(6i+2)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(6i+4)b_1{c}_3)(1+(6i+6)a_1{b}_3)}{c_1{a}_3{\prod }_{i=0}^{n-1}(1+(6i+1)b_1{c}_3)(1+(6i+3)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(6i+5)c_1{a}_3)}},\hfill \\ \hfill Y_{12n-7}& ={\displaystyle \frac{c_3{c}_1{a}_3{\prod }_{i=0}^{n-1}(1+(6i+2)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(6i+4)c_1{a}_3)(1+(6i+6)b_1{c}_3)}{a_1{b}_3{\prod }_{i=0}^{n-1}(1+(6i+1)c_1{a}_3)(1+(6i+3)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(6i+5)a_1{b}_3)}},\hfill \\ \hfill Z_{12n-7}& ={\displaystyle \frac{a_3{a}_1{b}_3{\prod }_{i=0}^{n-1}(1+(6i+2)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(6i+4)a_1{b}_3)(1+(6i+6)c_1{a}_3)}{b_1{c}_3{\prod }_{i=0}^{n-1}(1+(6i+1)a_1{b}_3)(1+(6i+3)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(6i+5)b_1{c}_3)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-6}& ={\displaystyle \frac{b_2{b}_0{c}_2{\prod }_{i=0}^{n-1}(1+(6i+2)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(6i+4)b_0{c}_2)(1+(6i+6)a_0{b}_2)}{c_0{a}_2{\prod }_{i=0}^{n-1}(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(6i+5)c_0{a}_2)}},\hfill \\ \hfill Y_{12n-6}& ={\displaystyle \frac{c_2{c}_0{a}_2{\prod }_{i=0}^{n-1}(1+(6i+2)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(6i+4)c_0{a}_2)(1+(6i+6)b_0{c}_2)}{a_0{b}_2{\prod }_{i=0}^{n-1}(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(6i+5)a_0{b}_2)}},\hfill \\ \hfill Z_{12n-6}& ={\displaystyle \frac{a_2{a}_0{b}_2{\prod }_{i=0}^{n-1}(1+(6i+2)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(6i+4)a_0{b}_2)(1+(6i+6)c_0{a}_2)}{b_0{c}_2{\prod }_{i=0}^{n-1}(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(6i+5)b_0{c}_2)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-5}& ={\displaystyle \frac{a_1{b}_3{\prod }_{i=0}^{n-1}(1+(6i+1)c_1{a}_3)(1+(6i+3)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(6i+5)a_1{b}_3)}{c_3{\prod }_{i=0}^{n-1}(1+(6i+2)a_1{b}_3)(1+(6i+4)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(6i+6)b_1{c}_3)}},\hfill \\ \hfill Y_{12n-5}& ={\displaystyle \frac{b_1{c}_3{\prod }_{i=0}^{n-1}(1+(6i+1)a_1{b}_3)(1+(6i+3)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(6i+5)b_1{c}_3)}{a_3{\prod }_{i=0}^{n-1}(1+(6i+2)b_1{c}_3)(1+(6i+4)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(6i+6)c_1{a}_3)}},\hfill \\ \hfill Z_{12n-5}& ={\displaystyle \frac{c_1{a}_3{\prod }_{i=0}^{n-1}(1+(6i+1)b_1{c}_3)(1+(6i+3)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(6i+5)c_1{a}_3)}{b_3{\prod }_{i=0}^{n-1}(1+(6i+2)c_1{a}_3)(1+(6i+4)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(6i+6)a_1{b}_3)}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-4}& ={\displaystyle \frac{a_0{b}_2{\prod }_{i=0}^{n-1}(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(6i+5)a_0{b}_2)}{c_2{\prod }_{i=0}^{n-1}(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(6i+6)b_0{c}_2)}},\hfill \\ \hfill Y_{12n-4}& ={\displaystyle \frac{b_0{c}_2{\prod }_{i=0}^{n-1}(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(6i+5)b_0{c}_2)}{a_2{\prod }_{i=0}^{n-1}(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(6i+6)c_0{a}_2)}},\hfill \\ \hfill Z_{12n-4}& ={\displaystyle \frac{c_0{a}_2{\prod }_{i=0}^{n-1}(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(6i+5)c_0{a}_2)}{b_2{\prod }_{i=0}^{n-1}(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(6i+6)a_0{b}_2)}}.\hfill \end{array}$$

Now, we will demonstrate the relations.

Substituting $12n$ into system (4), we obtain

$$\begin{array}{cc}\hfill X_{12n}& ={\displaystyle \frac{Z_{12n-2}{X}_{12n-4}}{Y_{12n-2}(1+Z_{12n-2}{X}_{12n-4})}},\hfill \\ \hfill & ={\displaystyle \frac{\frac{a_0{b}_2}{(1+(6i+5)a_0{b}_2)}}{\frac{b_2{\prod }_{i=0}^{n-1}(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(6i+6)a_0{b}_2)}{{\prod }_{i=0}^{n-1}(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}(1+\frac{a_0{b}_2}{(1+(6i+5)a_0{b}_2)})}}.\hfill \end{array}$$

Therefore, we obtain

$$\begin{array}{cc}\hfill X_{12n}& =a_0\prod _{i=0}^{n-1}{\displaystyle \frac{(1+(6i+1)b_0{c}_2)(1+(6i+3)a_0{b}_2)(1+(6i+5)c_0{a}_2)}{(1+(6i+2)c_0{a}_2)(1+(6i+4)b_0{c}_2)(1+(6i+6)a_0{b}_2)}}.\hfill \end{array}$$

$$\begin{array}{cc}\hfill Y_{12n}& ={\displaystyle \frac{X_{12n-2}{Y}_{12n-4}}{Z_{12n-2}(1+X_{12n-2}{Y}_{12n-4})}},\hfill \\ \hfill & ={\displaystyle \frac{\frac{b_0{c}_2}{(1+(6i+5)b_0{c}_2)}}{\frac{c_2{\prod }_{i=0}^{n-1}(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(6i+6)b_0{c}_2)}{{\prod }_{i=0}^{n-1}(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}(1+\frac{b_0{c}_2}{(1+(6i+5)b_0{c}_2)})}}.\hfill \end{array}$$

Hence, we obtain

$$\begin{array}{cc}\hfill Y_{12n}& =b_0\prod _{i=0}^{n-1}{\displaystyle \frac{(1+(6i+1)c_0{a}_2)(1+(6i+3)b_0{c}_2)(1+(6i+5)a_0{b}_2)}{(1+(6i+2)a_0{b}_2)(1+(6i+4)c_0{a}_2)(1+(6i+6)b_0{c}_2)}}.\hfill \end{array}$$

Also,

$$\begin{array}{cc}\hfill Z_{12n}& ={\displaystyle \frac{Y_{12n-2}{Z}_{12n-4}}{X_{12n-2}(1+Y_{12n-2}{Z}_{12n-4})}},\hfill \\ \hfill & ={\displaystyle \frac{\frac{c_0{a}_2}{(1+(6i+5)c_0{a}_2)}}{\frac{a_2{\prod }_{i=0}^{n-1}(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(6i+6)c_0{a}_2)}{{\prod }_{i=0}^{n-1}(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}(1+\frac{c_0{a}_2}{(1+(6i+5)c_0{a}_2)})}},\hfill \\ \hfill & =c_0\prod _{i=0}^{n-1}{\displaystyle \frac{(1+(6i+1)a_0{b}_2)(1+(6i+3)c_0{a}_2)(1+(6i+5)b_0{c}_2)}{(1+(6i+2)b_0{c}_2)(1+(6i+4)a_0{b}_2)(1+(6i+6)c_0{a}_2)}}.\hfill \end{array}$$

Similarly, we can prove the remaining relations. The proof is complete. □

Lemma 1. 

Every positive solution of the system (4) is bounded and approaches zero.

Proof. 

Let the sequence $\{X_n,Y_n,Z_n\}$ be solutions of system (4) in which the initial conditions are arbitrary positive numbers. We have from (4)

$$\begin{array}{cc}\hfill X_{n+1}& \le {\displaystyle \frac{1}{Y_{n-1}}},\hfill \\ \hfill Y_{n+1}& \le {\displaystyle \frac{1}{Z_{n-1}}},\hfill \\ \hfill Z_{n+1}& \le {\displaystyle \frac{1}{X_{n-1}}}.\hfill \end{array}$$

(17)

So, for $k=1,2,...,$ we obtain

$$X_{n+1}\le {\displaystyle \frac{1}{Y_{n-1}}}\le Z_{n-3}\le {\displaystyle \frac{1}{X_{n-5}}}\le Y_{n-7}\le {\displaystyle \frac{1}{Z_{n-9}}}\le X_{n-11}\le ...\le {\displaystyle \frac{1}{{\left(X_{n-6K+1}\right)}^{(-1)(k+1)}}},$$

setting $n=n+1,n+2,n+3,n+4$ and $n+5$, we obtain

$$X_{n+2}\le {\displaystyle \frac{1}{Y_n}}\le Z_{n-2}\le {\displaystyle \frac{1}{X_{n-4}}}\le Y_{n-6}\le {\displaystyle \frac{1}{Z_{n-8}}}\le X_{n-10}\le ...\le {\displaystyle \frac{1}{{\left(X_{n-6K+2}\right)}^{(-1)(k+1)}}},$$

$$X_{n+3}\le {\displaystyle \frac{1}{Y_n}}\le Z_{n-2}\le {\displaystyle \frac{1}{X_{n-4}}}\le Y_{n-6}\le {\displaystyle \frac{1}{Z_{n-8}}}\le X_{n-10}\le ...\le {\displaystyle \frac{1}{{\left(X_{n-6K+3}\right)}^{(-1)(k+1)}}},$$

$$X_{n+4}\le {\displaystyle \frac{1}{Y_n}}\le Z_{n-2}\le {\displaystyle \frac{1}{X_{n-4}}}\le Y_{n-6}\le {\displaystyle \frac{1}{Z_{n-8}}}\le X_{n-10}\le ...\le {\displaystyle \frac{1}{{\left(X_{n-6K+4}\right)}^{(-1)(k+1)}}},$$

$$X_{n+5}\le {\displaystyle \frac{1}{Y_n}}\le Z_{n-2}\le {\displaystyle \frac{1}{X_{n-4}}}\le Y_{n-6}\le {\displaystyle \frac{1}{Z_{n-8}}}\le X_{n-10}\le ...\le {\displaystyle \frac{1}{{\left(X_{n-6K+5}\right)}^{(-1)(k+1)}}},$$

$$X_{n+6}\le {\displaystyle \frac{1}{Y_n}}\le Z_{n-2}\le {\displaystyle \frac{1}{X_{n-4}}}\le Y_{n-6}\le {\displaystyle \frac{1}{Z_{n-8}}}\le X_{n-10}\le ...\le {\displaystyle \frac{1}{{\left(X_{n-6K+6}\right)}^{(-1)(k+1)}}}.$$

Likewise, for the sequences $Y_{n+1}$ and $Z_{n+1}$, we observe that all subsequences of $X_n$, $Y_n$, $Z_n$ are decreasing. Therefore, they are bounded from above by

$$\begin{array}{cc}\hfill X_{n+1}& =max\{X_{-3},X_{-2},X_{-1},X_0\},\hfill \\ \hfill Y_{n+1}& =max\{Y_{-3},Y_{-2},Y_{-1},Y_0\},\hfill \\ \hfill Z_{n+1}& =max\{Z_{-3},Z_{-2},Z_{-1},Z_0\}.\hfill \end{array}$$

(18)

□

2.2. The Second Case

The derivation of the solution’s form is demonstrated in this subsection for the following system:

$$\begin{array}{cc}\hfill X_{n+1}& ={\displaystyle \frac{Z_{n-1}{X}_{n-3}}{Y_{n-1}(1+Z_{n-1}{X}_{n-3})}},\hfill \\ \hfill Y_{n+1}& ={\displaystyle \frac{X_{n-1}{Y}_{n-3}}{Z_{n-1}(1+X_{n-1}{Y}_{n-3})}},\hfill \\ \hfill Z_{n+1}& ={\displaystyle \frac{Y_{n-1}{Z}_{n-3}}{X_{n-1}(-1+Y_{n-1}{Z}_{n-3})}},\hfill \end{array}$$

(19)

where $n=0,1,2,...$ and the initial conditions are arbitrary nonzero real numbers.

Theorem 2. 

