Solvability of the Class of Two-Dimensional Product-Type Systems of Difference Equations of Delay-Type (1, 3, 1, 1)

Abstract

This paper essentially presents the last and important steps in the study of (practical) solvability of two-dimensional product-type systems of difference equations of the following form $z_n=\alpha z_{n-k}^a{w}_{n-l}^b,$ $w_n=\beta w_{n-m}^c{z}_{n-s}^d,$ $n\in {\mathbb{N}}_0,$ where $k,l,m,s\in \mathbb{N}$, $a,b,c,d\in \mathbb{Z}$, and where $\alpha ,\beta$ and the initial values are complex numbers. It is devoted to the most complex case which has not been considered so far (the case $k=l=s=1$ and $m=3$). Closed form formulas for solutions to the system are found in all possible cases. The structure of the solutions to the system is considered in detail. The following five cases: (1) $b=0$; (2) $c=0$; (3) $d=0$; (4) $ac\ne 0$; (5) $a=0$, $bcd\ne 0$, are considered separately. Some of the situations appear for the first time in the literature.

1. Introduction

There has been a growing interest in difference equations and systems of difference equations (see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43]). Among several subfields of recent interest we mention here two, on whose intersection is the study in this paper. The first one is the classical subfield of finding solutions to the equations and systems in closed form. Books [6,12,13,14,15,16] contain some old information in the subfield. Some recent ones can be found, for example, in the following papers: [5,27,31,32,33,34,35,36,37,38,39,40,41,42,43] (see also numerous references therein). For some related results and applications of solvable difference equations and systems, see also [1,2,3,4]. The recent interest has been considerably motivated by the fact that some new interesting classes of nonlinear difference equations and systems have been solved by transforming them to known solvable ones. One of our transformations has had some impact in the recent interest. For more information, see, for example, [5,27,34], and the related references therein. Beside the line of investigation, there have been some other several ones which also use some related ideas. The reader can consult the representative paper [34] and find many other related ones in its list of references. Generally speaking, above mentioned lines of investigations use the method of transformation in solving the equations and systems therein. In many of these papers obtained formulas for the solutions to the equations and systems studied therein are used in describing their long-term behavior (for example, in [5,39]).

The second one is the subfield on concrete systems of difference equations. Some of the papers which have had some impact on the growing interest in the subfield are [19,20,21] by Papaschinopoulos and Schinas. One of their main ideas is to consider symmetric systems of difference equations obtained from the following scalar one

$$x_n=f(x_{n-k},x_{n-l}),n\in {\mathbb{N}}_0,$$

where $k,l\in \mathbb{N}$, $k\ne l$, for concrete values of function f, that is, to study some concrete systems of difference equations of one of the following forms

$$\begin{array}{cc}\hfill x_n& =f(x_{n-k},y_{n-l}),y_n=f(y_{n-k},x_{n-l}),\hfill \\ \hfill x_n& =f(y_{n-k},x_{n-l}),y_n=f(x_{n-k},y_{n-l}),\hfill \\ \hfill x_n& =f(y_{n-k},y_{n-l}),y_n=f(x_{n-k},x_{n-l}),n\in {\mathbb{N}}_0.\hfill \end{array}$$

This, among other things, has motivated us to study the solvability of some concrete systems of the form, such as the ones in [5,39,40] (see also the related references therein). The idea naturally evolved into the investigation of more general symmetric systems of difference equations. For example, a few symmetric systems with three variables were studied in [23], while in [22] were studied, among others, the invariants of the following system of nonlinear difference equations

$$\begin{array}{c}\hfill x_{n+1}=\frac{a+b(y_n+x_{n-1}+\cdots +y_{n-k+1}+x_{n-k})}{y_{n-k-1}}\\ \hfill y_{n+1}=\frac{a+b(x_n+y_{n-1}+\cdots +x_{n-k+1}+y_{n-k})}{x_{n-k-1}}\end{array}$$

where k is an odd number and $a,b\in (0,\infty )$, which means that more complex symmetric systems were studied therein. In fact, the study of the invariants of some equations and systems can be regarded as a kind of the study of their solvability, so that paper [22], as well as [21] essentially belong to both subfields.

It is highly expected that the methods and ideas used in the study of symmetric systems of difference equations can produce the same or related results for the systems which are not symmetric, but are close to them. For example, the following max-type system of difference equations

$$x_{n+1}=\frac{\max \{a_n{y}_n,b_n\}}{y_n{x}_{n-1}},y_{n+1}=\frac{\max \{c_n{x}_n,d_n\}}{x_n{y}_{n-1}},n\in {\mathbb{N}}_0,$$

where $a_n,b_n,c_n,d_n$ are sequences of positive numbers and $x_{-1},x_0,y_{-1},y_0$ are positive numbers, is such one, and was studied in [25] (for another max-type system see also [28]). Such system are called close-to-symmetric systems of difference equations and are frequently studied (see, for example, [8,17,24,29,31,33] and the related references therein). In paper [11] was initiated study of cyclic systems of difference equations, which naturally evolved into the study of some close to cyclic systems of difference equations (see, for example, [7,18]).

Studying positive solutions to some classes of equations and systems, such as the ones of the special cases of the following equation

$$x_n=\alpha +\frac{x_{n-k}^p}{x_{n-l}^q},n\in {\mathbb{N}}_0,$$

where $\alpha ,p,q\in (0,\infty )$, $k,l\in \mathbb{N}$, $k\ne l$ (see, for example, [30], as well as [9,10,26] and the related references therein), as well as of the corresponding two-dimensional symmetric systems of difference equations, such as

$$x_n=\alpha +\frac{y_{n-k}^p}{x_{n-l}^q},y_n=\alpha +\frac{x_{n-k}^p}{y_{n-l}^q},n\in {\mathbb{N}}_0,$$

we came across some product-type equations and systems.

The solvability of the product-type equations and systems in this case is something which should be known to any expert. Namely, if all the initial values are positive a simple inductive argument shows that all other terms are also positive, so that by using the logarithm the equation/system is transformed to a linear one with constant coefficients, which is one of the most known solvable equation/system. If the domain is changed, then several problems occur which prevent using the standard method for solving the equations and systems for the case of positive solutions. Our study of the systems on the complex plane was set off in [39]. It turned out that finding solutions to some related systems is not so simple problem. The next product-type system was solved in [40], but without detailed analysis of the structure of its solutions. The forms of the systems studied in [39,40] are similar, which suggested an investigation of the extensions which include both of them. On the other hand, the occurrence of some multipliers in some cases of the one-dimensional equation in [34] has suggested an investigation of the related systems with additional constant multipliers, which has been done for the first time in [31].

Somewhat later, a detailed analysis has shown that complete lists of formulas can be given for some of concrete systems of difference equations of the following form

$$\begin{array}{c}\hfill z_n=\alpha z_{n-k}^a{w}_{n-l}^b,w_n=\beta w_{n-m}^c{z}_{n-s}^d,n\in \mathbb{N},\end{array}$$

(1)

with “small" values of delays $k,l,m$ and s, which means that they are solvable. Since we have studied so far a number of the systems of type (1), to facilitate classification and terminology, from now on we will say that the system is of delay-type $(k,m,l,s)$.

The corresponding lists of formulas for solutions are given first for the systems in [33,43], unlike the systems in above mentioned paper [40] and for the system in [42]. For some systems such as the ones in [35,41] the solutions were obtained more easily, so the analysis was simpler and it was of a different character. Some technical problems in dealing with the systems of the sort in (1) lead us to devising another method for solving them in [32], which has been recently also used in [37]. Recently, we have done, for the first time a detailed analysis of the structure of solutions to a class of product-type systems with an associated polynomial (to the system) of the fourth order in [38], and quite recently in [36].

The main goal of the whole project is to classify solvable product-type systems of difference equations of the form in (1) and present their solutions in closed form in terms of the involved parameters and initial values. Here we continue the project. This paper is a natural continuation of our research in [31,32,33,35,36,37,38,39,40,41,42,43], and essentially presents the last and important steps in the finishing of the project.

Our task here is to show the solvability of system (1) of delay-type $(1,3,1,1)$ (the case $k=l=s=1$ and $m=3$), that is, of

$$\begin{array}{c}\hfill z_{n+1}=\alpha z_n^a{w}_n^b,w_{n+1}=\beta w_{n-2}^c{z}_n^d,n\in {\mathbb{N}}_0,\end{array}$$

(2)

where $a,b,c,d\in \mathbb{Z}$, $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. The case when some of the quantities $\alpha ,\beta ,z_0,$$w_{-2},w_{-1},w_0$ is zero we do not take into consideration because in the case are obtained solutions which are either not defined or trivial, so of not special interest.

The following five cases are considered separately in this paper: (1) $b=0$; (2) $c=0$; (3) $d=0$; (4) $ac\ne 0$; (5) $a=0$, $bcd\ne 0$. We would like to point out that the fifth case is not covered with the fourth one, since the condition $a=0$ changes the order of an associated polynomial appearing in the study. If $k,l\in \mathbb{Z}$, then the notation $\overline{k,l}$, denotes the set of all $j\in \mathbb{Z}$ such that $k\le j\le l,$ whereas we regard that ${\sum }_{j=l}^{l-1}{c}_j=0$ for each $l\in \mathbb{Z}$.

2. Auxiliary Results

The following three lemmas are useful tools in our investigation and have been already used in some of our previous papers devoted to the project on product-type systems. The first one is a consequence of the Langrage formula applied to the functions $f_s(t)=t^s$, $s\in \mathbb{N}$ (see, for example, [13] or [44], as well as [40] for a proof based on complex analysis).

Lemma 1.

Let

$$p_k(t)=c_k\prod _{j=1}^k(t-t_j),$$

$c_k\ne 0$ and $t_i\ne t_j$, $i\ne j$. Then

$$\sum _{j=1}^k\frac{t_j^s}{p_k^{\prime }(t_j)}=0,0\le s\le k-2,$$

and

$$\sum _{j=1}^k\frac{t_j^{k-1}}{p_k^{\prime }(t_j)}=\frac{1}{c_k}.$$

Further, we need several closed form formulas for some sums which can be found in numerous books (see, for example, [13] or [16]). For a general method for calculating this type of sums consult our recent paper [33] where a recurrent formula for this type of sums is presented, and by using it the sums can be calculated.

Lemma 2.

Let

$$\begin{array}{c}\hfill s_n^{(i)}(z)=\sum _{j=1}^n{j}^i{z}^{j-1},n\in \mathbb{N},\end{array}$$

(3)

where $i\in {\mathbb{N}}_0$ and $z\in \mathbb{C}.$

Then

$$\begin{array}{cc}\hfill s_n^{(0)}(z)& =\frac{1-z^n}{1-z},\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill s_n^{(1)}(z)& =\frac{1-(n+1)z^n+n{z}^{n+1}}{{(1-z)}^2},\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill s_n^{(2)}(z)& =\frac{1+z-{(n+1)}^2{z}^n+(2n^2+2n-1)z^{n+1}-n^2{z}^{n+2}}{{(1-z)}^3},\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill s_n^{(3)}(z)& =\frac{n^3{z}^n{(z-1)}^3-3n^2{z}^n{(z-1)}^2+3n{z}^n(z^2-1)-(z^n-1)(z^2+4z+1)}{{(1-z)}^4},\hfill \end{array}$$

(7)

for every $z\in \mathbb{C}\backslash \left\{1\right\}$ and $n\in \mathbb{N}$.

The following lemma describes the nature/type of the zeros of an arbitrary fourth order polynomial equation. The results in the lemma are certainly folklore and were essentially obtained, for example, in [45] (the lemma formulates the results appearing therein in a unified way, although the notation and some quantities are different).

Lemma 3.

Let

$$P_4(t)=t^4+b{t}^3+c{t}^2+dt+e,$$

where $b,c,d,e$ are real numbers,

$$\begin{array}{c}{\Delta }_0=c^2-3bd+12e,{\Delta }_1=2c^3-9bcd+27b^{2}e+27d^2-72ce,\hfill \\ \Delta =\frac{1}{27}(4{\Delta }_0^3-{\Delta }_1^2),P=8c-3b^2,\hfill \\ Q=b^3+8d-4bc,D=64e-16c^2+16b^{2}c-16bd-3b^4.\hfill \end{array}$$

Then, the following statements hold.

(a) 

If $\Delta <0$, then two zeros of $P_4$ are real and different, and two are complex conjugate.

(b) 

If $\Delta >0$, then all the zeros of $P_4$ are real or none is. More precisely,

$1^{\circ }$

if $P<0$ and $D<0$, then all four zeros of $P_4$ are real and different;

$2^{\circ }$

if $P>0$ or $D>0$, then there are two pairs of complex conjugate zeros of $P_4$.

(c) 

If $\Delta =0$, then and only then $P_4$ has a multiple zero. The following cases can occur:

$1^{\circ }$

if $P<0$, $D<0$ and ${\Delta }_0\ne 0$, then two zeros of $P_4$ are real and equal and two are real and simple;

$2^{\circ }$

if $D>0$ or ($P>0$ and ($D\ne 0$ or $Q\ne 0$)), then two zeros of $P_4$ are real and equal and two are complex conjugate;

$3^{\circ }$

if ${\Delta }_0=0$ and $D\ne 0$, there is a triple zero of $P_4$ and one simple, all real;

$4^{\circ }$

if $D=0$, then

${4.1}^{\circ }$

if $P<0$ there are two double real zeros of $P_4$;

${4.2}^{\circ }$

if $P>0$ and $Q=0$ there are two double complex conjugate zeros of $P_4$;

${4.3}^{\circ }$

if ${\Delta }_0=0$, then all four zeros of $P_4$ are real and equal to $-b/4$.

3. Main Results

In this section we state and prove our main results and by using them and some further analysis we get several corollaries. Before this we give a list of first several members of the sequences $z_n$ and $w_n$, whose values will be used in the proofs of some of the results.

We have

$$\begin{array}{cc}\hfill z_1& =\alpha z_0^a{w}_0^b,\hfill \\ \hfill w_1& =\beta w_{-2}^c{z}_0^d,\hfill \\ \hfill z_2& =\alpha {(\alpha z_0^a{w}_0^b)}^a{(\beta w_{-2}^c{z}_0^d)}^b={\alpha }^{1+a}{\beta }^b{z}_0^{a^2+bd}{w}_{-2}^{bc}{w}_0^{ab},\hfill \\ \hfill w_2& =\beta w_{-1}^c{z}_1^d=\beta w_{-1}^c{(\alpha z_0^a{w}_0^b)}^d={\alpha }^d\beta z_0^{ad}{w}_{-1}^c{w}_0^{bd}\hfill \\ \hfill z_3& =\alpha z_2^a{w}_2^b=\alpha {({\alpha }^{1+a}{\beta }^b{z}_0^{a^2+bd}{w}_{-2}^{bc}{w}_0^{ab})}^a{({\alpha }^d\beta z_0^{ad}{w}_{-1}^c{w}_0^{bd})}^b\hfill \\ & ={\alpha }^{1+a+a^2+bd}{\beta }^{b+ab}{z}_0^{a^3+2abd}{w}_{-2}^{abc}{w}_{-1}^{bc}{w}_0^{b(a^2+bd)}.\hfill \end{array}$$

(8)

As we have already mentioned we will consider the following five cases separately: (1) $b=0$; (2) $c=0$; (3) $d=0$; (4) $ac\ne 0$; (5) $a=0$, $bcd\ne 0$. Hence, we will prove five results on the solvability and by further analysis we will get several consequences from them.

Theorem 1.

Assume that $a,c,d\in \mathbb{Z}$, $b=0,$ $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then system (2) is solvable in closed form.

Proof. 

Since $b=0$, we have

$$\begin{array}{c}\hfill z_{n+1}=\alpha z_n^a,w_{n+1}=\beta w_{n-2}^c{z}_n^d,n\in {\mathbb{N}}_0,\end{array}$$

(9)

from which it follows that

$$\begin{array}{c}\hfill z_n={\alpha }^{{\sum }_{j=0}^{n-1}{a}^j}{z}_0^{a^n},\end{array}$$

(10)

for $n\in \mathbb{N}.$

From (10) we have that

$$\begin{array}{c}\hfill z_n={\alpha }^{\frac{1-a^n}{1-a}}{z}_0^{a^n},n\in \mathbb{N},\end{array}$$

(11)

when $a\ne 1$, and

$$\begin{array}{c}\hfill z_n={\alpha }^n{z}_0,n\in \mathbb{N},\end{array}$$

(12)

when $a=1$.

