\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 285, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/285\hfil Systems of difference equations]
{Third-order product-type systems of difference equations solvable in closed form}

\author[S. Stevi\'c \hfil EJDE-2016/285\hfilneg]
{Stevo Stevi\'c}

\address{Stevo Stevi\'c \newline
 Mathematical Institute of the Serbian Academy of Sciences,
 Knez Mihailova 36/III,  11000 Beograd, Serbia.\newline
Operator Theory and  Applications Research Group,
Department of Mathematics, Faculty of Science,
King Abdulaziz University, P.O. Box 80203,
Jeddah 21589, Saudi Arabia}
\email{sstevic@ptt.rs}

\thanks{Submitted September 15, 2016. Published October 26, 2016.}
\subjclass[2010]{39A20, 39A45}
\keywords{System of difference equations; product-type system;
\hfill\break\indent  solvable in closed form}

\begin{abstract}
 It is shown that a class of third order product-type systems of
 difference equations is solvable in closed form if initial values
 and multipliers are complex numbers, whereas the exponents are integers,
 by finding the formulas for the general solution in all possible cases.
 The main results complement some quite recent ones in the literature.
 The presented class of systems is the last one for whose investigation
 is not needed use of some associated polynomials of degree three or more,
 completing the investigation of such product-type systems.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks

\section{Introduction}

Many recent publications are devoted to the study of nonlinear difference 
equations and systems of difference equations; see
for example \cite{al0}-\cite{amc218-sde}, \cite{k}-\cite{kk}, 
\cite{mb}-\cite{ejde-2016}.
Papaschinopoulos and Schinas  essentially initiated a serious study of some 
classes of concrete systems of difference equations in \cite{ps1, ps2, ps3}, 
which was later continued by several authors in numerous other papers;
see for example
 \cite{amc218-sde, prs1, ps4, ps5, p-ejq, sps, ps-cana, ejde-fodce, ejqtde2015, 
ejde2015, ejqtde-maxsde, 508523, 541761, jdea205, jia2015-pts, ejde-2016} 
and the related references therein. 
The study of solvability of difference  equations and systems, which is a 
classical topic  \cite{ch, j, ll, mk}, has re-attracted some recent 
interest; see for example \cite{al0}-\cite{amc218-sde}, \cite{pst2},
 \cite{ejde-fodce}-\cite{ejde2015}, \cite{508523}-\cite{ejde-2016}, 
where several methods have been used. 
One of them is transforming the original equation,  which have been used and developed 
in several directions; see for example 
\cite{al0, amc218-sde, pst2, ana1, 508523, 541761, jdea205, ejde1} 
and the related references therein.

Having studied real-valued difference equations and systems whose right-hand 
sides are essentially obtained by acting with translations or some operators
with maximum on product-type expressions \cite{na4, ejqtde-maxsde}, 
we started studying some systems of difference equations in the complex domain 
(with complex initial values and/or parameters). One of the basic classes 
of difference equations and systems are product-type ones. 
The main obstacle in studying the equations and systems on the complex domain is 
the fact that many complex-valued functions are not single valued. 
Hence, we need to pose some conditions to prevent such a situation 
to obtain uniquely defined solutions. Also, the transformation method or
its modifications \cite{amc218-sde, 508523, 541761, jdea205, ejde1}
cannot be directly applied to product-type systems on the complex domain.

In \cite{ejde2015} we studied the following two-dimensional class of product-type
 systems of difference equations
$$
z_{n+1}=\frac{w_n^a}{z_{n-1}^b},\quad w_{n+1}=\frac{z_n^c}{w_{n-1}^d},\quad 
n\in\mathbb{N}_0,
$$
where $a, b, c, d\in\mathbb{Z}$, $z_{-1}, z_0, w_{-1}, w_0\in \mathbb{C}\setminus\{0\}$,
and showed that it is solvable in closed form (a three dimensional extension 
of the system was investigated in \cite{ejqtde2015}). A related product-type 
system was studied later in \cite{jia2015-pts}, whereas in paper \cite{ana1} 
appeared some product-type equations during the study of a general difference 
equation. Soon after the publication of \cite{ejqtde2015, ejde2015, jia2015-pts} 
we realized that some multipliers could be added in product-type systems 
so that the solvability of the systems is preserved. The first system of 
this type was studied in our paper \cite{ejde-fodce}. Quite recently we have 
presented another such a system in \cite{ejde-2016}. Another thing that we have 
realized is that there is only a few product-type systems of difference equations 
which are solvable in closed form, which is connected to the inability of 
solving the polynomial equations of the degree five or more by radicals. 
This means that finding all the product-type systems of difference equations 
which are solvable in closed form is of some interest and importance.

The purpose of this paper is to continue this research, by presenting another 
solvable product-type system of difference equations. More precisely, we will 
investigate the solvability of the  system
\begin{equation}
z_n=\alpha z_{n-2}^a w_{n-1}^b,\quad w_n=\beta w_{n-2}^cz_{n-3}^d,\quad n\in\mathbb{N}_0,
\label{ms}
\end{equation}
where $a,b,c,d\in\mathbb{Z}$, $\alpha, \beta\in\mathbb{C}\setminus\{0\}$,
$z_{-3}, z_{-2}, z_{-1}, w_{-2}, w_{-1}\in \mathbb{C}\setminus\{0\}$.
Cases when $\alpha=0$ or $\beta=0$ or if some of the initial values
$z_{-3}, z_{-2}, z_{-1}, w_{-2}, w_{-1}$ is equal to zero are quite
simple or produce not well-defined solutions, which is why they are excluded
from our consideration. The presented class of systems is the last one for
whose investigation is not needed use of some associated polynomials of
degree three or more, completing the investigation of such product-type systems.