Assume that $\{X_n,Y_n,Z_n\}$ are solutions of system (19). Then, for $n\ge 0$, the solutions of system (19) are formed as follows:

$$\begin{array}{cc}\hfill X_{12n-3}& ={\displaystyle \frac{{(-1)}^n{a}_3{(1-2a_1{b}_3)}^n{(-1+2b_1{c}_3)}^n}{{(1-b_1{c}_3)}^n{(-1-c_1{a}_3)}^n{(1+a_1{b}_3)}^n}},\hfill \\ \hfill Y_{12n-3}& ={\displaystyle \frac{b_3{(1-2b_1{c}_3)}^n{(1+2c_1{a}_3)}^n}{{(1-c_1{a}_3)}^n{(-1+b_1{c}_3)}^n{(1-a_1{b}_3)}^n}},\hfill \\ \hfill Z_{12n-3}& ={\displaystyle \frac{{(-1)}^n{c}_3}{{(-1+3b_1{c}_3)}^n{(1+c_1{a}_3)}^n{(-1+a_1{b}_3)}^n}},\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill X_{12n-2}& ={\displaystyle \frac{{(-1)}^n{a}_2{(1-2a_0{b}_2)}^n{(-1+2b_0{c}_2)}^n}{{(1-b_0{c}_2)}^n{(-1-c_0{a}_2)}^n{(1+a_0{b}_2)}^n}},\hfill \\ \hfill Y_{12n-2}& ={\displaystyle \frac{b_2{(1-2b_0{c}_2)}^n{(1+2c_0{a}_2)}^n}{{(1-c_0{a}_2)}^n{(-1+b_0{c}_2)}^n{(1-a_0{b}_2)}^n}},\hfill \\ \hfill Z_{12n-2}& ={\displaystyle \frac{{(-1)}^n{c}_2}{{(-1+3b_0{c}_2)}^n{(1+c_0{a}_2)}^n{(-1+a_0{b}_2)}^n}},\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill X_{12n-1}& ={\displaystyle \frac{a_1{(-1+b_1{c}_3)}^n{(1-a_1{b}_3)}^n{(1-c_1{a}_3)}^n}{{(1-2b_1{c}_3)}^n{(1+2c_1{a}_3)}^n}},\hfill \\ \hfill Y_{12n-1}& ={(-1)}^n{b}_1{(1+c_1{a}_3)}^n{(-1+3b_1{c}_3)}^n{(-1+a_1{b}_3)}^n,\hfill \\ \hfill Z_{12n-1}& ={\displaystyle \frac{{(-1)}^n{c}_1{(1+a_1{b}_3)}^n{(-1-c_1{a}_3)}^n{(1-b_1{c}_3)}^n}{{(1-2a_1{b}_3)}^n{(-1+2b_1{c}_3)}^n}},\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill X_{12n}& ={\displaystyle \frac{a_0{(-1+b_0{c}_2)}^n{(1-a_0{b}_2)}^n{(1-c_0{a}_2)}^n}{{(1-2b_0{c}_2)}^n{(1+2c_0{a}_2)}^n}},\hfill \\ \hfill Y_{12n}& ={(-1)}^n{b}_0{(1+c_0{a}_2)}^n{(-1+3b_0{c}_2)}^n{(-1+a_0{b}_2)}^n,\hfill \\ \hfill Z_{12n}& ={\displaystyle \frac{{(-1)}^n{c}_0{(1+a_0{b}_2)}^n{(-1-c_0{a}_2)}^n{(1-b_0{c}_2)}^n}{{(1-2a_0{b}_2)}^n{(-1+2b_0{c}_2)}^n}},\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill X_{12n+1}& ={\displaystyle \frac{{(-1)}^n{c}_1{a}_3}{b_1{(1+c_1{a}_3)}^{n+1}{(-1+3b_1{c}_3)}^n{(-1+a_1{b}_3)}^n}},\hfill \\ \hfill Y_{12n+1}& ={\displaystyle \frac{{(-1)}^n{a}_1{b}_3{(-1+2b_1{c}_3)}^n{(1-2a_1{b}_3)}^n}{c_1{(-1-c_1{a}_3)}^n{(1-b_1{c}_3)}^n{(1+a_1{b}_3)}^{n+1}}},\hfill \\ \hfill Z_{12n+1}& ={\displaystyle \frac{b_1{c}_3{(1+2c_1{a}_3)}^n{(1-2b_1{c}_3)}^n}{a_1{(1-a_1{b}_3)}^n{(1-c_1{a}_3)}^n{(-1+b_1{c}_3)}^{n+1}}},\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill X_{12n+2}& ={\displaystyle \frac{{(-1)}^n{c}_0{a}_2}{b_0{(1+c_0{a}_2)}^{n+1}{(-1+3b_0{c}_2)}^n{(-1+a_0{b}_2)}^n}},\hfill \\ \hfill Y_{12n+2}& ={\displaystyle \frac{{(-1)}^n{a}_0{b}_2{(-1+2b_0{c}_2)}^n{(1-2a_0{b}_2)}^n}{c_0{(-1-c_0{a}_2)}^n{(1-b_0{c}_2)}^n{(1+a_1{b}_3)}^{n+1}}},\hfill \\ \hfill Z_{12n+2}& ={\displaystyle \frac{b_0{c}_2{(1+2c_0{a}_2)}^n{(1-2b_0{c}_2)}^n}{a_1{(1-a_0{b}_2)}^n{(1-c_0{a}_2)}^n{(-1+b_0{c}_2)}^{n+1}}},\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill X_{12n+3}& ={\displaystyle \frac{{(-1)}^n{c}_1{c}_3{b}_1{(1+a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^n{(1-b_1{c}_3)}^n}{a_1{b}_3{(-1+2b_1{c}_3)}^{n+1}{(1-2a_1{b}_3)}^n}},\hfill \\ \hfill Y_{12n+3}& ={\displaystyle \frac{a_1{a}_3{c}_1{(-1+b_1{c}_3)}^{n+1}{(1-a_1{b}_3)}^n{(1-c_1{a}_3)}^n}{b_1{c}_3{(1-2b_1{c}_3)}^n{(1+2c_1{a}_3)}^{n+1}}},\hfill \\ \hfill Z_{12n+3}& ={\displaystyle \frac{{(-1)}^{n+1}{b}_1{b}_3{a}_1{(1+c_1{a}_3)}^{n+1}{(-1+3b_1{c}_3)}^n{(-1+a_1{b}_3)}^n}{c_1{a}_3}},\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill X_{12n+4}& ={\displaystyle \frac{{(-1)}^n{c}_0{c}_2{b}_0{(1+a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^n{(1-b_0{c}_2)}^n}{a_0{b}_2{(-1+2b_0{c}_2)}^{n+1}{(1-2a_0{b}_2)}^n}},\hfill \\ \hfill Y_{12n+4}& ={\displaystyle \frac{a_0{a}_2{c}_0{(-1+b_0{c}_2)}^{n+1}{(1-a_0{b}_2)}^n{(1-c_0{a}_2)}^n}{b_0{c}_2{(1-2b_0{c}_2)}^n{(1+2c_0{a}_2)}^{n+1}}},\hfill \\ \hfill Z_{12n+4}& ={\displaystyle \frac{{(-1)}^{n+1}{b}_0{b}_2{a}_0{(1+c_0{a}_2)}^{n+1}{(-1+3b_0{c}_2)}^n{(-1+a_0{b}_2)}^n}{c_0{a}_2}},\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill X_{12n+5}& ={\displaystyle \frac{{(-1)}^n{b}_1{b}_3{c}_3{(1+2c_1{a}_3)}^{n+1}{(1-2b_1{c}_3)}^n}{c_1{a}_3{(1-c_1{a}_3)}^n{(-1+b_1{c}_3)}^{n+1}{(1-a_1{b}_3)}^{n+1}}},\hfill \\ \hfill Y_{12n+5}& ={\displaystyle \frac{{(-1)}^n{c}_1{c}_3{a}_3}{a_1{b}_3{(1+c_1{a}_3)}^{n+1}{(-1+3b_1{c}_3)}^{n+1}{(-1+a_1{b}_3)}^n}},\hfill \\ \hfill Z_{12n+5}& ={\displaystyle \frac{{(-1)}^n{a}_1{a}_3{b}_3{(1-2a_1{b}_3)}^n{(-1+2b_1{c}_3)}^{n+1}}{b_1{c}_3{(1-b_1{c}_3)}^n{(-1-c_1{a}_3)}^{n+1}{(1+a_1{b}_3)}^{n+1}}},\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill X_{12n+6}& ={\displaystyle \frac{{(-1)}^n{b}_0{b}_2{c}_2{(1+2c_0{a}_2)}^{n+1}{(1-2b_0{c}_2)}^n}{c_0{a}_2{(1-c_0{a}_2)}^n{(-1+b_0{c}_2)}^{n+1}{(1-a_0{b}_2)}^{n+1}}},\hfill \\ \hfill Y_{12n+6}& ={\displaystyle \frac{{(-1)}^n{c}_0{c}_2{a}_2}{a_0{b}_2{(1+c_0{a}_2)}^{n+1}{(-1+3b_0{c}_2)}^{n+1}{(-1+a_0{b}_2)}^n}},\hfill \\ \hfill Z_{12n+6}& ={\displaystyle \frac{{(-1)}^n{a}_0{a}_2{b}_2{(1-2a_0{b}_2)}^n{(-1+2b_0{c}_2)}^{n+1}}{b_0{c}_2{(1-b_0{c}_2)}^n{(-1-c_0{a}_2)}^{n+1}{(1+a_0{b}_2)}^{n+1}}},\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill X_{12n+7}& ={\displaystyle \frac{{(-1)}^n{a}_1{b}_3{(-1+3b_1{c}_3)}^{n+1}{(1+c_1{a}_3)}^{n+1}{(-1+a_1{b}_3)}^n}{c_3}},\hfill \\ \hfill Y_{12n+7}& ={\displaystyle \frac{{(-1)}^{n+1}{b}_1{c}_3{(1+a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^{n+1}{(1-b_1{c}_3)}^n}{a_3{(1-2a_1{b}_3)}^{n+1}{(-1+2a_1{b}_3)}^{n+1}}},\hfill \\ \hfill Z_{12n+7}& ={\displaystyle \frac{-c_1{a}_3{(1-c_1{a}_3)}^n{(-1+b_1{c}_3)}^{n+1}{(1-a_1{b}_3)}^{n+1}}{b_3{(1-2b_1{c}_3)}^{n+1}{(1+2c_1{a}_3)}^{n+1}}},\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill X_{12n+8}& ={\displaystyle \frac{{(-1)}^n{a}_0{b}_2{(-1+3b_0{c}_2)}^{n+1}{(1+c_0{a}_2)}^{n+1}{(-1+a_0{b}_2)}^n}{c_2}},\hfill \\ \hfill Y_{12n+8}& ={\displaystyle \frac{{(-1)}^{n+1}{b}_0{c}_2{(1+a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^{n+1}{(1-b_0{c}_2)}^n}{a_2{(1-2a_0{b}_2)}^{n+1}{(-1+2a_0{b}_2)}^{n+1}}},\hfill \\ \hfill Z_{12n+8}& ={\displaystyle \frac{-c_0{a}_2{(1-c_0{a}_2)}^n{(-1+b_0{c}_2)}^{n+1}{(1-a_0{b}_2)}^{n+1}}{b_2{(1-2b_0{c}_2)}^{n+1}{(1+2c_0{a}_2)}^{n+1}}}.\hfill \end{array}$$

(31)

Proof. 

For $n=0$, it is obvious that the results are true. Now for $n>0$, assume that the results hold for $n-1$ and they are provided in the form of