Using (10) in the second equation in (9), is obtained

$$w_n={\alpha }^{d{\sum }_{j=0}^{n-2}{a}^j}\beta z_0^{d{a}^{n-1}}{w}_{n-3}^c,n\ge 2,$$

which is equivalent to

$$\begin{array}{c}\hfill w_{3m+i}={\alpha }^{d{\sum }_{j=0}^{3m+i-2}{a}^j}\beta z_0^{d{a}^{3m+i-1}}{w}_{3(m-1)+i}^c,\end{array}$$

(13)

for $m\in \mathbb{N}$ and $i=-1,0,1.$

From (13) and by induction it is proved that

$$w_{3m+i}={\alpha }^{d{\sum }_{l=0}^{k-1}{c}^l{\sum }_{j=0}^{3(m-l)+i-2}{a}^j}{\beta }^{{\sum }_{l=0}^{k-1}{c}^l}{z}_0^{d{\sum }_{l=0}^{k-1}{c}^l{a}^{3(m-l)+i-1}}{w}_{3(m-k)+i}^{c^k},$$

for $m\ge k$ and $i=-1,0,1,$ from which for $k=m$ is obtained

$$\begin{array}{c}\hfill w_{3m+i}={\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l{\sum }_{j=0}^{3(m-l)+i-2}{a}^j}{\beta }^{{\sum }_{l=0}^{m-1}{c}^l}{z}_0^{d{\sum }_{l=0}^{m-1}{c}^l{a}^{3(m-l)+i-1}}{w}_i^{c^m},\end{array}$$

(14)

for $m\in \mathbb{N}$ and $i=-1,0,1.$

From (8) and (14), we have

$$w_{3m-1}={\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l{\sum }_{j=0}^{3(m-l-1)}{a}^j}{\beta }^{{\sum }_{l=0}^{m-1}{c}^l}{z}_0^{d{\sum }_{l=0}^{m-1}{c}^l{a}^{3(m-l)-2}}{w}_{-1}^{c^m},$$

(15)

$$w_{3m}={\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l{\sum }_{j=0}^{3(m-l)-2}{a}^j}{\beta }^{{\sum }_{l=0}^{m-1}{c}^l}{z}_0^{d{\sum }_{l=0}^{m-1}{c}^l{a}^{3(m-l)-1}}{w}_0^{c^m},$$

(16)

$$\begin{array}{cc}\hfill w_{3m+1}& ={\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l{\sum }_{j=0}^{3(m-l)-1}{a}^j}{\beta }^{{\sum }_{l=0}^{m-1}{c}^l}{z}_0^{d{\sum }_{l=0}^{m-1}{c}^l{a}^{3(m-l)}}{(\beta w_{-2}^c{z}_0^d)}^{c^m}\hfill \\ & ={\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l{\sum }_{j=0}^{3(m-l)-1}{a}^j}{\beta }^{{\sum }_{l=0}^m{c}^l}{z}_0^{d{\sum }_{l=0}^m{c}^l{a}^{3(m-l)}}{w}_{-2}^{c^{m+1}},\hfill \end{array}$$

(17)

for $m\in \mathbb{N}$.

Now we use the formulas in (15)–(17) in five subcases separately.

Case $a\ne 1\ne c\ne a^3.$ From (15)–(17) and by Lemma 2, we obtain

$$\begin{array}{cc}\hfill w_{3m-1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l\frac{1-a^{3(m-l)-2}}{1-a}}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{ad\frac{a^{3m}-c^m}{a^3-c}}{w}_{-1}^{c^m}\hfill \\ \hfill =& {\alpha }^{\frac{d}{1-a}\left(\frac{1-c^m}{1-c}-a\frac{a^{3m}-c^m}{a^3-c}\right)}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{ad\frac{a^{3m}-c^m}{a^3-c}}{w}_{-1}^{c^m}\hfill \\ \hfill =& {\alpha }^{\frac{d(a^3-c+(1-a)(a+a^2+c)c^m+(c-1)a^{3m+1})}{(1-a)(1-c)(a^3-c)}}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{ad\frac{a^{3m}-c^m}{a^3-c}}{w}_{-1}^{c^m},\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill w_{3m}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l\frac{1-a^{3(m-l)-1}}{1-a}}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{d{a}^2\frac{a^{3m}-c^m}{a^3-c}}{w}_0^{c^m}\hfill \\ \hfill =& {\alpha }^{\frac{d}{1-a}\left(\frac{1-c^m}{1-c}-a^2\frac{a^{3m}-c^m}{a^3-c}\right)}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{d{a}^2\frac{a^{3m}-c^m}{a^3-c}}{w}_0^{c^m}\hfill \\ \hfill =& {\alpha }^{\frac{d(a^3-c+(1-a)(c+ca+a^2)c^m+(c-1)a^{3m+2})}{(1-a)(1-c)(a^3-c)}}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{d{a}^2\frac{a^{3m}-c^m}{a^3-c}}{w}_0^{c^m},\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill w_{3m+1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l\frac{1-a^{3(m-l)}}{1-a}}{\beta }^{\frac{1-c^{m+1}}{1-c}}{z}_0^{d\frac{a^{3m+3}-c^{m+1}}{a^3-c}}{w}_{-2}^{c^{m+1}}\hfill \\ \hfill =& {\alpha }^{\frac{d}{1-a}\left(\frac{1-c^m}{1-c}-a^3\frac{a^{3m}-c^m}{a^3-c}\right)}{\beta }^{\frac{1-c^{m+1}}{1-c}}{z}_0^{d\frac{a^{3m+3}-c^{m+1}}{a^3-c}}{w}_{-2}^{c^{m+1}}\hfill \\ \hfill =& {\alpha }^{\frac{d(a^3-c+(1-a^3)c^{m+1}+(c-1)a^{3m+3})}{(1-a)(1-c)(a^3-c)}}{\beta }^{\frac{1-c^{m+1}}{1-c}}{z}_0^{d\frac{a^{3m+3}-c^{m+1}}{a^3-c}}{w}_{-2}^{c^{m+1}},\hfill \end{array}$$

(20)

for $m\in \mathbb{N}$.

Case $a\ne 1\ne c=a^3.$ From (15)–(17) and by Lemma 2, we obtain

$$\begin{array}{cc}\hfill w_{3m-1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{a}^{3l}{\sum }_{j=0}^{3(m-l-1)}{a}^j}{\beta }^{{\sum }_{l=0}^{m-1}{a}^{3l}}{z}_0^{d{\sum }_{l=0}^{m-1}{a}^{3l}{a}^{3(m-l)-2}}{w}_{-1}^{a^{3m}}\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}{a}^{3l}\frac{1-a^{3(m-l)-2}}{1-a}}{\beta }^{\frac{1-a^{3m}}{1-a^3}}{z}_0^{dm{a}^{3m-2}}{w}_{-1}^{a^{3m}}\hfill \\ \hfill =& {\alpha }^{\frac{d}{1-a}\left(\frac{1-a^{3m}}{1-a^3}-m{a}^{3m-2}\right)}{\beta }^{\frac{1-a^{3m}}{1-a^3}}{z}_0^{dm{a}^{3m-2}}{w}_{-1}^{a^{3m}}\hfill \\ \hfill =& {\alpha }^{\frac{d(m{a}^{3m+1}-a^{3m}-m{a}^{3m-2}+1)}{(1-a)(1-a^3)}}{\beta }^{\frac{1-a^{3m}}{1-a^3}}{z}_0^{dm{a}^{3m-2}}{w}_{-1}^{a^{3m}},\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill w_{3m}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{a}^{3l}{\sum }_{j=0}^{3(m-l)-2}{a}^j}{\beta }^{{\sum }_{l=0}^{m-1}{a}^{3l}}{z}_0^{d{\sum }_{l=0}^{m-1}{a}^{3l}{a}^{3(m-l)-1}}{w}_0^{a^{3m}}\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}{a}^{3l}\frac{1-a^{3(m-l)-1}}{1-a}}{\beta }^{\frac{1-a^{3m}}{1-a^3}}{z}_0^{dm{a}^{3m-1}}{w}_0^{a^{3m}}\hfill \\ \hfill =& {\alpha }^{\frac{d}{1-a}\left(\frac{1-a^{3m}}{1-a^3}-m{a}^{3m-1}\right)}{\beta }^{\frac{1-a^{3m}}{1-a^3}}{z}_0^{md{a}^{3m-1}}{w}_0^{a^{3m}}\hfill \\ \hfill =& {\alpha }^{\frac{d(m{a}^{3m+2}-a^{3m}-m{a}^{3m-1}+1)}{(1-a)(1-a^3)}}{\beta }^{\frac{1-a^{3m}}{1-a^3}}{z}_0^{dm{a}^{3m-1}}{w}_0^{a^{3m}},\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill w_{3m+1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{a}^{3l}{\sum }_{j=0}^{3(m-l)-1}{a}^j}{\beta }^{{\sum }_{l=0}^m{a}^{3l}}{z}_0^{d{\sum }_{l=0}^m{a}^{3l}{a}^{3(m-l)}}{w}_{-2}^{a^{3m+3}}\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}{a}^{3l}\frac{1-a^{3(m-l)}}{1-a}}{\beta }^{\frac{1-a^{3m+3}}{1-a^3}}{z}_0^{d(m+1)a^{3m}}{w}_{-2}^{a^{3m+3}}\hfill \\ \hfill =& {\alpha }^{\frac{d}{1-a}\left(\frac{1-a^{3m}}{1-a^3}-m{a}^{3m}\right)}{\beta }^{\frac{1-a^{3m+3}}{1-a^3}}{z}_0^{d(m+1)a^{3m}}{w}_{-2}^{a^{3m+3}}\hfill \\ \hfill =& {\alpha }^{\frac{d(m{a}^{3m+3}-(m+1)a^{3m}+1)}{(1-a)(1-a^3)}}{\beta }^{\frac{1-a^{3m+3}}{1-a^3}}{z}_0^{d(m+1)a^{3m}}{w}_{-2}^{a^{3m+3}},\hfill \end{array}$$

(23)

for $m\in \mathbb{N}$.

Case $a\ne 1=c.$ From (15)–(17) and by Lemma 2, we obtain

$$\begin{array}{cc}\hfill w_{3m-1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{\sum }_{j=0}^{3(m-l-1)}{a}^j}{\beta }^{{\sum }_{l=0}^{m-1}1}{z}_0^{d{\sum }_{l=0}^{m-1}{a}^{3(m-l)-2}}{w}_{-1}\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}\frac{1-a^{3(m-l)-2}}{1-a}}{\beta }^m{z}_0^{ad\frac{a^{3m}-1}{a^3-1}}{w}_{-1}\hfill \\ \hfill =& {\alpha }^{\frac{d}{1-a}\left(m-a\frac{a^{3m}-1}{a^3-1}\right)}{\beta }^m{z}_0^{ad\frac{a^{3m}-1}{a^3-1}}{w}_{-1}\hfill \\ \hfill =& {\alpha }^{\frac{d(a^{3m+1}+m(1-a^3)-a)}{(1-a)(1-a^3)}}{\beta }^m{z}_0^{ad\frac{a^{3m}-1}{a^3-1}}{w}_{-1},\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill w_{3m}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{\sum }_{j=0}^{3(m-l)-2}{a}^j}{\beta }^{{\sum }_{l=0}^{m-1}1}{z}_0^{d{\sum }_{l=0}^{m-1}{a}^{3(m-l)-1}}{w}_0\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}\frac{1-a^{3(m-l)-1}}{1-a}}{\beta }^m{z}_0^{d{a}^2\frac{a^{3m}-1}{a^3-1}}{w}_0\hfill \\ \hfill =& {\alpha }^{\frac{d}{1-a}\left(m-a^2\frac{a^{3m}-1}{a^3-1}\right)}{\beta }^m{z}_0^{d{a}^2\frac{a^{3m}-1}{a^3-1}}{w}_0\hfill \\ \hfill =& {\alpha }^{\frac{d(a^{3m+2}+m(1-a^3)-a^2)}{(1-a)(1-a^3)}}{\beta }^m{z}_0^{d{a}^2\frac{a^{3m}-1}{a^3-1}}{w}_0,\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill w_{3m+1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{\sum }_{j=0}^{3(m-l)-1}{a}^j}{\beta }^{{\sum }_{l=0}^{m}1}{z}_0^{d{\sum }_{l=0}^m{a}^{3(m-l)}}{w}_{-2}\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}\frac{1-a^{3(m-l)}}{1-a}}{\beta }^{m+1}{z}_0^{d\frac{a^{3m+3}-1}{a^3-1}}{w}_{-2}\hfill \\ \hfill =& {\alpha }^{\frac{d}{1-a}\left(m-a^3\frac{a^{3m}-1}{a^3-1}\right)}{\beta }^{m+1}{z}_0^{d\frac{a^{3m+3}-1}{a^3-1}}{w}_{-2}\hfill \\ \hfill =& {\alpha }^{\frac{d(a^{3m+3}-(m+1)a^3+m)}{(1-a)(1-a^3)}}{\beta }^{m+1}{z}_0^{d\frac{a^{3m+3}-1}{a^3-1}}{w}_{-2},\hfill \end{array}$$

(26)

for $m\in \mathbb{N}$.

Case $a=1\ne c.$ From (15)–(17) and by Lemma 2, we obtain

$$\begin{array}{cc}\hfill w_{3m-1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l{\sum }_{j=0}^{3(m-l-1)}1}{\beta }^{{\sum }_{l=0}^{m-1}{c}^l}{z}_0^{d{\sum }_{l=0}^{m-1}{c}^l}{w}_{-1}^{c^m}\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}(3(m-l-1)+1)c^l}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{d\frac{1-c^m}{1-c}}{w}_{-1}^{c^m}\hfill \\ \hfill =& {\alpha }^{d\left((3m-2)\frac{1-c^m}{1-c}-3c\frac{1-m{c}^{m-1}+(m-1)c^m}{{(1-c)}^2}\right)}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{d\frac{1-c^m}{1-c}}{w}_{-1}^{c^m}\hfill \\ \hfill =& {\alpha }^{\frac{d(3m-2-(3m+1)c+2c^m+c^{m+1})}{{(1-c)}^2}}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{d\frac{1-c^m}{1-c}}{w}_{-1}^{c^m},\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill w_{3m}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l{\sum }_{j=0}^{3(m-l)-2}1}{\beta }^{{\sum }_{l=0}^{m-1}{c}^l}{z}_0^{d{\sum }_{l=0}^{m-1}{c}^l}{w}_0^{c^m}\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}(3(m-l)-1)c^l}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{d\frac{1-c^m}{1-c}}{w}_0^{c^m}\hfill \\ \hfill =& {\alpha }^{d\left((3m-1)\frac{1-c^m}{1-c}-3c\frac{1-m{c}^{m-1}+(m-1)c^m}{{(1-c)}^2}\right)}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{d\frac{1-c^m}{1-c}}{w}_0^{c^m}\hfill \\ \hfill =& {\alpha }^{\frac{d(3m-1-(3m+2)c+c^m+2c^{m+1})}{{(1-c)}^2}}{\beta }^{\frac{1-c^m}{1-c}}{z}_0^{d\frac{1-c^m}{1-c}}{w}_0^{c^m},\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill w_{3m+1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{c}^l{\sum }_{j=0}^{3(m-l)-1}1}{\beta }^{{\sum }_{l=0}^m{c}^l}{z}_0^{d{\sum }_{l=0}^m{c}^l}{w}_{-2}^{c^{m+1}}\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}3(m-l)c^l}{\beta }^{\frac{1-c^{m+1}}{1-c}}{z}_0^{d\frac{1-c^{m+1}}{1-c}}{w}_{-2}^{c^{m+1}}\hfill \\ \hfill =& {\alpha }^{d\left(3m\frac{1-c^m}{1-c}-3c\frac{1-m{c}^{m-1}+(m-1)c^m}{{(1-c)}^2}\right)}{\beta }^{\frac{1-c^{m+1}}{1-c}}{z}_0^{d\frac{1-c^{m+1}}{1-c}}{w}_{-2}^{c^{m+1}}\hfill \\ \hfill =& {\alpha }^{\frac{3d(m-(m+1)c+c^{m+1})}{{(1-c)}^2}}{\beta }^{\frac{1-c^{m+1}}{1-c}}{z}_0^{d\frac{1-c^{m+1}}{1-c}}{w}_{-2}^{c^{m+1}},\hfill \end{array}$$

(29)

for $m\in \mathbb{N}$.