\section{Main results}

In this section we prove our main results, which concern the solvability of
 system \eqref{ms}. Essentially there are six different results, but we 
incorporate them all in a theorem. Some of the formulas presented in the 
theorem hold on set $\mathbb{N}_0$, some hold on set $\mathbb{N}$, or even on the set
$\mathbb{N}\setminus\{1\}$ (such a situation appears, for example, if we have an
expression of the form $x^{n-1}$ and $x$ can be equal to zero and $n=1$). 
We will not specify which formula holds on which set and leave the minor 
observatory problem to the reader. What is interesting is that closed form 
formulas for solutions to the system of difference equations, although 
relatively complicated, are obtained in a more or less compact form, 
which is a rare case (one can note that it was not the case for the equation 
treated in \cite{jia2015-pts}).


\begin{theorem} \label{thm1} 
 Consider system of difference equation \eqref{ms} where $a,b,c,d\in\mathbb{Z}$,
 $\alpha,\beta\in\mathbb{C}\setminus\{0\}$ and
$z_{-3}, z_{-2}, z_{-1}, w_{-2}, w_{-1}\in\mathbb{C}\setminus\{0\}$.
Then the following statements hold.

(a) If $ac=bd$, $a+c\ne 1$, then the general solution to system
 \eqref{ms} is given by
\begin{gather} 
z_{2n} =\alpha^{\frac{1-c-a(a+c)^n}{1-a-c}}\beta^{b\frac{1-(a+c)^n}{1-a-c}}
 w_{-1}^{b(a+c)^n}z_{-2}^{a(a+c)^n},\label{d21}\\
z_{2n+1} =\alpha^{\frac{1-c-a(a+c)^n}{1-a-c}}\beta^{b\frac{1-(a+c)^{n+1}}{1-a-c}}
 w_{-2}^{bc(a+c)^n}z_{-3}^{bd(a+c)^n}z_{-1}^{a(a+c)^n},\label{d22}\\
w_{2n} =\alpha^{d\frac{1-(a+c)^{n-1}}{1-a-c}}\beta^{\frac{1-a-c(a+c)^n}{1-a-c}}
 w_{-2}^{c^2(a+c)^{n-1}}z_{-3}^{cd(a+c)^{n-1}}z_{-1}^{d(a+c)^{n-1}},\label{e12}\\
w_{2n-1} =\alpha^{d\frac{1-(a+c)^{n-1}}{1-a-c}}\beta^{\frac{1-a-c(a+c)^{n-1}}{1-a-c}}
w_{-1}^{c(a+c)^{n-1}}z_{-2}^{d(a+c)^{n-1}}.\label{e14}
\end{gather}


(b) If $ac=bd$, $a+c=1$, then the general solution to system \eqref{ms} is
given by
\begin{gather} 
z_{2n} =\alpha^{an+1}\beta^{bn}w_{-1}^bz_{-2}^a,\label{d23}\\
z_{2n+1} =\alpha^{an+1}\beta^{b(n+1)}w_{-2}^{bc}z_{-3}^{bd}z_{-1}^a,\label{d25}\\
w_{2n} =\alpha^{d(n-1)}\beta^{(1-a)n+1}w_{-2}^{c^2}z_{-3}^{cd}z_{-1}^d,\label{e12ab}\\
w_{2n-1}=\alpha^{d(n-1)}\beta^{(1-a)n+a}w_{-1}^cz_{-2}^d.\label{e14ab}
\end{gather}

