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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 159, pp. 1--13.\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-2016/159\hfil Close to symmetric system of difference equations]
{Solvability of a close to symmetric system of difference equations}

\author[S. Stevi\'c, B. Iri\v{c}anin, Z. \v{S}marda \hfil EJDE-2016/159\hfilneg]
{Stevo Stevi\'c, Bratislav Iri\v{c}anin, Zden\v{e}k \v{S}marda}

\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}

\address{Bratislav Iri\v{c}anin \newline
Faculty of Electrical Engineering,
Belgrade University, Bulevar Kralja Aleksandra 73,
11000 Beograd, Serbia}
\email{iricanin@etf.rs}

\address{Zden\v{e}k \v{S}marda \newline
 CEITEC - Central European Institute of Technology,
Brno University of Technology, Czech Republic}
\email{zdenek.smarda@ceitec.vutbr.cz}

\thanks{Submitted April 10, 2016. Published June 27, 2016.}
\subjclass[2010]{39A10, 39A20}
\keywords{Symmetric system of difference equations;
 product-type system}

\begin{abstract}
 The problem of solvability of a close to symmetric product-type system
 of difference equations of second order is investigated. Some recent results
 in the literature are extended.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

Various types of nonlinear difference equations and systems have been considerably 
studied recently 
(see, e.g., \cite{al0}-\cite{amc218-sde}, \cite{k1,kk,mj,mj1}, 
\cite{ps1}-\cite{508523}). 
Among other topics, there has been some renewed interest in
the equations and systems which can be solved (see, e.g.,
\cite{al0}-\cite{amc218-sde}, \cite{pst2,37264}, \cite{amc218-hde}-\cite{ejde2015}, 
\cite{508523}). Many of these papers essentially used a transformation method 
by  Stevi\'c (see, e.g., 
\cite{al0, al1, pst2, 37264, amc218-hde, amc-mtsde, amc219-dussde, 508523} 
where can be also found original sources and many other references). 
Some known  classes of difference equations and systems, including solvable ones, 
can be found, for example, in \cite{j, ll, mk, rv}. After the publication of 
some papers on concrete systems of difference equations by Papaschinopoulos and 
 Schinas almost two decades ago (see, e.g., \cite{ps1, ps2, ps3}), 
some interest in the area has also started 
(see, e.g., \cite{amc218-sde, ps4, p-ejq, sps, ps-cana, amc-mtsde, amc219-dussde, 
ejde-fodce, ejqtde-fib, ejqtde2015, ejde2015, ejqtde-maxsde, 508523}).

An investigation of the long-term behavior of solutions to some classes 
of difference  equations which are modifications/perturbations of 
product-type ones has been also started by  Stevi\'c 
(see, e.g., \cite{na4} and the references therein). 
The corresponding investigation of related systems of difference equations 
has been started  somewhat later. For example, in \cite{ejqtde-maxsde} 
the boundedness character of positive solutions of the following system
\begin{equation}
z_{n+1}=\max\big\{f,w_n^p/z_{n-1}^q\big\},\quad
w_{n+1}=\max\big\{f,z_n^p/w_{n-1}^q\big\},\quad
n\in\mathbb{N}_0,\label{a1}
\end{equation}
 with positive parameters $f, p$ and $q$, was investigated.
There are also some solvable max-type systems of difference equations
\cite{amc-mtsde}. The corresponding product-type system to \eqref{a1}
(system \eqref{a2} below with $\hat a=\hat c=p$ and $\hat b=\hat d=q$)
with positive initial values is solvable. However, the case of complex
initial values seems has not been studied in detail. These observations
motivated us to study product-type systems with such initial values.
One of the first papers on the problem is \cite{ejde2015}, where we have
studied the  product-type system
\begin{equation}
z_{n+1}=\frac{w_n^{\hat a}}{z_{n-1}^{\hat b}},\quad
w_{n+1}=\frac{z_n^{\hat c}}{w_{n-1}^{\hat d}},\quad n\in\mathbb{N}_0,\label{a2}
\end{equation}
where $\hat a,\hat b,\hat c,\hat d$ are integers (the condition is posed
to avoid multi-valued sequences).

Motivated by the close to symmetric systems in \cite{ejqtde-maxsde} 
and \cite{508523}, in \cite{ejde-fodce} S.~Stevi\'c has noticed that 
some complex parameters/coefficients can be included into a product-type 
system of difference equations so that the solvability of such obtained 
system is preserved. For some other results in the topic,
 see also \cite{ejqtde2015}.

Our aim is to investigate the solvability of the following close 
to symmetric system of difference equations
\begin{equation}
z_{n+1}=\alpha w_n^az_{n-1}^b,\quad
w_{n+1}=\beta z_n^cw_{n-1}^d,\quad n\in\mathbb{N}_0,\label{ms}
\end{equation}
where $a,b,c,d\in\mathbb{Z}$, $\alpha,\beta\in\mathbb{C}$ and $z_{-1}, z_0, w_{-1},
w_0\in\mathbb{C}$. Actually, since the cases $\alpha=0$ and $\beta=0$ are simple,
we will study only the case
when $\alpha,\beta\in\mathbb{C}\setminus\{0\}$ in detail.

We want to point out that system \eqref{ms} is not only an interesting and 
important extension of system \eqref{a2}, but also our approach in the 
paper will be different from the one in \cite{ejde2015}, but in the 
spirit of \cite{ejde-fodce}.

Note that the domain of undefinable
solutions (\cite{amc219-dussde}) to system \eqref{ms} is a subset of
$$
{\mathcal U}=\{(z_{-1}, z_0, w_{-1}, w_0)\in\mathbb{C}^4: z_{-1}=0 \text{ or }
 z_0=0\text{ or } w_{-1}=0\text{ or }w_0=0\}.
$$ 
This domain  is equal to ${\mathcal U}$ if $\min\{a,b,c,d\}<0$, but it can be also 
an empty set if $\min\{a,b,c,d\}>0$. To avoid some
quite simple and not so interesting discussions we will also assume that 
all the initial values belong to $\mathbb{C}\setminus\{0\}$. 
Throughout the paper we will frequently use the convention
$\sum_{j=l}^m a_j=0$, for $m<l$.


