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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 169, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/169\hfil
 Solvable product-type system of difference equations]
{Solvable product-type system of difference equations of second order}

\author[S. Stevi\'c, M. A. Alghamdi, A. Alotaibi, E. M. Elsayed
 \hfil EJDE-2015/169\hfilneg]
{Stevo Stevi\'c, Mohammed A. Alghamdi, \\ Abdullah Alotaibi, Elsayed M. Elsayed}

\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{Mohammed A. Alghamdi \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{proff-malghamdi@hotmail.com}

\address{Abdullah Alotaibi \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{aalotaibi@kau.edu.sa}

\address{Elsayed M. Elsayed \newline
Department of Mathematics, Faculty of Science,
King Abdulaziz University,  P.O. Box 80203,
Jeddah 21589, Saudi Arabia. \newline
Mathematics Department, Faculty of Science,
Mansoura University, Mansoura 35516, Egypt}
\email{emmelsayed@yahoo.com}

\thanks{Submitted February 7, 2015. Published June 18, 2015.}
\subjclass[2010]{39A10, 39A20}
\keywords{Solvable system of difference equations; second-order system;
\hfill\break\indent  product-type system; long-term behavior}

\begin{abstract}
 We show that the  system 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 $a,b,c,d\in\mathbb{Z}$, and initial values 
 $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}$,
 is solvable in closed form, and present a method for finding its solutions.
\end{abstract}

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

\section{Introduction}

Difference equations and systems not closely related to
differential equations is a topic of considerable recent interest
(see, e.g., \cite{al0}-\cite{amc218-sde}, \cite{c},
\cite{gl}-\cite{pst2}, \cite{sps}-\cite{tyt2}). Since the
appearance of paper \cite{amen}, in which was explained the
formula for solutions to the difference equation in \cite{c}, the
area of solving difference equations and systems of difference
equations reattracted some attention (see, e.g.,
\cite{al0}-\cite{bl}, \cite{amc218-sde}, \cite{c}, \cite{pst2},
\cite{37264}, \cite{amc218-sys1}, \cite{amc218-dcept},
\cite{amc219-srsy}, \cite{amc-thos},
\cite{amc-218solsys}-\cite{ejqtde1}, \cite{157943}-\cite{tyt2} and
the related references therein).

On the other hand, symmetric systems of difference equations and
systems of a similar appearance, whose investigation began by
Papaschinopoulos,  Schinas and their collaborators during
the mid of 1990's, is another area which has attracted some recent
attention (see, e.g., \cite{amc218-sde, ps1, ps2, ps3, ps4, sps,
ps-cana, ijpam1, amc218-sys1, amc219-srsy, amc-thos,
amc-218solsys, amc-mtsde, amc219-dussde, amc219-sdcescf,
amc219-sdeoos, amc219-ssktho, ejqtde-fib, ejqtde1, ejqtde-maxsde,
508523, jdea205, tyt2} and the related references therein).

The publication of \cite{tjm2} and \cite{jamc1} initiated a
considerable investigation of the boundedness character of some
classes of difference equations and systems containing non-integer
powers of their variables (see, e.g., \cite{bs2, jdea129, pss,
na4, jmaa376, ejqtde-maxsde} and the related references therein).
An interesting fact is that these equations and systems are
perturbations of some product-type equations and systems of
difference equations, usually obtained by using the translation
operator
\begin{equation}
\tau_a(s)=a+s,\quad a\in\mathbb{R},\label{to}
\end{equation}
or the following operator with maximum
\begin{equation}
m_a(s)=\max\{a,s\},\quad
a\in\mathbb{R}.\label{mxo}
\end{equation}
Note that operator \eqref{to} can act on
the space of complex sequences, unlike operator \eqref{mxo}, which
can act only on the space of real sequences. However, practically
there are no results which deal with the equations and systems
generated by operator \eqref{to} on the space of complex
sequences.

Properties of solutions to difference equations and systems
obtained by using operators \eqref{to} and \eqref{mxo} are
frequently closely related to the corresponding product-type ones.
For example, in \cite{ejqtde-maxsde}, it was studied the following
system of difference equations
\begin{equation}
x_{n+1}=\max\bigg\{a,\frac{y_n^p}{x_{n-1}^q}\bigg\},\quad
y_{n+1}=\max\bigg\{a,\frac{x_n^p}{y_{n-1}^q}\bigg\},\quad
n\in\mathbb{N}_0,\label{a1}
\end{equation}
with $\min\{a, p, q\}>0$, where the boundedness character of
their positive solutions was completely
characterized. Note that system \eqref{a1} can be regarded as a
perturbation of the following product-type system of difference
equations
\begin{equation}
x_{n+1}=\frac{y_n^p}{x_{n-1}^q},\quad
y_{n+1}=\frac{x_n^p}{y_{n-1}^q},\quad n\in\mathbb{N}_0.\label{a2}
\end{equation}

If only positive solutions to system \eqref{a2} are considered
then it can be solved in closed form. Generally speaking, a great
majority of papers on difference equations and systems consider
only their positive solutions. One of the reasons is that such
equations and systems can be frequently regarded as models of some
population or biological models (see, e.g., \cite{bc, 37264}). For
some other applications to difference equations, see, for example
\cite{gr, gl, rr1}. Beside this, their investigation is somewhat
simpler than in the general case. Hence, a natural problem, which
seems has been neglected so far, is to study behavior of solutions
to product-type equations and systems whose initial conditions
need not be positive numbers only. This paper is devoted to the
problem and can be regarded as a starting point in the
investigation.