(c) If $ac\ne bd$, $(a+c)^2\ne 4(ac-bd)$ and $bd\ne (a-1)(c-1)$, 
then the general solution to system \eqref{ms} is given by
\begin{gather} 
\begin{aligned}
z_{2n}&=\alpha^{\frac{(t_2-1)(t_1-c)t_1^{n+1}-(t_1-1)(t_2-c)t_2^{n+1}
 +(t_1-t_2)(1-c)}{(t_1-1)(t_2-1)(t_1-t_2)}} \\
&\quad \beta^{b\frac{(t_2-1)t_1^{n+1}-(t_1-1)t_2^{n+1}+t_1-t_2}{(t_1-1)(t_2-1)(t_1-t_2)}}
 \\
&\quad \times w_{-1}^{b\frac{t_1^{n+1}-t_2^{n+1}}{t_1-t_2}}
 z_{-2}^{\frac{(t_1-c)t_1^{n+1}-(t_2-c)t_2^{n+1}}{t_1-t_2}},
\end{aligned} \label{d12a}
\\
\begin{aligned}
z_{2n+1}&=\alpha^{\frac{(t_2-1)(t_1-c)t_1^{n+1}-(t_1-1)(t_2-c)t_2^{n+1}
 +(t_1-t_2)(1-c)}{(t_1-1)(t_2-1)(t_1-t_2)}}
\beta^{b\frac{(t_2-1)t_1^{n+2}-(t_1-1)t_2^{n+2}+t_1-t_2}{(t_1-1)(t_2-1)(t_1-t_2)}} \\
&\quad \times w_{-2}^{bc\frac{t_1^{n+1}-t_2^{n+1}}{t_1-t_2}}z_{-3}^{bd\frac{t_1^{n+1}
 -t_2^{n+1}}{t_1-t_2}}z_{-1}^{\frac{(t_1-c)t_1^{n+1}-(t_2-c)t_2^{n+1}}{t_1-t_2}},
\end{aligned}\label{d14a} \\
\begin{aligned}
w_{2n}&=\alpha^{d\frac{(t_2-1)t_1^n-(t_1-1)t_2^n+t_1-t_2}{(t_1-1)(t_2-1)(t_1-t_2)}} \\
&\quad \times \beta^{\frac{(t_2-1)(t_1-a)t_1^{n+1}-(t_1-1)(t_2-a)t_2^{n+1}
 +(t_1-t_2)(1-a)}{(t_1-1)(t_2-1)(t_1-t_2)}} \\
&\quad \times w_{-2}^{c\frac{(t_1-a)t_1^n-(t_2-a)t_2^n}{t_1-t_2}}z_{-3}^{d\frac{(t_1-a)
 t_1^n-(t_2-a)t_2^n}{t_1-t_2}}z_{-1}^{d\frac{t_1^n-t_2^n}{t_1-t_2}},
\end{aligned}\label{e12a}\\
\begin{aligned}
w_{2n-1}&=\alpha^{d\frac{(t_2-1)t_1^n-(t_1-1)t_2^n+t_1-t_2}{(t_1-1)(t_2-1)(t_1-t_2)}}
\beta^{\frac{(t_2-1)(t_1-a)t_1^n-(t_1-1)(t_2-a)t_2^n+(t_1-t_2)(1-a)}{(t_1-1)(t_2-1)
 (t_1-t_2)}} \\
&\quad \times w_{-1}^{\frac{(t_1-a)t_1^n-(t_2-a)t_2^n}{t_1-t_2}}
 z_{-2}^{d\frac{t_1^n-t_2^n}{t_1-t_2}},
\end{aligned}\label{e14a}
\end{gather}
where
\begin{equation}
t_{1,2}=\frac{a+c\pm\sqrt{(a+c)^2-4(ac-bd)}}2.\label{t12}
\end{equation}

(d) If $ac\ne bd$, $(a+c)^2=4(ac-bd)$, $bd\ne (a-1)(c-1)$, then the general 
solution to system \eqref{ms} is
given by
\begin{gather}
\begin{aligned} 
z_{2n}&=\alpha^{\frac{1-c+t_1^n((n+1)t_1^2-(n(c+1)+2)t_1+c(n+1))}{(1-t_1)^2}}
\beta^{b\frac{1-(n+1)t_1^n+nt_1^{n+1}}{(1-t_1)^2}} \\ 
&\quad \times w_{-1}^{b(n+1)t_1^n}z_{-2}^{(n(t_1-c)+2t_1-c)t_1^n},
\end{aligned} \label{d12b}\\
\begin{aligned}
z_{2n+1}&=\alpha^{\frac{1-c+t_1^n((n+1)t_1^2-(n(c+1)+2)t_1+c(n+1))}{(1-t_1)^2}}
 \beta^{b\frac{1-(n+2)t_1^{n+1}+(n+1)t_1^{n+2}}{(1-t_1)^2}} \\
&\quad \times w_{-2}^{bc(n+1)t_1^n}z_{-3}^{bd(n+1)t_1^n}z_{-1}^{(n(t_1-c)+2t_1-c)t_1^n},
\end{aligned}\label{d14b}\\
\begin{aligned}
w_{2n}&=\alpha^{d\frac{1-nt_1^{n-1}+(n-1)t_1^n}{(1-t_1)^2}}
\beta^{\frac{1-a+t_1^n((n+1)t_1^2-((1+a)n+2)t_1+a(n+1))}{(1-t_1)^2}} \\
&\quad \times w_{-2}^{c(n(t_1-a)+t_1) t_1^{n-1}}z_{-3}^{d(n(t_1-a)
 +t_1)t_1^{n-1}}z_{-1}^{dnt_1^{n-1}},
\end{aligned} \label{e12b}\\
\begin{aligned}
w_{2n-1}
&=\alpha^{d\frac{1-nt_1^{n-1}+(n-1)t_1^n}{(1-t_1)^2}}
 \beta^{\frac{1-a+t_1^{n-1}(nt_1^2-((1+a)n+1-a)t_1+an)}{(1-t_1)^2}} \\
&\quad \times w_{-1}^{(n(t_1-a)+t_1)t_1^{n-1}}z_{-2}^{dnt_1^{n-1}},
\end{aligned} \label{e14b}
\end{gather}
where
$$
t_1=\frac{a+c}2.
$$