\section{Main results}

The problem of solvability of system \eqref{ms} will be treated in this section. 
Three cases will be separately studied, namely, $a=0$, $c=0$ and $ac\ne 0$.

\begin{theorem} \label{thm1} 
 Assume that $b,c,d\in\mathbb{Z}$, $a=0$, $\alpha,\beta\in\mathbb{C}\setminus\{0\}$ and
initial values $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$.
Then system \eqref{ms} is solvable in closed form.
\end{theorem}

\begin{proof}
Since $a=0$ we have
\begin{equation}
z_{n+1}=\alpha z_{n-1}^b,\quad
w_{n+1}=\beta z_n^cw_{n-1}^d,\quad n\in\mathbb{N}_0.\label{f1}
\end{equation}
The first equation in \eqref{f1} easily yields
\begin{gather}
 z_{2n}=\alpha^{\sum_{j=0}^{n-1}b^j}z_0^{b^n},\quad n\in\mathbb{N},\label{f2}\\
 z_{2n-1}=\alpha^{\sum_{j=0}^{n-1}b^j}z_{-1}^{b^n},\quad n\in\mathbb{N}.\label{f3}
\end{gather}
From \eqref{f2} and \eqref{f3} we have
\begin{gather}
 z_{2n}=\alpha^{\frac{1-b^n}{1-b}}z_0^{b^n},\quad n\in\mathbb{N},\label{f5}\\
z_{2n-1}=\alpha^{\frac{1-b^n}{1-b}}z_{-1}^{b^n},\quad n\in\mathbb{N},\label{f6}
\end{gather}
when $b\ne 1$, while
\begin{gather}
 z_{2n}=\alpha^nz_0,\quad n\in\mathbb{N},\label{f7}\\
z_{2n-1}=\alpha^nz_{-1},\quad n\in\mathbb{N},\label{f8}
\end{gather}
when $b=1$ (note that \eqref{f2} and \eqref{f3} obviously hold for $n=0$,
when $b\ne 0$).

By using \eqref{f2} in the second equation in \eqref{f1} with $n\to 2n$, we obtain
\begin{equation}
\begin{aligned}
w_{2n+1}
&= \beta z_{2n}^cw_{2n-1}^d=\beta(\alpha^{\sum_{j=0}^{n-1}b^j}z_0^{b^n})^cw_{2n-1}^d \\
&= \beta\alpha^{c\sum_{j=0}^{n-1}b^j}z_0^{cb^n}w_{2n-1}^d,
\end{aligned}\label{f9}
\end{equation}
for $n\in\mathbb{N}$.

Suppose that for some $k\in\mathbb{N}$ we have proved
\begin{equation}
w_{2n+1}
= \beta^{\sum_{i=0}^{k-1}d^i}\alpha^{c\sum_{i=0}^{k-1}d^i\sum_{j=0}^{n-i-1}b^j}
z_0^{c\sum_{i=0}^{k-1}d^ib^{n-i}}w_{2(n-k)+1}^{d^k},\label{f11}
\end{equation}
for $n\ge k$.
By using \eqref{f9} with $n\to n-k$ into \eqref{f11} it follows that
\begin{equation}
\begin{aligned}
w_{2n+1}
&= \beta^{\sum_{i=0}^{k-1}d^i}\alpha^{c\sum_{i=0}^{k-1}d^i
\sum_{j=0}^{n-i-1}b^j}z_0^{c\sum_{i=0}^{k-1}d^ib^{n-i}} \\
&\quad\times (\beta\alpha^{c\sum_{j=0}^{n-k-1}b^j}z_0^{cb^{n-k}}w_{2(n-k-1)+1}^d)^{d^k} \\
&= \beta^{\sum_{i=0}^k d^i}\alpha^{c\sum_{i=0}^kd^i\sum_{j=0}^{n-i-1}b^j}
z_0^{c\sum_{i=0}^kd^ib^{n-i}}w_{2(n-k-1)+1}^{d^{k+1}},
\end{aligned} \label{f12}
\end{equation}
for $n\ge k+1$.

Formulas \eqref{f9}, \eqref{f12} along with the induction shows that 
\eqref{f11} holds for all natural numbers $k$ and $n$ such that $1\le k\le n$.
For $k=n$, \eqref{f11} becomes
\begin{equation}
w_{2n+1}=\beta^{\sum_{i=0}^{n-1} d^i}\alpha^{c\sum_{i=0}^{n-1}d^i
\sum_{j=0}^{n-i-1}b^j}z_0^{c\sum_{i=0}^{n-1}d^ib^{n-i}}w_1^{d^n},\quad
n\in\mathbb{N}.
\label{f14a}
\end{equation}
Using the relation $w_1=\beta z_0^cw_{-1}^d$ into \eqref{f14a}, we obtain
\begin{equation}
\begin{aligned}
w_{2n+1}
&= \beta^{\sum_{i=0}^{n-1} d^i}\alpha^{c\sum_{i=0}^{n-1}d^i
\sum_{j=0}^{n-i-1}b^j}z_0^{c\sum_{i=0}^{n-1}d^ib^{n-i}}(\beta z_0^cw_{-1}^d)^{d^n} \\
&= \beta^{\sum_{i=0}^n d^i}\alpha^{c\sum_{i=0}^{n-1}d^i\sum_{j=0}^{n-i-1}b^j}
z_0^{c\sum_{i=0}^n d^ib^{n-i}}w_{-1}^{d^{n+1}},\quad n\in\mathbb{N}_0.
\end{aligned}\label{f14}
\end{equation}
By using \eqref{f3} into the second equation in \eqref{f1} with $n\to 2n-1$,
we obtain
\begin{equation}
\begin{aligned}
w_{2n}
&= \beta z_{2n-1}^cw_{2n-2}^d=\beta(\alpha^{\sum_{j=0}^{n-1}b^j}z_{-1}^{b^n})^cw_{2n-2}^d \\
&= \beta\alpha^{c\sum_{j=0}^{n-1}b^j}z_{-1}^{cb^n}w_{2n-2}^d,
\end{aligned}\label{f15}
\end{equation}
for $n\in\mathbb{N}$.


Assume that for some $k\in\mathbb{N}$ we have proved that
\begin{equation}
w_{2n}
= \beta^{\sum_{i=0}^{k-1}d^i}\alpha^{c\sum_{i=0}^{k-1}d^i
\sum_{j=0}^{n-i-1}b^j}z_{-1}^{c\sum_{i=0}^{k-1}d^ib^{n-i}}w_{2(n-k)}^{d^k},\label{f17}
\end{equation}
for $n\ge k$.

By using \eqref{f15} with $n\to n-k$ into \eqref{f17} we obtain
\begin{equation}
\begin{aligned}
w_{2n}
&= \beta^{\sum_{i=0}^{k-1}d^i}\alpha^{c\sum_{i=0}^{k-1}d^i
 \sum_{j=0}^{n-i-1}b^j}z_{-1}^{c\sum_{i=0}^{k-1}d^ib^{n-i}}\\
&\quad\times (\beta\alpha^{c\sum_{j=0}^{n-k-1}b^j}z_{-1}^{cb^{n-k}}w_{2(n-k-1)}^d)^{d^k} \\
&= \beta^{\sum_{i=0}^kd^i}\alpha^{c\sum_{i=0}^kd^i\sum_{j=0}^{n-i-1}b^j}
 z_{-1}^{c\sum_{i=0}^kd^ib^{n-i}}w_{2(n-k-1)}^{d^{k+1}},
\end{aligned}\label{f18}
\end{equation}
for $n\ge k+1$.