The following second-order system of difference equations, which
is an extension of system \eqref{a2},
\begin{equation}
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,\label{ms}
\end{equation}
where $a,b,c,d\in\mathbb{R}$ and initial values $z_{-1}, z_0, w_{-1},
w_0$ are positive numbers can be solved in closed form. Namely, by
using the method of induction, it can be shown
$$
z_n>0,\quad w_n>0,\quad\text{for } n\ge -1,
$$
which enables us, by taking the logarithm to the both sides of both
equations in \eqref{ms}, to transform it to a linear second-order system of
difference equations with constant coefficients, which is solvable
in closed form. If $z_{-1}, z_0, w_{-1}, w_0$ are complex numbers,
then the method cannot be used, since in the case the sequences
$(z_n)_{n\ge -1}$ and  $(w_n)_{n\ge -1}$ need not be uniquely
defined.

Our aim here is to show that in some cases system \eqref{ms} can
be solved in closed form also when $z_{-1}, z_0, w_{-1}, w_0$ are
complex numbers. By the obtained formulas we will present some
results on the long-term behavior of solutions to system
\eqref{ms}.

A vector sequence $\vec z_n=(z_n^{(1)},\ldots,z_n^{(l)})$,
$n\ge -k$, is called \emph{periodic} (or \emph{eventually periodic}) with
period $p\in\mathbb{N}$ if there is $n_0\ge -k$, such that
$$
z^{(j)}_{n+p}=z^{(j)}_n,\quad\text{for } n\geq n_0,
$$
for every $j\in\{1,\ldots, l\}$.
Period $p$ is prime if there is
no $\hat p\in\mathbb{N}$, $\hat p<p$ which is a period for the vector
sequence. For $p=1$ the sequences are called eventually constant
or trivial (see, e.g., \cite{amc215-854}). The periodicity is also
one of the areas of considerable interest (see, e.g.,
\cite{bfs,lbs, gl, kkn, ph, ijpam1, 37264, 56813, um83-2, amc217-max,
amc-maxpsde}, and the related references therein). If we say that
a solution of system \eqref{ms} is periodic with period $p$ we
will tacitly regard that $p$ need not be its prime period.


\section{Main result}


In this section we prove the main result in this paper, which
presents formulas for solutions to system \eqref{ms}. Before we
formulate and prove it, note that the domain of undefinable
solutions (\cite{amc219-dussde}) to system \eqref{ms} is the set
$$
{\mathcal U}=\{(z_{-1}, z_0, w_{-1}, w_0)\in\mathbb{C}^4: z_{-1}=0\text{ or }
 z_0=0\text{ or } w_{-1}=0\text{ or }w_0=0\}.
 $$
 Hence, such solutions will be excluded from our
considerations.



\begin{theorem} \label{thm1}
 Assume that $a,b,c,d\in\mathbb{Z}$ 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}
 Let
 \begin{equation}
 a_1=a,\quad b_1=b,\quad c_1=c,\quad d_1=d.\label{ic}
 \end{equation}
By using the equations in \eqref{ms} we obtain
\begin{gather}
z_{n+1}=\frac{w_n^{a_1}}{z_{n-1}^{b_1}}
 =\frac{z_{n-1}^{ca_1-b_1}}{w_{n-2}^{da_1}}
  =\frac{z_{n-1}^{a_2}}{w_{n-2}^{b_2}},\label{a3} \\
w_{n+1}=\frac{z_n^{c_1}}{w_{n-1}^{d_1}}
 =\frac{w_{n-1}^{ac_1-d_1}}{z_{n-2}^{bc_1}}
 =\frac{w_{n-1}^{c_2}}{z_{n-2}^{d_2}},\label{a5}
\end{gather}
where we define $a_2, b_2, c_2$ and $d_2$ as follows
$$
a_2:=ca_1-b_1,\quad b_2:=da_1,\quad c_2:=ac_1-d_1,\quad
d_2:=bc_1.
$$

By using \eqref{a3}, \eqref{a5} and the equations in \eqref{ms} we
further obtain
\begin{gather}
z_{n+1}=\frac{z_{n-1}^{a_2}}{w_{n-2}^{b_2}}
=\frac{w_{n-2}^{aa_2-b_2}}{z_{n-3}^{ba_2}}
=\frac{w_{n-2}^{a_3}}{z_{n-3}^{b_3}},\label{a6}\\
w_{n+1}=\frac{w_{n-1}^{c_2}}{z_{n-2}^{d_2}}
=\frac{z_{n-2}^{cc_2-d_2}}{w_{n-3}^{dc_2}}
=\frac{z_{n-2}^{c_3}}{w_{n-3}^{d_3}},\label{a7}
\end{gather}
where we define $a_3, b_3, c_3$ and $d_3$ as follows
$$
a_3:=aa_2-b_2,\quad b_3:=ba_2,\quad c_3:=cc_2-d_2,\quad d_3:=dc_2.
$$
Assume that \begin{equation}
z_{n+1}=\frac{w_{n-2k+2}^{a_{2k-1}}}{z_{n-2k+1}^{b_{2k-1}}},\quad
w_{n+1}=\frac{z_{n-2k+2}^{c_{2k-1}}}{w_{n-2k+1}^{d_{2k-1}}},\label{p1}\end{equation}
where
\begin{gather*}
a_{2k-1}:=aa_{2k-2}-b_{2k-2},\quad b_{2k-1}:=ba_{2k-2},\\
c_{2k-1}:=cc_{2k-2}-d_{2k-2},\quad
d_{2k-1}:=dc_{2k-2},
\end{gather*}
and
\begin{equation}
z_{n+1}=\frac{z_{n-2k+1}^{a_{2k}}}{w_{n-2k}^{b_{2k}}},\quad
w_{n+1}=\frac{w_{n-2k+1}^{c_{2k}}}{z_{n-2k}^{d_{2k}}},\label{c1}
\end{equation}
where
\begin{gather*}
a_{2k}:=ca_{2k-1}-b_{2k-1},\quad b_{2k}:=da_{2k-1}, \\
c_{2k}:=ac_{2k-1}-d_{2k-1},\quad d_{2k}:=bc_{2k-1},
\end{gather*}
for some $k\in\mathbb{N}$ such that $n\ge 2k-2$.