$$\begin{array}{cc}\hfill X_{12n-15}& ={\displaystyle \frac{{(-1)}^{n-1}{a}_3{(1-2a_1{b}_3)}^{n-1}{(-1+2b_1{c}_3)}^{n-1}}{{(1-b_1{c}_3)}^{n-1}{(-1-c_1{a}_3)}^{n-1}{(1+a_1{b}_3)}^{n-1}}},\hfill \\ \hfill Y_{12n-15}& ={\displaystyle \frac{b_3{(1-2b_1{c}_3)}^{n-1}{(1+2c_1{a}_3)}^{n-1}}{{(1-c_1{a}_3)}^{n-1}{(-1+b_1{c}_3)}^{n-1}{(1-a_1{b}_3)}^{n-1}}},\hfill \\ \hfill Z_{12n-15}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_3}{{(-1+3b_1{c}_3)}^{n-1}{(1+c_1{a}_3)}^{n-1}{(-1+a_1{b}_3)}^{n-1}}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-14}& ={\displaystyle \frac{{(-1)}^{n-1}{a}_2{(1-2a_0{b}_2)}^{n-1}{(-1+2b_0{c}_2)}^{n-1}}{{(1-b_0{c}_2)}^{n-1}{(-1-c_0{a}_2)}^{n-1}{(1+a_0{b}_2)}^{n-1}}},\hfill \\ \hfill Y_{12n-14}& ={\displaystyle \frac{b_2{(1-2b_0{c}_2)}^{n-1}{(1+2c_0{a}_2)}^{n-1}}{{(1-c_0{a}_2)}^{n-1}{(-1+b_0{c}_2)}^{n-1}{(1-a_0{b}_2)}^{n-1}}},\hfill \\ \hfill Z_{12n-14}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_2}{{(-1+3b_0{c}_2)}^{n-1}{(1+c_0{a}_2)}^{n-1}{(-1+a_0{b}_2)}^{n-1}}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-13}& ={\displaystyle \frac{a_1{(-1+b_1{c}_3)}^{n-1}{(1-a_1{b}_3)}^{n-1}{(1-c_1{a}_3)}^{n-1}}{{(1-2b_1{c}_3)}^{n-1}{(1+2c_1{a}_3)}^{n-1}}},\hfill \\ \hfill Y_{12n-13}& ={(-1)}^{n-1}{b}_1{(1+c_1{a}_3)}^{n-1}{(-1+3b_1{c}_3)}^{n-1}{(-1+a_1{b}_3)}^{n-1},\hfill \\ \hfill Z_{12n-13}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_1{(1+a_1{b}_3)}^{n-1}{(-1-c_1{a}_3)}^{n-1}{(1-b_1{c}_3)}^{n-1}}{{(1-2a_1{b}_3)}^{n-1}{(-1+2b_1{c}_3)}^{n-1}}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-12}& ={\displaystyle \frac{a_0{(-1+b_0{c}_2)}^{n-1}{(1-a_0{b}_2)}^{n-1}{(1-c_0{a}_2)}^{n-1}}{{(1-2b_0{c}_2)}^{n-1}{(1+2c_0{a}_2)}^{n-1}}},\hfill \\ \hfill Y_{12n-12}& ={(-1)}^{n-1}{b}_0{(1+c_0{a}_2)}^{n-1}{(-1+3b_0{c}_2)}^{n-1}{(-1+a_0{b}_2)}^{n-1},\hfill \\ \hfill Z_{12n-12}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_0{(1+a_0{b}_2)}^{n-1}{(-1-c_0{a}_2)}^{n-1}{(1-b_0{c}_2)}^{n-1}}{{(1-2a_0{b}_2)}^{n-1}{(-1+2b_0{c}_2)}^{n-1}}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-11}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_1{a}_3}{b_1{(1+c_1{a}_3)}^n{(-1+3b_1{c}_3)}^{n-1}{(-1+a_1{b}_3)}^{n-1}}},\hfill \\ \hfill Y_{12n-11}& ={\displaystyle \frac{{(-1)}^{n-1}{a}_1{b}_3{(-1+2b_1{c}_3)}^{n-1}{(1-2a_1{b}_3)}^{n-1}}{c_1{(-1-c_1{a}_3)}^{n-1}{(1-b_1{c}_3)}^{n-1}{(1+a_1{b}_3)}^n}},\hfill \\ \hfill Z_{12n-11}& ={\displaystyle \frac{b_1{c}_3{(1+2c_1{a}_3)}^{n-1}{(1-2b_1{c}_3)}^{n-1}}{a_1{(1-a_1{b}_3)}^{n-1}{(1-c_1{a}_3)}^{n-1}{(-1+b_1{c}_3)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-10}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_0{a}_2}{b_0{(1+c_0{a}_2)}^n{(-1+3b_0{c}_2)}^{n-1}{(-1+a_0{b}_2)}^{n-1}}},\hfill \\ \hfill Y_{12n-10}& ={\displaystyle \frac{{(-1)}^{n-1}{a}_0{b}_2{(-1+2b_0{c}_2)}^{n-1}{(1-2a_0{b}_2)}^{n-1}}{c_0{(-1-c_0{a}_2)}^{n-1}{(1-b_0{c}_2)}^{n-1}{(1+a_1{b}_3)}^n}},\hfill \\ \hfill Z_{12n-10}& ={\displaystyle \frac{b_0{c}_2{(1+2c_0{a}_2)}^{n-1}{(1-2b_0{c}_2)}^{n-1}}{a_1{(1-a_0{b}_2)}^{n-1}{(1-c_0{a}_2)}^{n-1}{(-1+b_0{c}_2)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-9}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_1{c}_3{b}_1{(1+a_1{b}_3)}^n{(-1-c_1{a}_3)}^{n-1}{(1-b_1{c}_3)}^{n-1}}{a_1{b}_3{(-1+2b_1{c}_3)}^n{(1-2a_1{b}_3)}^{n-1}}},\hfill \\ \hfill Y_{12n-9}& ={\displaystyle \frac{a_1{a}_3{c}_1{(-1+b_1{c}_3)}^n{(1-a_1{b}_3)}^{n-1}{(1-c_1{a}_3)}^{n-1}}{b_1{c}_3{(1-2b_1{c}_3)}^{n-1}{(1+2c_1{a}_3)}^n}},\hfill \\ \hfill Z_{12n-9}& ={\displaystyle \frac{{(-1)}^n{b}_1{b}_3{a}_1{(1+c_1{a}_3)}^n{(-1+3b_1{c}_3)}^{n-1}{(-1+a_1{b}_3)}^{n-1}}{c_1{a}_3}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-8}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_0{c}_2{b}_0{(1+a_0{b}_2)}^n{(-1-c_0{a}_2)}^{n-1}{(1-b_0{c}_2)}^{n-1}}{a_0{b}_2{(-1+2b_0{c}_2)}^n{(1-2a_0{b}_2)}^{n-1}}},\hfill \\ \hfill Y_{12n-8}& ={\displaystyle \frac{a_0{a}_2{c}_0{(-1+b_0{c}_2)}^n{(1-a_0{b}_2)}^{n-1}{(1-c_0{a}_2)}^{n-1}}{b_0{c}_2{(1-2b_0{c}_2)}^{n-1}{(1+2c_0{a}_2)}^n}},\hfill \\ \hfill Z_{12n-8}& ={\displaystyle \frac{{(-1)}^n{b}_0{b}_2{a}_0{(1+c_0{a}_2)}^n{(-1+3b_0{c}_2)}^{n-1}{(-1+a_0{b}_2)}^{n-1}}{c_0{a}_2}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-7}& ={\displaystyle \frac{{(-1)}^{n-1}{b}_1{b}_3{c}_3{(1+2c_1{a}_3)}^n{(1-2b_1{c}_3)}^{n-1}}{c_1{a}_3{(1-c_1{a}_3)}^{n-1}{(-1+b_1{c}_3)}^n{(1-a_1{b}_3)}^n}},\hfill \\ \hfill Y_{12n-7}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_1{c}_3{a}_3}{a_1{b}_3{(1+c_1{a}_3)}^n{(-1+3b_1{c}_3)}^n{(-1+a_1{b}_3)}^{n-1}}},\hfill \\ \hfill Z_{12n-7}& ={\displaystyle \frac{{(-1)}^{n-1}{a}_1{a}_3{b}_3{(1-2a_1{b}_3)}^{n-1}{(-1+2b_1{c}_3)}^n}{b_1{c}_3{(1-b_1{c}_3)}^{n-1}{(-1-c_1{a}_3)}^n{(1+a_1{b}_3)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-6}& ={\displaystyle \frac{{(-1)}^{n-1}{b}_0{b}_2{c}_2{(1+2c_0{a}_2)}^n{(1-2b_0{c}_2)}^{n-1}}{c_0{a}_2{(1-c_0{a}_2)}^{n-1}{(-1+b_0{c}_2)}^n{(1-a_0{b}_2)}^n}},\hfill \\ \hfill Y_{12n-6}& ={\displaystyle \frac{{(-1)}^{n-1}{c}_0{c}_2{a}_2}{a_0{b}_2{(1+c_0{a}_2)}^n{(-1+3b_0{c}_2)}^n{(-1+a_0{b}_2)}^{n-1}}},\hfill \\ \hfill Z_{12n-6}& ={\displaystyle \frac{{(-1)}^{n-1}{a}_0{a}_2{b}_2{(1-2a_0{b}_2)}^{n-1}{(-1+2b_0{c}_2)}^n}{b_0{c}_2{(1-b_0{c}_2)}^{n-1}{(-1-c_0{a}_2)}^n{(1+a_0{b}_2)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-5}& ={\displaystyle \frac{{(-1)}^{n-1}{a}_1{b}_3{(-1+3b_1{c}_3)}^n{(1+c_1{a}_3)}^n{(-1+a_1{b}_3)}^{n-1}}{c_3}},\hfill \\ \hfill Y_{12n-5}& ={\displaystyle \frac{{(-1)}^n{b}_1{c}_3{(1+a_1{b}_3)}^n{(-1-c_1{a}_3)}^n{(1-b_1{c}_3)}^{n-1}}{a_3{(1-2a_1{b}_3)}^n{(-1+2a_1{b}_3)}^n}},\hfill \\ \hfill Z_{12n-5}& ={\displaystyle \frac{-c_1{a}_3{(1-c_1{a}_3)}^{n-1}{(-1+b_1{c}_3)}^n{(1-a_1{b}_3)}^n}{b_3{(1-2b_1{c}_3)}^n{(1+2c_1{a}_3)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{12n-4}& ={\displaystyle \frac{{(-1)}^{n-1}{a}_0{b}_2{(-1+3b_0{c}_2)}^n{(1+c_0{a}_2)}^n{(-1+a_0{b}_2)}^{n-1}}{c_2}},\hfill \\ \hfill Y_{12n-4}& ={\displaystyle \frac{{(-1)}^n{b}_0{c}_2{(1+a_0{b}_2)}^n{(-1-c_0{a}_2)}^n{(1-b_0{c}_2)}^{n-1}}{a_2{(1-2a_0{b}_2)}^n{(-1+2a_0{b}_2)}^n}},\hfill \\ \hfill Z_{12n-4}& ={\displaystyle \frac{-c_0{a}_2{(1-c_0{a}_2)}^{n-1}{(-1+b_0{c}_2)}^n{(1-a_0{b}_2)}^n}{b_2{(1-2b_0{c}_2)}^n{(1+2c_0{a}_2)}^n}}.\hfill \end{array}$$

Now, we will demonstrate the relation.

Substituting $12n-1$ into the system (19), we obtain

$$\begin{array}{cc}\hfill X_{12n-1}& ={\displaystyle \frac{Z_{12n-3}{X}_{12n-5}}{Y_{12n-3}(1+Z_{12n-3}{X}_{12n-5})}},\hfill \\ \hfill & ={\displaystyle \frac{\frac{a_1{b}_3}{(1-a_1{b}_3)}}{\frac{b_3{(1-2b_1{c}_3)}^n{(1+2c_1{a}_3)}^n}{{(1-c_1{a}_3)}^n{(-1+b_1{c}_3)}^n{(1-a_1{b}_3)}^n}(1+\frac{a_1{b}_3}{(1-a_1{b}_3)})}},\hfill \end{array}$$

and therefore,

$$\begin{array}{cc}\hfill X_{12n-1}& ={\displaystyle \frac{a_1{(-1+b_1{c}_3)}^n{(1-a_1{b}_3)}^n{(1-c_1{a}_3)}^n}{{(1-2b_1{c}_3)}^n{(1+2c_1{a}_3)}^n}}.\hfill \end{array}$$

Also,

$$\begin{array}{cc}\hfill Y_{12n-1}& ={\displaystyle \frac{X_{12n-3}{Y}_{12n-5}}{Z_{12n-3}(1+X_{12n-3}{Y}_{12n-5})}},\hfill \\ \hfill & ={\displaystyle \frac{\frac{b_1{c}_3}{(1-b_1{c}_3)}}{\frac{{(-1)}^n{c}_3}{{(-1+3b_1{c}_3)}^n{(1+c_1{a}_3)}^n{(-1+a_1{b}_3)}^n}(1+\frac{b_1{c}_3}{(1-b_1{c}_3)})}},\hfill \end{array}$$

and therefore,

$$\begin{array}{cc}\hfill Y_{12n-1}& ={(-1)}^n{b}_1{(1+c_1{a}_3)}^n{(-1+3b_1{c}_3)}^n{(-1+a_1{b}_3)}^n.\hfill \end{array}$$

Also, for equation $Y_{12n-1}$, we obtain from (19)

$$\begin{array}{cc}\hfill Z_{12n-1}& ={\displaystyle \frac{Y_{12n-3}{Z}_{12n-5}}{X_{12n-3}(-1+Y_{12n-3}{Z}_{12n-5})}},\hfill \\ \hfill & ={\displaystyle \frac{\frac{c_1{a}_3}{(-1+c_1{a}_3)}}{\frac{{(-1)}^n{a}_3{(1-2a_1{b}_3)}^n{(-1+2b_1{c}_3)}^n}{{(1-b_1{c}_3)}^n{(-1-c_1{a}_3)}^n{(1+a_1{b}_3)}^n}(-1+\frac{c_1{a}_3}{(-1+c_1{a}_3)})}},\hfill \end{array}$$

and hence,

$$\begin{array}{cc}\hfill Z_{12n-1}& ={\displaystyle \frac{{(-1)}^n{c}_1{(1+a_1{b}_3)}^n{(-1-c_1{a}_3)}^n{(1-b_1{c}_3)}^n}{{(1-2a_1{b}_3)}^n{(-1+2b_1{c}_3)}^n}}.\hfill \end{array}$$

Likewise, we can show the other relations. The proof is complete. □

2.3. The Third Case

In this section, we examine the solution of the SDEs

$$\begin{array}{cc}\hfill X_{n+1}& ={\displaystyle \frac{Z_{n-1}{X}_{n-3}}{Y_{n-1}(-1-Z_{n-1}{X}_{n-3})}},\hfill \\ \hfill Y_{n+1}& ={\displaystyle \frac{X_{n-1}{Y}_{n-3}}{Z_{n-1}(-1-X_{n-1}{Y}_{n-3})}},\hfill \\ \hfill Z_{n+1}& ={\displaystyle \frac{Y_{n-1}{Z}_{n-3}}{X_{n-1}(-1-Y_{n-1}{Z}_{n-3})}},\hfill \end{array}$$

(32)

where $n=0,1,2,...$ and the initial conditions are arbitrary nonzero real numbers.

Theorem 3. 