Case $a=c=1.$ From (15)–(17), we obtain

$$\begin{array}{cc}\hfill w_{3m-1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{\sum }_{j=0}^{3(m-l-1)}1}{\beta }^{{\sum }_{l=0}^{m-1}1}{z}_0^{d{\sum }_{l=0}^{m-1}1}{w}_{-1}\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}(3(m-l-1)+1)}{\beta }^m{z}_0^{dm}{w}_{-1}\hfill \\ \hfill =& {\alpha }^{d\frac{m(3m-1)}{2}}{\beta }^m{z}_0^{dm}{w}_{-1},\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill w_{3m}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{\sum }_{j=0}^{3(m-l)-2}1}{\beta }^{{\sum }_{l=0}^{m-1}1}{z}_0^{d{\sum }_{l=0}^{m-1}1}{w}_0\hfill \\ \hfill =& {\alpha }^{d{\sum }_{l=0}^{m-1}(3(m-l)-1)}{\beta }^m{z}_0^{dm}{w}_0\hfill \\ \hfill =& {\alpha }^{d\frac{m(3m+1)}{2}}{\beta }^m{z}_0^{dm}{w}_0,\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill w_{3m+1}=& {\alpha }^{d{\sum }_{l=0}^{m-1}{\sum }_{j=0}^{3(m-l)-1}1}{\beta }^{m+1}{z}_0^{d(m+1)}{w}_{-2}\hfill \\ \hfill =& {\alpha }^{3d{\sum }_{l=0}^{m-1}(m-l)}{\beta }^{m+1}{z}_0^{d(m+1)}{w}_{-2}\hfill \\ \hfill =& {\alpha }^{3d\frac{m(m+1)}{2}}{\beta }^{m+1}{z}_0^{d(m+1)}{w}_{-2},\hfill \end{array}$$

(32)

for $m\in \mathbb{N}$.

From (11), (12), (18)–(32) the theorem follows. ☐

The following corollary follows from Theorem 1.

Corollary 1.

Assume that $a,c,d\in \mathbb{Z}$, $b=0,$ $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then the following statements are true.

(a) 

If $a\ne 1\ne c\ne a^3$, then the general solution to system (2) is given by formulas (11), (18)–(20).

(b) 

If $a\ne 1\ne c=a^3$, then the general solution to system (2) is given by formulas (11), (21)–(23).

(c) 

If $a\ne 1=c$, then the general solution to system (2) is given by formulas (11), (24)–(26).

(d) 

If $a=1\ne c$, then the general solution to system (2) is given by formulas (12), (27)–(29).

(e) 

If $a=c=1$, then the general solution to system (2) is given by formulas (12), (30)–(32).

Theorem 2.

Assume that $a,b,c\in \mathbb{Z}$, $d=0,$ $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then system (2) is solvable in closed form.

Proof. 

In this case we have

$$\begin{array}{c}\hfill z_{n+1}=\alpha z_n^a{w}_n^b,w_{n+1}=\beta w_{n-2}^c,n\in {\mathbb{N}}_0.\end{array}$$

(33)

Hence

$$\begin{array}{c}\hfill w_{3n+i}=\beta w_{3(n-1)+i}^c,\end{array}$$

(34)

for $n\in \mathbb{N}$ and $i=\overline{-2,0},$ and consequently

$$\begin{array}{c}\hfill w_{3n+i}={\beta }^{{\sum }_{j=0}^{n-1}{c}^j}{w}_i^{c^n},\end{array}$$

(35)

for $n\in \mathbb{N}$ and $i=\overline{-2,0}.$

When $c\ne 1$ relation (35) implies

$$\begin{array}{c}\hfill w_{3n+i}={\beta }^{\frac{1-c^n}{1-c}}{w}_i^{c^n},\end{array}$$

(36)

for $n\in \mathbb{N}$ and $i=\overline{-2,0},$ and

$$\begin{array}{c}\hfill w_{3n+i}={\beta }^n{w}_i,\end{array}$$

(37)

for $n\in \mathbb{N}$ and $i=\overline{-2,0},$ when $c=1$.

Further, we have

$$\begin{array}{cc}\hfill z_n& =\alpha {(\alpha z_{n-2}^a{w}_{n-2}^b)}^a{w}_{n-1}^b\hfill \\ & ={\alpha }^{1+a}{z}_{n-2}^{a^2}{w}_{n-2}^{ba}{w}_{n-1}^b\hfill \\ & ={\alpha }^{1+a}{(\alpha z_{n-3}^a{w}_{n-3}^b)}^{a^2}{w}_{n-2}^{ba}{w}_{n-1}^b\hfill \\ & ={\alpha }^{1+a+a^2}{w}_{n-3}^{b{a}^2}{w}_{n-2}^{ba}{w}_{n-1}^b{z}_{n-3}^{a^3},\hfill \end{array}$$

for $n\ge 3$, which can be written as

$$\begin{array}{cc}\hfill z_{3n}& ={\alpha }^{1+a+a^2}{w}_{3n-3}^{b{a}^2}{w}_{3n-2}^{ba}{w}_{3n-1}^b{z}_{3(n-1)}^{a^3}\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill z_{3n+1}& ={\alpha }^{1+a+a^2}{w}_{3n-2}^{b{a}^2}{w}_{3n-1}^{ba}{w}_{3n}^b{z}_{3(n-1)+1}^{a^3}\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill z_{3n+2}& ={\alpha }^{1+a+a^2}{w}_{3n-1}^{b{a}^2}{w}_{3n}^{ba}{w}_{3n+1}^b{z}_{3(n-1)+2}^{a^3},\hfill \end{array}$$

(40)

for $n\in \mathbb{N}.$

Employing the method of induction we get

$$\begin{array}{cc}\hfill z_{3n}& =z_0^{a^{3n}}\prod _{j=0}^{n-1}{\left({\alpha }^{1+a+a^2}{w}_{3j}^{b{a}^2}{w}_{3j+1}^{ba}{w}_{3j+2}^b\right)}^{a^{3(n-j-1)}}\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill z_{3n+1}& =z_1^{a^{3n}}\prod _{j=0}^{n-1}{\left({\alpha }^{1+a+a^2}{w}_{3j+1}^{b{a}^2}{w}_{3j+2}^{ba}{w}_{3j+3}^b\right)}^{a^{3(n-j-1)}}\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill z_{3n+2}& =z_2^{a^{3n}}\prod _{j=0}^{n-1}{\left({\alpha }^{1+a+a^2}{w}_{3j+2}^{b{a}^2}{w}_{3j+3}^{ba}{w}_{3j+4}^b\right)}^{a^{3(n-j-1)}},\hfill \end{array}$$

(43)

for $n\in {\mathbb{N}}_0$.

Using formula (35), along with the first equation in (33) with $n=0,1$, into (41)–(43), we get

$$\begin{array}{cc}\hfill z_{3n}=& z_0^{a^{3n}}\prod _{j=0}^{n-1}{\left({\alpha }^{1+a+a^2}{\beta }^{b{a}^2{\sum }_{l=0}^{j-1}{c}^l}{w}_0^{b{a}^2{c}^j}{\beta }^{ba{\sum }_{l=0}^j{c}^l}{w}_{-2}^{ba{c}^{j+1}}{\beta }^{b{\sum }_{l=0}^j{c}^l}{w}_{-1}^{b{c}^{j+1}}\right)}^{a^{3(n-j-1)}}\hfill \\ \hfill =& {\alpha }^{{\sum }_{j=0}^{3n-1}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^{j-1}{c}^l+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l}\hfill \\ & \times z_0^{a^{3n}}{w}_{-2}^{bac{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_{-1}^{bc{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_0^{b{a}^2{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}},\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill z_{3n+1}=& {(\alpha z_0^a{w}_0^b)}^{a^{3n}}\prod _{j=0}^{n-1}{\left({\alpha }^{1+a+a^2}{\beta }^{b{a}^2{\sum }_{l=0}^j{c}^l}{w}_{-2}^{b{a}^2{c}^{j+1}}{\beta }^{ba{\sum }_{l=0}^j{c}^l}{w}_{-1}^{ba{c}^{j+1}}{\beta }^{b{\sum }_{l=0}^j{c}^l}{w}_0^{b{c}^{j+1}}\right)}^{a^{3(n-j-1)}}\hfill \\ \hfill =& {\alpha }^{{\sum }_{j=0}^{3n}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l}\hfill \\ & \times z_0^{a^{3n+1}}{w}_{-2}^{b{a}^{2}c{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_{-1}^{bac{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_0^{b{\sum }_{j=0}^n{c}^j{a}^{3(n-j)}},\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill z_{3n+2}=& {({\alpha }^{1+a}{\beta }^b{z}_0^{a^2}{w}_{-2}^{bc}{w}_0^{ab})}^{a^{3n}}\hfill \\ & \times \prod _{j=0}^{n-1}{\left({\alpha }^{1+a+a^2}{\beta }^{b{a}^2{\sum }_{l=0}^j{c}^l}{w}_{-1}^{b{a}^2{c}^{j+1}}{\beta }^{ba{\sum }_{l=0}^j{c}^l}{w}_0^{ba{c}^{j+1}}{\beta }^{b{\sum }_{l=0}^{j+1}{c}^l}{w}_{-2}^{b{c}^{j+2}}\right)}^{a^{3(n-j-1)}}\hfill \\ \hfill =& {\alpha }^{{\sum }_{j=0}^{3n+1}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^n{a}^{3(n-j)}{\sum }_{l=0}^j{c}^l}\hfill \\ & \times z_0^{a^{3n+2}}{w}_{-2}^{bc{\sum }_{j=0}^n{c}^j{a}^{3(n-j)}}{w}_{-1}^{b{a}^{2}c{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_0^{ab{\sum }_{j=0}^n{c}^j{a}^{3(n-j)}},\hfill \end{array}$$

(46)

for $n\in {\mathbb{N}}_0$.

As in the proof of the previous theorem, there are five subcases, depending on the values of parameters a and c, for which we get closed form formulas for solutions to system (2).

Case $a\ne 1\ne c\ne a^3$. From (44)–(46) and by Lemma 2, we have

$$\begin{array}{cc}\hfill z_{3n}=& {\alpha }^{{\sum }_{j=0}^{3n-1}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^{j-1}{c}^l+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l}\hfill \\ & \times z_0^{a^{3n}}{w}_{-2}^{bac{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_{-1}^{bc{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_0^{b{a}^2{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n}}{1-a}}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-c^j}{1-c}+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-c^{j+1}}{1-c}+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-c^{j+1}}{1-c}}\hfill \\ & \times z_0^{a^{3n}}{w}_{-2}^{bac\frac{a^{3n}-c^n}{a^3-c}}{w}_{-1}^{bc\frac{a^{3n}-c^n}{a^3-c}}{w}_0^{b{a}^2\frac{a^{3n}-c^n}{a^3-c}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n}}{1-a}}{\beta }^{\frac{b(a^2+a+1)}{1-c}\frac{1-a^{3n}}{1-a^3}-\frac{b(a^2+ac+c)}{1-c}\frac{a^{3n}-c^n}{a^3-c}}{z}_0^{a^{3n}}{w}_{-2}^{bac\frac{a^{3n}-c^n}{a^3-c}}{w}_{-1}^{bc\frac{a^{3n}-c^n}{a^3-c}}{w}_0^{b{a}^2\frac{a^{3n}-c^n}{a^3-c}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n}}{1-a}}{\beta }^{\frac{b(a^3-c+(c-1)a^{3n+2}-(a^2+ac+c)(a-1)c^n)}{(1-c)(a^3-c)(1-a)}}{z}_0^{a^{3n}}{w}_{-2}^{bac\frac{a^{3n}-c^n}{a^3-c}}{w}_{-1}^{bc\frac{a^{3n}-c^n}{a^3-c}}{w}_0^{b{a}^2\frac{a^{3n}-c^n}{a^3-c}},\hfill \end{array}$$

(47)

$$\begin{array}{cc}\hfill z_{3n+1}=& {\alpha }^{{\sum }_{j=0}^{3n}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l}\hfill \\ & \times z_0^{a^{3n+1}}{w}_{-2}^{b{a}^{2}c{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_{-1}^{bac{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_0^{b{\sum }_{j=0}^n{c}^j{a}^{3(n-j)}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+1}}{1-a}}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-c^{j+1}}{1-c}+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-c^{j+1}}{1-c}+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-c^{j+1}}{1-c}}\hfill \\ & \times z_0^{a^{3n+1}}{w}_{-2}^{b{a}^{2}c\frac{a^{3n}-c^n}{a^3-c}}{w}_{-1}^{bac\frac{a^{3n}-c^n}{a^3-c}}{w}_0^{b\frac{a^{3n+3}-c^{n+1}}{a^3-c}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+1}}{1-a}}{\beta }^{\frac{b(a^2+a+1)}{1-c}\frac{1-a^{3n}}{1-a^3}-\frac{bc(a^2+a+1)}{1-c}\frac{a^{3n}-c^n}{a^3-c}}{z}_0^{a^{3n+1}}{w}_{-2}^{b{a}^{2}c\frac{a^{3n}-c^n}{a^3-c}}{w}_{-1}^{bac\frac{a^{3n}-c^n}{a^3-c}}{w}_0^{b\frac{a^{3n+3}-c^{n+1}}{a^3-c}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+1}}{1-a}}{\beta }^{\frac{b(a^3-c+(c-1)a^{3n+3}+(1-a^3)c^{n+1})}{(1-c)(1-a)(a^3-c)}}{z}_0^{a^{3n+1}}{w}_{-2}^{b{a}^{2}c\frac{a^{3n}-c^n}{a^3-c}}{w}_{-1}^{bac\frac{a^{3n}-c^n}{a^3-c}}{w}_0^{b\frac{a^{3n+3}-c^{n+1}}{a^3-c}},\hfill \end{array}$$

(48)

$$\begin{array}{cc}\hfill z_{3n+2}=& {\alpha }^{{\sum }_{j=0}^{3n+1}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^n{a}^{3(n-j)}{\sum }_{l=0}^j{c}^l}\hfill \\ & \times z_0^{a^{3n+2}}{w}_{-2}^{bc{\sum }_{j=0}^n{c}^j{a}^{3(n-j)}}{w}_{-1}^{b{a}^{2}c{\sum }_{j=0}^{n-1}{c}^j{a}^{3(n-j-1)}}{w}_0^{ab{\sum }_{j=0}^n{c}^j{a}^{3(n-j)}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+2}}{1-a}}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-c^{j+1}}{1-c}+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-c^{j+1}}{1-c}+b{\sum }_{j=0}^n{a}^{3(n-j)}\frac{1-c^{j+1}}{1-c}}\hfill \\ & \times z_0^{a^{3n+2}}{w}_{-2}^{bc\frac{a^{3n+3}-c^{n+1}}{a^3-c}}{w}_{-1}^{b{a}^{2}c\frac{a^{3n}-c^n}{a^3-c}}{w}_0^{ab\frac{a^{3n+3}-c^{n+1}}{a^3-c}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+2}}{1-a}}{\beta }^{\frac{b(a^2+a)}{1-c}\frac{1-a^{3n}}{1-a^3}+\frac{b(1-a^{3n+3})}{(1-c)(1-a^3)}-\frac{bc(a^2+a)}{1-c}\frac{a^{3n}-c^n}{a^3-c}-\frac{bc}{1-c}\frac{a^{3n+3}-c^{n+1}}{a^3-c}}\hfill \\ & \times z_0^{a^{3n+2}}{w}_{-2}^{bc\frac{a^{3n+3}-c^{n+1}}{a^3-c}}{w}_{-1}^{b{a}^{2}c\frac{a^{3n}-c^n}{a^3-c}}{w}_0^{ab\frac{a^{3n+3}-c^{n+1}}{a^3-c}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+2}}{1-a}}{\beta }^{\frac{b(a^3-c+(c-1)a^{3n+4}+(1-a)(a^2+a+c)c^{n+1})}{(1-c)(1-a)(a^3-c)}}{z}_0^{a^{3n+2}}\hfill \\ & \times w_{-2}^{bc\frac{a^{3n+3}-c^{n+1}}{a^3-c}}{w}_{-1}^{b{a}^{2}c\frac{a^{3n}-c^n}{a^3-c}}{w}_0^{ab\frac{a^{3n+3}-c^{n+1}}{a^3-c}},\hfill \end{array}$$

(49)

for $n\in {\mathbb{N}}_0$.