(e) If $ac\ne bd$, $(a+c)^2\ne 4(ac-bd)$, $bd=(a-1)(c-1)$, $a+c\ne 2$, 
then the general solution to system \eqref{ms} is
given by
\begin{gather}
\begin{aligned} 
z_{2n}&=\alpha^{\frac{(t_1-c)t_1^{n+1}+((c-1)n+c-2)t_1+(1-c)n+1}{(1-t_1)^2}}
\beta^{b\frac{t_1^{n+1}-(n+1)t_1+n}{(1-t_1)^2}} \\
&\quad \times w_{-1}^{b\frac{t_1^{n+1}-1}{t_1-1}}z_{-2}^{\frac{(t_1-c)t_1^{n+1}
+c-1}{t_1-1}},
\end{aligned} \label{d12o}\\
\begin{aligned}
z_{2n+1}
&=\alpha^{\frac{(t_1-c)t_1^{n+1}+((c-1)n+c-2)t_1+(1-c)n+1}{(1-t_1)^2}}
\beta^{b\frac{t_1^{n+2}-(n+2)t_1+n+1}{(1-t_1)^2}} \\
&\quad \times w_{-2}^{bc\frac{t_1^{n+1}-1}{t_1-1}}z_{-3}^{bd\frac{t_1^{n+1}-1}{t_1-1}}
 z_{-1}^{\frac{(t_1-c)t_1^{n+1}+c-1}{t_1-1}},
\end{aligned} \label{d14o}\\
\begin{aligned}
w_{2n}
&=\alpha^{d\frac{t_1^n-nt_1+n-1}{(1-t_1)^2}}\beta^{\frac{(t_1-a)t_1^{n+1}
 +((a-1)n+a-2)t_1+(1-a)n+1}{(1-t_1)^2}} \\
&\quad \times w_{-2}^{c\frac{(t_1-a)t_1^n+a-1}{t_1-1}}
 z_{-3}^{d\frac{(t_1-a)t_1^n+a-1}{t_1-1}}z_{-1}^{d\frac{t_1^n-1}{t_1-1}},
\end{aligned} \label{e12o}\\
\begin{aligned}
w_{2n-1}&= \alpha^{d\frac{t_1^n-nt_1+n-1}{(1-t_1)^2}}\beta^{\frac{(t_1-a)t_1^n+((a-1)n-1)t_1+(1-a)n+a}{(1-t_1)^2}} \\
&\times w_{-1}^{\frac{(t_1-a)t_1^n+a-1}{t_1-1}}z_{-2}^{d\frac{t_1^n-1}{t_1-1}},
\end{aligned} \label{e14o}
\end{gather}
where $t_1=a+c-1$.


(f) If $ac\ne bd$, $(a+c)^2=4(ac-bd)$, $bd=(a-1)(c-1)$, and $a+c=2$, 
then the general solution to system \eqref{ms} is
given by
\begin{gather} 
z_{2n}=\alpha^{\frac{(n+1)((1-c)n+2)}2}\beta^{b\frac{n(n+1)}2}w_{-1}^{b(n+1)}
 z_{-2}^{(1-c)n+2-c},\label{d12p}\\
z_{2n+1}=\alpha^{\frac{(n+1)((1-c)n+2)}2}\beta^{b\frac{(n+1)(n+2)}2}w_{-2}^{bc(n+1)}
 z_{-3}^{bd(n+1)}z_{-1}^{(1-c)n+2-c},\label{d14p}\\
w_{2n}=\alpha^{d\frac{(n-1)n}2}\beta^{\frac{(n+1)((1-a)n+2)}2}w_{-2}^{c((1-a)n+1)}
 z_{-3}^{d((1-a)n+1)}z_{-1}^{dn},\label{e12p}\\
w_{2n-1}=\alpha^{d\frac{(n-1)n}2}\beta^{\frac{n((1-a)n+1+a)}2} w_{-1}^{(1-a)n+1}
 z_{-2}^{dn}.\label{e14p}
\end{gather}
\end{theorem}

\begin{proof} 
Since $\alpha,\beta\in\mathbb{C}\setminus\{0\}$ and
$z_{-3}, z_{-2}, z_{-1}, w_{-2}, w_{-1}\in\mathbb{C}\setminus\{0\}$, using \eqref{ms}
and induction we easily get
$$
z_n\ne 0\qquad \text{for }n\ge -3, \quad\text{and}\quad 
 w_n\ne 0\quad\text{for }n\ge -2.
$$
 Hence, from \eqref{ms} we have
\begin{gather}
w_{n-1}^b=\frac{z_n}{\alpha z_{n-2}^a},\quad n\in\mathbb{N}_0,\label{d1} \\
w_n^b=\beta^b w_{n-2}^{bc}z_{n-3}^{bd},\quad n\in\mathbb{N}_0.\label{d2}
\end{gather}
From \eqref{d1} and \eqref{d2} it follows that
\begin{equation}
z_{n+1}=\alpha^{1-c}\beta^bz_{n-1}^{a+c}z_{n-3}^{bd-ac},\quad n\in\mathbb{N}.\label{d3}
\end{equation}
Let $\eta:=\alpha^{1-c}\beta^b$,
\begin{equation}
 u_1=1,\quad a_1=a+c,\quad b_1=bd-ac.\label{ic}
\end{equation}
From \eqref{d3} we have
\begin{equation}
z_{2(n+1)+i}=\eta^{u_1}z_{2n+i}^{a_1}z_{2(n-1)+i}^{b_1},\label{d5}
\end{equation}
for $n\in\mathbb{N}_0$ and $i=0,1$.
From \eqref{d5} it follows that
\begin{equation}
\begin{aligned}
 z_{2(n+1)+i}
&=\eta^{u_1}(\eta z_{2(n-1)+i}^{a_1}z_{2(n-2)+i}^{b_1})^{a_1}z_{2(n-1)+i}^{b_1} \\
&=\eta^{u_1+a_1}z_{2(n-1)+i}^{a_1a_1+b_1}z_{2(n-2)+i}^{b_1a_1} \\
&=\eta^{u_2}z_{2(n-1)+i}^{a_2}z_{2(n-2)+i}^{b_2},
\end{aligned}\label{d6}
\end{equation}
for $n\in\mathbb{N}$ and $i=0,1$, where
\begin{equation}
u_2:=u_1+a_1,\quad a_2:=a_1a_1+b_1,\quad b_2:=b_1a_1.\label{d7}
\end{equation}
Assume that for a $k\ge 2$ it  holds
\begin{equation}
z_{2(n+1)+i}=\eta^{u_k}z_{2(n-k+1)+i}^{a_k}z_{2(n-k)+i}^{b_k},\label{d8}
\end{equation}
for $n\ge k-1$ and $i=0,1$, where
\begin{equation}
u_k:=u_{k-1}+a_{k-1},\quad a_k:=a_1a_{k-1}+b_{k-1},\quad b_k:=b_1a_{k-1}.\label{d9}
\end{equation}