From \eqref{f15}, \eqref{f18} and the induction we have that \eqref{f17} 
holds for all natural numbers $k$ and $n$ such that $1\le k\le n$.
For $k=n$, \eqref{f17} becomes
\begin{equation}
 w_{2n}=\beta^{\sum_{i=0}^{n-1}d^i}\alpha^{c\sum_{i=0}^{n-1}d^i
\sum_{j=0}^{n-i-1}b^j}z_{-1}^{c\sum_{i=0}^{n-1}d^ib^{n-i}}w_0^{d^n},\label{f19}
\end{equation}
for $n\in\mathbb{N}$.
\smallskip

\noindent\textbf{Case $b\ne d$.} 
From \eqref{f14} and \eqref{f19} we have that
\begin{equation}
 w_{2n+1}=\beta^{\sum_{i=0}^nd^i}\alpha^{c\sum_{i=0}^{n-1}d^i
\sum_{j=0}^{n-i-1}b^j}z_0^{c\frac{d^{n+1}-b^{n+1}}{d-b}}w_{-1}^{d^{n+1}},
\label{f20}
\end{equation}
for $n\in\mathbb{N}_0$, and
\begin{equation}
 w_{2n}=\beta^{\sum_{i=0}^{n-1}d^i}\alpha^{c\sum_{i=0}^{n-1}d^i
\sum_{j=0}^{n-i-1}b^j}z_{-1}^{bc\frac{d^n-b^n}{d-b}}w_0^{d^n},\quad
n\in\mathbb{N}.\label{f21}
\end{equation}
\smallskip

\noindent\textbf{Subcase $b\ne 1\ne d.$} 
From \eqref{f20} we have
\begin{equation}
\begin{aligned}
w_{2n+1}&= \beta^{\frac{1-d^{n+1}}{1-d}}\alpha^{c\sum_{i=0}^{n-1}
 d^i\frac{1-b^{n-i}}{1-b}}z_0^{c\frac{d^{n+1}-b^{n+1}}{d-b}}w_{-1}^{d^{n+1}} \\
&= \beta^{\frac{1-d^{n+1}}{1-d}} \alpha^{\frac{c}{1-b}
 \big(\frac{1-d^n}{1-d}-b\frac{b^n-d^n}{b-d}\big)}
z_0^{c\frac{d^{n+1}-b^{n+1}}{d-b}}w_{-1}^{d^{n+1}} \\
&= \beta^{\frac{1-d^{n+1}}{1-d}} \alpha^{\frac{c(b-d+d^{n+1}
 -b^{n+1}+db^{n+1}-bd^{n+1})}{(1-b)(1-d)(b-d)}}
z_0^{c\frac{d^{n+1}-b^{n+1}}{d-b}}w_{-1}^{d^{n+1}},
\end{aligned} \label{f22}
\end{equation}
for $n\in\mathbb{N}_0$.
From \eqref{f21} and by employing a formula used in getting \eqref{f22}, we have
\begin{equation}
\begin{aligned}
w_{2n}&= \beta^{\sum_{i=0}^{n-1}d^i}\alpha^{c\sum_{i=0}^{n-1}d^i
\frac{1-b^{n-i}}{1-b}}z_{-1}^{bc\frac{d^n-b^n}{d-b}}w_0^{d^n} \\
&= \beta^{\frac{1-d^n}{1-d}}\alpha^{\frac{c(b-d+d^{n+1}-b^{n+1}
+db^{n+1}-bd^{n+1})}{(1-b)(1-d)(b-d)}}z_{-1}^{bc\frac{d^n-b^n}{d-b}}
w_0^{d^n},
\end{aligned} \label{f28}
\end{equation}
for $n\in\mathbb{N}$.
\smallskip

\noindent\textbf{Subcase $b\ne 1=d.$} 
From \eqref{f20} we have
\begin{equation}
\begin{aligned}
w_{2n+1}
&= \beta^{n+1}\alpha^{c\sum_{i=0}^{n-1}\frac{1-b^{n-i}}{1-b}}
 z_0^{c\frac{1-b^{n+1}}{1-b}}w_{-1} \\
&= \beta^{n+1} \alpha^{\frac{c}{1-b}\big(n-b\frac{1-b^n}{1-b}\big)}
z_0^{c\frac{1-b^{n+1}}{1-b}}w_{-1} \\
&= \beta^{n+1} \alpha^{\frac{c(n-(n+1)b+b^{n+1})}{(1-b)^2}}
z_0^{c\frac{1-b^{n+1}}{1-b}}w_{-1},
\end{aligned} \label{f23}
\end{equation}
for $n\in\mathbb{N}_0$.
From \eqref{f21} and by employing a formula used in getting \eqref{f23},
we have
\begin{equation}
\begin{aligned}
w_{2n}&= \beta^n\alpha^{c\sum_{i=0}^{n-1}\frac{1-b^{n-i}}{1-b}}
 z_{-1}^{bc\frac{1-b^n}{1-b}}w_0 \\
&= \beta^n\alpha^{\frac{c(n-(n+1)b+b^{n+1})}{(1-b)^2}}
 z_{-1}^{bc\frac{1-b^n}{1-b}}w_0,
\end{aligned} \label{f29}
\end{equation}
for $n\in\mathbb{N}$.
\smallskip

\noindent\textbf{Subcase $b=1\ne d.$} 
From \eqref{f20} we have
\begin{equation}
\begin{aligned}
w_{2n+1}&= \beta^{\frac{1-d^{n+1}}{1-d}}\alpha^{c\sum_{i=0}^{n-1}
 d^i(n-i)}z_0^{c\frac{d^{n+1}-1}{d-1}}w_{-1}^{d^{n+1}} \\
&= \beta^{\frac{1-d^{n+1}}{1-d}} \alpha^{c\big(n\frac{1-d^n}{1-d}
 -d\frac{1-nd^{n-1}+(n-1)d^n}{(1-d)^2}\big)}
z_0^{c\frac{d^{n+1}-1}{d-1}}w_{-1}^{d^{n+1}} \\
&= \beta^{\frac{1-d^{n+1}}{1-d}} \alpha^{\frac{c(n-(n+1)d+d^{n+1})}{(1-d)^2}}
z_0^{c\frac{d^{n+1}-1}{d-1}}w_{-1}^{d^{n+1}},
\end{aligned}\label{f25}
\end{equation}
for $n\in\mathbb{N}_0$.
From \eqref{f21} and by employing a formula used in getting \eqref{f25}, we have
\begin{equation}
\begin{aligned}
w_{2n}&= \beta^{\frac{1-d^n}{1-d}}\alpha^{c\sum_{i=0}^{n-1}
 d^i(n-i)}z_{-1}^{c\frac{1-d^n}{1-d}}w_0^{d^n} \\
&= \beta^{\frac{1-d^n}{1-d}}\alpha^{\frac{c(n-(n+1)d+d^{n+1})}{(1-d)^2}}
 z_{-1}^{c\frac{d^n-1}{d-1}}w_0^{d^n},
\end{aligned} \label{f30}
\end{equation}
for $n\in\mathbb{N}$.
\smallskip