Case $a\ne 1\ne c=a^3$. From (44)–(46) and by Lemma 2, we have

$$\begin{array}{cc}\hfill z_{3n}=& {\alpha }^{{\sum }_{j=0}^{3n-1}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^{j-1}{a}^{3l}+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{a}^{3l}+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{a}^{3l}}\hfill \\ & \times z_0^{a^{3n}}{w}_{-2}^{b{a}^4{\sum }_{j=0}^{n-1}{a}^{3j}{a}^{3(n-j-1)}}{w}_{-1}^{b{a}^3{\sum }_{j=0}^{n-1}{a}^{3j}{a}^{3(n-j-1)}}{w}_0^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3j}{a}^{3(n-j-1)}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n}}{1-a}}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-a^{3j}}{1-a^3}+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-a^{3j+3}}{1-a^3}+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-a^{3j+3}}{1-a^3}}\hfill \\ & \times z_0^{a^{3n}}{w}_{-2}^{bn{a}^{3n+1}}{w}_{-1}^{bn{a}^{3n}}{w}_0^{bn{a}^{3n-1}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n}}{1-a}}{\beta }^{\frac{b{a}^2}{1-a^3}\left(\frac{1-a^{3n}}{1-a^3}-n{a}^{3n-3}\right)+\frac{ba}{1-a^3}\left(\frac{1-a^{3n}}{1-a^3}-n{a}^{3n}\right)+\frac{b}{1-a^3}\left(\frac{1-a^{3n}}{1-a^3}-n{a}^{3n}\right)}\hfill \\ & \times z_0^{a^{3n}}{w}_{-2}^{bn{a}^{3n+1}}{w}_{-1}^{bn{a}^{3n}}{w}_0^{bn{a}^{3n-1}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n}}{1-a}}{\beta }^{b\frac{n{a}^{3n+2}-a^{3n}-n{a}^{3n-1}+1}{(1-a)(1-a^3)}}{z}_0^{a^{3n}}{w}_{-2}^{bn{a}^{3n+1}}{w}_{-1}^{bn{a}^{3n}}{w}_0^{bn{a}^{3n-1}},\hfill \end{array}$$

(50)

$$\begin{array}{cc}\hfill z_{3n+1}=& {\alpha }^{{\sum }_{j=0}^{3n}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{a}^{3l}+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{a}^{3l}+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{a}^{3l}}\hfill \\ & \times z_0^{a^{3n+1}}{w}_{-2}^{b{a}^5{\sum }_{j=0}^{n-1}{a}^{3j}{a}^{3(n-j-1)}}{w}_{-1}^{b{a}^4{\sum }_{j=0}^{n-1}{a}^{3j}{a}^{3(n-j-1)}}{w}_0^{b{\sum }_{j=0}^n{a}^{3j}{a}^{3(n-j)}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+1}}{1-a}}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-a^{3j+3}}{1-a^3}+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-a^{3j+3}}{1-a^3}+b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-a^{3j+3}}{1-a^3}}\hfill \\ & \times z_0^{a^{3n+1}}{w}_{-2}^{bn{a}^{3n+2}}{w}_{-1}^{bn{a}^{3n+1}}{w}_0^{b(n+1)a^{3n}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+1}}{1-a}}{\beta }^{\frac{b{a}^2}{1-a^3}\left(\frac{1-a^{3n}}{1-a^3}-n{a}^{3n}\right)+\frac{ba}{1-a^3}\left(\frac{1-a^{3n}}{1-a^3}-n{a}^{3n}\right)+\frac{b}{1-a^3}\left(\frac{1-a^{3n}}{1-a^3}-n{a}^{3n}\right)}\hfill \\ & \times z_0^{a^{3n+1}}{w}_{-2}^{bn{a}^{3n+2}}{w}_{-1}^{bn{a}^{3n+1}}{w}_0^{b(n+1)a^{3n}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+1}}{1-a}}{\beta }^{\frac{b(n{a}^{3n+3}-(n+1)a^{3n}+1)}{(1-a)(1-a^3)}}{z}_0^{a^{3n+1}}{w}_{-2}^{bn{a}^{3n+2}}{w}_{-1}^{bn{a}^{3n+1}}{w}_0^{b(n+1)a^{3n}},\hfill \end{array}$$

(51)

$$\begin{array}{cc}\hfill z_{3n+2}=& {\alpha }^{{\sum }_{j=0}^{3n+1}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{a}^{3l}+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}{\sum }_{l=0}^j{a}^{3l}+b{\sum }_{j=0}^n{a}^{3(n-j)}{\sum }_{l=0}^j{a}^{3l}}\hfill \\ & \times z_0^{a^{3n+2}}{w}_{-2}^{b{a}^3{\sum }_{j=0}^n{a}^{3j}{a}^{3(n-j)}}{w}_{-1}^{b{a}^5{\sum }_{j=0}^{n-1}{a}^{3j}{a}^{3(n-j-1)}}{w}_0^{ab{\sum }_{j=0}^n{a}^{3j}{a}^{3(n-j)}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+2}}{1-a}}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-a^{3j+3}}{1-a^3}+ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}\frac{1-a^{3j+3}}{1-a^3}+b{\sum }_{j=0}^n{a}^{3(n-j)}\frac{1-a^{3j+3}}{1-a^3}}\hfill \\ & \times z_0^{a^{3n+2}}{w}_{-2}^{b(n+1)a^{3n+3}}{w}_{-1}^{bn{a}^{3n+2}}{w}_0^{b(n+1)a^{3n+1}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+2}}{1-a}}{\beta }^{\frac{b{a}^2}{1-a^3}\left(\frac{1-a^{3n}}{1-a^3}-n{a}^{3n}\right)+\frac{ba}{1-a^3}\left(\frac{1-a^{3n}}{1-a^3}-n{a}^{3n}\right)+\frac{b}{1-a^3}\left(\frac{1-a^{3n+3}}{1-a^3}-(n+1)a^{3n+3}\right)}\hfill \\ & \times z_0^{a^{3n+2}}{w}_{-2}^{b(n+1)a^{3n+3}}{w}_{-1}^{bn{a}^{3n+2}}{w}_0^{b(n+1)a^{3n+1}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+2}}{1-a}}{\beta }^{\frac{b((n+1)a^{3n+4}-a^{3n+3}-(n+1)a^{3n+1}+1)}{(1-a)(1-a^3)}}{z}_0^{a^{3n+2}}{w}_{-2}^{b(n+1)a^{3n+3}}{w}_{-1}^{bn{a}^{3n+2}}{w}_0^{b(n+1)a^{3n+1}},\hfill \end{array}$$

(52)

for $n\in {\mathbb{N}}_0$.

Case If $a\ne 1=c$. From (44)–(46) and by Lemma 2, we have

$$\begin{array}{cc}\hfill z_{3n}=& {\alpha }^{{\sum }_{j=0}^{3n-1}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}j{a}^{3(n-j-1)}+ba{\sum }_{j=0}^{n-1}(j+1)a^{3(n-j-1)}+b{\sum }_{j=0}^{n-1}(j+1)a^{3(n-j-1)}}\hfill \\ & \times z_0^{a^{3n}}{w}_{-2}^{ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}}{w}_{-1}^{b{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}}{w}_0^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n}}{1-a}}{\beta }^{b{a}^2\frac{a^{3n}-n{a}^3+n-1}{{(a^3-1)}^2}+ba\frac{a^{3n+3}-(n+1)a^3+n}{{(a^3-1)}^2}+b\frac{a^{3n+3}-(n+1)a^3+n}{{(a^3-1)}^2}}\hfill \\ & \times z_0^{a^{3n}}{w}_{-2}^{ba\frac{1-a^{3n}}{1-a^3}}{w}_{-1}^{b\frac{1-a^{3n}}{1-a^3}}{w}_0^{b{a}^2\frac{1-a^{3n}}{1-a^3}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n}}{1-a}}{\beta }^{b\frac{(a^2+a+1)a^{3n+2}-n{a}^5-(n+1)a^4-(n+1)a^3+(n-1)a^2+na+n}{{(a^3-1)}^2}}\hfill \\ & \times z_0^{a^{3n}}{w}_{-2}^{ba\frac{1-a^{3n}}{1-a^3}}{w}_{-1}^{b\frac{1-a^{3n}}{1-a^3}}{w}_0^{b{a}^2\frac{1-a^{3n}}{1-a^3}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n}}{1-a}}{\beta }^{b\frac{a^{3n+2}-n{a}^3+n-a^2}{(a-1)(a^3-1)}}{z}_0^{a^{3n}}{w}_{-2}^{ba\frac{1-a^{3n}}{1-a^3}}{w}_{-1}^{b\frac{1-a^{3n}}{1-a^3}}{w}_0^{b{a}^2\frac{1-a^{3n}}{1-a^3}},\hfill \end{array}$$

(53)

$$\begin{array}{cc}\hfill z_{3n+1}=& {\alpha }^{{\sum }_{j=0}^{3n}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}(j+1)a^{3(n-j-1)}+ba{\sum }_{j=0}^{n-1}(j+1)a^{3(n-j-1)}+b{\sum }_{j=0}^{n-1}(j+1)a^{3(n-j-1)}}\hfill \\ & \times z_0^{a^{3n+1}}{w}_{-2}^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}}{w}_{-1}^{ba{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}}{w}_0^{b{\sum }_{j=0}^n{a}^{3(n-j)}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+1}}{1-a}}{\beta }^{b\frac{a^{3n+3}-(n+1)a^3+n}{(a-1)(a^3-1)}}{z}_0^{a^{3n+1}}{w}_{-2}^{b{a}^2\frac{1-a^{3n}}{1-a^3}}{w}_{-1}^{ba\frac{1-a^{3n}}{1-a^3}}{w}_0^{b\frac{1-a^{3n+3}}{1-a^3}},\hfill \end{array}$$

(54)

$$\begin{array}{cc}\hfill z_{3n+2}=& {\alpha }^{{\sum }_{j=0}^{3n+1}{a}^j}{\beta }^{b{a}^2{\sum }_{j=0}^{n-1}(j+1)a^{3(n-j-1)}+ba{\sum }_{j=0}^{n-1}(j+1)a^{3(n-j-1)}+b{\sum }_{j=0}^n(j+1)a^{3(n-j)}}\hfill \\ & \times z_0^{a^{3n+2}}{w}_{-2}^{b{\sum }_{j=0}^n{a}^{3(n-j)}}{w}_{-1}^{b{a}^2{\sum }_{j=0}^{n-1}{a}^{3(n-j-1)}}{w}_0^{ab{\sum }_{j=0}^n{a}^{3(n-j)}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+2}}{1-a}}{\beta }^{b{a}^2\frac{a^{3n+3}-(n+1)a^3+n}{{(a^3-1)}^2}+ba\frac{a^{3n+3}-(n+1)a^3+n}{{(a^3-1)}^2}+b\frac{a^{3n+6}-(n+2)a^3+n+1}{{(a^3-1)}^2}}\hfill \\ & \times z_0^{a^{3n+2}}{w}_{-2}^{b\frac{1-a^{3n+3}}{1-a^3}}{w}_{-1}^{b{a}^2\frac{1-a^{3n}}{1-a^3}}{w}_0^{ab\frac{1-a^{3n+3}}{1-a^3}}\hfill \\ \hfill =& {\alpha }^{\frac{1-a^{3n+2}}{1-a}}{\beta }^{b\frac{a^{3n+4}+(1-a^3)n-(a^3+a-1)}{(a-1)(a^3-1)}}{z}_0^{a^{3n+2}}{w}_{-2}^{b\frac{1-a^{3n+3}}{1-a^3}}{w}_{-1}^{b{a}^2\frac{1-a^{3n}}{1-a^3}}{w}_0^{ab\frac{1-a^{3n+3}}{1-a^3}},\hfill \end{array}$$

(55)

for $n\in {\mathbb{N}}_0$.

Case If $a=1\ne c$. From (44)–(46) and by Lemma 2, we have

$$\begin{array}{cc}\hfill z_{3n}=& {\alpha }^{3n}{\beta }^{b{\sum }_{j=0}^{n-1}{\sum }_{l=0}^{j-1}{c}^l+b{\sum }_{j=0}^{n-1}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^{n-1}{\sum }_{l=0}^j{c}^l}{z}_0{w}_{-2}^{bc{\sum }_{j=0}^{n-1}{c}^j}{w}_{-1}^{bc{\sum }_{j=0}^{n-1}{c}^j}{w}_0^{b{\sum }_{j=0}^{n-1}{c}^j}\hfill \\ \hfill =& {\alpha }^{3n}{\beta }^{b{\sum }_{j=0}^{n-1}\frac{1-c^j}{1-c}+b{\sum }_{j=0}^{n-1}\frac{1-c^{j+1}}{1-c}+b{\sum }_{j=0}^{n-1}\frac{1-c^{j+1}}{1-c}}{z}_0{w}_{-2}^{bc\frac{1-c^n}{1-c}}{w}_{-1}^{bc\frac{1-c^n}{1-c}}{w}_0^{b\frac{1-c^n}{1-c}}\hfill \\ \hfill =& {\alpha }^{3n}{\beta }^{\frac{b(2c^{n+1}+c^n-3(c-1)n-2c-1)}{{(1-c)}^2}}{z}_0{w}_{-2}^{bc\frac{1-c^n}{1-c}}{w}_{-1}^{bc\frac{1-c^n}{1-c}}{w}_0^{b\frac{1-c^n}{1-c}},\hfill \end{array}$$

(56)

$$\begin{array}{cc}\hfill z_{3n+1}=& {\alpha }^{3n+1}{\beta }^{b{\sum }_{j=0}^{n-1}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^{n-1}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^{n-1}{\sum }_{l=0}^j{c}^l}{z}_0{w}_{-2}^{bc{\sum }_{j=0}^{n-1}{c}^j}{w}_{-1}^{bc{\sum }_{j=0}^{n-1}{c}^j}{w}_0^{b{\sum }_{j=0}^n{c}^j}\hfill \\ \hfill =& {\alpha }^{3n+1}{\beta }^{3b{\sum }_{j=0}^{n-1}\frac{1-c^{j+1}}{1-c}}{z}_0{w}_{-2}^{bc\frac{1-c^n}{1-c}}{w}_{-1}^{bc\frac{1-c^n}{1-c}}{w}_0^{b\frac{1-c^{n+1}}{1-c}}\hfill \\ \hfill =& {\alpha }^{3n+1}{\beta }^{3b\frac{c^{n+1}-(c-1)n-c}{{(1-c)}^2}}{z}_0{w}_{-2}^{bc\frac{1-c^n}{1-c}}{w}_{-1}^{bc\frac{1-c^n}{1-c}}{w}_0^{b\frac{1-c^{n+1}}{1-c}},\hfill \end{array}$$

(57)

$$\begin{array}{cc}\hfill z_{3n+2}=& {\alpha }^{3n+2}{\beta }^{b{\sum }_{j=0}^{n-1}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^{n-1}{\sum }_{l=0}^j{c}^l+b{\sum }_{j=0}^n{\sum }_{l=0}^j{c}^l}\hfill \\ & \times z_0{w}_{-2}^{bc{\sum }_{j=0}^n{c}^j}{w}_{-1}^{bc{\sum }_{j=0}^{n-1}{c}^j}{w}_0^{b{\sum }_{j=0}^n{c}^j}\hfill \\ \hfill =& {\alpha }^{3n+2}{\beta }^{b{\sum }_{j=0}^{n-1}\frac{1-c^{j+1}}{1-c}+b{\sum }_{j=0}^{n-1}\frac{1-c^{j+1}}{1-c}+b{\sum }_{j=0}^n\frac{1-c^{j+1}}{1-c}}\hfill \\ & \times z_0{w}_{-2}^{bc\frac{1-c^{n+1}}{1-c}}{w}_{-1}^{bc\frac{1-c^n}{1-c}}{w}_0^{b\frac{1-c^{n+1}}{1-c}}\hfill \\ \hfill =& {\alpha }^{3n+2}{\beta }^{\frac{b(c^{n+2}+2c^{n+1}-3(c-1)n+1-4c)}{{(1-c)}^2}}{z}_0{w}_{-2}^{bc\frac{1-c^{n+1}}{1-c}}{w}_{-1}^{bc\frac{1-c^n}{1-c}}{w}_0^{b\frac{1-c^{n+1}}{1-c}},\hfill \end{array}$$

(58)

for $n\in {\mathbb{N}}_0$.

Case If $a=c=1$. From (44)–(46), we have

$$\begin{array}{cc}\hfill z_{3n}=& {\alpha }^{3n}{\beta }^{b{\sum }_{j=0}^{n-1}j+b{\sum }_{j=0}^{n-1}(j+1)+b{\sum }_{j=0}^{n-1}(j+1)}{z}_0{w}_{-2}^{bn}{w}_{-1}^{bn}{w}_0^{bn}\hfill \\ \hfill =& {\alpha }^{3n}{\beta }^{b\frac{n(3n+1)}{2}}{z}_0{w}_{-2}^{bn}{w}_{-1}^{bn}{w}_0^{bn},\hfill \end{array}$$

(59)

$$\begin{array}{cc}\hfill z_{3n+1}=& {\alpha }^{3n+1}{\beta }^{b{\sum }_{j=0}^{n-1}(j+1)+b{\sum }_{j=0}^{n-1}(j+1)+b{\sum }_{j=0}^{n-1}(j+1)}{z}_0{w}_{-2}^{bn}{w}_{-1}^{bn}{w}_0^{b(n+1)}\hfill \\ \hfill =& {\alpha }^{3n+1}{\beta }^{3b\frac{n(n+1)}{2}}{z}_0{w}_{-2}^{bn}{w}_{-1}^{bn}{w}_0^{b(n+1)},\hfill \end{array}$$

(60)

$$\begin{array}{cc}\hfill z_{3n+2}=& {\alpha }^{3n+2}{\beta }^{b{\sum }_{j=0}^{n-1}(j+1)+b{\sum }_{j=0}^{n-1}(j+1)+b{\sum }_{j=0}^n(j+1)}{z}_0{w}_{-2}^{b(n+1)}{w}_{-1}^{bn}{w}_0^{b(n+1)}\hfill \\ \hfill =& {\alpha }^{3n+2}{\beta }^{b\frac{(n+1)(3n+2)}{2}}{z}_0{w}_{-2}^{b(n+1)}{w}_{-1}^{bn}{w}_0^{b(n+1)},\hfill \end{array}$$

(61)

for $n\in {\mathbb{N}}_0$.