Using \eqref{d5} in \eqref{d8}, it follows that
\begin{equation}
\begin{aligned}
z_{2(n+1)+i}&=\eta^{u_k}z_{2(n-k+1)+i}^{a_k}z_{2(n-k)+i}^{b_k} \\
&=\eta^{u_k}(\eta z_{2(n-k)+i}^{a_1}z_{2(n-k-1)+i}^{b_1})^{a_k}z_{2(n-k)+i}^{b_k} \\
&=\eta^{u_k+a_k}z_{2(n-k)+i}^{a_1a_k+b_k}z_{2(n-k-1)+i}^{b_1a_k} \\
&=\eta^{u_{k+1}}z_{2(n-k)+i}^{a_{k+1}}z_{2(n-k-1)+i}^{b_{k+1}},
\end{aligned} \label{d10}
\end{equation}
for $n\ge k$ and $i=0,1$, where
\begin{equation}
u_{k+1}:=u_k+a_k,\quad a_{k+1}:=a_1a_k+b_k,\quad b_{k+1}:=b_1a_k.\label{d11}
\end{equation}
Equalities \eqref{d6}, \eqref{d7}, \eqref{d10}, \eqref{d11} along with induction
show that \eqref{d8} and \eqref{d9} hold for
all $k, n\in\mathbb{N}$ such that $2\le k\le n+1$.

From \eqref{d8} we have
$$
z_{2n+i}=\eta^{u_n}z_i^{a_n}z_{i-2}^{b_n},
$$
for $n\in\mathbb{N}$ and $i=0,1$, from which along with
$$
z_0=\alpha z_{-2}^aw_{-1}^b,\quad z_1=\alpha z_{-1}^aw_0^b
=\alpha z_{-1}^a(\beta w_{-2}^cz_{-3}^d)^b
=\alpha \beta^bw_{-2}^{bc}z_{-3}^{bd}z_{-1}^a,
$$
it follows that
\begin{gather}
\begin{aligned}
 z_{2n}
&= \eta^{u_n}z_0^{a_n}z_{-2}^{b_n}=(\alpha^{1-c}\beta^b)^{u_n}(\alpha z_{-2}^a
 w_{-1}^b)^{a_n}z_{-2}^{b_n} \\
&= \alpha^{(1-c)u_n+a_n}\beta^{bu_n}w_{-1}^{ba_n}z_{-2}^{aa_n+b_n} \\
&= \alpha^{u_{n+1}-cu_n}\beta^{bu_n}w_{-1}^{ba_n}z_{-2}^{a_{n+1}-ca_n},
\end{aligned} \label{d12}\\
\begin{aligned}
z_{2n+1}
&= \eta^{u_n}z_1^{a_n}z_{-1}^{b_n}=(\alpha^{1-c}\beta^b)^{u_n}
(\alpha \beta^bw_{-2}^{bc}z_{-3}^{bd}z_{-1}^a)^{a_n}z_{-1}^{b_n} \\
&= \alpha^{(1-c)u_n+a_n}\beta^{bu_n+ba_n}w_{-2}^{bca_n}z_{-3}^{bda_n}z_{-1}^{aa_n+b_n} \\
&= \alpha^{u_{n+1}-cu_n}\beta^{bu_{n+1}}w_{-2}^{bca_n}z_{-3}^{bda_n}z_{-1}^{a_{n+1}-ca_n},
\end{aligned} \label{d14}
\end{gather}
for $n\in\mathbb{N}$.

From \eqref{d9} and since $u_1=1$, we have
\begin{gather} 
a_k=a_1a_{k-1}+b_1a_{k-2},\quad k\ge 3,\label{d15} \\
 u_k=1+\sum_{j=1}^{k-1}a_j,\quad k\in\mathbb{N}.\label{d16}
\end{gather}
\smallskip


\noindent\textbf{Case $ac=bd$.} 
Since $b_1=bd-ac=0$ equation \eqref{d15} is reduced to
$$
a_k=a_1a_{k-1}=(a+c)a_{k-1},\quad k\ge 3,
$$
which implies 
\begin{equation}
a_k=a_2(a+c)^{k-2}=(a+c)^k,\label{d17}
\end{equation}
for $k\in\mathbb{N}$ (for $k=1,2$ this is directly verified).