Assume that the sequence $\{X_n,Y_n,Z_n\}$ are solutions of system (32). Then, for $n=0,1,2,3,...$

$$\begin{array}{cc}\hfill X_{12n-3}& ={\displaystyle \frac{a_3}{{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n}},\hfill \\ \hfill Y_{12n-3}& ={\displaystyle \frac{b_3}{{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n}},\hfill \\ \hfill Z_{12n-3}& ={\displaystyle \frac{c_3}{{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n}},\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill X_{12n-2}& ={\displaystyle \frac{a_2}{{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n}},\hfill \\ \hfill Y_{12n-2}& ={\displaystyle \frac{b_2}{{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n}},\hfill \\ \hfill Z_{12n-2}& ={\displaystyle \frac{c_2}{{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n}},\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill X_{12n-1}& =a_1{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n,\hfill \\ \hfill Y_{12n-1}& =b_1{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n,\hfill \\ \hfill Z_{12n-1}& =c_1{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n,\hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill X_{12n}& =a_0{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n,\hfill \\ \hfill Y_{12n}& =b_0{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n,\hfill \\ \hfill Z_{12n}& =c_0{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n,\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill X_{12n+1}& ={\displaystyle \frac{c_1{a}_3}{b_1{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n}},\hfill \\ \hfill Y_{12n+1}& ={\displaystyle \frac{a_1{b}_3}{c_1{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n}},\hfill \\ \hfill Z_{12n+1}& ={\displaystyle \frac{b_1{c}_3}{a_1{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n}},\hfill \end{array}$$

(37)

$$\begin{array}{cc}\hfill X_{12n+2}& ={\displaystyle \frac{c_0{a}_2}{b_0{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n}},\hfill \\ \hfill Y_{12n+2}& ={\displaystyle \frac{a_0{b}_2}{c_0{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n}},\hfill \\ \hfill Z_{12n+2}& ={\displaystyle \frac{b_0{c}_2}{a_0{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n}},\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill X_{12n+3}& ={\displaystyle \frac{c_1{c}_3{b}_1{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n}{a_1{b}_3}},\hfill \\ \hfill Y_{12n+3}& ={\displaystyle \frac{a_1{a}_3{c}_1{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n}{b_1{c}_3}},\hfill \\ \hfill Z_{12n+3}& ={\displaystyle \frac{b_1{b}_3{a}_1{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n}{c_1{a}_3}},\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill X_{12n+4}& ={\displaystyle \frac{c_0{c}_2{b}_0{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n}{a_0{b}_2}},\hfill \\ \hfill Y_{12n+4}& ={\displaystyle \frac{a_0{a}_2{c}_0{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n}{b_0{c}_2}},\hfill \\ \hfill Z_{12n+4}& ={\displaystyle \frac{b_0{b}_2{a}_0{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n}{c_0{a}_2}},\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill X_{12n+5}& ={\displaystyle \frac{b_1{b}_3{c}_3}{c_1{a}_3{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^n}},\hfill \\ \hfill Y_{12n+5}& ={\displaystyle \frac{c_1{c}_3{a}_3}{a_1{b}_3{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^n}},\hfill \\ \hfill Z_{12n+5}& ={\displaystyle \frac{a_1{a}_3{b}_3}{b_1{c}_3{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^n}},\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill X_{12n+6}& ={\displaystyle \frac{b_0{b}_2{c}_2}{c_0{a}_2{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^n}},\hfill \\ \hfill Y_{12n+6}& ={\displaystyle \frac{c_0{c}_2{a}_2}{a_0{b}_2{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^n}},\hfill \\ \hfill Z_{12n+6}& ={\displaystyle \frac{a_0{a}_2{b}_2}{b_0{c}_2{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^n}},\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill X_{12n+7}& ={\displaystyle \frac{a_1{b}_3{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^n}{c_3}},\hfill \\ \hfill Y_{12n+7}& ={\displaystyle \frac{b_1{c}_3{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^n}{a_3}},\hfill \\ \hfill Z_{12n+7}& ={\displaystyle \frac{c_1{a}_3{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^n}{b_3}},\hfill \end{array}$$

(43)

$$\begin{array}{cc}\hfill X_{12n+8}& ={\displaystyle \frac{a_0{b}_2{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^n}{c_3}},\hfill \\ \hfill Y_{12n+8}& ={\displaystyle \frac{b_0{c}_2{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^n}{a_2}},\hfill \\ \hfill Z_{12n+8}& ={\displaystyle \frac{c_0{a}_2{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^n}{b_2}},\hfill \end{array}$$

(44)

where $X_0=a_0$, $X_{-1}=a_1$, $X_{-2}=a_2$, $X_{-3}=a_3$, $Y_0=b_0$, $Y_{-1}=b_1$, $Y_{-2}=b_2$, $Y_{-3}=b_3$, $Z_0=c_0$, $Z_{-1}=c_1$, $Z_{-2}=c_2$ and $Z_{-3}=c_3$.

Proof. 

We apply the same approach used in Theorem 2. See Appendix A. □

Theorem 4. 

The system of nonlinear difference Equation (32) has a periodic solution of period twelve iff $a_1{b}_3=c_1{a}_3=b_1{c}_3=a_0{b}_2=c_0{a}_2=b_0{c}_2=-2$, and it will take the following form:

$$\left\{X_n\right\}=\{a_3,a_2,a_1,a_0,{\displaystyle \frac{-2}{b_1}},{\displaystyle \frac{-2}{b_0}},c_1,c_0,b_3,b_2,{\displaystyle \frac{-2}{c_3}},{\displaystyle \frac{-2}{c_2}},...\},$$

(45)

$$\left\{Y_n\right\}=\{b_3,b_2,b_1,b_0,{\displaystyle \frac{-2}{c_1}},{\displaystyle \frac{-2}{c_0}},a_1,a_0,c_3,c_2,{\displaystyle \frac{-2}{a_3}},{\displaystyle \frac{-2}{a_2}},...\},$$

(46)

$$\left\{Z_n\right\}=\{c_3,c_2,c_1,c_0,{\displaystyle \frac{-2}{a_1}},{\displaystyle \frac{-2}{a_0}},b_1,b_0,a_3,a_2,{\displaystyle \frac{-2}{b_3}},{\displaystyle \frac{-2}{b_2}},...\}.$$

(47)

Proof. 

Suppose that a prime period twelve solution exists of system (32):

$$\left\{X_n\right\}=\{a_3,a_2,a_1,a_0,{\displaystyle \frac{-2}{b_1}},{\displaystyle \frac{-2}{b_0}},c_1,c_0,b_3,b_2,{\displaystyle \frac{-2}{c_3}},{\displaystyle \frac{-2}{c_2}},...\},$$

$$\left\{Y_n\right\}=\{b_3,b_2,b_1,b_0,{\displaystyle \frac{-2}{c_1}},{\displaystyle \frac{-2}{c_0}},a_1,a_0,c_3,c_2,{\displaystyle \frac{-2}{a_3}},{\displaystyle \frac{-2}{a_2}},...\},$$

$$\left\{Z_n\right\}=\{c_3,c_2,c_1,c_0,{\displaystyle \frac{-2}{a_1}},{\displaystyle \frac{-2}{a_0}},b_1,b_0,a_3,a_2,{\displaystyle \frac{-2}{b_3}},{\displaystyle \frac{-2}{b_2}},...\}.$$

Then, we can recognize from the form of the solution of system (32) that

$$\begin{array}{cc}\hfill a_3& ={\displaystyle \frac{a_3}{{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n}},\hfill \\ \hfill b_3& ={\displaystyle \frac{b_3}{{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n}},\hfill \\ \hfill c_3& ={\displaystyle \frac{c_3}{{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill a_2& ={\displaystyle \frac{a_2}{{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n}},\hfill \\ \hfill b_2& ={\displaystyle \frac{b_2}{{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n}},\hfill \\ \hfill c_2& ={\displaystyle \frac{c_2}{{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill a_1& =a_1{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n,\hfill \\ \hfill b_1& =b_1{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n,\hfill \\ \hfill c_1& =c_1{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n,\hfill \end{array}$$

$$\begin{array}{cc}\hfill a_0& =a_0{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n,\hfill \\ \hfill b_0& =b_0{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n,\hfill \\ \hfill c_0& =c_0{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n,\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle \frac{-2}{b_1}}& ={\displaystyle \frac{c_1{a}_3}{b_1{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n}},\hfill \\ \hfill {\displaystyle \frac{-2}{c_1}}& ={\displaystyle \frac{a_1{b}_3}{c_1{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n}},\hfill \\ \hfill {\displaystyle \frac{-2}{a_1}}& ={\displaystyle \frac{b_1{c}_3}{a_1{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle \frac{-2}{b_0}}& ={\displaystyle \frac{c_0{a}_2}{b_0{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n}},\hfill \\ \hfill {\displaystyle \frac{-2}{c_0}}& ={\displaystyle \frac{a_0{b}_2}{c_0{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n}},\hfill \\ \hfill {\displaystyle \frac{-2}{a_0}}& ={\displaystyle \frac{b_0{c}_2}{a_0{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill c_1& ={\displaystyle \frac{c_1{c}_3{b}_1{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^n{(-1-b_1{c}_3)}^n}{a_1{b}_3}},\hfill \\ \hfill a_1& ={\displaystyle \frac{a_1{a}_3{c}_1{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^n{(-1-c_1{a}_3)}^n}{b_1{c}_3}},\hfill \\ \hfill b_1& ={\displaystyle \frac{b_1{b}_3{a}_1{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^n{(-1-a_1{b}_3)}^n}{c_1{a}_3}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill c_0& ={\displaystyle \frac{c_0{c}_2{b}_0{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^n{(-1-b_0{c}_2)}^n}{a_0{b}_2}},\hfill \\ \hfill a_0& ={\displaystyle \frac{a_0{a}_2{c}_0{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^n{(-1-c_0{a}_2)}^n}{b_0{c}_2}},\hfill \\ \hfill b_0& ={\displaystyle \frac{b_0{b}_2{a}_0{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^n{(-1-a_0{b}_2)}^n}{c_0{a}_2}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill b_3& ={\displaystyle \frac{b_1{b}_3{c}_3}{c_1{a}_3{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^n}},\hfill \\ \hfill c_3& ={\displaystyle \frac{c_1{c}_3{a}_3}{a_1{b}_3{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^n}},\hfill \\ \hfill a_3& ={\displaystyle \frac{a_1{a}_3{b}_3}{b_1{c}_3{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill b_2& ={\displaystyle \frac{b_0{b}_2{c}_2}{c_0{a}_2{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^n}},\hfill \\ \hfill c_2& ={\displaystyle \frac{c_0{c}_2{a}_2}{a_0{b}_2{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^n}},\hfill \\ \hfill a_2& ={\displaystyle \frac{a_0{a}_2{b}_2}{b_0{c}_2{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^n}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle \frac{-2}{c_3}}& ={\displaystyle \frac{a_1{b}_3{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^n}{c_3}},\hfill \\ \hfill {\displaystyle \frac{-2}{a_3}}& ={\displaystyle \frac{b_1{c}_3{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^{n+1}{(-1-b_1{c}_3)}^n}{a_3}},\hfill \\ \hfill {\displaystyle \frac{-2}{b_3}}& ={\displaystyle \frac{c_1{a}_3{(-1-b_1{c}_3)}^{n+1}{(-1-a_1{b}_3)}^{n+1}{(-1-c_1{a}_3)}^n}{b_3}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle \frac{-2}{c_2}}& ={\displaystyle \frac{a_0{b}_2{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^n}{c_3}},\hfill \\ \hfill {\displaystyle \frac{-2}{a_2}}& ={\displaystyle \frac{b_0{c}_2{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^{n+1}{(-1-b_0{c}_2)}^n}{a_2}},\hfill \\ \hfill {\displaystyle \frac{-2}{b_2}}& ={\displaystyle \frac{c_0{a}_2{(-1-b_0{c}_2)}^{n+1}{(-1-a_0{b}_2)}^{n+1}{(-1-c_0{a}_2)}^n}{b_2}}.\hfill \end{array}$$

Then, assume that $a_1{b}_3=c_1{a}_3=b_1{c}_3=a_0{b}_2=c_0{a}_2=b_0{c}_2=-2$.

So, we see from the form of the solution of system (32) that

$$\begin{array}{cc}\hfill X_{12n-3}& =a_3,Y_{12n-3}=b_3,Z_{12n-3}=c_3,\hfill \\ \hfill X_{12n-2}& =a_2,Y_{12n-2}=b_2,Z_{12n-2}=c_2,\hfill \\ \hfill X_{12n-1}& =a_1,Y_{12n-1}=b_1,Z_{12n-1}=c_1,\hfill \\ \hfill X_{12n}& =a_0,Y_{12n}=b_0,Z_{12n}=c_0,\hfill \\ \hfill X_{12n+1}& ={\displaystyle \frac{-2}{b_1}},Y_{12n+1}={\displaystyle \frac{-2}{c_1}},Z_{12n+1}={\displaystyle \frac{-2}{a_1}},\hfill \\ \hfill X_{12n+2}& ={\displaystyle \frac{-2}{b_0}},Y_{12n+2}={\displaystyle \frac{-2}{c_0}},Z_{12n+2}={\displaystyle \frac{-2}{a_0}},\hfill \\ \hfill X_{12n+3}& =c_1,Y_{12n+3}=a_1,Z_{12n+3}=b_1,\hfill \\ \hfill X_{12n+4}& =c_0,Y_{12n+4}=a_0,Z_{12n+4}=b_0,\hfill \\ \hfill X_{12n+5}& =b_3,Y_{12n+5}=c_3,Z_{12n+5}=a_3,\hfill \\ \hfill X_{12n+6}& =b_2,Y_{12n+6}=c_2,Z_{12n+6}=a_2,\hfill \\ \hfill X_{12n+7}& ={\displaystyle \frac{-2}{c_3}},Y_{12n+7}={\displaystyle \frac{-2}{a_3}},Z_{12n+7}={\displaystyle \frac{-2}{b_3}},\hfill \\ \hfill X_{12n+8}& ={\displaystyle \frac{-2}{c_2}},Y_{12n+8}={\displaystyle \frac{-2}{a_2}},Z_{12n+8}={\displaystyle \frac{-2}{b_2}}.\hfill \end{array}$$

Therefore, we have a periodic solution of period twelve. The proof is complete. □

2.4. The Fourth Case

This section is designed to discuss the form of solutions of the following system

$$\begin{array}{cc}\hfill X_{n+1}& ={\displaystyle \frac{Z_{n-1}{X}_{n-3}}{Y_{n-1}(1+Z_{n-1}{X}_{n-3})}},\hfill \\ \hfill Y_{n+1}& ={\displaystyle \frac{X_{n-1}{Y}_{n-3}}{Z_{n-1}(1+X_{n-1}{Y}_{n-3})}},\hfill \\ \hfill Z_{n+1}& ={\displaystyle \frac{Y_{n-1}{Z}_{n-3}}{X_{n-1}(1-Y_{n-1}{Z}_{n-3})}},\hfill \end{array}$$

(48)

where $n=0,1,2,...$ and the initial conditions are arbitrary nonzero real numbers.