\noindent\textbf{Case $b=d\ne 1$.} 
From \eqref{f14} we have
\begin{equation}
\begin{aligned} w_{2n+1}
&= \beta^{\frac{1-b^{n+1}}{1-b}}\alpha^{c\sum_{i=0}^{n-1}b^i\frac{1-b^{n-i}}{1-b}}
 z_0^{cb^n(n+1)}w_{-1}^{b^{n+1}} \\
&= \beta^{\frac{1-b^{n+1}}{1-b}}\alpha^{\frac{c}{1-b}\big(\frac{1-b^n}{1-b}-nb^n\big)}
 z_0^{cb^n(n+1)}w_{-1}^{b^{n+1}} \\
&= \beta^{\frac{1-b^{n+1}}{1-b}}\alpha^{\frac{c(1-(n+1)b^n+nb^{n+1})}{(1-b)^2}}
 z_0^{cb^n(n+1)}w_{-1}^{b^{n+1}},
\end{aligned}\label{f26}
\end{equation}
for $n\in\mathbb{N}_0$.

From \eqref{f19} and by employing a formula used in getting \eqref{f26}, we have
\begin{equation}
\begin{aligned} w_{2n}
&= \beta^{\frac{1-b^n}{1-b}}\alpha^{c\sum_{i=0}^{n-1}b^i\frac{1-b^{n-i}}{1-b}}
z_{-1}^{cnb^n}w_0^{b^n} \\
&= \beta^{\frac{1-b^n}{1-b}}\alpha^{\frac{c(1-(n+1)b^n+nb^{n+1})}{(1-b)^2}}
z_{-1}^{cnb^n}w_0^{b^n},
\end{aligned}\label{f31}
\end{equation}
for $n\in\mathbb{N}_0$.
\smallskip

\noindent\textbf{Case $b=d=1$.} 
From \eqref{f14} we have
\begin{equation}
\begin{aligned} w_{2n+1}
&= \beta^{n+1}\alpha^{c\sum_{i=0}^{n-1}(n-i)}z_0^{c(n+1)}w_{-1} \\
&= \beta^{n+1}\alpha^{\frac{cn(n+1)}2}z_0^{c(n+1)}w_{-1},
\end{aligned}\label{f27}
\end{equation}
for $n\in\mathbb{N}_0$.
From \eqref{f19} we have
\begin{equation}
\begin{aligned}
w_{2n}&= \beta^n\alpha^{c\sum_{i=0}^{n-1}(n-i)}z_{-1}^{cn}w_0 \\
&= \beta^n\alpha^{\frac{cn(n+1)}2}z_{-1}^{cn}w_0,
\end{aligned}\label{f32}
\end{equation}
for $n\in\mathbb{N}_0$. This completes the proof.
\end{proof}


\begin{corollary} \label{coro1} 
 Consider system \eqref{ms}. Assume that $b,c,d\in\mathbb{Z}$, $a=0$,
 $\alpha,\beta\in\mathbb{C}\setminus\{0\}$ and
 $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$.
Then the following statements are true.
\begin{itemize}

\item[(a)] If $b\ne 1\ne d\ne b$, then the general solution of system
\eqref{ms} is given by \eqref{f5}, \eqref{f6}, \eqref{f22} and \eqref{f28}.

\item[(b)] If $b\ne 1=d$, then the general solution of system \eqref{ms}
is given by \eqref{f5}, \eqref{f6}, \eqref{f23} and \eqref{f29}.

\item[(c)] If $b=1\ne d$, then the general solution of system \eqref{ms}
is given by \eqref{f7}, \eqref{f8}, \eqref{f25} and \eqref{f30}.

\item[(d)] If $b=d\ne 1$, then the general solution of system \eqref{ms}
is given by \eqref{f5}, \eqref{f6}, \eqref{f26} and \eqref{f31}.

\item[(e)] If $b=d=1$, then the general solution of system \eqref{ms}
is given by \eqref{f7}, \eqref{f8}, \eqref{f27} and \eqref{f32}.

\end{itemize}
\end{corollary}


\begin{theorem} \label{thm2} 
 Assume that $a,b,d\in\mathbb{Z}$, $c=0$, $\alpha,\beta\in\mathbb{C}\setminus\{0\}$ and
$z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$.
Then system \eqref{ms} is solvable in closed form.
\end{theorem}

\begin{proof} This theorem follows from the proof of Theorem \ref{thm1},
 since essentially the same system is obtained in this case. 
Namely, it is enough to change letter $a$ to $c$, letter $b$ to $d$, 
letter $z$ to $w$, and letter $\alpha$ to $\beta$, in the system
$$
z_{n+1}=\alpha w_n^az_{n-1}^b,\quad
w_{n+1}=\beta w_{n-1}^d,\quad n\in\mathbb{N}_0,
$$
and it will become system \eqref{f1}.
\end{proof}

\begin{theorem} \label{thm3} 
 Assume that $a,b,c,d\in\mathbb{Z}$, $a\ne 0\ne c$, $\alpha,\beta\in\mathbb{C}\setminus\{0\}$ and
$z_{-1}$, $z_0$, $w_{-1}$,$ w_0$ $\in\mathbb{C}\setminus\{0\}$.
Then system \eqref{ms} is solvable in closed form.
\end{theorem}

\begin{proof} 
Using the conditions $\alpha,\beta\in\mathbb{C}\setminus\{0\}$ and
$z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$ into the equations 
in \eqref{ms} it is easy to see by using the induction that 
$z_n\ne 0\ne w_n$ for every $n\ge -1$. Hence, from the first equation 
in \eqref{ms} for any such solution we have
\begin{equation}
w_n^a=\frac{z_{n+1}}{\alpha z_{n-1}^b},\quad n\in\mathbb{N}_0.\label{d1}
\end{equation}
By taking the second equation in \eqref{ms} to the $a$-th power is obtained
\begin{equation}
w_{n+1}^a=\beta^a z_n^{ac}w_{n-1}^{ad},\quad n\in\mathbb{N}_0.\label{d2}
\end{equation}
If we use \eqref{d1} into \eqref{d2} we easily get
\begin{equation}
z_{n+2}=\alpha^{1-d}\beta^az_n^{ac+b+d}z_{n-2}^{-bd},\quad n\in\mathbb{N}.\label{d3}
\end{equation}
Let $\gamma:=\alpha^{1-d}\beta^a$,
\begin{equation}
x_1=1,\quad a_1=ac+b+d,\quad b_1=-bd.\label{ic}
\end{equation}
From \eqref{d3} we have that
\begin{equation}
z_{2(n+1)+i}=\gamma^{x_1}z_{2n+i}^{a_1}z_{2(n-1)+i}^{b_1},\quad n\in\mathbb{N},\label{d5}
\end{equation}
for $i=-1,0$.