From all above presented formulas we see that system of difference equations (2) is solvable in this case, as claimed. ☐

The statements in the following corollary are direct consequences of the formulas presented in the proof of Theorem 2.

Corollary 2.

Assume that $a,b,c\in \mathbb{Z}$, $d=0,$ $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then the following statements are true.

(a) 

If $a\ne 1\ne c\ne a^3$, then the general solution to system (2) is given by (36), (47)–(49).

(b) 

If $a\ne 1\ne c=a^3$, then the general solution to system (2) is given by (36), (50)–(52).

(c) 

If $a\ne 1=c$, then the general solution to system (2) is given by (37), (53)–(55).

(d) 

If $a=1\ne c$, then the general solution to system (2) is given by (36), (56)–(58).

(e) 

If $a=c=1$, then the general solution to system (2) is given by (37), (59)–(61).

The following result concerns the case $c=0$. Note that under the condition system (2) of difference equations is

$$\begin{array}{c}\hfill z_{n+1}=\alpha z_n^a{w}_n^b,w_{n+1}=\beta z_n^d,n\in {\mathbb{N}}_0,\end{array}$$

(62)

which is system (3.15) in [33].

From (62) it follows that

$$z_{n+1}=\alpha {\beta }^b{z}_n^a{z}_{n-1}^{bd},$$

for $n\in \mathbb{N}.$

How this product-type difference equation can be solved was explained in the proof of Theorem 3.3 in [33]. From the closed-form formulas obtained in the proof of Theorem 3.3 therein it follows directly that the following result holds. Hence, to avoid repeating, the proof of the theorem is omitted here.

Theorem 3.

Assume that $a,b,d\in \mathbb{Z}$, $c=0,$ $\alpha ,\beta ,z_0,w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then the following statements are true.

(a) 

If $bd\ne 0$, $a^2+4bd\ne 0$ and $a+bd\ne 1$, then the general solution to system (2) is given by

$$\begin{array}{cc}\hfill z_n=& {\alpha }^{\frac{({\lambda }_2-1){\lambda }_1^{n+1}-({\lambda }_1-1){\lambda }_2^{n+1}+{\lambda }_1-{\lambda }_2}{({\lambda }_1-{\lambda }_2)({\lambda }_1-1)({\lambda }_2-1)}}{\beta }^{b\frac{({\lambda }_2-1){\lambda }_1^n-({\lambda }_1-1){\lambda }_2^n+{\lambda }_1-{\lambda }_2}{({\lambda }_1-{\lambda }_2)({\lambda }_1-1)({\lambda }_2-1)}}{z}_0^{\frac{{\lambda }_1^{n+1}-{\lambda }_2^{n+1}}{{\lambda }_1-{\lambda }_2}}{w}_0^{b\frac{{\lambda }_1^n-{\lambda }_2^n}{{\lambda }_1-{\lambda }_2}},\hfill \\ \hfill w_n=& {\alpha }^{d\frac{({\lambda }_2-1){\lambda }_1^n-({\lambda }_1-1){\lambda }_2^n+{\lambda }_1-{\lambda }_2}{({\lambda }_1-{\lambda }_2)({\lambda }_1-1)({\lambda }_2-1)}}{\beta }^{1+bd\frac{({\lambda }_2-1){\lambda }_1^{n-1}-({\lambda }_1-1){\lambda }_2^{n-1}+{\lambda }_1-{\lambda }_2}{({\lambda }_1-{\lambda }_2)({\lambda }_1-1)({\lambda }_2-1)}}{z}_0^{d\frac{{\lambda }_1^n-{\lambda }_2^n}{{\lambda }_1-{\lambda }_2}}{w}_0^{bd\frac{{\lambda }_1^{n-1}-{\lambda }_2^{n-1}}{{\lambda }_1-{\lambda }_2}},\hfill \end{array}$$

for $n\in \mathbb{N}$, where

$${\lambda }_{1,2}=\frac{a\pm \sqrt{a^2+4bd}}{2}.$$

(b) 

If $bd\ne 0$, $a^2+4bd=0$ and $a+bd\ne 1$, then the general solution to system (2) is given by

$$\begin{array}{cc}\hfill z_n=& {\alpha }^{\frac{1-(n+1){\lambda }_1^n+n{\lambda }_1^{n+1}}{{(1-{\lambda }_1)}^2}}{\beta }^{b\frac{1-n{\lambda }_1^{n-1}+(n-1){\lambda }_1^n}{{(1-{\lambda }_1)}^2}}{z}_0^{(n+1){\lambda }_1^n}{w}_0^{bn{\lambda }_1^{n-1}}\hfill \\ \hfill w_n=& {\alpha }^{d\frac{1-n{\lambda }_1^{n-1}+(n-1){\lambda }_1^n}{{(1-{\lambda }_1)}^2}}{\beta }^{\frac{1-2{\lambda }_1+(n-1){\lambda }_1^n-(n-2){\lambda }_1^{n+1}}{{(1-{\lambda }_1)}^2}}{z}_0^{dn{\lambda }_1^{n-1}}{w}_0^{(1-n){\lambda }_1^n}\hfill \end{array}$$

for $n\in \mathbb{N},$ where

$${\lambda }_1=\frac{a}{2}.$$

(c) 

If $bd\ne 0$, $a^2+4bd\ne 0$ and $a+bd=1$, then the general solution to system (2) is given by

$$\begin{array}{cc}\hfill z_n=& {\alpha }^{\frac{{(-bd)}^{n+1}+(n+1)bd+n}{{(1+bd)}^2}}{\beta }^{b\frac{{(-bd)}^n+nbd+n-1}{{(1+bd)}^2}}{z}_0^{\frac{1-{(-bd)}^{n+1}}{1+bd}}{w}_0^{b\frac{1-{(-bd)}^n}{1+bd}},\hfill \\ \hfill w_n=& {\alpha }^{d\frac{{(-bd)}^n+nbd+n-1}{{(1+bd)}^2}}{\beta }^{1+bd\frac{{(-bd)}^{n-1}+(n-1)bd+n-2}{{(1+bd)}^2}}{z}_0^{d\frac{1-{(-bd)}^n}{1+bd}}{w}_0^{bd\frac{1-{(-bd)}^{n-1}}{1+bd}},\hfill \end{array}$$

for $n\in \mathbb{N}$.

(d) 

If $bd\ne 0$, $a^2+4bd=0$ and $a+bd=1$, then the general solution to system (2) is given by

$$\begin{array}{cc}\hfill z_n=& {\alpha }^{\frac{n(n+1)}{2}}{\beta }^{b\frac{(n-1)n}{2}}{z}_0^{n+1}{w}_0^{bn}\hfill \\ \hfill w_n=& {\alpha }^{d\frac{(n-1)n}{2}}{\beta }^{\frac{(3-n)n}{2}}{z}_0^{dn}{w}_0^{1-n}\hfill \end{array}$$

for $n\in \mathbb{N}.$

(e) 

If $bd=0$ and $a\ne 1$, then the general solution to system (2) is given by

$$\begin{array}{cc}\hfill z_n=& {\alpha }^{\frac{1-a^n}{1-a}}{\beta }^{b\frac{1-a^{n-1}}{1-a}}{z}_0^{a^n}{w}_0^{b{a}^{n-1}}\hfill \\ \hfill w_n=& {\alpha }^{d\frac{1-a^{n-1}}{1-a}}\beta z_0^{d{a}^{n-1}},\hfill \end{array}$$

for $n\in \mathbb{N}$.

(f) 

If $bd=0$ and $a=1$, then the general solution to system (2) is given by

$$\begin{array}{cc}\hfill z_n=& {\alpha }^n{\beta }^{b(n-1)}{z}_0{w}_0^b\hfill \\ \hfill w_n=& {\alpha }^{d(n-1)}\beta z_0^d,\hfill \end{array}$$

for $n\in \mathbb{N}$.

Theorem 4.

Assume that $a,b,c,d\in \mathbb{Z}$, $ac\ne 0,$ $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then system (2) is solvable in closed form.

Proof. 

To deal with the case we use and modify our method previously used, for example, in [40,43]. Since $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$, we have $z_n{w}_n\ne 0$ for $n\in {\mathbb{N}}_0.$ Hence,

$$\begin{array}{c}\hfill w_n^b=\frac{z_{n+1}}{\alpha z_n^a},n\in {\mathbb{N}}_0,\end{array}$$

(63)

$$\begin{array}{c}\hfill w_{n+1}^b={\beta }^b{w}_{n-2}^{bc}{z}_n^{bd},n\in {\mathbb{N}}_0,\end{array}$$

(64)

from which we get

$$\begin{array}{c}\hfill z_{n+2}={\alpha }^{1-c}{\beta }^b{z}_{n+1}^a{z}_n^{bd}{z}_{n-1}^c{z}_{n-2}^{-ac},\end{array}$$

(65)

for $n\ge 2$.

Let $\delta ={\alpha }^{1-c}{\beta }^b$,

$$\begin{array}{c}\hfill a_1=a,b_1=bd,c_1=c,d_1=-ac,y_1=1,\end{array}$$

(66)

then

$$\begin{array}{c}\hfill z_{n+2}={\delta }^{y_1}{z}_{n+1}^{a_1}{z}_n^{b_1}{z}_{n-1}^{c_1}{z}_{n-2}^{d_1},n\ge 2.\end{array}$$

(67)

From (67) is obtained

$$\begin{array}{cc}\hfill z_{n+2}& ={\delta }^{y_1}{(\delta z_n^{a_1}{z}_{n-1}^{b_1}{z}_{n-2}^{c_1}{z}_{n-3}^{d_1})}^{a_1}{z}_n^{b_1}{z}_{n-1}^{c_1}{z}_{n-2}^{d_1},\hfill \\ & ={\delta }^{y_1+a_1}{z}_n^{a_1{a}_1+b_1}{z}_{n-1}^{b_1{a}_1+c_1}{z}_{n-2}^{c_1{a}_1+d_1}{z}_{n-3}^{d_1{a}_1}\hfill \\ & ={\delta }^{y_2}{z}_n^{a_2}{z}_{n-1}^{b_2}{z}_{n-2}^{c_2}{z}_{n-3}^{d_2},\hfill \end{array}$$

(68)

for $n\ge 3,$ where

$$\begin{array}{c}\hfill a_2:=a_1{a}_1+b_1,b_2:=b_1{a}_1+c_1,c_2:=c_1{a}_1+d_1,d_2:=d_1{a}_1,y_2:=y_1+a_1.\end{array}$$

(69)

Suppose

$$\begin{array}{c}\hfill z_{n+2}={\delta }^{y_k}{z}_{n+2-k}^{a_k}{z}_{n+1-k}^{b_k}{z}_{n-k}^{c_k}{z}_{n-k-1}^{d_k},\end{array}$$

(70)

for a $k\ge 2$ and $n\ge k+1$, and

$$\begin{array}{cc}\hfill a_k& =a_1{a}_{k-1}+b_{k-1},b_k=b_1{a}_{k-1}+c_{k-1},\hfill \\ \hfill c_k& =c_1{a}_{k-1}+d_{k-1},d_k=d_1{a}_{k-1},\hfill \end{array}$$

(71)

$$\begin{array}{c}\hfill y_k=y_{k-1}+a_{k-1}.\end{array}$$

(72)

Using (67) in (70), we get

$$\begin{array}{cc}\hfill z_{n+2}& ={\delta }^{y_k}{(\delta z_{n+1-k}^{a_1}{z}_{n-k}^{b_1}{z}_{n-k-1}^{c_1}{z}_{n-k-2}^{d_1})}^{a_k}{z}_{n+1-k}^{b_k}{z}_{n-k}^{c_k}{z}_{n-k-1}^{d_k}\hfill \\ & ={\delta }^{y_k+a_k}{z}_{n+1-k}^{a_1{a}_k+b_k}{z}_{n-k}^{b_1{a}_k+c_k}{z}_{n-k-1}^{c_1{a}_k+d_k}{z}_{n-k-2}^{d_1{a}_k}\hfill \\ & ={\delta }^{y_{k+1}}{z}_{n+1-k}^{a_{k+1}}{z}_{n-k}^{b_{k+1}}{z}_{n-k-1}^{c_{k+1}}{z}_{n-k-2}^{d_{k+1}},\hfill \end{array}$$

(73)

for $n\ge k+2$, where

$$\begin{array}{cc}\hfill a_{k+1}& :=a_1{a}_k+b_k,b_{k+1}:=b_1{a}_k+c_k,\hfill \\ \hfill c_{k+1}& :=c_1{a}_k+d_k,d_{k+1}:=d_1{a}_k,\hfill \end{array}$$

(74)

$$\begin{array}{c}\hfill y_{k+1}:=y_k+a_k.\end{array}$$

(75)

Hence, by the induction we get that (70)–(72) hold.

From (70) with $k=n-1$ and (8), it follows that

$$\begin{array}{cc}\hfill z_{n+2}=& {\delta }^{y_{n-1}}{z}_3^{a_{n-1}}{z}_2^{b_{n-1}}{z}_1^{c_{n-1}}{z}_0^{d_{n-1}}\hfill \\ \hfill =& {({\alpha }^{1-c}{\beta }^b)}^{y_{n-1}}{({\alpha }^{1+a+a^2+bd}{\beta }^{b+ab}{z}_0^{a^3+2abd}{w}_{-2}^{abc}{w}_{-1}^{bc}{w}_0^{b(a^2+bd)})}^{a_{n-1}}\hfill \\ & \times {({\alpha }^{1+a}{\beta }^b{z}_0^{a^2+bd}{w}_{-2}^{bc}{w}_0^{ab})}^{b_{n-1}}{(\alpha z_0^a{w}_0^b)}^{c_{n-1}}{z}_0^{d_{n-1}}\hfill \\ \hfill =& {\alpha }^{(1-c)y_{n-1}+(1+a+a^2+bd)a_{n-1}+(1+a)b_{n-1}+c_{n-1}}{\beta }^{b{y}_{n-1}+b(1+a)a_{n-1}+b{b}_{n-1}}\hfill \\ & \times z_0^{(a^3+2abd)a_{n-1}+(a^2+bd)b_{n-1}+a{c}_{n-1}+d_{n-1}}{w}_{-2}^{abc{a}_{n-1}+bc{b}_{n-1}}{w}_{-1}^{bc{a}_{n-1}}\hfill \\ & \times w_0^{b(a^2+bd)a_{n-1}+ab{b}_{n-1}+b{c}_{n-1}}\hfill \\ \hfill =& {\alpha }^{y_{n+2}-c{y}_{n-1}}{\beta }^{b{y}_{n+1}}{z}_0^{a_{n+2}-c{a}_{n-1}}{w}_{-2}^{bc{a}_n}{w}_{-1}^{bc{a}_{n-1}}{w}_0^{b{a}_{n+1}},\hfill \end{array}$$

(76)

for $n\in {\mathbb{N}}_0.$

From (71) we have

$$\begin{array}{c}\hfill a_k=a_1{a}_{k-1}+b_1{a}_{k-2}+c_1{a}_{k-3}+d_1{a}_{k-4},k\ge 5,\end{array}$$

(77)

and that $b_k$, $c_k$ and $d_k$ also satisfy the equation, and using (74) and (75) for $k=0,-1,-2,-3$ is obtained

$$\begin{array}{cc}\hfill & a_{-3}=0,a_{-2}=0,a_{-1}=0,a_0=1,\hfill \end{array}$$

(78)

$$\begin{array}{c}\hfill y_{-3}=y_{-2}=y_{-1}=y_0=0,y_1=1,\end{array}$$

(79)

and

$$\begin{array}{c}\hfill y_k=\sum _{j=0}^{k-1}{a}_j,\end{array}$$

(80)

(see, for example, [33] for more details).

The solvability of (77) is a classical thing. Hence, by finding closed form formula for $a_k$, employing it in (80), then using Lemma 2, is calculated $y_k$. These two formulas and (76) give a closed-form formula for solution to (65).