By using \eqref{c1} and the equations in \eqref{ms} we obtain
\begin{gather}
z_{n+1}=\frac{z_{n-2k+1}^{a_{2k}}}{w_{n-2k}^{b_{2k}}}
=\frac{w_{n-2k}^{aa_{2k}-b_{2k}}}{z_{n-2k-1}^{ba_{2k}}}
=\frac{w_{n-2k}^{a_{2k+1}}}{z_{n-2k-1}^{b_{2k+1}}},\label{a9a}\\
w_{n+1}=\frac{w_{n-2k+1}^{c_{2k}}}{z_{n-2k}^{d_{2k}}}
=\frac{z_{n-2k}^{cc_{2k}-d_{2k}}}{w_{n-2k-1}^{dc_{2k}}}
=\frac{z_{n-2k}^{c_{2k+1}}}{w_{n-2k-1}^{d_{2k+1}}},\label{a10a}
\end{gather}
where we define $a_{2k+1}$, $b_{2k+1}$, $c_{2k+1}$ and $d_{2k+1}$ as
follows
$$
a_{2k+1}:=aa_{2k}-b_{2k},\quad b_{2k+1}:=ba_{2k},\quad
c_{2k+1}:=cc_{2k}-d_{2k},\quad d_{2k+1}:=dc_{2k}.
$$

From \eqref{a9a}, \eqref{a10a} and by using the equations in
\eqref{ms} we obtain
\begin{gather}
z_{n+1}=\frac{w_{n-2k}^{a_{2k+1}}}{z_{n-2k-1}^{b_{2k+1}}}
=\frac{z_{n-2k-1}^{ca_{2k+1}-b_{2k+1}}}{w_{n-2k-2}^{da_{2k+1}}}
=\frac{z_{n-2k-1}^{a_{2k+2}}}{w_{n-2k-2}^{b_{2k+2}}},\label{a9}\\
w_{n+1}=\frac{z_{n-2k}^{c_{2k+1}}}{w_{n-2k-1}^{d_{2k+1}}}
=\frac{w_{n-2k-1}^{ac_{2k+1}-d_{2k+1}}}{z_{n-2k-2}^{bc_{2k+1}}}
=\frac{w_{n-2k-1}^{c_{2k+2}}}{z_{n-2k-2}^{d_{2k+2}}},\label{a10}
\end{gather}
where we define $a_{2k+2}, b_{2k+2}, c_{2k+2}$ and $d_{2k+2}$ as
follows
\begin{gather*}
 a_{2k+2}:=ca_{2k+1}-b_{2k+1},\quad b_{2k+2}:=da_{2k+1},\\
 c_{2k+2}:=ac_{2k+1}-d_{2k+1},\quad d_{2k+2}:=bc_{2k+1}.
\end{gather*}

Hence, this inductive argument shows that relations \eqref{p1} and
\eqref{c1} hold for every $k\in\mathbb{N}$ 
and $n \geq 2k-1$,  and that above defined
sequences $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ satisfy the
following recurrent relations
\begin{gather}
a_{2k}=ca_{2k-1}-b_{2k-1},\quad b_{2k}=da_{2k-1},\quad\label{a21}\\
a_{2k+1}=aa_{2k}-b_{2k},\qquad\quad b_{2k+1}=ba_{2k},\label{a22}\\
c_{2k}=ac_{2k-1}-d_{2k-1},\quad d_{2k}=bc_{2k-1},\label{a23}\\
c_{2k+1}=cc_{2k}-d_{2k},\qquad\quad d_{2k+1}=dc_{2k},\label{a25}
\end{gather}
for $k\in\mathbb{N}$.

From \eqref{a9a}-\eqref{a10} we obtain
\begin{gather}
z_{2n+1}=\frac{w_0^{a_{2n+1}}}{z_{-1}^{b_{2n+1}}},\quad
z_{2n+2}=\frac{z_0^{a_{2n+2}}}{w_{-1}^{b_{2n+2}}}\label{z1}\\
w_{2n+1}=\frac{z_0^{c_{2n+1}}}{w_{-1}^{d_{2n+1}}},\quad
w_{2n+2}=\frac{w_0^{c_{2n+2}}}{z_{-1}^{d_{2n+2}}},\label{z2}
\end{gather}
for $n\in\mathbb{N}_0$.