Using \eqref{d5} with $n\to n-1$ into itself, we obtain
\begin{equation}
\begin{aligned} z_{2(n+1)+i}
&=\gamma^{x_1}(\gamma z_{2(n-1)+i}^{a_1}z_{2(n-2)+i}^{b_1})^{a_1}z_{2(n-1)+i}^{b_1} \\
&=\gamma^{x_1+a_1}z_{2(n-1)+i}^{a_1a_1+b_1}z_{2(n-2)+i}^{b_1a_1} \\
&=\gamma^{x_2}z_{2(n-1)+i}^{a_2}z_{2(n-2)+i}^{b_2},\label{d6}
\end{aligned}
\end{equation}
for $n\ge 2$ and $i=-1,0$, where
\begin{equation}
x_2:=x_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 $k\ge 2$ have been proved the following equality
\begin{equation}
z_{2(n+1)+i}=\gamma^{x_k}z_{2(n-k+1)+i}^{a_k}z_{2(n-k)+i}^{b_k},\label{d8}
\end{equation}
for $n\ge k$ and $i=-1,0$, where
\begin{equation}
x_k:=x_{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}
Then, by using \eqref{d5} with $n\to n-k$ into \eqref{d8}, we have
\begin{equation}
\begin{aligned}
z_{2(n+1)+i}
&=\gamma^{x_k}z_{2(n-k+1)+i}^{a_k}z_{2(n-k)+i}^{b_k} \\
&=\gamma^{x_k}(\gamma z_{2(n-k)+i}^{a_1}z_{2(n-k-1)+i}^{b_1})^{a_k}z_{2(n-k)+i}^{b_k} \\
&=\gamma^{x_k+a_k}z_{2(n-k)+i}^{a_1a_k+b_k}z_{2(n-k-1)+i}^{b_1a_k} \\
&=\gamma^{x_{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 $n\ge k+1$ and $i=-1,0$, where
\begin{equation}
x_{k+1}:=x_k+a_k,\quad a_{k+1}:=a_1a_k+b_k,\quad b_{k+1}:=b_1a_k.\label{d11}
\end{equation}
From \eqref{d6}, \eqref{d7}, \eqref{d10}, \eqref{d11} and the induction we
obtain that \eqref{d8} and \eqref{d9} hold for
all natural numbers $k$ and $n$ such that $2\le k\le n$.
Note that \eqref{d8} also holds for $1\le k\le n$.

For $k=n$, \eqref{d8} becomes
$$
z_{2(n+1)+i}=\gamma^{x_n}z_{2+i}^{a_n}z_{i}^{b_n},
$$
for $n\in\mathbb{N}$ and $i=-1,0$, from which along with
$$
z_1=\alpha w_0^az_{-1}^b,\quad z_2=\alpha w_1^az_0^b=\alpha\beta^az_0^{ac+b}w_{-1}^{ad},
$$
it follows that
\begin{gather}
\begin{aligned} 
z_{2n}&=\gamma^{x_{n-1}}z_2^{a_{n-1}}z_0^{b_{n-1}}\\
&=(\alpha^{1-d}\beta^a)^{x_{n-1}}(\alpha\beta^az_0^{ac+b}w_{-1}^{ad})^{a_{n-1}}z_0^{b_{n-1}} \\
&=\alpha^{(1-d)x_{n-1}+a_{n-1}}\beta^{ax_{n-1}+aa_{n-1}}z_0^{(ac+b)a_{n-1}
 +b_{n-1}}w_{-1}^{ada_{n-1}},
\end{aligned}\label{d12}\\
\begin{aligned}
z_{2n-1}
&=\gamma^{x_{n-1}}z_1^{a_{n-1}}z_{-1}^{b_{n-1}}\\
&=(\alpha^{1-d}\beta^a)^{x_{n-1}}(\alpha w_0^az_{-1}^b)^{a_{n-1}}z_{-1}^{b_{n-1}} \\
&=\alpha^{(1-d)x_{n-1}+a_{n-1}}\beta^{ax_{n-1}}w_0^{aa_{n-1}}z_{-1}^{ba_{n-1}+b_{n-1}},
\end{aligned} \label{d14}
\end{gather}
for $n\ge 2$.
From \eqref{d9} and since $x_1=1$, we have that
\begin{gather}
 a_k=a_1a_{k-1}+b_1a_{k-2},\quad k\ge 3,\label{d15} \\
x_k=1+\sum_{j=1}^{k-1}a_j,\quad k\in\mathbb{N}.\label{d16}
\end{gather}

In what follows we  consider three cases separately, that is,
$b=0$, $d=0$ and $bd\ne 0$.
\smallskip

\noindent\textbf{Case $b=0$.}
 In this case \eqref{d15} is
$$
a_k=a_1a_{k-1}=(ac+d)a_{k-1},\quad k\in\mathbb{N},
$$
from which it follows that 
\begin{equation}
\begin{aligned}
a_k&=a_1(ac+d)^{k-1}=(ac+d)^k,\quad k\in\mathbb{N},
\end{aligned} \label{d17}
\end{equation}
and which along with $b_1=0$ and $b_k=b_1a_{k-1}$, $k\ge 2$, implies that
\begin{equation}
b_k=0,\quad k\in\mathbb{N}.\label{d18}
\end{equation}
From \eqref{d16} and \eqref{d17} we have
$$
x_k=1+\sum_{j=1}^{k-1}(ac+d)^j,\quad k\in\mathbb{N},
$$
from which it follows that
\begin{equation}
x_k=\frac{1-(ac+d)^k}{1-ac-d},\quad k\in\mathbb{N},\label{d19}
\end{equation}
if $ac+d\ne 1$, while
\begin{equation}
x_k=k,\quad k\in\mathbb{N},\label{d20}
\end{equation}
if $ac+d=1$.

From \eqref{d12}, \eqref{d14}, \eqref{d17}, \eqref{d18} and \eqref{d19}, we have that
\begin{align}
z_{2n}
&= \alpha^{\frac{1-d-ac(ac+d)^{n-1}}{1-ac-d}}
 \beta^{a\frac{1-(ac+d)^n}{1-ac-d}} z_0^{ac(ac+d)^{n-1}}w_{-1}^{ad(ac+d)^{n-1}},
 \label{d21}\\
z_{2n-1}
&= \alpha^{\frac{1-d-ac(ac+d)^{n-1}}{1-ac-d}}\beta^{a\frac{1-(ac+d)^{n-1}}{1-ac-d}}
 w_0^{a(ac+d)^{n-1}},\label{d22}
\end{align}
for $n\ge 2$, if $ac+d\ne 1$, while from \eqref{d12}, \eqref{d14},
\eqref{d17}, \eqref{d18} and \eqref{d20}, we have
\begin{align}
 z_{2n} &=\alpha^{(1-d)n+d}\beta^{an}z_0^{ac}w_{-1}^{ad},\label{d23}\\
z_{2n-1} &=\alpha^{(1-d)n+d}\beta^{a(n-1)}w_0^{a},\label{d25}
\end{align}
for $n\ge 2$, if $ac+d=1$.
\smallskip

\noindent\textbf{Case $d=0$.} 
In this case \eqref{d15} is
$$
a_k=a_1a_{k-1}=(ac+b)a_{k-1},\quad k\in\mathbb{N},
$$
from which it follows that 
\begin{equation}
a_k=a_1(ac+b)^{k-1}=(ac+b)^k,\quad k\in\mathbb{N},\label{d27}
\end{equation}
and which along with $b_1=0$ and $b_k=b_1a_{k-1}$, $k\ge 2$, implies that
\eqref{d18} holds.