Now note that

$$\begin{array}{c}\hfill z_n^d=\frac{w_{n+1}}{\beta w_{n-2}^c},n\in {\mathbb{N}}_0,\end{array}$$

(81)

$$\begin{array}{c}\hfill z_{n+1}^d={\alpha }^d{z}_n^{ad}{w}_n^{bd},n\in {\mathbb{N}}_0,\end{array}$$

(82)

from which we get

$$\begin{array}{c}\hfill w_{n+2}={\alpha }^d{\beta }^{1-a}{w}_{n+1}^a{w}_n^{bd}{w}_{n-1}^c{w}_{n-2}^{-ac},n\in {\mathbb{N}}_0.\end{array}$$

(83)

As above is get

$$\begin{array}{c}\hfill w_{n+2}={\eta }^{y_k}{w}_{n+2-k}^{a_k}{w}_{n+1-k}^{b_k}{w}_{n-k}^{c_k}{w}_{n-k-1}^{d_k},\end{array}$$

(84)

for every $k,n\in \mathbb{N},$ such that $n\ge k-1$, where $\eta ={\alpha }^d{\beta }^{1-a}$, ${(a_k)}_{k\in \mathbb{N}}$, ${(b_k)}_{k\in \mathbb{N}}$, ${(c_k)}_{k\in \mathbb{N}}$ and ${(d_k)}_{k\in \mathbb{N}}$ are defined by (66) and (71), while ${(y_k)}_{k\in \mathbb{N}}$ is defined by (72) and (79).

From (84) with $k=n+1$ and (8), we have

$$\begin{array}{cc}\hfill w_{n+2}& ={\eta }^{y_{n+1}}{w}_1^{a_{n+1}}{w}_0^{b_{n+1}}{w}_{-1}^{c_{n+1}}{w}_{-2}^{d_{n+1}}\hfill \\ & ={({\alpha }^d{\beta }^{1-a})}^{y_{n+1}}{(\beta w_{-2}^c{z}_0^d)}^{a_{n+1}}{w}_0^{b_{n+1}}{w}_{-1}^{c_{n+1}}{w}_{-2}^{d_{n+1}}\hfill \\ & ={\alpha }^{d{y}_{n+1}}{\beta }^{(1-a)y_{n+1}+a_{n+1}}{z}_0^{d{a}_{n+1}}{w}_{-2}^{c{a}_{n+1}+d_{n+1}}{w}_{-1}^{c_{n+1}}{w}_0^{b_{n+1}}\hfill \\ & ={\alpha }^{d{y}_{n+1}}{\beta }^{y_{n+2}-a{y}_{n+1}}{z}_0^{d{a}_{n+1}}{w}_{-2}^{c(a_{n+1}-a{a}_n)}{w}_{-1}^{c(a_n-a{a}_{n-1})}{w}_0^{a_{n+2}-a{a}_{n+1}},\hfill \end{array}$$

(85)

for $n\in {\mathbb{N}}_0.$

Recall that closed form formulas for $a_k$ and $y_k$ can be found. Applying them into (85) we show the solvability of equation (83). It can be checked that (76) and (85) present a solution to (2), from which the theorem follows. ☐

Corollary 3.

Consider system (2) with $a,b,c,d\in \mathbb{Z}$, $ac\ne 0,$ $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then the general solution to (2) is given by (76) and (85), where ${(a_k)}_{k\in \mathbb{N}}$ is defined by (77) and (78), while ${(y_k)}_{k\in \mathbb{N}}$ is defined by (79) and (80)

Detailed Form of Solutions Given in (76) and (85)

Equation (77) is not only theoretically but also practically solvable. The reason for this is that its characteristic polynomial

$$\begin{array}{c}\hfill p_4(\lambda )={\lambda }^4-a{\lambda }^3-bd{\lambda }^2-c\lambda +ac,\end{array}$$

(86)

for the case $ac\ne 0$, is of the forth degree, thus, solvable by radicals.

Note that the equation $p_4(\lambda )=0$ can be written as follows [44].

$$\begin{array}{c}\hfill {\left({\lambda }^2-\frac{a}{2}\lambda +\frac{s}{2}\right)}^2-\left(\left(\frac{a^2}{4}+s+bd\right){\lambda }^2-\left(\frac{as}{2}-c\right)\lambda +\frac{s^2}{4}-ac\right)=0,\end{array}$$

(87)

Now, choose s so that the expression in the second bracket in (87) is a perfect square. Thus, it must be

$${(as-2c)}^2=(a^2+4s+4bd)(s^2-4ac),$$

that is,

$$\begin{array}{c}\hfill s^3+bd{s}^2-3acs-a^{3}c-c^2-4abcd=0.\end{array}$$

(88)

Hence, (87) can be written as

$$\begin{array}{c}\hfill {\left({\lambda }^2-\frac{a}{2}\lambda +\frac{s}{2}\right)}^2-{\left(\frac{\sqrt{a^2+4s+4bd}}{2}\lambda -\frac{as-2c}{2\sqrt{a^2+4s+4bd}}\right)}^2=0,\end{array}$$

(89)

which is equivalent to

$$\begin{array}{c}{\lambda }^2-\left(\frac{a}{2}+\frac{\sqrt{a^2+4s+4bd}}{2}\right)\lambda +\frac{s}{2}+\frac{as-2c}{2\sqrt{a^2+4s+4bd}}=0,\hfill \end{array}$$

(90)

$$\begin{array}{c}{\lambda }^2-\left(\frac{a}{2}-\frac{\sqrt{a^2+4s+4bd}}{2}\right)\lambda +\frac{s}{2}-\frac{as-2c}{2\sqrt{a^2+4s+4bd}}=0.\hfill \end{array}$$

(91)

Using the change of variables $s=t-\frac{bd}{3}$ in (88), it follows that

$$\begin{array}{c}\hfill t^3-\left(\frac{b^2{d}^2}{3}+3ac\right)t+\frac{2b^3{d}^3}{27}-a^{3}c-c^2-3abcd=0.\end{array}$$

(92)

Let

$$p=-\left(\frac{b^2{d}^2}{3}+3ac\right)\mathrm{and}q=-\frac{27a^{3}c+27c^2+81abcd-2b^3{d}^3}{27}.$$

As usual, a solution to (92) is found in the form $t=u+v$. Putting it into (92) and requesting $uv=-p/3$, is get $u^3+v^3=-q$ and $u^3{v}^3=-p^3/27.$ Hence, $u^3$ and $v^3$ are solutions to the equation $z^2+qz-p^3/27=0$, so, they must be $(-q\pm \sqrt{q^2+4p^3/27})/2.$

Hence

$$\begin{array}{c}\hfill t=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}},\end{array}$$

(93)

where any of the three possible values of the right-hand side can be chosen. If $p=-{\Delta }_0/3$ and $q=-{\Delta }_1/27$, then it can be written as

$$\begin{array}{c}\hfill t=\frac{1}{3\sqrt[3]{2}}\left(\sqrt[3]{{\Delta }_1-\sqrt{{\Delta }_1^2-4{\Delta }_0^3}}+\sqrt[3]{{\Delta }_1+\sqrt{{\Delta }_1^2-4{\Delta }_0^3}}\right).\end{array}$$

(94)

For such chosen s, that is, t, (90) and (91) can be solved and by some calculation it is obtained that the zeros of $p_4$ are

$$\begin{array}{cc}\hfill {\lambda }_1=& \frac{a}{4}+\frac{1}{2}\sqrt{\frac{a^2}{4}+\frac{2bd}{3}+t}+\frac{1}{2}\sqrt{\frac{a^2}{2}+\frac{4bd}{3}-t-\frac{Q}{4\sqrt{\frac{a^2}{4}+\frac{2bd}{3}+t}}},\hfill \end{array}$$

(95)

$$\begin{array}{cc}\hfill {\lambda }_2=& \frac{a}{4}+\frac{1}{2}\sqrt{\frac{a^2}{4}+\frac{2bd}{3}+t}-\frac{1}{2}\sqrt{\frac{a^2}{2}+\frac{4bd}{3}-t-\frac{Q}{4\sqrt{\frac{a^2}{4}+\frac{2bd}{3}+t}}},\hfill \end{array}$$

(96)

$$\begin{array}{cc}\hfill {\lambda }_3=& \frac{a}{4}-\frac{1}{2}\sqrt{\frac{a^2}{4}+\frac{2bd}{3}+t}+\frac{1}{2}\sqrt{\frac{a^2}{2}+\frac{4bd}{3}-t+\frac{Q}{4\sqrt{\frac{a^2}{4}+\frac{2bd}{3}+t}}},\hfill \end{array}$$

(97)

$$\begin{array}{cc}\hfill {\lambda }_4=& \frac{a}{4}-\frac{1}{2}\sqrt{\frac{a^2}{4}+\frac{2bd}{3}+t}-\frac{1}{2}\sqrt{\frac{a^2}{2}+\frac{4bd}{3}-t+\frac{Q}{4\sqrt{\frac{a^2}{4}+\frac{2bd}{3}+t}}},\hfill \end{array}$$

(98)

where

$$\begin{array}{cc}\hfill Q:=& -a^3-8c-4abd.\hfill \end{array}$$

(99)

Recall, that the nature of these ${\lambda }_j$’s depends on the sign of the discriminant

$$\begin{array}{c}\hfill \Delta :=\frac{1}{27}(4{\Delta }_0^3-{\Delta }_1^2),\end{array}$$

(100)

where

$$\begin{array}{cc}\hfill {\Delta }_0:=& b^2{d}^2+9ac,\hfill \end{array}$$

(101)

$$\begin{array}{cc}\hfill {\Delta }_1:=& -2b^3{d}^3+81acbd+27a^{3}c+27c^2,\hfill \end{array}$$

(102)

and the signs of

$$\begin{array}{c}\hfill P:=-8bd-3a^2\end{array}$$

(103)

and

$$\begin{array}{c}\hfill D:=48ac-16b^2{d}^2-16a^{2}bd-3a^4.\end{array}$$

(104)

Zeros of $p_4$ are different and none of them is 1. If a, b, c and d are chosen such that

$${\Delta }_0=b^2{d}^2+9ac<0,$$

then it will be $\Delta <0$, from which by Lemma 3 we have in this case that $p_4$ has four different zeros. Moreover, since $\Delta <0$ two zeros are real and two are complex-conjugate.

Zeros of $p_4$ are different and one of them is 1. Polynomial $p_4$ has a zero equal to 1 if $p_4(1)=1-a-bd-c+ac=0,$ that is, if

$$\begin{array}{c}\hfill (a-1)(c-1)=bd,\end{array}$$

(105)

so that

$$\begin{array}{c}\hfill p_4(\lambda )={\lambda }^4-a{\lambda }^3-(a-1)(c-1){\lambda }^2-c\lambda +ac.\end{array}$$

(106)

Thus, if we choose a and c such that

$$p_4^{\prime }(1)=4-3a-2(a-1)(c-1)-c\ne 0,$$

that is, $(2a-1)(2c+1)\ne 3$, then $p_4$ will be such a polynomial if $\Delta \ne 0$. For example, if $a=3$ and $c=2$, then $bd=2\ne 0$, $\Delta \ne 0$, which means that polynomial $p_4$ has all zeros mutually different and exactly one of them is equal to 1

$$\begin{array}{c}\hfill p_4(\lambda )={\lambda }^4-3{\lambda }^3-2{\lambda }^2-2\lambda +6=(\lambda -1)({\lambda }^3-2{\lambda }^2-4\lambda -6).\end{array}$$

(107)

Since in these two cases ${\lambda }_j\ne {\lambda }_i$, $i\ne j$, then the general solution to (77) is

$$\begin{array}{c}\hfill a_n={\gamma }_1{\lambda }_1^n+{\gamma }_2{\lambda }_2^n+{\gamma }_3{\lambda }_3^n+{\gamma }_4{\lambda }_4^n,n\in \mathbb{N},\end{array}$$

(108)

where ${\gamma }_i$, $i=\overline{1,4}$, are arbitrary constants.

Lemma 1 implies

$$\begin{array}{c}\hfill \sum _{j=1}^4\frac{{\lambda }_j^l}{p_4^{\prime }({\lambda }_j)}=0\mathrm{for}l=\overline{0,2},\mathrm{and}\sum _{j=1}^4\frac{{\lambda }_j^3}{p_4^{\prime }({\lambda }_j)}=1.\end{array}$$

(109)

From initial conditions (78) and (109), it is obtained

$$\begin{array}{cc}\hfill a_n=& \sum _{j=1}^4\frac{{\lambda }_j^{n+3}}{p_4^{\prime }({\lambda }_j)}=\frac{{\lambda }_1^{n+3}}{({\lambda }_1-{\lambda }_2)({\lambda }_1-{\lambda }_3)({\lambda }_1-{\lambda }_4)}+\frac{{\lambda }_2^{n+3}}{({\lambda }_2-{\lambda }_1)({\lambda }_2-{\lambda }_3)({\lambda }_2-{\lambda }_4)}\hfill \\ & +\frac{{\lambda }_3^{n+3}}{({\lambda }_3-{\lambda }_1)({\lambda }_3-{\lambda }_2)({\lambda }_3-{\lambda }_4)}+\frac{{\lambda }_4^{n+3}}{({\lambda }_4-{\lambda }_1)({\lambda }_4-{\lambda }_2)({\lambda }_4-{\lambda }_3)},\hfill \end{array}$$

(110)

for $n\ge -3$ ([40]).

Combining (80) and (110), we get

$$\begin{array}{c}\hfill y_n=\sum _{j=0}^{n-1}\sum _{i=1}^4\frac{{\lambda }_i^{j+3}}{p_4^{\prime }({\lambda }_i)}=\sum _{i=1}^4\frac{{\lambda }_i^3({\lambda }_i^n-1)}{p_4^{\prime }({\lambda }_i)({\lambda }_i-1)},n\in \mathbb{N},\end{array}$$

(111)

when ${\lambda }_i\ne 1$, $i=\overline{1,4}$, and

$$\begin{array}{c}\hfill y_n=\frac{n}{2-a+c-2ac}+\sum _{i=2}^4\frac{{\lambda }_i^3({\lambda }_i^n-1)}{p_4^{\prime }({\lambda }_i)({\lambda }_i-1)},n\in \mathbb{N},\end{array}$$

(112)

when one of the zeros is 1, say ${\lambda }_1$, is equal to 1.

Note that if one of the zeros is equal to 1, then we have

$$\begin{array}{cc}\hfill p_4(\lambda )=& {\lambda }^4-a{\lambda }^3-(a-1)(c-1){\lambda }^2-c\lambda +ac\hfill \\ \hfill =& (\lambda -1)({\lambda }^3+(1-a){\lambda }^2+(1-a)c\lambda -ac).\hfill \end{array}$$

(113)

By using the change of variables $\lambda =t+\frac{a-1}{3}$ in (113) we get that the equation

$$\begin{array}{c}\hfill {\lambda }^3+(1-a){\lambda }^2+(1-a)c\lambda -ac=0,\end{array}$$

(114)

is transformed to the following one

$$t^3+\tilde{p}t+\tilde{q}=0,$$

with

$$\begin{array}{c}\hfill \tilde{p}=-\frac{(a-1)(a+3c-1)}{3}\mathrm{and}\tilde{q}=-\left(\frac{2{(a-1)}^3}{27}+\frac{c{(a-1)}^2}{3}+ac\right),\end{array}$$

(115)

whose solutions are given by

$$t_j={\epsilon }^{j-1}\sqrt[3]{-\frac{\tilde{q}}{2}-\sqrt{\frac{{\tilde{q}}^2}{4}+\frac{{\tilde{p}}^3}{27}}}+{\epsilon }^{j-1}\sqrt[3]{-\frac{\tilde{q}}{2}+\sqrt{\frac{{\tilde{q}}^2}{4}+\frac{{\tilde{p}}^3}{27}}},j=\overline{1,3},$$

where ${\epsilon }^3=1$ and $\epsilon \ne 1$.

Hence,

$$\begin{array}{c}\hfill {\lambda }_j=\frac{a-1}{3}+{\epsilon }^{j-2}\sqrt[3]{-\frac{\tilde{q}}{2}-\sqrt{\frac{q^2}{4}+\frac{{\tilde{p}}^3}{27}}}+{\epsilon }^{j-2}\sqrt[3]{-\frac{\tilde{q}}{2}+\sqrt{\frac{{\tilde{q}}^2}{4}+\frac{{\tilde{p}}^3}{27}}},\end{array}$$

(116)

for $j=\overline{2,4},$ are the other three zeros of the equation $p_4(\lambda )=0$, in this case.

A simple calculation along with (109) shows that (111) holds also for $n=-j$, $j=\overline{0,3}.$

The previous analysis along with Corollary 3 implies the following corollary.

Corollary 4.

Assume that $a,b,c,d\in \mathbb{Z}$, $ac\ne 0$, $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$ and $\Delta \ne 0$. Then the following statements are true.

(a) 

If $(a-1)(c-1)\ne bd$, then the general solution to (2) is given by (76) and (85), where ${(a_n)}_{n\ge -3}$ is given by (110), ${(y_n)}_{n\ge -3}$ is given by (111), while ${\lambda }_j$’s, $j=\overline{1,4}$, are given by (95)–(98).