Using \eqref{a21} and \eqref{a22} we have
$$
a_{2k+1}=aa_{2k}-da_{2k-1},\quad
a_{2k+2}=ca_{2k+1}-ba_{2k},\quad k\in\mathbb{N},
$$
from which it follows that
\begin{gather}
 a_{2k+3}-(ac-b-d)a_{2k+1}+bda_{2k-1}=0,\quad k\in\mathbb{N},\label{a11} \\
a_{2k+2}-(ac-b-d)a_{2k}+bda_{2k-2}=0,\quad k\ge 2.\label{a12}
\end{gather}
From \eqref{a23} and \eqref{a25} we have
$$
c_{2k+1}=cc_{2k}-bc_{2k-1},\quad
c_{2k+2}=ac_{2k+1}-dc_{2k},\quad k\in\mathbb{N},
$$ from which it follows that
\begin{gather}
c_{2k+3}-(ac-b-d)c_{2k+1}+bdc_{2k-1}=0,\quad
k\in\mathbb{N},\label{a26}\\
c_{2k+2}-(ac-b-d)c_{2k}+bdc_{2k-2}=0,\quad k\ge 2.\label{a27}
\end{gather}

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

\noindent\textit{Case $b=0$.} In this case equations \eqref{a11}-\eqref{a27}
become
\begin{gather*}
a_{2k+3}=(ac-d)a_{2k+1},\quad a_{2k+2}=(ac-d)a_{2k},\quad k\in\mathbb{N},\\
c_{2k+3}=(ac-d)c_{2k+1},\quad c_{2k+2}=(ac-d)c_{2k},\quad k\in\mathbb{N},
\end{gather*}
from which it follows that
\begin{gather}
a_{2k+1}=a_1(ac-d)^k=a(ac-d)^k,\label{c14}\\
a_{2k+2}=a_2(ac-d)^k=ac(ac-d)^k,\label{a14}\\
c_{2k+1}=c_1(ac-d)^k=c(ac-d)^k,\label{c14a}\\
c_{2k+2}=c_2(ac-d)^k=(ac-d)^{k+1},\label{a14a}
\end{gather}
for $k\in\mathbb{N}_0$.


Using \eqref{c14}-\eqref{a14a} in \eqref{a21}-\eqref{a25}, as
well as the condition $b=0$, it follows that
\begin{gather}
b_{2k+1}=0,\quad
b_{2k+2}=ad(ac-d)^k,\quad k\in\mathbb{N}_0,\label{a14b}\\
d_{2k+1}=d(ac-d)^k,\quad d_{2k+2}=0,\quad
k\in\mathbb{N}_0.\label{a14c}
\end{gather}

Employing \eqref{c14}-\eqref{a14c} in \eqref{z1} and \eqref{z2}
we obtain that well-defined solutions to system \eqref{ms} in this
case are given by the following formulas
\begin{gather}
z_{2n+1}=w_0^{a(ac-d)^n},\quad
z_{2n+2}=\frac{z_0^{ac(ac-d)^n}}{w_{-1}^{ad(ac-d)^n}}\label{z1b}\\
w_{2n+1}=\frac{z_0^{c(ac-d)^n}}{w_{-1}^{d(ac-d)^n}},\quad
w_{2n+2}=w_0^{(ac-d)^{n+1}},\quad n\in\mathbb{N}_0.\label{z2b}
\end{gather}
\smallskip

\noindent\emph{Case $d=0$.} In this case equations \eqref{a11}-\eqref{a27}
become
\begin{gather*}
a_{2k+3}=(ac-b)a_{2k+1},\quad a_{2k+2}=(ac-b)a_{2k},\quad k\in\mathbb{N},\\
c_{2k+3}=(ac-b)c_{2k+1},\quad c_{2k+2}=(ac-b)c_{2k},\quad k\in\mathbb{N},
\end{gather*}
from which it follows that
\begin{gather}
a_{2k+1}=a_1(ac-b)^k=a(ac-b)^k,\label{c15}\\
a_{2k+2}=a_2(ac-b)^k=(ac-b)^{k+1},\label{a15}\\
c_{2k+1}=c_1(ac-b)^k=c(ac-b)^k,\label{c16}\\
c_{2k+2}=c_2(ac-b)^k=ac(ac-b)^k,\label{a15a}
\end{gather}
for $k\in\mathbb{N}_0$.

Using \eqref{c15}-\eqref{a15a} in \eqref{a21}-\eqref{a25}, as
well as the condition $d=0$, it follows that
\begin{gather}
b_{2k+1}=b(ac-b)^k,\quad
b_{2k+2}=0,\quad k\in\mathbb{N}_0,\label{a15b}\\
d_{2k+1}=0,\quad d_{2k+2}=bc(ac-b)^k,\quad
k\in\mathbb{N}_0.\label{a15c}
\end{gather}

Employing \eqref{c15}-\eqref{a15c} in \eqref{z1} and \eqref{z2}
we obtain that well-defined solutions to system \eqref{ms} in this
case are given by the following formulas
\begin{gather}
z_{2n+1}=\frac{w_0^{a(ac-b)^n}}{z_{-1}^{b(ac-b)^n}},\quad
z_{2n+2}=z_0^{(ac-b)^{n+1}}\label{z1d}\\
w_{2n+1}=z_0^{c(ac-b)^n},\quad
w_{2n+2}=\frac{w_0^{ac(ac-b)^n}}{z_{-1}^{bc(ac-b)^n}},\quad
n\in\mathbb{N}_0.\label{z2d}
\end{gather}
\smallskip