Equalities \eqref{d16} and \eqref{d17} yield
$$
u_k=1+\sum_{j=1}^{k-1}(a+c)^j,\quad k\in\mathbb{N},
$$
so that
\begin{equation}
u_k=\frac{1-(a+c)^k}{1-a-c},\quad k\in\mathbb{N},\label{d19}
\end{equation}
if $a+c\ne 1$, whereas
\begin{equation}
 u_k=k,\quad k\in\mathbb{N},\label{d20}
\end{equation}
if $a+c=1$.

If $a+c\ne 1$, then from \eqref{d12}, \eqref{d14}, \eqref{d17}, \eqref{d19} 
and since
\begin{gather*}
u_{n+1}-cu_n= \frac{1-c-a(a+c)^n}{1-a-c}\\
a_{n+1}-ca_n= a(a+c)^n
\end{gather*}
we obtain  formulas \eqref{d21} and \eqref{d22},
for $n\ge 2$.

If $a+c=1$, then from \eqref{d12}, \eqref{d14}, \eqref{d17}, \eqref{d20} and since
$$u_{n+1}-cu_n=(1-c)n+1=an+1$$
we obtain formulas \eqref{d23} and \eqref{d25}, for $n\in \mathbb{N}$.
\smallskip


\noindent\textbf{Case $ac\ne bd$.} 
Let $t_{1,2}$ be the roots of the characteristic polynomial 
\begin{equation}
P(t)=t^2-(a+c)t+ac-bd,\label{q28}
\end{equation}
associated with  difference equation \eqref{d15}.
 Note that they are given by the formulas in \eqref{t12}.
We have
$$
a_n=c_1t_1^n+c_2t_2^n,\quad n\in\mathbb{N},
$$
where $c_1, c_2\in \mathbb{R}$, if $(a+c)^2\ne 4(ac-bd)$, whereas
$$
u_n=(d_1n+d_2)t_1^n,\quad n\in\mathbb{N},
$$
where $d_1, d_2\in\mathbb{R}$, if $(a+c)^2=4(ac-bd)$.

Since $a_1=t_1+t_2$ and $a_2=(t_1+t_2)^2-t_1t_2=t_1^2+t_1t_2+t_2^2$, 
it is easily obtained that
\begin{equation}
a_k=\frac{t_1^{k+1}-t_2^{k+1}}{t_1-t_2},\quad k\in\mathbb{N},\label{d35}
\end{equation}
if $(a+c)^2\ne 4(ac-bd)$, whereas
\begin{equation} a_k=(k+1)t_1^k\label{d36},\quad k\in\mathbb{N},\end{equation}
if $(a+c)^2=4(ac-bd)$.

From  \eqref{d16} and \eqref{d35} we have
\begin{equation}
u_k=1+\sum_{j=1}^{k-1}\frac{t_1^{j+1}-t_2^{j+1}}{t_1-t_2}
=\frac{(t_2-1)t_1^{k+1}-(t_1-1)t_2^{k+1}+t_1-t_2}{(t_1-1)(t_2-1)(t_1-t_2)},
\label{d39}
\end{equation}
for $k\in\mathbb{N}$, if $(a+c)^2\ne 4(ac-bd)$ and $bd\ne (a-1)(c-1)$.

From \eqref{d16} and \eqref{d36} it follows that
\begin{equation}
u_k=1+\sum_{j=1}^{k-1}(j+1)t_1^j=\frac{1-(k+1)t_1^k+kt_1^{k+1}}{(1-t_1)^2},
\label{d40}
\end{equation}
for $k\in\mathbb{N}$, if $(a+c)^2=4(ac-bd)$ and $bd\ne (a-1)(c-1)$.

If $(a+c)^2\ne 4(ac-bd)$, $bd=(a-1)(c-1)$ and $a+c\ne 2$, then polynomial 
\eqref{q28} has exactly one zero equal to one, say $t_2$. 
From  \eqref{d16} and \eqref{d35} we have
\begin{equation}
u_k=1+\sum_{j=1}^{k-1}\frac{t_1^{j+1}-1}{t_1-1}
=\frac{t_1^{k+1}-(k+1)t_1+k}{(t_1-1)^2},\label{d39w}
\end{equation}
for $k\in\mathbb{N}$.

Finally, if $(a+c)^2=4(ac-bd)$, $bd=(a-1)(c-1)$ and $a+c=2$, then both 
zeros of polynomial \eqref{q28} are equal to one. From \eqref{d16} and 
\eqref{d36} it follows that
\begin{equation}
u_k=1+\sum_{j=1}^{k-1}(j+1)=\frac{k(k+1)}2,\label{d40w}
\end{equation}
for $k\in\mathbb{N}$.