(b) 

If $(a-1)(c-1)=bd$ and $(2a-1)(2c+1)\ne 3$, then the general solution to (2) is given by (76) and (85), where ${(a_n)}_{n\ge -3}$ is given by (110) with ${\lambda }_1=1$, ${(y_n)}_{n\ge -3}$ is given by (112), ${\lambda }_1=1$, while ${\lambda }_j$’s, $j=\overline{2,4}$, are given by (116) and (115).

$p_4$ has only one double zero which is equal to 1. Polynomial (86) has a double zero equal to 1 if (105) holds and

$$\begin{array}{c}\hfill (2a-1)(2c+1)=3.\end{array}$$

(117)

From (117) we have that it must be $a=2$ and $c=0$, or $a=1$ and $c=1$, or $a=0$ and $c=-2$, or $a=-1$ and $c=-1$. If $a=0$ or $c=0$, then $ac=0$, which contradicts to the assumption $ac\ne 0$.

If $a=c=1$, then $bd=0$, from which it follows that

$$p_4(\lambda )={\lambda }^4-{\lambda }^3-\lambda +1={(\lambda -1)}^2({\lambda }^2+\lambda +1),$$

and consequently

$$\begin{array}{c}\hfill {\lambda }_{3,4}=\frac{-1\pm i\sqrt{3}}{2},\end{array}$$

(118)

are the non-unit zeros of polynomial $p_4$ in the case.

If $a=c=-1$, then $bd=4$, from which it follows that

$$p_4(\lambda )={\lambda }^4+{\lambda }^3-4{\lambda }^2+\lambda +1={(\lambda -1)}^2({\lambda }^2+3\lambda +1),$$

and consequently

$$\begin{array}{c}\hfill {\lambda }_{3,4}=\frac{-3\pm \sqrt{5}}{2},\end{array}$$

(119)

are the non-unit zeros of polynomial $p_4$ in the case.

From this, we have proved in passing, that there are no $a,c\in \mathbb{Z}\backslash \left\{0\right\}$, such that 1 is a triple zero of $p_4$ or that it has two pairs of double zeros one of which is 1.

In these two cases we have (see, for example, [38])

$$\begin{array}{cc}\hfill a_n=& \frac{n(1-{\lambda }_3)(1-{\lambda }_4)+3{\lambda }_3{\lambda }_4-2{\lambda }_3-2{\lambda }_4+1}{{(1-{\lambda }_3)}^2{(1-{\lambda }_4)}^2}+\frac{{\lambda }_3^{n+3}}{{({\lambda }_3-1)}^2({\lambda }_3-{\lambda }_4)}+\frac{{\lambda }_4^{n+3}}{{({\lambda }_4-1)}^2({\lambda }_4-{\lambda }_3)},\hfill \end{array}$$

(120)

and

$$\begin{array}{cc}\hfill y_n=& \sum _{j=0}^{n-1}\left(\frac{j(1-{\lambda }_3)(1-{\lambda }_4)+3{\lambda }_3{\lambda }_4-2{\lambda }_3-2{\lambda }_4+1}{{(1-{\lambda }_3)}^2{(1-{\lambda }_4)}^2}+\frac{{\lambda }_3^{j+3}}{{({\lambda }_3-1)}^2({\lambda }_3-{\lambda }_4)}+\frac{{\lambda }_4^{j+3}}{{({\lambda }_4-1)}^2({\lambda }_4-{\lambda }_3)}\right)\hfill \\ \hfill =& \frac{(n-1)n}{2(1-{\lambda }_3)(1-{\lambda }_4)}+\frac{n(3{\lambda }_3{\lambda }_4-2{\lambda }_3-2{\lambda }_4+1)}{{(1-{\lambda }_3)}^2{(1-{\lambda }_4)}^2}+\frac{{\lambda }_3^3({\lambda }_3^n-1)}{{({\lambda }_3-1)}^3({\lambda }_3-{\lambda }_4)}+\frac{{\lambda }_4^3({\lambda }_4^n-1)}{{({\lambda }_4-1)}^3({\lambda }_4-{\lambda }_3)}.\hfill \end{array}$$

(121)

$p_4$ has a double zero different from 1. Let $bd=0$, then

$$p_4(\lambda )={\lambda }^4-a{\lambda }^3-c\lambda +ac=(\lambda -a)({\lambda }^3-c).$$

If we take $c=a^3$, then it is obtained

$$p_4(\lambda )={(\lambda -a)}^2({\lambda }^2+a\lambda +a^2),$$

which for $a\in \mathbb{Z}\backslash \{0,1\}$ is a polynomial with a double zero different from 1 and two non-real complex-conjugate zeros.

Since, in the case ${\lambda }_1={\lambda }_2$, ${\lambda }_i\ne {\lambda }_j$, $2\le i,j\le 4,$ we have

$$\begin{array}{c}\hfill a_n=({\gamma }_1+{\gamma }_{2}n){\lambda }_2^n+{\gamma }_3{\lambda }_3^n+{\gamma }_4{\lambda }_4^n,n\in \mathbb{N},\end{array}$$

(122)

where ${\gamma }_i$, $i=\overline{1,4}$, are arbitrary constants, and the solution satisfying (78) can be obtained, for example, by letting ${\lambda }_1\to {\lambda }_2$ in (110) [38]

$$\begin{array}{cc}\hfill a_n=& \frac{{\lambda }_2^{n+2}((n+3)({\lambda }_2-{\lambda }_3)({\lambda }_2-{\lambda }_4)-{\lambda }_2(2{\lambda }_2-{\lambda }_3-{\lambda }_4))}{{({\lambda }_2-{\lambda }_3)}^2{({\lambda }_2-{\lambda }_4)}^2}\hfill \\ & +\frac{{\lambda }_3^{n+3}}{{({\lambda }_3-{\lambda }_2)}^2({\lambda }_3-{\lambda }_4)}+\frac{{\lambda }_4^{n+3}}{{({\lambda }_4-{\lambda }_2)}^2({\lambda }_4-{\lambda }_3)}.\hfill \end{array}$$

(123)

From (80), (123) and by Lemma 2, we get

$$\begin{array}{cc}\hfill y_n=& \sum _{j=0}^{n-1}(\frac{{\lambda }_2^{j+2}((j+3)({\lambda }_2-{\lambda }_3)({\lambda }_2-{\lambda }_4)-{\lambda }_2(2{\lambda }_2-{\lambda }_3-{\lambda }_4))}{{({\lambda }_2-{\lambda }_3)}^2{({\lambda }_2-{\lambda }_4)}^2}\hfill \\ & +\frac{{\lambda }_3^{j+3}}{{({\lambda }_3-{\lambda }_2)}^2({\lambda }_3-{\lambda }_4)}+\frac{{\lambda }_4^{j+3}}{{({\lambda }_4-{\lambda }_2)}^2({\lambda }_4-{\lambda }_3)})\hfill \\ \hfill =& \frac{{\lambda }_2^3-n{\lambda }_2^{n+2}+(n-1){\lambda }_2^{n+3}}{({\lambda }_2-{\lambda }_3)({\lambda }_2-{\lambda }_4){(1-{\lambda }_2)}^2}+\frac{({\lambda }_2^4-2{\lambda }_2^3{\lambda }_3-2{\lambda }_2^3{\lambda }_4+3{\lambda }_2^2{\lambda }_3{\lambda }_4)({\lambda }_2^n-1)}{{({\lambda }_2-{\lambda }_3)}^2{({\lambda }_2-{\lambda }_4)}^2({\lambda }_2-1)}\hfill \\ & +\frac{{\lambda }_3^3({\lambda }_3^n-1)}{{({\lambda }_3-{\lambda }_2)}^2({\lambda }_3-{\lambda }_4)({\lambda }_3-1)}+\frac{{\lambda }_4^3({\lambda }_4^n-1)}{{({\lambda }_4-{\lambda }_2)}^2({\lambda }_4-{\lambda }_3)({\lambda }_4-1)}.\hfill \end{array}$$

(124)

From the previous analysis and Corollary 3 we obtain the following result.

Corollary 5.

Assume that $a,b,c,d\in \mathbb{Z}$, $ac\ne 0$ and $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then the following statements are true.

(a) 

If only one of the zeros of $p_4$ is double and different from 1, then the general solution to (2) is given by (76) and (85), where ${(a_n)}_{n\ge -3}$ is given by (123), while ${(y_n)}_{n\ge -3}$ is given by (124).

(b) 

If 1 is a unique double zero of polynomial $p_4$, say ${\lambda }_1={\lambda }_2=1$, then the general solution to (2) is given by (76) and (85), where ${(a_n)}_{n\ge -3}$ is given by (120), ${(y_n)}_{n\ge -3}$ is given by (121), while ${\lambda }_{3,4}$ are given by (118) if $a=c=1$ or by (119) if $a=c=-1$.

Two pairs of different double zeros. In this case it must be $D=0$ which implies that

$$\begin{array}{c}\hfill 16t^2+16a^{2}t+3a^4-48ac=0,\end{array}$$

(125)

where $t=bd.$ On the other hand, it must be $\Delta =0$, which is equivalent to $4{\Delta }_0^3={\Delta }_1^2,$ that is,

$${(-2b^3{d}^3+81acbd+27a^{3}c+27c^2)}^2=4{(b^2{d}^2+9ac)}^3,$$

from which it follows that

$$\begin{array}{c}\hfill c\left(16a{t}^4+4(a^3+c)t^3-207a^{2}c{t}^2-162ac(a^3+c)t-27c{(a^3-c)}^2)\right)=0.\end{array}$$

(126)

The problem of the existence of a joint zero of the polynomials in (125) and (126) for some integers $a,$b, c and d, such that $ac\ne 0$, seems quite technical, so we leave it to the reader.

Solutions to (77) in this case are

$$\begin{array}{c}\hfill a_n=({\gamma }_1+{\gamma }_{2}n){\lambda }_1^n+({\gamma }_3+{\gamma }_{4}n){\lambda }_3^n,n\in \mathbb{N},\end{array}$$

(127)

where ${\gamma }_i$, $i=\overline{1,4}$ are constants. The solution to (127) such that (78) holds is [38]

$$\begin{array}{cc}\hfill a_n=& \frac{{\lambda }_2^{n+2}(n{({\lambda }_2-{\lambda }_4)}^2+{\lambda }_2^2-4{\lambda }_2{\lambda }_4+3{\lambda }_4^2)}{{({\lambda }_2-{\lambda }_4)}^4}+\frac{{\lambda }_4^{n+2}(n{({\lambda }_4-{\lambda }_2)}^2+{\lambda }_4^2-4{\lambda }_2{\lambda }_4+3{\lambda }_2^2)}{{({\lambda }_4-{\lambda }_2)}^4}.\hfill \end{array}$$

(128)

From (80), (128) and Lemma 2, we get

$$\begin{array}{cc}\hfill y_n=& \sum _{j=0}^{n-1}\left(\frac{{\lambda }_2^{j+2}(j{({\lambda }_2-{\lambda }_4)}^2+{\lambda }_2^2-4{\lambda }_2{\lambda }_4+3{\lambda }_4^2)}{{({\lambda }_2-{\lambda }_4)}^4}+\frac{{\lambda }_4^{j+2}(j{({\lambda }_4-{\lambda }_2)}^2+{\lambda }_4^2-4{\lambda }_2{\lambda }_4+3{\lambda }_2^2)}{{({\lambda }_4-{\lambda }_2)}^4}\right)\hfill \\ \hfill =& \frac{{\lambda }_2^3-n{\lambda }_2^{n+2}+(n-1){\lambda }_2^{n+3}}{{({\lambda }_2-{\lambda }_4)}^2{(1-{\lambda }_2)}^2}+\frac{({\lambda }_2^4-4{\lambda }_2^3{\lambda }_4+3{\lambda }_2^2{\lambda }_4^2)({\lambda }_2^n-1)}{{({\lambda }_2-{\lambda }_4)}^4({\lambda }_2-1)}\hfill \\ & +\frac{{\lambda }_4^3-n{\lambda }_4^{n+2}+(n-1){\lambda }_4^{n+3}}{{({\lambda }_4-{\lambda }_2)}^2{(1-{\lambda }_4)}^2}+\frac{({\lambda }_4^4-4{\lambda }_2{\lambda }_4^3+3{\lambda }_2^2{\lambda }_4^2)({\lambda }_4^n-1)}{{({\lambda }_4-{\lambda }_2)}^4({\lambda }_4-1)}.\hfill \end{array}$$

(129)

Corollary 6.

Assume that $a,b,c,d\in \mathbb{Z}$, $ac\ne 0$ and $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then the following statements are true.

(a) 

If polynomial $p_4$ has two pairs of double zeros both different from 1, then the general solution to (2) is given by (76) and (85), where ${(a_n)}_{n\ge -3}$ is given by (128), while ${(y_n)}_{n\ge -3}$ is given by (129).

(b) 

The polynomial in (86) cannot have two pairs of double zeros such that one of them is equal to 1.

Triple zero case. In this case it must be $\Delta ={\Delta }_0=0$, or equivalently, ${\Delta }_0={\Delta }_1=0$. Hence,

$$ac=-b^2{d}^2/9\mathrm{and}2b^3{d}^3-81acbd-27a^{3}c-27c^2=0,$$

from which it follows that

$$\begin{array}{c}\hfill 11{(bd)}^3+3a^2{(bd)}^2-\frac{{(bd)}^4}{3a^2}=0.\end{array}$$

(130)

Since $bd=0$ implies $ac=0$, which contradicts to the assumption $ac\ne 0$, from (130) it follows that

$$\begin{array}{c}\hfill s^2-33s-9=0,\end{array}$$

(131)

where $s=bd/a^2$. Hence, it must be

$$bd/a^2=\frac{33+\sqrt{1125}}{2}\mathrm{or}bd/a^2=\frac{33-\sqrt{1125}}{2},$$

which is not possible since $bd/a^2$ is a rational number, whereas $(33\pm \sqrt{1125})/2$ are both irrational numbers. Hence, $p_4$ cannot have a triple, and consequently cannot have a quadruple zero.

Corollary 7.

Assume that $a,b,c,d\in \mathbb{Z}$, $ac\ne 0$ and $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then polynomial (86) cannot have a triple zero.

Theorem 5.

Assume that $b,c,d\in \mathbb{Z}$, $a=0,$ $bcd\ne 0$, $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$. Then system (2) is solvable in closed form.

Proof. 

In this case (2) is

$$\begin{array}{c}\hfill z_{n+1}=\alpha w_n^b,w_{n+1}=\beta w_{n-2}^c{z}_n^d,n\in {\mathbb{N}}_0.\end{array}$$

(132)

Using the first equation in (132) into the second one it follows that

$$\begin{array}{c}\hfill w_{n+1}={\alpha }^d\beta w_{n-1}^{bd}{w}_{n-2}^c,n\in \mathbb{N}.\end{array}$$

(133)

Let $\eta ={\alpha }^d\beta$,

$$\begin{array}{c}\hfill {\beta }_1=bd,{\gamma }_1=c,{\delta }_1=0,x_1=1,\end{array}$$

(134)

then, we have

$$\begin{array}{c}\hfill w_{n+1}=\eta w_{n-1}^{{\beta }_1}{w}_{n-2}^{{\gamma }_1}{w}_{n-3}^{{\delta }_1},n\in \mathbb{N}.\end{array}$$

(135)

Hence

$$\begin{array}{cc}\hfill w_{n+1}=& {\eta }^{x_1}{(\eta w_{n-3}^{{\beta }_1}{w}_{n-4}^{{\gamma }_1})}^{{\beta }_1}{w}_{n-2}^{{\gamma }_1}{w}_{n-3}^{{\delta }_1}\hfill \\ \hfill =& {\eta }^{x_1+{\beta }_1}{w}_{n-2}^{{\gamma }_1}{w}_{n-3}^{{\beta }_1{\beta }_1+{\delta }_1}{w}_{n-4}^{{\gamma }_1{\beta }_1}\hfill \\ \hfill =& {\eta }^{x_2}{w}_{n-2}^{{\beta }_2}{w}_{n-3}^{{\gamma }_2}{w}_{n-4}^{{\delta }_2},\hfill \end{array}$$

for $n\ge 3$, where

$$x_2:=x_1+{\beta }_1,{\beta }_2:={\gamma }_1,{\gamma }_2:={\beta }_1{\beta }_1+{\delta }_1,{\delta }_2:={\gamma }_1{\beta }_1.$$

Suppose that

$$\begin{array}{cc}\hfill w_{n+1}=& {\eta }^{x_k}{w}_{n-k}^{{\beta }_k}{w}_{n-k-1}^{{\gamma }_k}{w}_{n-k-2}^{{\delta }_k},\hfill \end{array}$$

(136)

for a $k\in \mathbb{N}\backslash \left\{1\right\}$ and every $n\ge k+1$, and

$$\begin{array}{c}\hfill {\beta }_k={\gamma }_{k-1},{\gamma }_k={\beta }_1{\beta }_{k-1}+{\delta }_{k-1},{\delta }_k={\gamma }_1{\beta }_{k-1},\end{array}$$

(137)

$$\begin{array}{c}\hfill x_k=x_{k-1}+{\beta }_{k-1}.\end{array}$$

(138)

Using (135) in (136), we have

$$\begin{array}{cc}\hfill w_{n+1}=& {\eta }^{x_k}{(\eta w_{n-k-2}^{{\beta }_1}{w}_{n-k-3}^{{\gamma }_1})}^{{\beta }_k}{w}_{n-k-1}^{{\gamma }_k}{w}_{n-k-2}^{{\delta }_k}\hfill \\ \hfill =& {\eta }^{x_k+{\beta }_k}{w}_{n-k-1}^{{\gamma }_k}{w}_{n-k-2}^{{\beta }_1{\beta }_k+{\delta }_k}{w}_{n-k-3}^{{\gamma }_1{\beta }_k}\hfill \\ \hfill =& {\eta }^{x_{k+1}}{w}_{n-k-1}^{{\beta }_{k+1}}{w}_{n-k-2}^{{\gamma }_{k+1}}{w}_{n-k-3}^{{\delta }_{k+1}},\hfill \end{array}$$

for a $k\ge 2$ and $n\ge k+2$, and where

$${\beta }_{k+1}:={\gamma }_k,{\gamma }_{k+1}:={\beta }_1{\beta }_k+{\delta }_k,{\delta }_{k+1}:={\gamma }_1{\beta }_k,$$

$$x_{k+1}:=x_k+{\beta }_k.$$

From this we see that hypotheses (136)–(138) are true.