\noindent\emph{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}
of the difference equation
\begin{equation}
u_{n+2}-(ac-b-d)u_{n+1}+bdu_n=0,\quad
n\in\mathbb{N}.\label{ge}
\end{equation}
 From \eqref{a11}-\eqref{a27} it is clear
that the sequences $(a_{2k+1})_{k\in\mathbb{N}_0}$, $(a_{2k})_{k\in\mathbb{N}}$,
$(c_{2k+1})_{k\in\mathbb{N}_0}$, $(c_{2k})_{k\in\mathbb{N}}$, are solutions to
equation \eqref{ge}.

It is known that the general solution of \eqref{ge} has the form
$$
u_n=\alpha_1\lambda_1^n+\alpha_2\lambda_2^n,\quad n\in\mathbb{N},
$$
if $(ac-b-d)^2\ne 4bd$, where $\alpha_1$ and $\alpha_2$ are arbitrary
constants, while in the case $(ac-b-d)^2=4bd$, the general
solution has the following form
$$
u_n=(\beta_1n+\beta_2)\lambda_1^n,\quad n\in\mathbb{N},
$$
 where $\beta_1$ and $\beta_2$ are arbitrary constants.

By some calculation and using the values for $a_i, b_i, c_i, d_i$,
for $i\in\{1,2,3,4\}$, if $(ac-b-d)^2\ne 4bd$, we obtain
\begin{gather}
a_{2k+1}=a\frac{\lambda_1^{k+1}-\lambda_2^{k+1}}{\lambda_1-\lambda_2},\label{s1}\\
a_{2k+2}=\frac{(ac-b-\lambda_2)\lambda_1^{k+1}-(ac-b-\lambda_1)\lambda_2^{k+1}}{\lambda_1-\lambda_2},\label{s2}\\
b_{2k+1}=b\frac{(ac-b-\lambda_2)\lambda_1^k-(ac-b-\lambda_1)\lambda_2^k}{\lambda_1-\lambda_2},\label{s4}\\
b_{2k+2}=ad\frac{\lambda_1^{k+1}-\lambda_2^{k+1}}{\lambda_1-\lambda_2},\label{s3}\\
c_{2k+1}=c\frac{\lambda_1^{k+1}-\lambda_2^{k+1}}{\lambda_1-\lambda_2},\label{s5}\\
c_{2k+2}=\frac{(ac-d-\lambda_2)\lambda_1^{k+1}-(ac-d-\lambda_1)\lambda_2^{k+1}}{\lambda_1-\lambda_2},\label{s6}\\
d_{2k+1}=d\frac{(ac-d-\lambda_2)\lambda_1^k-(ac-d-\lambda_1)\lambda_2^k}{\lambda_1-\lambda_2},\label{s8}\\
d_{2k+2}=bc\frac{\lambda_1^{k+1}-\lambda_2^{k+1}}{\lambda_1-\lambda_2}\label{s7},
\end{gather}
for $k\in\mathbb{N}_0$.

By using \eqref{s1}-\eqref{s7} into \eqref{z1} and \eqref{z2} we
obtain that well-defined solutions to system \eqref{ms} in this
case are given by the following formulas
\begin{gather}
z_{2n+1}=w_0^{a\frac{\lambda_1^{n+1}-\lambda_2^{n+1}}{\lambda_1-\lambda_2}}
 z_{-1}^{-b\frac{(ac-b-\lambda_2)\lambda_1^n-(ac-b-\lambda_1)\lambda_2^n}{\lambda_1-\lambda_2}},\label{b1}\\
z_{2n+2}=z_0^{\frac{(ac-b-\lambda_2)\lambda_1^{n+1}-(ac-b-\lambda_1)\lambda_2^{n+1}}{\lambda_1-\lambda_2}}
 w_{-1}^{-ad\frac{\lambda_1^{n+1}-\lambda_2^{n+1}}{\lambda_1-\lambda_2}},\label{b2}\\
w_{2n+1}=z_0^{c\frac{\lambda_1^{n+1}-\lambda_2^{n+1}}{\lambda_1-\lambda_2}}w_{-1}
 ^{-d\frac{(ac-d-\lambda_2)\lambda_1^n-(ac-d-\lambda_1)\lambda_2^n}{\lambda_1-\lambda_2}},\label{b3}\\
w_{2n+2}=w_0^{\frac{(ac-d-\lambda_2)\lambda_1^{n+1}-(ac-d-\lambda_1)\lambda_2^{n+1}}{\lambda_1-\lambda_2}}
 z_{-1}^{-bc\frac{\lambda_1^{n+1}-\lambda_2^{n+1}}{\lambda_1-\lambda_2}},\label{b4}
\end{gather}
for $n\in\mathbb{N}_0$.