If $(a+c)^2\ne 4(ac-bd)$ and $bd\ne(a-1)(c-1)$, then from 
\eqref{d12}, \eqref{d14}, \eqref{d35}, \eqref{d39} and since
\begin{gather*}
u_{n+1}-cu_n= \frac{(t_2-1)(t_1-c)t_1^{n+1}-(t_1-1)(t_2-c)t_2^{n+1}
 +(t_1-t_2)(1-c)}{(t_1-1)(t_2-1)(t_1-t_2)}\\
a_{n+1}-ca_n= \frac{(t_1-c)t_1^{n+1}-(t_2-c)t_2^{n+1}}{t_1-t_2},
\end{gather*}
we obtain formulas \eqref{d12a} and \eqref{d14a}.

If $(a+c)^2=4(ac-bd)$ and $bd\ne (a-1)(c-1)$, then from \eqref{d12}, \eqref{d14}, 
\eqref{d36}, \eqref{d40} and since
\begin{gather*}
u_{n+1}-cu_n= \frac{1-c+t_1^n((n+1)t_1^2-((c+1)n+2)t_1+c(n+1))}{(1-t_1)^2}\\
a_{n+1}-ca_n= ((t_1-c)n+2t_1-c)t_1^n,
\end{gather*}
we obtain formulas \eqref{d12b} and \eqref{d14b}.

If $(a+c)^2\ne 4(ac-bd)$, $bd=(a-1)(c-1)$ and $a+c\ne 2$, then 
from \eqref{d12}, \eqref{d14}, \eqref{d35} with $t_2=1$, \eqref{d39w} and since
\begin{gather*}
u_{n+1}-cu_n= \frac{(t_1-c)t_1^{n+1}+((c-1)n+c-2)t_1+(1-c)n+1}{(1-t_1)^2}\\
a_{n+1}-ca_n= \frac{(t_1-c)t_1^{n+1}+c-1}{t_1-1},
\end{gather*}
we obtain formulas \eqref{d12o} and \eqref{d14o}, where $t_1=ac-bd=a+c-1$.

If $(a+c)^2= 4(ac-bd)$, $bd=(a-1)(c-1)$ and $a+c=2$, then from 
\eqref{d12}, \eqref{d14}, \eqref{d36} with $t_1=1$, \eqref{d40w} and since
\begin{gather*}
u_{n+1}-cu_n= \frac{(n+1)((1-c)n+2)}2\\
a_{n+1}-ca_n= (1-c)n+2-c,
\end{gather*}
we obtain formulas \eqref{d12p} and \eqref{d14p}.

From \eqref{ms} we also have 
\begin{gather}
z_{n-3}^d=\frac{w_n}{\beta w_{n-2}^c},\quad n\in\mathbb{N}_0,\label{e1}\\
z_n^d=\alpha^d z_{n-2}^{ad}w_{n-1}^{bd},\quad n\in\mathbb{N}_0.\label{e2}
\end{gather}
Thus, from \eqref{e1} and \eqref{e2}  it follows that
\begin{equation}
w_{n+3}=\alpha^d\beta^{1-a}w_{n+1}^{a+c}w_{n-1}^{bd-ac},\quad n\in\mathbb{N}_0.\label{e3}
\end{equation}
Note that difference equations \eqref{d3} and \eqref{e3} have only
different constant multipliers.

Let $\nu:=\alpha^d\beta^{1-a}$,
\begin{equation}
 \hat u_1=1,\quad \hat a_1=a+c,\quad \hat b_1=bd-ac.\label{ic1}
\end{equation}
As above it is proved that for any
$k\in\mathbb{N}$ it  holds
\begin{equation}
w_{2(n+1)+i}=\nu^{\hat u_k}w_{2(n-k+1)+i}^{\hat a_k}w_{2(n-k)+i}^{\hat b_k},
\label{e8}
\end{equation}
for $n\ge k$ and $i=-1,0$, where
\begin{equation}
\hat u_k:=\hat u_{k-1}+\hat a_{k-1},\quad
\hat a_k:=\hat a_1\hat a_{k-1}+\hat b_{k-1},\quad
\hat b_k=\hat b_1\hat a_{k-1}.\label{e9}
\end{equation}