Let $k=n-1$, then from (136)–(138), it follows that

$$\begin{array}{cc}\hfill w_{n+1}=& {\eta }^{x_{n-1}}{w}_1^{{\beta }_{n-1}}{w}_0^{{\gamma }_{n-1}}{w}_{-1}^{{\delta }_{n-1}}\hfill \\ \hfill =& {({\alpha }^d\beta )}^{x_{n-1}}{(\beta w_{-2}^c{z}_0^d)}^{{\beta }_{n-1}}{w}_0^{{\gamma }_{n-1}}{w}_{-1}^{{\delta }_{n-1}}\hfill \\ \hfill =& {\alpha }^{d{x}_{n-1}}{\beta }^{x_{n-1}+{\beta }_{n-1}}{z}_0^{d{\beta }_{n-1}}{w}_{-2}^{c{\beta }_{n-1}}{w}_{-1}^{{\delta }_{n-1}}{w}_0^{{\gamma }_{n-1}}\hfill \\ \hfill =& {\alpha }^{d{x}_{n-1}}{\beta }^{x_n}{z}_0^{d{\beta }_{n-1}}{w}_{-2}^{c{\beta }_{n-1}}{w}_{-1}^{c{\beta }_{n-2}}{w}_0^{{\beta }_n},\hfill \end{array}$$

(139)

for $n\ge 3.$

From (137) it follows that

$$\begin{array}{c}\hfill {\beta }_k={\beta }_1{\beta }_{k-2}+{\gamma }_1{\beta }_{k-3},k\ge 4,\end{array}$$

(140)

and that ${\gamma }_k$ and ${\delta }_k$ also satisfy (140).

From (140) and since ${\gamma }_1=c\ne 0$, we have that

$$\begin{array}{c}\hfill {\beta }_{k-3}=\frac{{\beta }_k-{\beta }_1{\beta }_{k-2}}{{\gamma }_1},\end{array}$$

(141)

From (138) and (141) with $k=3,2,1$ and some calculation is obtained

$$\begin{array}{cc}& {\beta }_{-3}=0,{\beta }_{-2}=0,{\beta }_{-1}=1,{\beta }_0=0,\hfill \end{array}$$

(142)

$$\begin{array}{c}\hfill x_{-3}=x_{-2}=x_{-1}=0,x_0=x_1=1,\end{array}$$

(143)

and

$$\begin{array}{c}\hfill x_k=1+\sum _{j=1}^{k-1}{\beta }_j.\end{array}$$

(144)

Since equation (140) is solvable, we can calculate ${\beta }_k$, from which along with (144) and Lemma 2, $y_k$ is calculated. These facts along with (139) gives a closed form formula for (133).

Using (139) in the first equation in (132) we get

$$\begin{array}{c}\hfill z_{n+1}={\alpha }^{1+bd{x}_{n-2}}{\beta }^{b{x}_{n-1}}{z}_0^{bd{\beta }_{n-2}}{w}_{-2}^{bc{\beta }_{n-2}}{w}_{-1}^{bc{\beta }_{n-3}}{w}_0^{b{\beta }_{n-1}},n\in {\mathbb{N}}_0.\end{array}$$

(145)

It is not difficult to see that (139) and (145) are solutions to (2) in the case. ☐

Theorem 5 solves theoretically system (2) when $a=0$ and $bcd\ne 0$. Now we will practically solve it in terms of the parameters and initial values. The following polynomial

$$\begin{array}{c}\hfill p_3(\lambda )={\lambda }^3-bd\lambda -c,\end{array}$$

(146)

is the characteristic one associated to Equation (140), and its solutions are

$$\begin{array}{c}\hfill {\lambda }_j=\frac{1}{3\sqrt[3]{2}}\left({\epsilon }^j\sqrt[3]{{\widehat{\Delta }}_1-\sqrt{{\widehat{\Delta }}_1^2-4{\widehat{\Delta }}_0^3}}+{\overline{\epsilon }}^j\sqrt[3]{{\widehat{\Delta }}_1+\sqrt{{\widehat{\Delta }}_1^2-4{\widehat{\Delta }}_0^3}}\right),j=\overline{0,2},\end{array}$$

(147)

where

$$\begin{array}{c}\hfill {\widehat{\Delta }}_0=3bd=:-3p\mathrm{and}{\widehat{\Delta }}_1=27c=:-27q,\end{array}$$

(148)

and ${\epsilon }^3=1$, $\epsilon \ne 1$.

Zeros of $p_3$ are different and none of them is 1. In this case it must be ${\widehat{\Delta }}_1^2\ne 4{\widehat{\Delta }}_0^3$, which can written as

$$27c^2\ne 4{(bd)}^3.$$

Hence, in the case all the zeros of $p_3$ are different. If additionally $p_3(1)\ne 1$, that is, $c+bd\ne 1,$ then none of the zeros is 1. For example, if $c=bd=k\in \mathbb{N}$, then we have such a situation.

Zeros of $p_3$ are different and one of them is 1. Polynomial $p_3$ has a zero equal to 1 if $c+bd=1$. Then

$$p_3(\lambda )={\lambda }^3+(c-1)\lambda -c=(\lambda -1)({\lambda }^2+\lambda +c).$$

Hence

$$\begin{array}{c}\hfill {\lambda }_1=1,{\lambda }_{2,3}=\frac{1\pm \sqrt{1-4c}}{2},\end{array}$$

(149)

are the zeros of the equation $p_3(\lambda )=0$, in this case. Since $p_3^{\prime }(1)=2+c$, it follows that $p_3$ can have 1 as a double zero only if $c=-2$.

The general solution to (140) in these two cases is

$$\begin{array}{c}\hfill {\beta }_n={\alpha }_1{\lambda }_1^n+{\alpha }_2{\lambda }_2^n+{\alpha }_3{\lambda }_3^n,n\in \mathbb{N},\end{array}$$

(150)

for some constants ${\alpha }_i$, $i=\overline{1,3}$, which due to ${\gamma }_1=c\ne 0$ can be prolonged for every non-positive index.

From Lemma 1 with $Q(t)=p_3(t)={\prod }_{j=1}^3(t-{\lambda }_j)$, we have

$$\begin{array}{c}\hfill \sum _{j=1}^3\frac{{\lambda }_j^l}{p_3^{\prime }({\lambda }_j)}=0,\mathrm{for}l=0,1,\mathrm{and}\sum _{j=1}^3\frac{{\lambda }_j^2}{p_3^{\prime }({\lambda }_j)}=1.\end{array}$$

(151)

From (142), (150) and (151), we get

$$\begin{array}{cc}\hfill {\beta }_n=& \frac{{\lambda }_1^{n+3}}{({\lambda }_1-{\lambda }_2)({\lambda }_1-{\lambda }_3)}+\frac{{\lambda }_2^{n+3}}{({\lambda }_2-{\lambda }_1)({\lambda }_2-{\lambda }_3)}+\frac{{\lambda }_3^{n+3}}{({\lambda }_3-{\lambda }_1)({\lambda }_3-{\lambda }_2)},\hfill \end{array}$$

(152)

for $n\ge -2$.

From (144) and (152), it follows that

$$\begin{array}{c}\hfill x_n=1+\sum _{k=1}^{n-1}\sum _{j=1}^3\frac{{\lambda }_j^{k+3}}{p_3^{\prime }({\lambda }_j)},\end{array}$$

(153)

for $n\in \mathbb{N}.$

If ${\lambda }_j\ne 1$, $j=\overline{1,3}$, then from formula (153), it follows that

$$\begin{array}{cc}\hfill x_n=& 1+\frac{{\lambda }_1^4({\lambda }_1^{n-1}-1)}{({\lambda }_1-{\lambda }_2)({\lambda }_1-{\lambda }_3)({\lambda }_1-1)}+\frac{{\lambda }_2^4({\lambda }_2^{n-1}-1)}{({\lambda }_2-{\lambda }_1)({\lambda }_2-{\lambda }_3)({\lambda }_2-1)}\hfill \\ & +\frac{{\lambda }_3^4({\lambda }_3^{n-1}-1)}{({\lambda }_3-{\lambda }_1)({\lambda }_3-{\lambda }_2)({\lambda }_3-1)},\hfill \end{array}$$

(154)

for $n\in \mathbb{N}$, moreover, (154) holds for every $n\ge -2$.

If one of the zeros is 1, say ${\lambda }_3$, then $1\ne {\lambda }_1\ne {\lambda }_2\ne 1$, and we have

$$\begin{array}{cc}\hfill x_n=& 1+\frac{{\lambda }_1^4({\lambda }_1^{n-1}-1)}{({\lambda }_1-{\lambda }_2){({\lambda }_1-1)}^2}+\frac{{\lambda }_2^4({\lambda }_2^{n-1}-1)}{({\lambda }_2-{\lambda }_1){({\lambda }_2-1)}^2}+\frac{n-1}{({\lambda }_1-1)({\lambda }_2-1)}.\hfill \end{array}$$

(155)

for $n\in \mathbb{N}$. Moreover, due to (151), (155) holds for every $n\ge -2$.

Corollary 8.

Assume that $a,b,c,d\in \mathbb{Z}$, $a=0$, $bcd\ne 0$, $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$ and ${\widehat{\Delta }}_1^2\ne 4{\widehat{\Delta }}_0^3$. Then the following statements are true.

(a) 

If $c+bd\ne 1$, then the general solution to (2) is given by (139) and (145), where ${({\beta }_n)}_{n\ge -2}$ is given by (152), ${(x_n)}_{n\ge -2}$ is given by (154), while ${\lambda }_j$’s, $j=\overline{1,3}$ are given by (147) and (148).

(b) 

If $c+bd=1$, then $p_3$ has a unique zero equal to 1, say ${\lambda }_3$, and the general solution to (2) is given by formulas (139) and (145), where ${({\beta }_n)}_{n\ge -2}$ is given by (152) with ${\lambda }_3=1$, ${(x_n)}_{n\ge -2}$ is given by (155), while ${\lambda }_j$’s, $j=\overline{1,3}$ are given by (149).

One of the zeros is double. In this case it must be ${\widehat{\Delta }}_1^2=4{\widehat{\Delta }}_0^3$, that is, ${(bd)}^3=27c^2/4.$ Assume that m is a double zero of $p_3$, then it must be

$$m^3-bdm-c=0\mathrm{and}3m^2-bd=0,$$

from which it follows that

$$\begin{array}{c}\hfill p_3(\lambda )={\lambda }^3-3m^2\lambda +2m^3={(\lambda -m)}^2(\lambda +2m).\end{array}$$

(156)

Since $bd\ne 0$, we have $m\ne 0$. From this and since $c\ne 0$, from (156) we see that $p_3$ cannot have a triple zero. It also cannot have a unique zero equal to 1, since otherwise we would have $2m=-1$, from which it would be $bd=3/4\notin \mathbb{Z}$, which would be a contradiction. Note also that the polynomial can have 1 as a double zero when $bd=3$ and $c=-2$.

If ${\lambda }_1\ne {\lambda }_2={\lambda }_3,$ then the general solution to (140) has the following form

$$\begin{array}{c}\hfill {\beta }_n={\widehat{\alpha }}_1{\lambda }_1^n+({\widehat{\alpha }}_2+{\widehat{\alpha }}_{3}n){\lambda }_2^n,n\in \mathbb{N},\end{array}$$

(157)

where ${\widehat{\alpha }}_i$, $i=\overline{1,3}$ are constants. Since, in our case condition (142) must be satisfied, the solution ${({\beta }_n)}_{n\ge -2}$ to (140) can be found by letting ${\lambda }_3\to {\lambda }_2$ in (152), so that

$$\begin{array}{c}\hfill {\beta }_n=\frac{{\lambda }_1^{n+3}+\left(2{\lambda }_2-3{\lambda }_1+n({\lambda }_2-{\lambda }_1)\right){\lambda }_2^{n+2}}{{({\lambda }_2-{\lambda }_1)}^2},\end{array}$$

(158)

for $n\ge -2$.

From (144) and (158), we have

$$\begin{array}{cc}\hfill x_n=& 1+\sum _{j=1}^{n-1}{\beta }_j=1+\sum _{j=1}^{n-1}\frac{{\lambda }_1^{j+3}+\left(2{\lambda }_2-3{\lambda }_1+j({\lambda }_2-{\lambda }_1)\right){\lambda }_2^{j+2}}{{({\lambda }_2-{\lambda }_1)}^2},\hfill \end{array}$$

(159)

for every $n\in \mathbb{N}.$

From (159) and Lemma 2, we get

$$\begin{array}{cc}\hfill x_n=& 1+\frac{{\lambda }_1^4({\lambda }_1^{n-1}-1)}{{({\lambda }_2-{\lambda }_1)}^2({\lambda }_1-1)}+\frac{(2{\lambda }_2-3{\lambda }_1){\lambda }_2^3({\lambda }_2^{n-1}-1)}{{({\lambda }_2-{\lambda }_1)}^2({\lambda }_2-1)}+\frac{{\lambda }_2^3(1-n{\lambda }_2^{n-1}+(n-1){\lambda }_2^n)}{({\lambda }_2-{\lambda }_1){({\lambda }_2-1)}^2},\hfill \end{array}$$

(160)

for $n\in \mathbb{N}$ (in fact, (160) hold also for every $n\ge -2$).

If we assume that ${\lambda }_1\ne 1$ and ${\lambda }_2={\lambda }_3=1$, then from (159) it follows that

$$\begin{array}{cc}\hfill x_n=& 1+\frac{{\lambda }_1^4({\lambda }_1^{n-1}-1)}{{({\lambda }_1-1)}^3}+\frac{(2-3{\lambda }_1)(n-1)}{{({\lambda }_1-1)}^2}+\frac{(n-1)n}{2(1-{\lambda }_1)}.\hfill \end{array}$$

(161)

A direct calculation shows that (161) holds also for every $n\ge -2.$

Corollary 9.

Assume that $b,c,d\in \mathbb{Z}$, $a=0$, $bcd\ne 0$, $\alpha ,\beta ,z_0,w_{-2},w_{-1},w_0\in \mathbb{C}\backslash \left\{0\right\}$ and ${\widehat{\Delta }}_1^2=4{\widehat{\Delta }}_0^3$. Then the following statements are true.

(a) 

If $c+bd\ne 1,$ then the general solution to (2) is given by (139) and (145), where ${({\beta }_n)}_{n\ge -2}$ is given by (158), while ${(x_n)}_{n\ge -2}$ is given by (160).

(b) 

If $bd=3$ and $c=-2$, then 1 is a double zero of $p_3$, say, ${\lambda }_2={\lambda }_3=1$, then the general solution to system (2) is given by (139) and (145), where ${({\beta }_n)}_{n\ge -2}$ is given by (158) with ${\lambda }_2=1$, ${(x_n)}_{n\ge -2}$ is given by (161), while ${\lambda }_3=-2$.

(c) 

It is not possible that 1 is a simple zero of $p_3$.

Triple zero case. Since in this case $p_3$ must have the form in (156), we see that the only possibility that this polynomial has a triple zero is if $m=0$, which is impossible due to the condition $c\ne 0$.