Theorem 5. 

Assume that $\{X_n,Y_n,Z_n\}$ are solutions of system (48). Then, for $n\ge 0$, the solutions of system (48) can take the following form:

$$\begin{array}{cc}\hfill X_{12n-3}& ={\displaystyle \frac{a_3{\prod }_{i=0}^{n-1}(1+(2i+2)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(2i+2)c_1{a}_3)(1+(2i+2)b_1{c}_3)}{{\prod }_{i=0}^{n-1}(1+(2i+1)a_1{b}_3)(1+(2i+1)c_1{a}_3)(1+(2i+1)b_1{c}_3)}},\hfill \\ \hfill Y_{12n-3}& ={\displaystyle \frac{b_3{\prod }_{i=0}^{n-1}(1+(2i+2)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(2i+2)a_1{b}_3)(1+(2i+2)b_1{c}_3)}{{\prod }_{i=0}^{n-1}(1+(2i-1)b_1{c}_3)(1+(2i+1)a_1{b}_3)(1+(2i+3)c_1{a}_3)}},\hfill \\ \hfill Z_{12n-3}& ={\displaystyle \frac{c_3{\prod }_{i=0}^{n-1}(1+(2i+2)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(2i+2)a_1{b}_3)(1+(2i+2)b_1{c}_3)}{{\prod }_{i=0}^{n-1}(1+(2i+1)c_1{a}_3)(1+(2i+1)b_1{c}_3)(1+(2i+1)a_1{b}_3)}},\hfill \end{array}$$

(49)

$$\begin{array}{cc}\hfill X_{12n-2}& ={\displaystyle \frac{a_2{\prod }_{i=0}^{n-1}(1+(2i+2)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(2i+2)c_0{a}_2)(1+(2i+2)b_0{c}_2)}{{\prod }_{i=0}^{n-1}(1+(2i+1)a_0{b}_2)(1+(2i+1)c_0{a}_2)(1+(2i+1)b_0{c}_2)}},\hfill \\ \hfill Y_{12n-2}& ={\displaystyle \frac{b_2{\prod }_{i=0}^{n-1}(1+(2i+2)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(2i+2)a_0{b}_2)(1+(2i+2)b_0{c}_2)}{{\prod }_{i=0}^{n-1}(1+(2i-1)b_0{c}_2)(1+(2i+1)a_0{b}_2)(1+(2i+3)c_0{a}_2)}},\hfill \\ \hfill Z_{12n-2}& ={\displaystyle \frac{c_2{\prod }_{i=0}^{n-1}(1+(2i+2)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(2i+2)a_0{b}_2)(1+(2i+2)b_0{c}_2)}{{\prod }_{i=0}^{n-1}(1+(2i+1)c_0{a}_2)(1+(2i+1)b_0{c}_2)(1+(2i+1)a_0{b}_2)}},\hfill \end{array}$$

(50)

$$\begin{array}{cc}\hfill X_{12n-1}& ={\displaystyle \frac{a_1{\prod }_{i=0}^{n-1}(1+(2i-1)b_1{c}_3)(1+(2i+1)a_1{b}_3)(1+(2i+3)c_1{a}_3)}{{\prod }_{i=0}^{n-1}(1+(2i+2)c_1{a}_3)(1+(2i+2)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(2i+2)b_1{c}_3)}},\hfill \\ \hfill Y_{12n-1}& ={\displaystyle \frac{b_1{\prod }_{i=0}^{n-1}(1+(2i+1)c_1{a}_3)(1+(2i+1)b_1{c}_3)(1+(2i+1)a_1{b}_3)}{{\prod }_{i=0}^{n-1}(1+(2i+2)c_1{a}_3)(1+(2i+2)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(2i+2)a_1{b}_3)}},\hfill \\ \hfill Z_{12n-1}& ={\displaystyle \frac{c_1{\prod }_{i=0}^{n-1}(1+(2i+1)a_1{b}_3)(1+(2i+1)c_1{a}_3)(1+(2i+1)b_1{c}_3)}{{\prod }_{i=0}^{n-1}(1+(2i+2)a_1{b}_3)(1+(2i+2)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(2i+2)b_1{c}_3)}},\hfill \end{array}$$

(51)

$$\begin{array}{cc}\hfill X_{12n}& ={\displaystyle \frac{a_0{\prod }_{i=0}^{n-1}(1+(2i-1)b_0{c}_2)(1+(2i+1)a_0{b}_2)(1+(2i+3)c_0{a}_2)}{{\prod }_{i=0}^{n-1}(1+(2i+2)c_0{a}_2)(1+(2i+2)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(2i+2)b_0{c}_2)}},\hfill \\ \hfill Y_{12n}& ={\displaystyle \frac{b_0{\prod }_{i=0}^{n-1}(1+(2i+1)c_0{a}_2)(1+(2i+1)b_0{c}_2)(1+(2i+1)a_0{b}_2)}{{\prod }_{i=0}^{n-1}(1+(2i+2)c_0{a}_2)(1+(2i+2)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(2i+2)a_0{b}_2)}},\hfill \\ \hfill Z_{12n}& ={\displaystyle \frac{c_0{\prod }_{i=0}^{n-1}(1+(2i+1)a_0{b}_2)(1+(2i+1)c_0{a}_2)(1+(2i+1)b_0{c}_2)}{{\prod }_{i=0}^{n-1}(1+(2i+2)a_0{b}_2)(1+(2i+2)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(2i+2)b_0{c}_2)}},\hfill \end{array}$$

(52)

$$\begin{array}{cc}\hfill X_{12n+1}& ={\displaystyle \frac{c_1{a}_3{\prod }_{i=0}^{n-1}(1+(2i+2)c_1{a}_3)(1+(2i+2)b_1{c}_3){\prod }_{i=0}^{n-2}(1+(2i+2)a_1{b}_3)}{b_1{\prod }_{i=0}^n(1+(2i+1)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(2i+1)b_1{c}_3)(1+(2i+1)a_1{b}_3)}},\hfill \\ \hfill Y_{12n+1}& ={\displaystyle \frac{a_1{b}_3{\prod }_{i=0}^{n-1}(1+(2i+2)a_1{b}_3)(1+(2i+2)c_1{a}_3){\prod }_{i=0}^{n-2}(1+(2i+2)b_1{c}_3)}{c_1{\prod }_{i=0}^n(1+(2i+1)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(2i+1)c_1{a}_3)(1+(2i+1)b_1{c}_3)}},\hfill \\ \hfill Z_{12n+1}& ={\displaystyle \frac{b_1{c}_3{\prod }_{i=0}^{n-1}(1+(2i+2)c_1{a}_3)(1+(2i+2)a_1{b}_3){\prod }_{i=0}^{n-2}(1+(2i+2)b_1{c}_3)}{a_1{\prod }_{i=0}^n(1+(2i+1)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(2i+1)a_1{b}_3)(1+(2i+1)c_1{a}_3)}},\hfill \end{array}$$

(53)

$$\begin{array}{cc}\hfill X_{12n+2}& ={\displaystyle \frac{c_0{a}_2{\prod }_{i=0}^{n-1}(1+(2i+2)c_0{a}_2)(1+(2i+2)b_0{c}_2){\prod }_{i=0}^{n-2}(1+(2i+2)a_0{b}_2)}{b_0{\prod }_{i=0}^n(1+(2i+1)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(2i+1)b_0{c}_2)(1+(2i+1)a_0{b}_2)}},\hfill \\ \hfill Y_{12n+2}& ={\displaystyle \frac{a_0{b}_2{\prod }_{i=0}^{n-1}(1+(2i+2)a_0{b}_2)(1+(2i+2)c_0{a}_2){\prod }_{i=0}^{n-2}(1+(2i+2)b_0{c}_2)}{c_0{\prod }_{i=0}^n(1+(2i+1)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(2i+1)c_0{a}_2)(1+(2i+1)b_0{c}_2)}},\hfill \\ \hfill Z_{12n+2}& ={\displaystyle \frac{b_0{c}_2{\prod }_{i=0}^{n-1}(1+(2i+2)c_0{a}_2)(1+(2i+2)a_0{b}_2){\prod }_{i=0}^{n-2}(1+(2i+2)b_0{c}_2)}{a_0{\prod }_{i=0}^n(1+(2i+1)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(2i+1)a_0{b}_2)(1+(2i+1)c_0{a}_2)}},\hfill \end{array}$$

(54)

$$\begin{array}{cc}\hfill X_{12n+3}& ={\displaystyle \frac{c_1{c}_3{b}_1{\prod }_{i=0}^n(1+(2i+1)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(2i+1)c_1{a}_3)(1+(2i+1)b_1{c}_3)}{a_1{b}_3{\prod }_{i=0}^{n-1}(1+(2i+2)a_1{b}_3)(1+(2i+2)c_1{a}_3)(1+(2i+2)b_1{c}_3)}},\hfill \\ \hfill Y_{12n+3}& ={\displaystyle \frac{a_1{a}_3{c}_1{\prod }_{i=0}^n(1+(2i-1)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(2i+3)c_1{a}_3)(1+(2i+1)a_1{b}_3)}{b_1{a}_3{\prod }_{i=0}^n(1+(2i+2)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(2i+2)a_1{b}_3)}},\hfill \\ \hfill Z_{12n+3}& ={\displaystyle \frac{b_1{b}_3{a}_1{\prod }_{i=0}^n(1+(2i+1)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(2i+1)b_1{c}_3)(1+(2i+1)a_1{b}_3)}{c_1{a}_3{\prod }_{i=0}^{n-1}(1+(2i+2)c_1{a}_3)(1+(2i+2)b_1{c}_3)(1+(2i+2)a_1{b}_3)}},\hfill \end{array}$$

(55)

$$\begin{array}{cc}\hfill X_{12n+4}& ={\displaystyle \frac{c_0{c}_2{b}_0{\prod }_{i=0}^n(1+(2i+1)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(2i+1)c_0{a}_2)(1+(2i+1)b_0{c}_2)}{a_0{b}_2{\prod }_{i=0}^{n-1}(1+(2i+2)a_0{b}_2)(1+(2i+2)c_0{a}_2)(1+(2i+2)b_0{c}_2)}},\hfill \\ \hfill Y_{12n+4}& ={\displaystyle \frac{a_0{a}_2{c}_0{\prod }_{i=0}^n(1+(2i-1)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(2i+3)c_0{a}_2)(1+(2i+1)a_0{b}_2)}{b_0{a}_2{\prod }_{i=0}^n(1+(2i+2)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(2i+2)a_0{b}_2)}},\hfill \\ \hfill Z_{12n+4}& ={\displaystyle \frac{b_0{b}_2{a}_0{\prod }_{i=0}^n(1+(2i+1)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(2i+1)b_0{c}_2)(1+(2i+1)a_0{b}_2)}{c_0{a}_2{\prod }_{i=0}^{n-1}(1+(2i+2)c_0{a}_2)(1+(2i+2)b_0{c}_2)(1+(2i+2)a_0{b}_2)}},\hfill \end{array}$$

(56)

$$\begin{array}{cc}\hfill X_{12n+5}& ={\displaystyle \frac{b_1{b}_3{c}_3{\prod }_{i=0}^n(1+(2i+2)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(2i+2)a_1{b}_3)}{c_1{a}_3{\prod }_{i=0}^n(1+(2i-1)b_1{c}_3)(1+(2i+1)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(2i+3)c_1{a}_3)}},\hfill \\ \hfill Y_{12n+5}& ={\displaystyle \frac{c_1{c}_3{a}_3{\prod }_{i=0}^{n-1}(1+(2i+2)c_1{a}_3)(1+(2i+2)b_1{c}_3)(1+(2i+2)a_1{b}_3)}{a_1{b}_3{\prod }_{i=0}^n(1+(2i+1)c_1{a}_3)(1+(2i+1)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(2i+1)a_1{b}_3)}},\hfill \\ \hfill Z_{12n+5}& ={\displaystyle \frac{a_1{a}_3{b}_3{\prod }_{i=0}^{n-1}(1+(2i+2)a_1{b}_3)(1+(2i+2)c_1{a}_3)(1+(2i+2)b_1{c}_3)}{b_1{c}_3{\prod }_{i=0}^n(1+(2i+1)a_1{b}_3)(1+(2i+1)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(2i+1)b_1{c}_3)}},\hfill \end{array}$$