If $(ac-b-d)^2=4bd$, that is, if
$$
\lambda_1=\lambda_2=\frac{ac-b-d}2,
$$
we have
\begin{gather}
a_{2k+1}=a(k+1)\lambda_1^k\label{s1a}\\
a_{2k+2}=\big((ac-b-\lambda_1)(k+1)+\lambda_1\big)\lambda_1^k,\label{s2a}\\
b_{2k+1}=b\big((ac-b-\lambda_1)k+\lambda_1\big)\lambda_1^{k-1},\label{s4a}\\
b_{2k+2}=ad(k+1)\lambda_1^k,\label{s3a}\\
c_{2k+1}=c(k+1)\lambda_1^k,\label{s5a}\\
c_{2k+2}=\big((ac-d-\lambda_1)(k+1)+\lambda_1\big)\lambda_1^k,\label{s6a}\\
d_{2k+1}=d\big((ac-d-\lambda_1)k+\lambda_1\big)\lambda_1^{k-1},\label{s8a}\\
d_{2k+2}=bc(k+1)\lambda_1^k,\label{s7a}
\end{gather}
for $k\in\mathbb{N}_0$.

By using \eqref{s1a}-\eqref{s7a} into \eqref{z1} and \eqref{z2} we
obtain that well-defined solutions to system \eqref{ms} in this
case are given by the following formulas
\begin{gather}
z_{2n+1}=w_0^{a(n+1)\lambda_1^n}{z_{-1}^{-b((ac-b-\lambda_1)n+\lambda_1)\lambda_1^{n-1}}},\label{b5}\\
z_{2n+2}=z_0^{((ac-b-\lambda_1)(n+1)+\lambda_1)\lambda_1^n}w_{-1}^{-ad(n+1)\lambda_1^n},\label{b6}\\
w_{2n+1}=z_0^{c(n+1)\lambda_1^n}w_{-1}^{-d((ac-d-\lambda_1)n+\lambda_1)\lambda_1^{n-1}},\label{b7}\\
w_{2n+2}=w_0^{((ac-d-\lambda_1)(n+1)+\lambda_1)\lambda_1^n}z_{-1}^{-bc(n+1)\lambda_1^n},\quad
n\in\mathbb{N}_0,\label{b8}
\end{gather}
finishing the proof of the theorem.
\end{proof}

From the proof of Theorem \ref{thm1} we obtain the following corollary.

\begin{corollary} \label{coro1}
Consider  \eqref{ms} with $a,b,c,d\in\mathbb{Z}$. Assume that
$z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$. Then the following
statements hold:
\begin{itemize}

\item[(a)] If $b=0$, then the general solution of system \eqref{ms} is
given by \eqref{z1b} and \eqref{z2b}.

\item[(b)] If $d=0$, then the general solution of system \eqref{ms} is
given by \eqref{z1d} and \eqref{z2d}.

\item[(c)] If $bd\ne 0$ and $(ac-b-d)^2\ne 4bd$, then the general
solution of system \eqref{ms} is given by \eqref{b1}-\eqref{b4}.

\item[(d)] If $bd\ne 0$ and $(ac-b-d)^2=4bd$, then the general solution
of system \eqref{ms} is given by \eqref{b5}-\eqref{b8}.

\end{itemize}
\end{corollary}

\begin{remark} \label{rmk1}\rm
The condition $a,b,c,d\in\mathbb{Z}$ is posed in
order to avoid multi-values of powers of complex numbers, that is,
we want that initial values $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}$
define unique solutions to system \eqref{ms}.
\end{remark}


\section{Applications}

In this section we give some applications of the formulas obtained
in the previous section. The long-term behavior of solutions to
system \eqref{ms} in several cases is described.

\begin{theorem} \label{thm2}
 Consider system \eqref{ms}. Assume that $a,c,d\in\mathbb{Z}$, $b=0$,
and initial values $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$.
Then the following statements hold:
\begin{itemize}

\item[(a)] If $ac=d$, then the solution $(z_n,w_n)_{n\ge -1}$ is
eventually constant.

\item[(b)] If $ac-d=1$, then the solution $(z_n,w_n)_{n\ge -1}$ is
two-periodic, while if $ac-d=-1$, then the solution
$(z_n,w_n)_{n\ge -1}$ is four-periodic.

\item[(c)] If $ac-d\ge 2$ and $0<|w_0^a|<1$, then $z_{2n+1}\to 0$, as
$n\to\infty$.

\item[(d)] If $ac-d\ge 2$ and $|w_0^a|>1$, then $z_{2n+1}\to \infty$,
as $n\to\infty$.

\item[(e)] If $w_0^a=1$, then $z_{2n+1}=1$, $n\in\mathbb{N}_0$.

\item[(f)] If $ac\ne d$, $a\ne0$, and $w_0=e^{i\theta}$,
$\theta=p\pi/q$, $q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $z_{2n+1}$ is
periodic with period $T\le 2q$.

\item[(g)] If $ac-d\le -2$ and $0<|w_0^a|<1$, then $z_{4n+1}\to 0$, as
$n\to\infty$, while if $|w_0^a|>1$, then $z_{4n+1}\to \infty$, as
$n\to\infty$.

\item[(h)] If $ac-d\le -2$ and $0<|w_0^a|<1$, then $z_{4n+3}\to
\infty$, as $n\to\infty$, while if $|w_0^a|>1$, then $z_{4n+3}\to
0$, as $n\to\infty$.


\item[(i)] If $ac-d\ge 2$ and $0<|z_0^{ac}/w_{-1}^{ad}|<1$, then
$z_{2n+2}\to 0$, as $n\to\infty$.

\item[(j)] If $ac-d\ge 2$ and $|z_0^{ac}/w_{-1}^{ad}|>1$, then
$z_{2n+2}\to \infty$, as $n\to\infty$.