From \eqref{d16} and \eqref{d27} we have
$$
x_k=1+\sum_{j=1}^{k-1}(ac+b)^j,\quad k\in\mathbb{N},
$$
from which it follows that
\begin{equation}
 x_k=\frac{1-(ac+b)^k}{1-ac-b},\quad k\in\mathbb{N},\label{d29}
\end{equation}
if $ac+b\ne 1$, while
\begin{equation}
x_k=k,\quad k\in\mathbb{N},\label{d30}
\end{equation}
if $ac+b=1$.
From \eqref{d12}, \eqref{d14}, \eqref{d18}, \eqref{d27}  and \eqref{d29},
we have
\begin{gather}
z_{2n} =\alpha^{\frac{1-(ac+b)^n}{1-ac-b}}\beta^{a\frac{1-(ac+b)^n}{1-ac-b}}z_0^{(ac+b)^n},
\label{d31}\\
z_{2n-1}
=\alpha^{\frac{1-(ac+b)^n}{1-ac-b}}\beta^{a\frac{1-(ac+b)^{n-1}}{1-ac-b}}
w_0^{a(ac+b)^{n-1}}z_{-1}^{b(ac+b)^{n-1}},\label{d32}
\end{gather}
for $n\ge 2$, if $ac+b\ne 1$, while from \eqref{d12}, \eqref{d14}, \eqref{d18},
\eqref{d27}  and \eqref{d30}, we have
\begin{align} z_{2n}
&=\alpha^n\beta^{an}z_0,\label{d33}\\
z_{2n-1}
&=\alpha^n\beta^{a(n-1)}w_0^az_{-1}^b,\label{d34}
\end{align}
for $n\in\mathbb{N}$, if $ac+b=1$.

\smallskip

\noindent\textbf{Case $b\ne 0\ne d$.} 
Let $\lambda_{1,2}$ be the roots of the characteristic polynomial 
\begin{equation}
P(\lambda)=\lambda^2-(ac+b+d)\lambda+bd,\label{q28}
\end{equation}
associate to difference equation \eqref{d15}.

Recall that then the general solution to equation \eqref{d15} is
$$
a_n=c_1\lambda_1^n+c_2\lambda_2^n,\quad n\in\mathbb{N},
$$
if $(ac+b+d)^2\ne 4bd$, where $c_1$ and $c_2$ are arbitrary
constants, while in the case $(ac+b+d)^2=4bd$, it has the following form
$$
u_n=(d_1n+d_2)\lambda_1^n,\quad n\in\mathbb{N},
$$
where $d_1$ and $d_2$ are arbitrary constants.

By some calculation and using the values for $a_1$ and $a_2$, 
if $(ac+b+d)^2\ne 4bd$, we obtain
\begin{equation}
a_k=\frac{\lambda_1^{k+1}-\lambda_2^{k+1}}{\lambda_1-\lambda_2},\quad k\in\mathbb{N},\label{d35}
\end{equation}
while if $(ac+b+d)^2=4bd$, we obtain
\begin{equation}
a_k=(k+1)\lambda_1^k\label{d36},\quad k\in\mathbb{N}.
\end{equation}
By using \eqref{d35} into the third equation in \eqref{d9} we obtain
\begin{equation}
b_k=-bda_{k-1}=-bd\frac{\lambda_1^k-\lambda_2^k}{\lambda_1-\lambda_2},\quad k\ge 2,\label{d37}
\end{equation}
if $(ac+b+d)^2\ne 4bd$, while if $(ac+b+d)^2=4bd$, by using  \eqref{d36}
into the third equation in \eqref{d9} we obtain
\begin{equation}
b_k=-bda_{k-1}=-bdk\lambda_1^{k-1},\quad k\ge 2.\label{d38}
\end{equation}
On the other hand, by using \eqref{d35} into \eqref{d16} we obtain
\begin{equation}
x_k=1+\sum_{j=1}^{k-1}\frac{\lambda_1^{j+1}-\lambda_2^{j+1}}{\lambda_1-\lambda_2}
=\frac{(\lambda_2-1)\lambda_1^{k+1}-(\lambda_1-1)\lambda_2^{k+1}
+\lambda_1-\lambda_2}{(\lambda_1-1)(\lambda_2-1)(\lambda_1-\lambda_2)},\label{d39}
\end{equation}
for $k\in\mathbb{N}$, if $(ac+b+d)^2\ne 4bd$, while if $(ac+b+d)^2=4bd$,
by using \eqref{d36} into \eqref{d16} we obtain
\begin{equation}
x_k=1+\sum_{j=1}^{k-1}(j+1)\lambda_1^j=\frac{1-(k+1)\lambda_1^k+k\lambda_1^{k+1}}{(1-\lambda_1)^2},
\quad k\in\mathbb{N}.\label{d40}
\end{equation}
From \eqref{d12}, \eqref{d14}, \eqref{d35}, \eqref{d37} and \eqref{d39},
we obtain formulas for $z_n$ in the case $(ac+b+d)^2\ne 4bd$, while from
\eqref{d12}, \eqref{d14}, \eqref{d36}, \eqref{d38} and \eqref{d40},
we obtain formulas for $z_n$ in the case $(ac+b+d)^2=4bd$.

From the second equation in \eqref{ms} we have 
\begin{equation}
z_n^c=\frac{w_{n+1}}{\beta w_{n-1}^d},\quad n\in\mathbb{N}_0.\label{e1}
\end{equation}
By taking the first equation in \eqref{ms} to the $c$-th power is obtained
\begin{equation}
z_{n+1}^c=\alpha^c w_n^{ac}z_{n-1}^{bc},\quad n\in\mathbb{N}_0.\label{e2}
\end{equation}
Using \eqref{e1} into \eqref{e2} we easily obtain
\begin{equation}
 w_{n+2}=\alpha^c\beta^{1-b}w_n^{ac+b+d}w_{n-2}^{-bd},\quad n\in\mathbb{N},\label{e3}
\end{equation}
which differs from \eqref{d3} only for the coefficient $\alpha^c\beta^{1-b}$.