(57)

$$\begin{array}{cc}\hfill X_{12n+6}& ={\displaystyle \frac{b_0{b}_2{c}_2{\prod }_{i=0}^n(1+(2i+2)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(2i+2)a_0{b}_2)}{c_0{a}_2{\prod }_{i=0}^n(1+(2i-1)b_0{c}_2)(1+(2i+1)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(2i+3)c_0{a}_2)}},\hfill \\ \hfill Y_{12n+6}& ={\displaystyle \frac{c_0{c}_2{a}_2{\prod }_{i=0}^{n-1}(1+(2i+2)c_0{a}_2)(1+(2i+2)b_0{c}_2)(1+(2i+2)a_0{b}_2)}{a_0{b}_2{\prod }_{i=0}^n(1+(2i+1)c_0{a}_2)(1+(2i+1)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(2i+1)a_0{b}_2)}},\hfill \\ \hfill Z_{12n+6}& ={\displaystyle \frac{a_0{a}_2{b}_2{\prod }_{i=0}^{n-1}(1+(2i+2)a_0{b}_2)(1+(2i+2)c_0{a}_2)(1+(2i+2)b_0{c}_2)}{b_0{c}_2{\prod }_{i=0}^n(1+(2i+1)a_0{b}_2)(1+(2i+1)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(2i+1)b_0{c}_2)}},\hfill \end{array}$$

(58)

$$\begin{array}{cc}\hfill X_{12n+7}& ={\displaystyle \frac{a_1{b}_3{\prod }_{i=0}^n(1+(2i+1)c_1{a}_3)(1+(2i+1)b_1{c}_3){\prod }_{i=0}^{n-1}(1+(2i+1)a_1{b}_3)}{c_3{\prod }_{i=0}^n(1+(2i+2)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(2i+2)b_1{c}_3)(1+(2i+2)a_1{b}_3)}},\hfill \\ \hfill Y_{12n+7}& ={\displaystyle \frac{b_1{c}_3{\prod }_{i=0}^n(1+(2i+1)a_1{b}_3)(1+(2i+1)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(2i+1)b_1{c}_3)}{a_3{\prod }_{i=0}^n(1+(2i+2)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(2i+2)c_1{a}_3)(1+(2i+2)b_1{c}_3)}},\hfill \\ \hfill Z_{12n+7}& ={\displaystyle \frac{c_1{a}_3{\prod }_{i=0}^n(1+(2i-1)b_1{c}_3)(1+(2i+1)a_1{b}_3){\prod }_{i=0}^{n-1}(1+(2i+1)c_1{a}_3)}{b_3{\prod }_{i=0}^n(1+(2i+2)c_1{a}_3){\prod }_{i=0}^{n-1}(1+(2i+2)a_1{b}_3)(1+(2i+2)b_1{c}_3)}},\hfill \end{array}$$

(59)

$$\begin{array}{cc}\hfill X_{12n+8}& ={\displaystyle \frac{a_0{b}_2{\prod }_{i=0}^n(1+(2i+1)c_0{a}_2)(1+(2i+1)b_0{c}_2){\prod }_{i=0}^{n-1}(1+(2i+1)a_0{b}_2)}{c_2{\prod }_{i=0}^n(1+(2i+2)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(2i+2)b_0{c}_2)(1+(2i+2)a_0{b}_2)}},\hfill \\ \hfill Y_{12n+8}& ={\displaystyle \frac{b_0{c}_2{\prod }_{i=0}^n(1+(2i+1)a_0{b}_2)(1+(2i+1)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(2i+1)b_0{c}_2)}{a_2{\prod }_{i=0}^n(1+(2i+2)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(2i+2)c_0{a}_2)(1+(2i+2)b_0{c}_2)}},\hfill \\ \hfill Z_{12n+8}& ={\displaystyle \frac{c_0{a}_2{\prod }_{i=0}^n(1+(2i-1)b_0{c}_2)(1+(2i+1)a_0{b}_2){\prod }_{i=0}^{n-1}(1+(2i+1)c_0{a}_2)}{b_2{\prod }_{i=0}^n(1+(2i+2)c_0{a}_2){\prod }_{i=0}^{n-1}(1+(2i+2)a_0{b}_2)(1+(2i+2)b_0{c}_2)}},\hfill \end{array}$$

(60)

where $X_0=a_0$, $X_{-1}=a_1$, $X_{-2}=a_2$, $Y_0=b_0$, $Y_{-1}=b_1$, $Y_{-2}=b_2$, $Z_0=c_0$, $Z_{-1}=c_1$ and $Z_{-2}=c_2$.

Proof. 

We apply the same approach used in Theorem 1. See Appendix B. □

2.5. The Fifth Case

In this section, we explore the solution formulas of the following system:

$$\begin{array}{cc}\hfill X_{n+1}& ={\displaystyle \frac{Z_{n-1}{X}_{n-3}}{Y_{n-1}(-1+Z_{n-1}{X}_{n-3})}},\hfill \\ \hfill Y_{n+1}& ={\displaystyle \frac{X_{n-1}{Y}_{n-3}}{Z_{n-1}(-1+X_{n-1}{Y}_{n-3})}},\hfill \\ \hfill Z_{n+1}& ={\displaystyle \frac{Y_{n-1}{Z}_{n-3}}{X_{n-1}(-1+Y_{n-1}{Z}_{n-3})}},\hfill \end{array}$$

(61)

where $n=0,1,2,...$ and the initial conditions are arbitrary nonzero real numbers.

Theorem 6. 

Assume that $\{X_n,Y_n,Z_n\}$ are solutions of system (61). Then, for $n\ge 0$, the solutions of system (61) can be formed as follows:

$$\begin{array}{cc}\hfill X_{12n-3}& ={\displaystyle \frac{a_3}{{(-1+a_1{b}_3)}^n{(-1+c_1{a}_3)}^n{(-1+b_1{c}_3)}^n}},\hfill \\ \hfill Y_{12n-3}& ={\displaystyle \frac{b_3}{{(-1+b_1{c}_3)}^n{(-1+a_1{b}_3)}^n{(-1+c_1{a}_3)}^n}},\hfill \\ \hfill Z_{12n-3}& ={\displaystyle \frac{c_3}{{(-1+c_1{a}_3)}^n{(-1+b_1{c}_3)}^n{(-1+a_1{b}_3)}^n}},\hfill \end{array}$$

(62)

$$\begin{array}{cc}\hfill X_{12n-2}& ={\displaystyle \frac{a_2}{{(-1+a_0{b}_2)}^n{(-1+c_0{a}_2)}^n{(-1+b_0{c}_2)}^n}},\hfill \\ \hfill Y_{12n-2}& ={\displaystyle \frac{b_2}{{(-1+b_0{c}_2)}^n{(-1+a_0{b}_2)}^n{(-1+c_0{a}_2)}^n}},\hfill \\ \hfill Z_{12n-2}& ={\displaystyle \frac{c_2}{{(-1+c_0{a}_2)}^n{(-1+b_0{c}_2)}^n{(-1+a_0{b}_2)}^n}},\hfill \end{array}$$

(63)

$$\begin{array}{cc}\hfill X_{12n-1}& =a_1{(-1+b_1{c}_3)}^n{(-1+a_1{b}_3)}^n{(-1+c_1{a}_3)}^n,\hfill \\ \hfill Y_{12n-1}& =b_1{(-1+c_1{a}_3)}^n{(-1+b_1{c}_3)}^n{(-1+a_1{b}_3)}^n,\hfill \\ \hfill Z_{12n-1}& =c_1{(-1+a_1{b}_3)}^n{(-1+c_1{a}_3)}^n{(-1+b_1{c}_3)}^n,\hfill \end{array}$$

(64)

$$\begin{array}{cc}\hfill X_{12n}& =a_0{(-1+b_0{c}_2)}^n{(-1+a_0{b}_2)}^n{(-1+c_0{a}_2)}^n,\hfill \\ \hfill Y_{12n}& =b_0{(-1+c_0{a}_2)}^n{(-1+b_0{c}_2)}^n{(-1+a_0{b}_2)}^n,\hfill \\ \hfill Z_{12n}& =c_0{(-1+a_0{b}_2)}^n{(-1+c_0{a}_2)}^n{(-1+b_0{c}_2)}^n,\hfill \end{array}$$

(65)

$$\begin{array}{cc}\hfill X_{12n+1}& ={\displaystyle \frac{c_1{a}_3}{b_1{(-1+c_1{a}_3)}^{n+1}{(-1+b_1{c}_3)}^n{(-1+a_1{b}_3)}^n}},\hfill \\ \hfill Y_{12n+1}& ={\displaystyle \frac{a_1{b}_3}{c_1{(-1+a_1{b}_3)}^{n+1}{(-1+c_1{a}_3)}^n{(-1+b_1{c}_3)}^n}},\hfill \\ \hfill Z_{12n+1}& ={\displaystyle \frac{b_1{c}_3}{a_1{(-1+b_1{c}_3)}^{n+1}{(-1+a_1{b}_3)}^n{(-1+c_1{a}_3)}^n}},\hfill \end{array}$$

(66)

$$\begin{array}{cc}\hfill X_{12n+2}& ={\displaystyle \frac{c_0{a}_2}{b_0{(-1+c_0{a}_2)}^{n+1}{(-1+b_0{c}_2)}^n{(-1+a_0{b}_2)}^n}},\hfill \\ \hfill Y_{12n+2}& ={\displaystyle \frac{a_0{b}_2}{c_0{(-1+a_0{b}_2)}^{n+1}{(-1+c_0{a}_2)}^n{(-1+b_0{c}_2)}^n}},\hfill \\ \hfill Z_{12n+2}& ={\displaystyle \frac{b_0{c}_2}{a_0{(-1+b_0{c}_2)}^{n+1}{(-1+a_0{b}_2)}^n{(-1+c_0{a}_2)}^n}},\hfill \end{array}$$

(67)

$$\begin{array}{cc}\hfill X_{12n+3}& ={\displaystyle \frac{c_1{c}_3{b}_1{(-1+a_1{b}_3)}^{n+1}{(-1+c_1{a}_3)}^n{(-1+b_1{c}_3)}^n}{a_1{b}_3}},\hfill \\ \hfill Y_{12n+3}& ={\displaystyle \frac{a_1{a}_3{c}_1{(-1+b_1{c}_3)}^{n+1}{(-1+a_1{b}_3)}^n{(-1+c_1{a}_3)}^n}{b_1{c}_3}},\hfill \\ \hfill Z_{12n+3}& ={\displaystyle \frac{b_1{b}_3{a}_1{(-1+c_1{a}_3)}^{n+1}{(-1+b_1{c}_3)}^n{(-1+a_1{b}_3)}^n}{c_1{a}_3}},\hfill \end{array}$$

(68)

$$\begin{array}{cc}\hfill X_{12n+4}& ={\displaystyle \frac{c_0{c}_2{b}_0{(-1+a_0{b}_2)}^{n+1}{(-1+c_0{a}_2)}^n{(-1+b_0{c}_2)}^n}{a_0{b}_2}},\hfill \\ \hfill Y_{12n+4}& ={\displaystyle \frac{a_0{a}_2{c}_0{(-1+b_0{c}_2)}^{n+1}{(-1+a_0{b}_2)}^n{(-1+c_0{a}_2)}^n}{b_0{c}_2}},\hfill \\ \hfill Z_{12n+4}& ={\displaystyle \frac{b_0{b}_2{a}_0{(-1+c_0{a}_2)}^{n+1}{(-1+b_0{c}_2)}^n{(-1+a_0{b}_2)}^n}{c_0{a}_2}},\hfill \end{array}$$

(69)

$$\begin{array}{cc}\hfill X_{12n+5}& ={\displaystyle \frac{b_1{b}_3{c}_3}{c_1{a}_3{(-1+b_1{c}_3)}^{n+1}{(-1+a_1{b}_3)}^{n+1}{(-1+c_1{a}_3)}^n}},\hfill \\ \hfill Y_{12n+5}& ={\displaystyle \frac{c_1{c}_3{a}_3}{a_1{b}_3{(-1+c_1{a}_3)}^{n+1}{(-1+b_1{c}_3)}^{n+1}{(-1+a_1{b}_3)}^n}},\hfill \\ \hfill Z_{12n+5}& ={\displaystyle \frac{a_1{a}_3{b}_3}{b_1{c}_3{(-1+a_1{b}_3)}^{n+1}{(-1+c_1{a}_3)}^{n+1}{(-1+b_1{c}_3)}^n}},\hfill \end{array}$$

(70)

$$\begin{array}{cc}\hfill X_{12n+6}& ={\displaystyle \frac{b_0{b}_2{c}_2}{c_0{a}_2{(-1+b_0{c}_2)}^{n+1}{(-1+a_0{b}_2)}^{n+1}{(-1+c_0{a}_2)}^n}},\hfill \\ \hfill Y_{12n+6}& ={\displaystyle \frac{c_0{c}_2{a}_2}{a_0{b}_2{(-1+c_0{a}_2)}^{n+1}{(-1+b_0{c}_2)}^{n+1}{(-1+a_0{b}_2)}^n}},\hfill \\ \hfill Z_{12n+6}& ={\displaystyle \frac{a_0{a}_2{b}_2}{b_0{c}_2{(-1+a_0{b}_2)}^{n+1}{(-1+c_0{a}_2)}^{n+1}{(-1+b_0{c}_2)}^n}},\hfill \end{array}$$