\item[(k)] If $z_0^{ac}=w_{-1}^{ad}$, then $z_{2n+2}=1$, $n\in\mathbb{N}_0$.

\item[(l)] If $ac\ne d$, $a\ne0$, $c\ne0$ or $d\ne 0$, and
$z_0^{ac}=w_{-1}^{ad}e^{i\theta}$, $\theta=p\pi/q$, $q\in\mathbb{N}$ and
$p\in\mathbb{Z}$, then $z_{2n+2}$ is periodic with period $T\le 2q$.


\item[(m)] If $ac-d\le -2$ and $0<|z_0^{ac}/w_{-1}^{ad}|<1$, then
$z_{4n+2}\to 0$, as $n\to\infty$, while if
$|z_0^{ac}/w_{-1}^{ad}|>1$, then $z_{4n+2}\to \infty$, as
$n\to\infty$.

\item[(n)] If $ac-d\le -2$ and $0<|z_0^{ac}/w_{-1}^{ad}|<1$, then
$z_{4n}\to \infty$, as $n\to\infty$, while if
$|z_0^{ac}/w_{-1}^{ad}|>1$, then $z_{4n}\to 0$, as $n\to\infty$.


\item[(o)] If $ac-d\ge 2$ and $0<|z_0^c/w_{-1}^d|<1$, then $w_{2n+1}\to
0$, as $n\to\infty$.

\item[(p)] If $ac-d\ge 2$, and $|z_0^c/w_{-1}^d|>1$, then $w_{2n+1}\to
\infty$, as $n\to\infty$.

\item[(q)] If $z_0^c=w_{-1}^d$, then $w_{2n+1}=1$, $n\in\mathbb{N}_0$.

\item[(r)] If $ac\ne d$, and $z_0^c=w_{-1}^de^{i\theta}$,
$\theta=p\pi/q$, $q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $w_{2n+1}$ is
periodic with period $T\le 2q$.


\item[(s)] If $ac-d\le -2$, $0<|z_0^c/w_{-1}^d|<1$, then $w_{4n+1}\to
0$, as $n\to\infty$, while if $|z_0^c/w_{-1}^d|>1$, then
$w_{4n+1}\to \infty$, as $n\to\infty$.

\item[(t)] If $ac-d\le -2$, and $0<|z_0^c/w_{-1}^d|<1$, then
$w_{4n+3}\to \infty$, as $n\to\infty$, while if
$|z_0^c/w_{-1}^d|>1$, then $w_{4n+3}\to 0$, as $n\to\infty$.



\item[(u)] If $ac-d\ge 2$ and $0<|w_0|<1$, then $w_{2n+2}\to 0$, as
$n\to\infty$.

\item[(v)] If $ac-d\ge 2$, and $|w_0|>1$, then $w_{2n+2}\to \infty$, as
$n\to\infty$.

\item[(w)] If $w_0=1$, then $w_{2n+2}=1$, $n\in\mathbb{N}_0$.

\item[(x)] If $ac\ne d$, and $w_0=e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $w_{2n+2}$ is periodic with period
$T\le 2q$.


\item[(y)] If $ac-d\le -2$, $|w_0|>1$, then $w_{4n+2}\to 0$, as
$n\to\infty$, while if $0<|w_0|<1$, then $w_{4n+2}\to \infty$, as
$n\to\infty$.

\item[(z)] If $ac-d\le -2$, and $0<|w_0|<1$, then $w_{4n+4}\to 0$, as
$n\to\infty$, while if $|w_0|>1$, then $w_{4n+4}\to \infty$, as
$n\to\infty$.

\end{itemize}
\end{theorem}


\begin{proof}  (a) Using the condition $ac=d$ in \eqref{z1b} and
\eqref{z2b} we obtain
$$
z_{2n+1}=z_{2n+2}=w_{2n+1}=w_{2n+2}=1,\quad n\in\mathbb{N},
$$
from which the statement follows.

 (b) Using the condition $ac-d=1$ in \eqref{z1b} and
\eqref{z2b} we obtain
$$
z_{2n+1}=w_0^a,\quad z_{2n+2}=\frac{z_0^{ac}}{w_{-1}^{ad}},\quad
w_{2n+1}=\frac{z_0^c}{w_{-1}^d},\quad w_{2n+2}=w_0,\quad
n\in\mathbb{N}_0,
$$
 which means that $(z_n,w_n)_{n\ge -1}$ is two-periodic.

Using the condition $ac-d=-1$ in \eqref{z1b} and \eqref{z2b} we obtain
$$
z_{2n+1}=w_0^{a(-1)^n},\quad
z_{2n+2}=\frac{z_0^{ac(-1)^n}}{w_{-1}^{ad(-1)^n}}, \quad
w_{2n+1}=\frac{z_0^{c(-1)^n}}{w_{-1}^{d(-1)^n}},\quad
w_{2n+2}=w_0^{(-1)^{n+1}},
$$
for $n\in\mathbb{N}_0$. From this we have
\begin{gather}
z_{4n+1}=w_0^{a},\quad z_{4n+2}=\frac{z_0^{ac}}{w_{-1}^{ad}},\quad
w_{4n+1}=\frac{z_0^{c}}{w_{-1}^{d}},\quad
w_{4n+2}=\frac1{w_0}, \\
z_{4n+3}=\frac1{w_0^a},\quad
z_{4n+4}=\frac{w_{-1}^{ad}}{z_0^{ac}},\quad
w_{4n+3}=\frac{w_{-1}^d}{z_0^c},\quad
w_{4n+4}=w_0,
\end{gather}
for $n\in\mathbb{N}_0$, which means
that $(z_n,w_n)_{n\ge -1}$ is four-periodic.