Let $\delta:=\alpha^c\beta^{1-b}$,
\begin{equation}
 y_1=1,\quad a_1=ac+b+d,\quad b_1=-bd.\label{ic1}
\end{equation}
By the above described procedure for $z_{2n+i}$, $n\in\mathbb{N}_0$, $i=-1,0$,
it can be shown that  for any
$k\in\mathbb{N}$ such that $1\le k\le n$, hold
\begin{equation}
w_{2(n+1)+i}=\delta^{y_k}w_{2(n-k+1)+i}^{a_k}w_{2(n-k)+i}^{b_k},\label{e8}
\end{equation}
for $n\ge k$ and $i=-1,0$, where
\begin{equation}
y_k:=y_{k-1}+a_{k-1},\quad a_k:=a_1a_{k-1}+b_{k-1},\quad
b_k=b_1a_{k-1}.\label{e9}
\end{equation}
For $k=n$, \eqref{e8} becomes
$$
w_{2(n+1)+i}=\delta^{y_n}w_{2+i}^{a_n}w_{i}^{b_n},
$$
for $n\in\mathbb{N}$ and $i=-1,0$, from which along with
$$
w_1=\beta z_0^cw_{-1}^d,\quad w_2=\beta z_1^cw_0^d=\alpha^c\beta w_0^{ac+d}z_{-1}^{bc},
$$
it follows that
\begin{gather}
\begin{aligned}
w_{2n}
&=\delta^{y_{n-1}}w_2^{a_{n-1}}w_0^{b_{n-1}}=(\alpha^c\beta^{1-b})^{y_{n-1}}
(\alpha^c\beta w_0^{ac+d}z_{-1}^{bc})^{a_{n-1}}w_0^{b_{n-1}} \\
&=\alpha^{cy_{n-1}+ca_{n-1}}\beta^{(1-b)y_{n-1}+a_{n-1}}z_{-1}^{bca_{n-1}}
 w_0^{(ac+d)a_{n-1}+b_{n-1}},
\end{aligned}\label{e12}\\
\begin{aligned}
w_{2n-1}&=\delta^{y_{n-1}}w_1^{a_{n-1}}w_{-1}^{b_{n-1}}=(\alpha^c\beta^{1-b})^{y_{n-1}}
(\beta z_0^cw_{-1}^d)^{a_{n-1}}w_{-1}^{b_{n-1}} \\
&=\alpha^{cy_{n-1}}\beta^{(1-b)y_{n-1}+a_{n-1}}w_{-1}^{da_{n-1}+b_{n-1}}z_0^{ca_{n-1}},
\end{aligned} \label{e14}
\end{gather}
for $n\ge 2$.
From \eqref{e9} and the fact that $y_1=1$, we see that the sequence
$(a_k)_{k\in\mathbb{N}}$ satisfies recurrent relation \eqref{d15}, while
\begin{equation} y_k=1+\sum_{j=1}^{k-1}a_j,\quad k\in\mathbb{N},\label{e16}\end{equation}
which means that
\begin{equation}
y_k=x_k,\quad k\in\mathbb{N}.\label{a3}
\end{equation}

Now we consider three cases separately, that is,
$b=0$, $d=0$ and $bd\ne 0$.
\smallskip

\noindent\textbf{Case $b=0$.} From the above consideration it is clear 
that for $a_k$ holds formula \eqref{d17}, for $b_k$ formula \eqref{d18}, while
for $y_k$, we have
\begin{equation}
y_k=\frac{1-(ac+d)^k}{1-ac-d},\quad k\in\mathbb{N},\label{e19}
\end{equation}
if $ac+d\ne 1$, while
\begin{equation}
y_k=k,\quad k\in\mathbb{N},\label{e20}
\end{equation}
if $ac+d=1$.

From \eqref{d17}, \eqref{d18}, \eqref{e12}, \eqref{e14} and \eqref{e19}, 
we have that
\begin{align}
 w_{2n}
&=\alpha^{c\frac{1-(ac+d)^n}{1-ac-d}}\beta^{\frac{1-(ac+d)^n}{1-ac-d}}
 w_0^{(ac+d)^n},\label{e21}\\
w_{2n-1}
&=\alpha^{c\frac{1-(ac+d)^{n-1}}{1-ac-d}}
\beta^{\frac{1-(ac+d)^n}{1-ac-d}}w_{-1}^{d(ac+d)^{n-1}}z_0^{c(ac+d)^{n-1}},\label{e22}
\end{align}
for $n\ge 2$, if $ac+d\ne 1$, while from \eqref{d17}, \eqref{d18}, 
 \eqref{e12}, \eqref{e14} and \eqref{e20}, we have that
\begin{align} w_{2n}
&=\alpha^{cn}\beta^nw_0,\label{e23}\\
w_{2n-1}
&=\alpha^{c(n-1)}\beta^n w_{-1}^dz_0^c,\label{e25}
\end{align}
for $n\in\mathbb{N}$, if $ac+d=1$.
\smallskip

\noindent\textbf{Case $d=0$.} 
From the above consideration it is clear that for $a_k$ holds formula 
\eqref{d27}, for $b_k$ formula \eqref{d18}, while
for $y_k$, we have
\begin{equation}
 y_k=\frac{1-(ac+b)^k}{1-ac-b},\quad k\in\mathbb{N},\label{e29}
\end{equation}
if $ac+b\ne 1$, while
\begin{equation}
y_k=k,\quad k\in\mathbb{N},\label{e30}
\end{equation}
if $ac+b=1$.

From \eqref{d18}, \eqref{d27}, \eqref{e12}, \eqref{e14} and \eqref{e29},
 we have that
\begin{align} 
w_{2n}
&=\alpha^{c\frac{1-(ac+b)^n}{1-ac-b}}\beta^{\frac{1-b-ac(ac+b)^{n-1}}{1-ac-b}}
z_{-1}^{bc(ac+b)^{n-1}}w_0^{ac(ac+b)^{n-1}},\label{e31}\\
w_{2n-1}
&=\alpha^{c\frac{1-(ac+b)^{n-1}}{1-ac-b}}\beta^{\frac{1-b-ac(ac+b)^{n-1}}
{1-ac-b}}z_0^{c(ac+b)^{n-1}},\label{e32}
\end{align}
for $n\ge 2$, if $ac+b\ne 1$, while from \eqref{d18}, \eqref{d27},
 \eqref{e12}, \eqref{e14} and \eqref{e30}, we have that
\begin{align} w_{2n}
&=\alpha^{cn}\beta^{(1-b)n+b}z_{-1}^{bc}w_0^{ac},\label{e33}\\
w_{2n-1}
&=\alpha^{c(n-1)}\beta^{(1-b)n+b}z_0^c,\label{e34}
\end{align}
for $n\in\mathbb{N}$, if $ac+b=1$.
\smallskip