(71)

$$\begin{array}{cc}\hfill X_{12n+7}& ={\displaystyle \frac{a_1{b}_3{(-1+c_1{a}_3)}^{n+1}{(-1+b_1{c}_3)}^{n+1}{(-1+a_1{b}_3)}^n}{c_3}},\hfill \\ \hfill Y_{12n+7}& ={\displaystyle \frac{b_1{c}_3{(-1+a_1{b}_3)}^{n+1}{(-1+c_1{a}_3)}^{n+1}{(-1+b_1{c}_3)}^n}{a_3}},\hfill \\ \hfill Z_{12n+7}& ={\displaystyle \frac{c_1{a}_3{(-1+b_1{c}_3)}^{n+1}{(-1+a_1{b}_3)}^{n+1}{(-1+c_1{a}_3)}^n}{b_3}},\hfill \end{array}$$

(72)

$$\begin{array}{cc}\hfill X_{12n+8}& ={\displaystyle \frac{a_0{b}_2{(-1+c_0{a}_2)}^{n+1}{(-1+b_0{c}_2)}^{n+1}{(-1+a_0{b}_2)}^n}{c_3}},\hfill \\ \hfill Y_{12n+8}& ={\displaystyle \frac{b_0{c}_2{(-1+a_0{b}_2)}^{n+1}{(-1+c_0{a}_2)}^{n+1}{(-1+b_0{c}_2)}^n}{a_2}},\hfill \\ \hfill Z_{12n+8}& ={\displaystyle \frac{c_0{a}_2{(-1+b_0{c}_2)}^{n+1}{(-1+a_0{b}_2)}^{n+1}{(-1+c_0{a}_2)}^n}{b_2}}.\hfill \end{array}$$

(73)

Proof. 

Readers can verify this proof by applying the same methodology as Theorem 2. □

Theorem 7. 

The system of nonlinear difference Equation (61) has a periodic solution of period twelve iff $a_1{b}_3=c_1{a}_3=b_1{c}_3=a_0{b}_2=c_0{a}_2=b_0{c}_2=2$ and it will take the following form

$$\left\{X_n\right\}=\{a_3,a_2,a_1,a_0,{\displaystyle \frac{2}{b_1}},{\displaystyle \frac{2}{b_0}},c_1,c_0,b_3,b_2,{\displaystyle \frac{2}{c_3}},{\displaystyle \frac{2}{c_2}},...\},$$

(74)

$$\left\{Y_n\right\}=\{b_3,b_2,b_1,b_0,{\displaystyle \frac{2}{c_1}},{\displaystyle \frac{2}{c_0}},a_1,a_0,c_3,c_2,{\displaystyle \frac{2}{a_3}},{\displaystyle \frac{2}{a_2}},...\},$$

(75)

$$\left\{Z_n\right\}=\{c_3,c_2,c_1,c_0,{\displaystyle \frac{2}{a_1}},{\displaystyle \frac{2}{a_0}},b_1,b_0,a_3,a_2,{\displaystyle \frac{2}{b_3}},{\displaystyle \frac{2}{b_2}},...\}.$$

(76)

Proof. 

We let the readers establish this and can be demonstrated by applying the same methodology as Theorem 4. □

2.6. The Sixth Case

This section is assigned to investigate the solution forms of the following system:

$$\begin{array}{cc}\hfill X_{n+1}& ={\displaystyle \frac{Z_{n-1}{X}_{n-3}}{Y_{n-1}(1-Z_{n-1}{X}_{n-3})}},\hfill \\ \hfill Y_{n+1}& ={\displaystyle \frac{X_{n-1}{Y}_{n-3}}{Z_{n-1}(1-X_{n-1}{Y}_{n-3})}},\hfill \\ \hfill Z_{n+1}& ={\displaystyle \frac{Y_{n-1}{Z}_{n-3}}{X_{n-1}(1-Y_{n-1}{Z}_{n-3})}},\hfill \end{array}$$

(77)

where $n=0,1,2,...$ and the initial conditions are arbitrary nonzero real numbers.

Theorem 8. 

Assume that $\{X_n,Y_n,Z_n\}$ are solutions of system (77). Then, for $n\ge 0$, the solutions of system (77) can be formed as follows:

$$\begin{array}{cc}\hfill X_{12n-3}& ={\displaystyle \frac{a_3{\prod }_{i=0}^{n-1}(1-(6i+2)b_1{c}_3)(1-(6i+4)a_1{b}_3){\prod }_{i=0}^{n-2}(1-(6i+6)c_1{a}_3)}{{\prod }_{i=0}^{n-1}(1-(6i+1)a_1{b}_3)(1-(6i+3)c_1{a}_3)(1-(6i+5)b_1{c}_3)}},\hfill \\ \hfill Y_{12n-3}& ={\displaystyle \frac{b_3{\prod }_{i=0}^{n-1}(1-(6i+2)c_1{a}_3)(1-(6i+4)b_1{c}_3){\prod }_{i=0}^{n-2}(1-(6i+6)a_1{b}_3)}{{\prod }_{i=0}^{n-1}(1-(6i+1)b_1{c}_3)(1-(6i+3)a_1{b}_3)(1-(6i+5)c_1{a}_3)}},\hfill \\ \hfill Z_{12n-3}& ={\displaystyle \frac{c_3{\prod }_{i=0}^{n-1}(1-(6i+2)a_1{b}_3)(1-(6i+4)c_1{a}_3){\prod }_{i=0}^{n-2}(1-(6i+6)b_1{c}_3)}{{\prod }_{i=0}^{n-1}(1-(6i+1)c_1{a}_3)(1-(6i+3)b_1{c}_3)(1-(6i+5)a_1{b}_3)}},\hfill \end{array}$$

(78)

$$\begin{array}{cc}\hfill X_{12n-2}& ={\displaystyle \frac{a_2{\prod }_{i=0}^{n-1}(1-(6i+2)b_0{c}_2)(1-(6i+4)a_0{b}_2){\prod }_{i=0}^{n-2}(1-(6i+6)c_0{a}_2)}{{\prod }_{i=0}^{n-1}(1-(6i+1)a_0{b}_2)(1-(6i+3)c_0{a}_2)(1-(6i+5)b_0{c}_2)}},\hfill \\ \hfill Y_{12n-2}& ={\displaystyle \frac{b_2{\prod }_{i=0}^{n-1}(1-(6i+2)c_0{a}_2)(1-(6i+4)b_0{c}_2){\prod }_{i=0}^{n-2}(1-(6i+6)a_0{b}_2)}{{\prod }_{i=0}^{n-1}(1-(6i+1)b_0{c}_2)(1-(6i+3)a_0{b}_2)(1-(6i+5)c_0{a}_2)}},\hfill \\ \hfill Z_{12n-2}& ={\displaystyle \frac{c_2{\prod }_{i=0}^{n-1}(1-(6i+2)a_0{b}_2)(1-(6i+4)c_0{a}_2){\prod }_{i=0}^{n-2}(1-(6i+6)b_0{c}_2)}{{\prod }_{i=0}^{n-1}(1-(6i+1)c_0{a}_2)(1-(6i+3)b_0{c}_2)(1-(6i+5)a_0{b}_2)}},\hfill \end{array}$$

(79)

$$\begin{array}{cc}\hfill X_{12n-1}& =a_1\prod _{i=0}^{n-1}{\displaystyle \frac{(1-(6i+1)b_1{c}_3)(1-(6i+3)a_1{b}_3)(1-(6i+5)c_1{a}_3)}{(1-(6i+2)c_1{a}_3)(1-(6i+4)b_1{c}_3)(1-(6i+6)a_1{b}_3)}},\hfill \\ \hfill Y_{12n-1}& =b_1\prod _{i=0}^{n-1}{\displaystyle \frac{(1-(6i+1)c_1{a}_3)(1-(6i+3)b_1{c}_3)(1-(6i+5)a_1{b}_3)}{(1-(6i+2)a_1{b}_3)(1-(6i+4)c_1{a}_3)(1-(6i+6)b_1{c}_3)}},\hfill \\ \hfill Z_{12n-1}& =c_1\prod _{i=0}^{n-1}{\displaystyle \frac{(1-(6i+1)a_1{b}_3)(1+(6i+3)c_1{a}_3)(1-(6i+5)b_1{c}_3)}{(1-(6i+2)b_1{c}_3)(1-(6i+4)a_1{b}_3)(1-(6i+6)c_1{a}_3)}},\hfill \end{array}$$

(80)

$$\begin{array}{cc}\hfill X_{12n}& =a_0\prod _{i=0}^{n-1}{\displaystyle \frac{(1-(6i+1)b_0{c}_2)(1-(6i+3)a_0{b}_2)(1-(6i+5)c_0{a}_2)}{(1-(6i+2)c_0{a}_2)(1-(6i+4)b_0{c}_2)(1-(6i+6)a_0{b}_2)}},\hfill \\ \hfill Y_{12n}& =b_0\prod _{i=0}^{n-1}{\displaystyle \frac{(1-(6i+1)c_0{a}_2)(1-(6i+3)b_0{c}_2)(1-(6i+5)a_0{b}_2)}{(1-(6i+2)a_0{b}_2)(1-(6i+4)c_0{a}_2)(1-(6i+6)b_0{c}_2)}},\hfill \\ \hfill Z_{12n}& =c_0\prod _{i=0}^{n-1}{\displaystyle \frac{(1-(6i+1)a_0{b}_2)(1-(6i+3)c_0{a}_2)(1-(6i+5)b_0{c}_2)}{(1-(6i+2)b_0{c}_2)(1-(6i+4)a_0{b}_2)(1-(6i+6)c_0{a}_2)}},\hfill \end{array}$$

(81)

$$\begin{array}{cc}\hfill X_{12n+1}& ={\displaystyle \frac{c_1{a}_3{\prod }_{i=0}^{n-1}(1-(6i+2)a_1{b}_3)(1-(6i+4)c_1{a}_3)(1-(6i+6)b_1{c}_3)}{b_1{\prod }_{i=0}^n(1-(6i+1)c_1{a}_3){\prod }_{i=0}^{n-1}(1-(6i+3)b_1{c}_3)(1-(6i+5)a_1{b}_3)}},\hfill \\ \hfill Y_{12n+1}& ={\displaystyle \frac{a_1{b}_3{\prod }_{i=0}^{n-1}(1-(6i+2)b_1{c}_3)(1-(6i+4)a_1{b}_3)(1-(6i+6)c_1{a}_3)}{c_1{\prod }_{i=0}^n(1-(6i+1)a_1{b}_3){\prod }_{i=0}^{n-1}(1-(6i+3)c_1{a}_3)(1-(6i+5)b_1{c}_3)}},\hfill \\ \hfill Z_{12n+1}& ={\displaystyle \frac{b_1{c}_3{\prod }_{i=0}^{n-1}(1-(6i+2)c_1{a}_3)(1-(6i+4)b_1{c}_3)(1-(6i+6)a_1{b}_3)}{a_1{\prod }_{i=0}^n(1-(6i+1)b_1{c}_3){\prod }_{i=0}^{n-1}(1-(6i+3)a_1{b}_3)(1-(6i+5)c_1{a}_3)}},\hfill \end{array}$$

(82)

$$\begin{array}{cc}\hfill X_{12n+2}& ={\displaystyle \frac{c_0{a}_2{\prod }_{i=0}^{n-1}(1-(6i+2)a_0{b}_2)(1-(6i+4)c_0{a}_2)(1-(6i+6)b_0{c}_2)}{b_0{\prod }_{i=0}^n(1-(6i+1)c_0{a}_2){\prod }_{i=0}^{n-1}(1-(6i+3)b_0{c}_2)(1-(6i+5)a_0{b}_2)}},\hfill \\ \hfill Y_{12n+2}& ={\displaystyle \frac{a_0{b}_2{\prod }_{i=0}^{n-1}(1-(6i+2)b_0{c}_2)(1-(6i+4)a_0{b}_2)(1-(6i+6)c_0{a}_2)}{c_0{\prod }_{i=0}^n(1-(6i+1)a_0{b}_2){\prod }_{i=0}^{n-1}(1-(6i+3)c_0{a}_2)(1-(6i+5)b_0{c}_2)}},\hfill \\ \hfill Z_{12n+2}& ={\displaystyle \frac{b_0{c}_2{\prod }_{i=0}^{n-1}(1-(6i+2)c_0{a}_2)(1-(6i+4)b_0{c}_2)(1-(6i+6)a_0{b}_2)}{a_0{\prod }_{i=0}^n(1-(6i+1)b_0{c}_2){\prod }_{i=0}^{n-1}(1-(6i+3)a_0{b}_2)(1-(6i+5)c_0{a}_2)}},\hfill \end{array}$$