Since initial conditions \eqref{ic1} and system \eqref{e9} are the same 
as those in \eqref{ic} and \eqref{d9}, it follows that
\begin{equation} 
\hat a_k=a_k, \quad \hat b_k=b_k,\quad \hat u_k=u_k,\label{e16}
\end{equation}
for every $k\in\mathbb{N}$.
From \eqref{e8} we have
$$
w_{2n+i}=\nu^{u_{n-1}}w_{2+i}^{a_{n-1}}w_{i}^{b_{n-1}},
$$
for $n\ge 2$ and $i=-1,0$, from which along with
\begin{gather*}
w_0=\beta w_{-2}^cz_{-3}^d,\qquad w_1=\beta w_{-1}^cz_{-2}^d, \\
w_2=\beta w_0^cz_{-1}^d=\beta(\beta w_{-2}^cz_{-3}^d)^cz_{-1}^d
=\beta^{1+c}w_{-2}^{c^2}z_{-3}^{cd}z_{-1}^d,
\end{gather*}
it follows that
\begin{gather}
\begin{aligned}
w_{2n}
&=\nu^{ u_{n-1}}w_2^{a_{n-1}}w_0^{b_{n-1}}\\
&=(\alpha^d\beta^{1-a})^{u_{n-1}}(\beta^{1+c}w_{-2}^{c^2}z_{-3}^{cd}z_{-1}^d)^{a_{n-1}}
 (\beta w_{-2}^cz_{-3}^d)^{b_{n-1}} \\
&=\alpha^{du_{n-1}}\beta^{(1-a) u_{n-1}+(1+c)a_{n-1}+b_{n-1}}w_{-2}^{c^2a_{n-1}
 +cb_{n-1}}z_{-3}^{cda_{n-1}+db_{n-1}}z_{-1}^{da_{n-1}} \\
&=\alpha^{du_{n-1}}\beta^{u_{n+1}-a u_n}w_{-2}^{c(a_n-aa_{n-1})}
 z_{-3}^{d(a_n-aa_{n-1})}z_{-1}^{da_{n-1}},
\end{aligned} \label{e12c}\\
\begin{aligned}
w_{2n-1}
&=\nu^{u_{n-1}}w_1^{a_{n-1}}w_{-1}^{b_{n-1}}\\
&=(\alpha^d\beta^{1-a})^{u_{n-1}}(\beta w_{-1}^cz_{-2}^d)^{a_{n-1}}w_{-1}^{b_{n-1}} \\
&=\alpha^{du_{n-1}}\beta^{(1-a)u_{n-1}+a_{n-1}}w_{-1}^{ca_{n-1}+b_{n-1}}z_{-2}^{da_{n-1}} \\
&=\alpha^{du_{n-1}}\beta^{u_n-au_{n-1}}w_{-1}^{a_n-aa_{n-1}}z_{-2}^{da_{n-1}},\label{e14c}
\end{aligned}
\end{gather}
for $n\ge 2$.
\smallskip

\noindent\textbf{Case $ac=bd$.} 
If $a+c\ne 1$, then from \eqref{d17}, \eqref{d19}, \eqref{e12c}, \eqref{e14c} 
and since
\begin{gather*}
u_n-au_{n-1}= \frac{1-a-c(a+c)^{n-1}}{1-a-c}\\
a_n-aa_{n-1}= c(a+c)^{n-1}
\end{gather*}
we obtain formulas \eqref{e12} and \eqref{e14}.

If $a+c=1$, then from \eqref{d17} with $a+c=1$, \eqref{d20}, \eqref{e12c}, 
\eqref{e14c} and since
$$
u_n-au_{n-1}=(1-a)n+a
$$
we obtain formulas \eqref{e12ab} and \eqref{e14ab}.
\smallskip

\noindent\textbf{Case $ac\ne bd$.} 
If $ac\ne bd$, $(a+c)^2\ne 4(ac-bd)$ and $bd\ne(a-1)(c-1)$, then from \eqref{d35}, 
\eqref{d39}, \eqref{e12c}, \eqref{e14c} and since
\begin{gather*}
u_n-au_{n-1}= \frac{(t_2-1)(t_1-a)t_1^n-(t_1-1)
 (t_2-a)t_2^n+(t_1-t_2)(1-a)}{(t_1-1)(t_2-1)(t_1-t_2)}\\
a_n-aa_{n-1}= \frac{(t_1-a)t_1^n-(t_2-a)t_2^n}{t_1-t_2}
\end{gather*}
we obtain formulas  \eqref{e12a} and \eqref{e14a}.

If $ac\ne bd$, $(a+c)^2=4(ac-bd)$, $bd\ne(a-1)(c-1)$, then from \eqref{d36}, \eqref{d40}, \eqref{e12c}, \eqref{e14c} and since
\begin{gather*}
u_n-au_{n-1}= \frac{1-a+t_1^{n-1}(nt_1^2-((1+a)n+1-a)t_1+an)}{(1-t_1)^2}\\
a_n-aa_{n-1}= (n(t_1-a)+t_1)t_1^{n-1}
\end{gather*}
we obtain formulas \eqref{e12b} and \eqref{e14b}.


If $ac\ne bd$, $(a+c)^2\ne 4(ac-bd)$, $bd=(a-1)(c-1)$ and $a+c\ne 2$, 
then from \eqref{d35} with $t_2=1$, \eqref{d39w}, \eqref{e12c}, \eqref{e14c} 
and since
\begin{gather*}
u_n-au_{n-1}= \frac{(t_1-a)t_1^n+((a-1)n-1)t_1+(1-a)n+a}{(1-t_1)^2}\\
a_n-aa_{n-1}= \frac{(t_1-a)t_1^n+a-1}{t_1-1}
\end{gather*}
we obtain formulas \eqref{e12o} and \eqref{e14o}, where $t_1=ac-bd=a+c-1$.

If $ac\ne bd$, $(a+c)^2=4(ac-bd)$, $bd=(a-1)(c-1)$ and $a+c=2$, then 
from \eqref{d36} with $t_1=1$, \eqref{d40w}, \eqref{e12c}, \eqref{e14c} and since
\begin{gather*}
u_n-au_{n-1}= \frac{n((1-a)n+1+a)}2\\
a_n-aa_{n-1}= (1-a)n+1
\end{gather*}
we obtain formulas \eqref{e12p} and \eqref{e14p}.

By some standard but tedious and time-consuming calculations it is 
checked that all the formulas in the theorem really present general 
solution to system \eqref{ms} (in each of these six cases), 
completing the proof of the theorem.
\end{proof}


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\end{document}