\noindent\textbf{Case $b\ne 0\ne d$.} 
Let $\lambda_{1,2}$ be the roots of
characteristic polynomial \eqref{q28} associated to difference equation
\eqref{d15}. From the above consideration we see that formulas
\eqref{d35}-\eqref{d38} hold and that
\begin{equation}
y_k=1+\sum_{j=1}^{k-1}\frac{\lambda_1^{j+1}-\lambda_2^{j+1}}{\lambda_1-\lambda_2}
=\frac{(\lambda_2-1)\lambda_1^{k+1}-(\lambda_1-1)\lambda_2^{k+1}
 +\lambda_1-\lambda_2}{(\lambda_1-1)(\lambda_2-1)(\lambda_1-\lambda_2)},\label{e39}
\end{equation}
for $k\in\mathbb{N}$, if $(ac+b+d)^2\ne 4bd$, while if $(ac+b+d)^2=4bd$,
then we have that the following formula holds
\begin{equation}
 y_k=1+\sum_{j=1}^{k-1}(j+1)\lambda_1^j=\frac{1-(k+1)
\lambda_1^k+k\lambda_1^{k+1}}{(1-\lambda_1)^2},\quad k\in\mathbb{N}.\label{d40a}
\end{equation}
Using formulas \eqref{d35}-\eqref{d38}, \eqref{e39} and \eqref{d40a}
into \eqref{e12} and \eqref{e14} are obtained closed form formulas
for sequence $w_n$ in this case, finishing the proof.
\end{proof}

From the proof of Theorem \ref{thm3} we obtain the following corollary.

\begin{corollary} \label{coro2}
 Consider system \eqref{ms} with $a,b,c,d\in\mathbb{Z}$, $ac\ne 0$. 
Assume that $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$. 
Then the following statements are true.
\begin{itemize}

\item[(a)] If $b=0$, $ac+d\ne 1$, then the general solution of system \eqref{ms} is
given by \eqref{d21}, \eqref{d22}, \eqref{e21} and \eqref{e22}.

\item[(b)] If $b=0$, $ac+d=1$, then the general solution of system \eqref{ms} is
given by \eqref{d23}, \eqref{d25}, \eqref{e23} and \eqref{e25}.

\item[(c)] If $d=0$, $ac+b\ne 1$, then the general solution of system \eqref{ms} is
given by \eqref{d31}, \eqref{d32}, \eqref{e31} and \eqref{e32}.

\item[(d)] If $d=0$, $ac+b=1$, then the general solution of system \eqref{ms} is
given by \eqref{d33}, \eqref{d34}, \eqref{e33} and \eqref{e34}.

\item[(e)] If $bd\ne 0$ and $(ac+b+d)^2\ne 4bd$, then the general
solution of system \eqref{ms} is given by
\begin{gather} 
z_{2n}=\alpha^{x_n-dx_{n-1}}\beta^{ax_n}z_0^{(ac+b)a_{n-1}-bda_{n-2}}w_{-1}^{ada_{n-1}},
 \label{a5}\\
z_{2n-1}=\alpha^{x_n-dx_{n-1}}\beta^{ax_{n-1}}w_0^{aa_{n-1}}z_{-1}^{ba_{n-1}-bda_{n-2}},
 \label{a6}\\
w_{2n}=\alpha^{cx_n}\beta^{x_n-bx_{n-1}}z_{-1}^{bca_{n-1}}w_0^{(ac+d)a_{n-1}-bda_{n-2}},
 \label{a7}\\
w_{2n-1}=\alpha^{cx_{n-1}}\beta^{x_n-bx_{n-1}}w_{-1}^{da_{n-1}-bda_{n-2}}z_0^{ca_{n-1}},
 \label{a8}
\end{gather}
where $(a_n)_{n\in\mathbb{N}}$ and $(x_n)_{n\in\mathbb{N}}$ are given by
\eqref{d35} and \eqref{d39} respectively.

\item[(f)] If $bd\ne 0$ and $(ac+b+d)^2=4bd$, then the general solution
of system \eqref{ms} is given by formulas \eqref{a5}-\eqref{a8},
 where $(a_n)_{n\in\mathbb{N}}$ and $(x_n)_{n\in\mathbb{N}}$ are given by \eqref{d36} 
and \eqref{d40} respectively.
\end{itemize}
\end{corollary}

\begin{proof}  Statements (a)--(d) follow directly from the proof 
of Theorem \ref{thm3}.

 (e), (f): By using \eqref{d9} and \eqref{a3} in \eqref{d12}, \eqref{d14}, 
\eqref{e12} and \eqref{e14}, and some calculation formulas 
\eqref{a5}-\eqref{a8} follow.
\end{proof}


\begin{remark} \label{rmk1}\rm
 A relatively long and tedious calculation shows that formulas 
\eqref{a5}-\eqref{a8} really present general solution to system 
\eqref{ms} in the ``main" case, that is, $abcd\ne 0$. 
The authors have verified this, but since such calculations 
are traditionally not quite suitable for publication we omit the 
calculation and left it to the reader as an exercise.
\end{remark}

\begin{remark} \label{rmk2}  \rm
Bearing in mind that sequence $a_n$ is defined for $n\in\mathbb{N}$, one may 
think that formulas \eqref{a5}-\eqref{a8} hold only for $n\ge 3$. 
However, since $b_1=bd\ne 0$, by using recurrent relation \eqref{d15}
 we see that sequence $a_n$ can be prolonged for $n\in \mathbb{Z}$. Indeed, 
for $k=2$ equation \eqref{d15} becomes $a_2=a_1a_1+b_1a_0$, from which 
it follows that $a_0=(a_2-a_1a_1)/b_1=1$. In general, if $a_{n-1}$ and $a_n$ 
are defined for some $n\in\mathbb{Z}$, then $a_{n-2}$ can be calculated/defined by 
using the following consequence of \eqref{d15}
\begin{equation}
a_{n-2}:=\frac1{b_1}(a_n-a_1a_{n-1}).\label{a9}
\end{equation}
By using \eqref{a9} for $n=1$ is obtained $a_{-1}=0$, from which along
with \eqref{a9} for $n=0$ is obtained $a_{-2}=-1/(bd)$. Consequently,
$x_n$ can be calculated/defined also for every $n\in\mathbb{Z}$, by using the
relation $x_{n-1}=x_n-a_{n-1}$. For $n=1$ is obtained $x_0=0$.
Using this ``prolongation" it is easy to verify that \eqref{a5}-\eqref{a8}
 hold for every $n\in\mathbb{N}$.
\end{remark}

\subsection*{Acknowledgements} 
 Stevo Stevi\'c was supported  by the Serbian Ministry of Education 
and Science projects III 41025 and III 44006.
 Bratislav Iri\v{c}anin was supported by
the Serbian Ministry of Education and Science projects III 41025
and OI 171007. The work of Zden\v ek \v Smarda was realized in
CEITEC - Central European Institute of Technology supported by 
the Czech Science Foundation under the project 16-08549S.


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\end{document}