(83)

$$\begin{array}{cc}\hfill X_{12n+3}& ={\displaystyle \frac{b_1{c}_3{c}_1{\prod }_{i=0}^n(1-(6i+1)a_1{b}_3){\prod }_{i=0}^{n-1}(1-(6i+3)c_1{a}_3)(1-(6i+5)b_1{c}_3)}{a_1{b}_3{\prod }_{i=0}^n(1-(6i+2)b_1{c}_3){\prod }_{i=0}^{n-1}(1-(6i+4)a_1{b}_3)(1-(6i+6)c_1{a}_3)}},\hfill \\ \hfill Y_{12n+3}& ={\displaystyle \frac{c_1{a}_3{a}_1{\prod }_{i=0}^n(1-(6i+1)b_1{c}_3){\prod }_{i=0}^{n-1}(1-(6i+3)a_1{b}_3)(1-(6i+5)c_1{a}_3)}{b_1{c}_3{\prod }_{i=0}^n(1-(6i+2)c_1{a}_3){\prod }_{i=0}^{n-1}(1-(6i+4)b_1{c}_3)(1-(6i+6)a_1{b}_3)}},\hfill \\ \hfill Z_{12n+3}& ={\displaystyle \frac{a_1{b}_3{b}_1{\prod }_{i=0}^n(1-(6i+1)c_1{a}_3){\prod }_{i=0}^{n-1}(1-(6i+3)b_1{c}_3)(1-(6i+5)a_1{b}_3)}{c_1{a}_3{\prod }_{i=0}^n(1-(6i+2)a_1{b}_3){\prod }_{i=0}^{n-1}(1-(6i+4)c_1{a}_3)(1-(6i+6)b_1{c}_3)}},\hfill \end{array}$$

(84)

$$\begin{array}{cc}\hfill X_{12n+4}& ={\displaystyle \frac{b_0{c}_2{c}_0{\prod }_{i=0}^n(1-(6i+1)a_0{b}_2){\prod }_{i=0}^{n-1}(1-(6i+3)c_0{a}_2)(1-(6i+5)b_0{c}_2)}{a_0{b}_2{\prod }_{i=0}^n(1-(6i+2)b_0{c}_2){\prod }_{i=0}^{n-1}(1-(6i+4)a_0{b}_2)(1-(6i+6)c_0{a}_2)}},\hfill \\ \hfill Y_{12n+4}& ={\displaystyle \frac{c_0{a}_2{a}_0{\prod }_{i=0}^n(1-(6i+1)b_0{c}_2){\prod }_{i=0}^{n-1}(1-(6i+3)a_0{b}_2)(1-(6i+5)c_0{a}_2)}{b_0{c}_2{\prod }_{i=0}^n(1-(6i+2)c_0{a}_2){\prod }_{i=0}^{n-1}(1-(6i+4)b_0{c}_2)(1-(6i+6)a_0{b}_2)}},\hfill \\ \hfill Z_{12n+4}& ={\displaystyle \frac{a_0{b}_2{b}_0{\prod }_{i=0}^n(1-(6i+1)c_0{a}_2){\prod }_{i=0}^{n-1}(1-(6i+3)b_0{c}_2)(1-(6i+5)a_0{b}_2)}{c_0{a}_2{\prod }_{i=0}^n(1-(6i+2)a_0{b}_2){\prod }_{i=0}^{n-1}(1-(6i+4)c_0{a}_2)(1-(6i+6)b_0{c}_2)}},\hfill \end{array}$$

(85)

$$\begin{array}{cc}\hfill X_{12n+5}& ={\displaystyle \frac{b_3{b}_1{c}_3{\prod }_{i=0}^n(1-(6i+2)c_1{a}_3){\prod }_{i=0}^{n-1}(1-(6i+4)b_1{c}_3)(1-(6i+6)a_1{b}_3)}{c_1{a}_3{\prod }_{i=0}^n(1-(6i+1)b_1{c}_3)(1-(6i+3)a_1{b}_3){\prod }_{i=0}^{n-1}(1-(6i+5)c_1{a}_3)}},\hfill \\ \hfill Y_{12n+5}& ={\displaystyle \frac{c_3{c}_1{a}_3{\prod }_{i=0}^n(1-(6i+2)a_1{b}_3){\prod }_{i=0}^{n-1}(1-(6i+4)c_1{a}_3)(1-(6i+6)b_1{c}_3)}{a_1{b}_3{\prod }_{i=0}^n(1-(6i+1)c_1{a}_3)(1-(6i+3)b_1{c}_3){\prod }_{i=0}^{n-1}(1-(6i+5)a_1{b}_3)}},\hfill \\ \hfill Z_{12n+5}& ={\displaystyle \frac{a_3{a}_1{b}_3{\prod }_{i=0}^n(1-(6i+2)b_1{c}_3){\prod }_{i=0}^{n-1}(1-(6i+4)a_1{b}_3)(1-(6i+6)c_1{a}_3)}{b_1{c}_3{\prod }_{i=0}^n(1-(6i+1)a_1{b}_3)(1-(6i+3)c_1{a}_3){\prod }_{i=0}^{n-1}(1-(6i+5)b_1{c}_3)}},\hfill \end{array}$$

(86)

$$\begin{array}{cc}\hfill X_{12n+6}& ={\displaystyle \frac{b_2{b}_0{c}_2{\prod }_{i=0}^n(1-(6i+2)c_0{a}_2){\prod }_{i=0}^{n-1}(1-(6i+4)b_0{c}_2)(1-(6i+6)a_0{b}_2)}{c_0{a}_2{\prod }_{i=0}^n(1-(6i+1)b_0{c}_2)(1-(6i+3)a_0{b}_2){\prod }_{i=0}^{n-1}(1-(6i+5)c_0{a}_2)}},\hfill \\ \hfill Y_{12n+6}& ={\displaystyle \frac{c_2{c}_0{a}_2{\prod }_{i=0}^n(1-(6i+2)a_0{b}_2){\prod }_{i=0}^{n-1}(1-(6i+4)c_0{a}_2)(1-(6i+6)b_0{c}_2)}{a_0{b}_2{\prod }_{i=0}^n(1-(6i+1)c_0{a}_2)(1-(6i+3)b_0{c}_2){\prod }_{i=0}^{n-1}(1-(6i+5)a_0{b}_2)}},\hfill \\ \hfill Z_{12n+6}& ={\displaystyle \frac{a_2{a}_0{b}_2{\prod }_{i=0}^n(1-(6i+2)b_0{c}_2){\prod }_{i=0}^{n-1}(1-(6i+4)a_0{b}_2)(1-(6i+6)c_0{a}_2)}{b_0{c}_2{\prod }_{i=0}^n(1-(6i+1)a_0{b}_2)(1-(6i+3)c_0{a}_2){\prod }_{i=0}^{n-1}(1-(6i+5)b_0{c}_2)}},\hfill \end{array}$$

(87)

$$\begin{array}{cc}\hfill X_{12n+7}& ={\displaystyle \frac{a_1{b}_3{\prod }_{i=0}^n(1-(6i+1)c_1{a}_3)(1-(6i+3)b_1{c}_3){\prod }_{i=0}^{n-1}(1-(6i+5)a_1{b}_3)}{c_3{\prod }_{i=0}^n(1-(6i+2)a_1{b}_3)(1-(6i+4)c_1{a}_3){\prod }_{i=0}^{n-1}(1-(6i+6)b_1{c}_3)}},\hfill \\ \hfill Y_{12n+7}& ={\displaystyle \frac{b_1{c}_3{\prod }_{i=0}^n(1-(6i+1)a_1{b}_3)(1-(6i+3)c_1{a}_3){\prod }_{i=0}^{n-1}(1-(6i+5)b_1{c}_3)}{a_3{\prod }_{i=0}^n(1-(6i+2)b_1{c}_3)(1-(6i+4)a_1{b}_3){\prod }_{i=0}^{n-1}(1-(6i+6)c_1{a}_3)}},\hfill \\ \hfill Z_{12n+7}& ={\displaystyle \frac{c_1{a}_3{\prod }_{i=0}^n(1-(6i+1)b_1{c}_3)(1-(6i+3)a_1{b}_3){\prod }_{i=0}^{n-1}(1-(6i+5)c_1{a}_3)}{b_3{\prod }_{i=0}^n(1-(6i+2)c_1{a}_3)(1-(6i+4)b_1{c}_3){\prod }_{i=0}^{n-1}(1-(6i+6)a_1{b}_3)}},\hfill \end{array}$$

(88)

$$\begin{array}{cc}\hfill X_{12n+8}& ={\displaystyle \frac{a_0{b}_2{\prod }_{i=0}^n(1-(6i+1)c_0{a}_2)(1-(6i+3)b_0{c}_2){\prod }_{i=0}^{n-1}(1-(6i+5)a_0{b}_2)}{c_2{\prod }_{i=0}^n(1-(6i+2)a_0{b}_2)(1-(6i+4)c_0{a}_2){\prod }_{i=0}^{n-1}(1-(6i+6)b_0{c}_2)}},\hfill \\ \hfill Y_{12n+8}& ={\displaystyle \frac{b_0{c}_2{\prod }_{i=0}^n(1-(6i+1)a_0{b}_2)(1-(6i+3)c_0{a}_2){\prod }_{i=0}^{n-1}(1-(6i+5)b_0{c}_2)}{a_2{\prod }_{i=0}^n(1-(6i+2)b_0{c}_2)(1-(6i+4)a_0{b}_2){\prod }_{i=0}^{n-1}(1-(6i+6)c_0{a}_2)}},\hfill \\ \hfill Z_{12n+8}& ={\displaystyle \frac{c_0{a}_2{\prod }_{i=0}^n(1-(6i+1)b_0{c}_2)(1-(6i+3)a_0{b}_2){\prod }_{i=0}^{n-1}(1-(6i+5)c_0{a}_2)}{b_2{\prod }_{i=0}^n(1-(6i+2)c_0{a}_2)(1-(6i+4)b_0{c}_2){\prod }_{i=0}^{n-1}(1-(6i+6)a_0{b}_2)}},\hfill \end{array}$$

(89)

where $X_0=a_0$, $X_{-1}=a_1$, $X_{-2}=a_2$, $Y_0=b_0$, $Y_{-1}=b_1$, $Y_{-2}=b_2$, $Z_0=c_0$, $Z_{-1}=c_1$ and $Z_{-2}=c_2$.

Proof. 

We let the readers prove this and can be demonstrated by applying the same methodology as Theorem 1. □

3. Numerical Simulations

In this section, we present several intriguing numerical examples that confirm our previous theoretical results and strengthen the validity of our proofs. Furthermore, these examples illustrate the behavior of different solutions to the nonlinear systems discussed in the previous sections. All plots (Figure 1, Figure 2, Figure 3, Figure 4, Figure 5, Figure 6 and Figure 7) in this section are created using MATLAB (R2024b).

Example 1. 

This example illustrates the boundedness of the solution for the first system (4) with the initial values $X_{-3}=0.2,X_{-2}=1.4,X_{-1}=0.3,X_0=0.5,Y_{-3}=0.2,Y_{-2}=0.9,Y_{-1}=1,Y_0=0.8,Z_{-3}=0.1,Z_{-2}=1,Z_{-1}=0.5$ and $Z_0=9.5$. It is clear that system (4) is bounded from above by two and zero from below.

Example 2. 

The solution of the second system (19) is depicted in this example with initial values $X_{-3}=1.5,X_{-2}=0.9,X_{-1}=2,X_0=1,Y_{-3}=7,Y_{-2}=3.8,Y_{-1}=0.33,Y_0=5,Z_{-3}=2,Z_{-2}=6,Z_{-1}=9$ and $Z_0=3.5.$

Example 3. 

We show the behavior of the third system (32) in this example with random initial values $X_{-3}=7,X_{-2}=5.2,X_{-1}=1,X_0=2,Y_{-3}=0.7,Y_{-2}=0.22,Y_{-1}=10,Y_0=6.5,Z_{-3}=8,Z_{-2}=6,Z_{-1}=1.9$ and $Z_0=9.5$.

Example 4. 

This example shows that the third system (32) has a periodic solution of period twelve when the condition of Theorem 4 is satisfied.

Example 5. 

The behavior of system (48) is represented in example (5) with initial conditions $X_{-3}=9,X_{-2}=4,X_{-1}=3,X_0=5,Y_{-3}=1,Y_{-2}=3,Y_{-1}=1,Y_0=6,Z_{-3}=4,Z_{-2}=2,Z_{-1}=7$ and $Z_0=2.$

Example 6. 

This example shows the dynamics of system (61) under the initial values $X_{-3}=0.22,X_{-2}=9,X_{-1}=2,X_0=1,Y_{-3}=7,Y_{-2}=3.8,Y_{-1}=1,Y_0=0.5,Z_{-3}=4,Z_{-2}=6,Z_{-1}=0.3$ and $Z_0=5$.

Example 7. 

For the sixth system (77), consider that the initial conditions are $X_{-3}=9,X_{-2}=4,X_{-1}=3,X_0=1,Y_{-3}=4,Y_{-2}=8,Y_{-1}=1,Y_0=5,Z_{-3}=3,Z_{-2}=6,Z_{-1}=1.5$ and $Z_0=1$.

4. Conclusions

The quest for solvability often revolves around deriving general formulas that can be applied. Although these formulas can sometimes become quite complex, we have used the iteration method to derive a solution formula for a nonlinear system involved in recursive relations. First, we obtained the general solution form for the system (4) and established that the solution is bounded. Subsequently, we formed the expressions of solutions for systems (19), (32), (48), (61) and (77). Furthermore, we discovered that nonlinear systems (32) and (61) exhibit periodic behavior with a period of twelve cycles. Finally, we present some illustrative numerical examples to corroborate our findings.