 (c), (d) If $ac-d\ge 2$, then $(ac-d)^n\to+\infty$ as
$n\to+\infty$. Hence, if $0<|w_0^a|<1$, by using the formula
\begin{equation}
z_{2n+1}=w_0^{a(ac-d)^n},\quad n\in\mathbb{N}_0,\label{q1}
\end{equation}
we obtain that $z_{2n+1}\to 0$, as $n\to\infty$, while if $|w_0^a|>1$, then
we obtain that $z_{2n+1}\to \infty$, as $n\to\infty$.

 (e) The statement directly follows by using the condition
$w_0^a=1$ in \eqref{q1}.

 (f) Using the conditions $w_0=e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, in \eqref{q1} we have \begin{equation}
z_{2n+1}=e^{i\pi \frac{pa(ac-d)^n}q},\quad
n\in\mathbb{N}_0.\label{q3}
\end{equation}
Now note that among the numbers
$$
pa,\quad pa(ac-d),\quad pa(ac-d)^2,\ldots, pa(ac-d)^{2q},
$$
there are two which have the same reminder by dividing by $2q$,
say, $pa(ac-d)^i$ and $pa(ac-d)^j$, $0\le i<j\le 2q$. This means
that there is a $k_0\in\mathbb{N}$ such that
$$
pa(ac-d)^j-pa(ac-d)^i=2k_0q,
$$
from which it follows that
$$
pa(ac-d)^{m+j}-pa(ac-d)^{m+i}=2k_0q(ac-d)^m,
$$
for every $m\in\mathbb{N}_0$. This means that the sequence $pa(ac-d)^n$
(mod $2q$), $n\in\mathbb{N}_0$, is eventually periodic with period
$T=j-i\le 2q$. Using this fact in \eqref{q3} the statement easily
follows.

 (g), (h) Since $ac-d\le -2$ from \eqref{q1} we have
\begin{equation}
z_{4n+1}=w_0^{a(ac-d)^{2n}},\quad
z_{4n+3}=w_0^{-a|ac-d|^{2n+1}},\quad n\in\mathbb{N}_0.\label{q5}
\end{equation}
From \eqref{q5} and the posed conditions these two statements
easily follow.

 (i)-(l) The proofs are the same as those ones of 
(c)-(f), when $w_0^a$ is replaced by $z_0^{ac}/w_{-1}^{ad}$, and
$z_{2n+1}$ is replaced by $z_{2n+2}$. Hence, we omit the detail.

 (m), (n) From \eqref{z1b} we have
 \begin{equation}
z_{4n+2}=\Big(\frac{z_0^{ac}}{w_{-1}^{ad}}\Big)^{(ac-d)^{2n}},\quad
z_{4n+4}=\Big(\frac{z_0^{ac}}{w_{-1}^{ad}}\Big)^{-|ac-d|^{2n+1}},\label{q6}
\end{equation}
for $n\in\mathbb{N}_0$. From \eqref{q6} and the posed conditions these
two statements easily follow.

 (o)-(r) Using the formula
\begin{equation}
w_{2n+1}=\Big(\frac{z_0^c}{w_{-1}^d}\Big)^{(ac-d)^n},\quad
n\in\mathbb{N}_0,\label{q7}
\end{equation}
statements  (o)-(q) easily follow,
while  (r) is proved similarly to  (f).

 (s), (t) From \eqref{q7} and since $ac-d\le -2$, it follows
that \
\begin{equation}
w_{4n+1}=\Big(\frac{z_0^c}{w_{-1}^d}\Big)^{(ac-d)^{2n}},\quad
w_{4n+3}=\Big(\frac{z_0^c}{w_{-1}^d}\Big)^{-|ac-d|^{2n+1}},\quad
n\in\mathbb{N}_0.\label{q17}
\end{equation}
Using the formulas in \eqref{q17} these
two statements easily follow.

 (u)-(x) Using the formula
\begin{equation}
w_{2n+2}=w_0^{(ac-d)^{n+1}},\quad n\in\mathbb{N}_0,\label{q8}
\end{equation}
statements  (u)-(w) easily follow, while  (x) is proved
similar to  (f).

 (y), (z) From \eqref{q8} and since $ac-d\le -2$, it follows
that
\begin{equation}
w_{4n+2}=w_0^{-|ac-d|^{2n+1}},\quad
w_{4n+4}=w_0^{(ac-d)^{2n+2}},\quad n\in\mathbb{N}_0.\label{q18}
\end{equation}
Using the formulas in \eqref{q18} these two statements easily
follow.
\end{proof}


\begin{remark} \label{rmk2} \rm
The long-term behavior of solutions to
system \eqref{ms} in the cases $w_0^a=e^{i\theta}$ or
$z_0^{ac}/w_{-1}^{ad}=e^{i\theta}$ or $z_0^c/w_{-1}^d=e^{i\theta}$
or $w_0=e^{i\theta}$, when $\theta/\pi\not\in \mathbb{Q}$, is more
complex and will be not treated here in detail. We can only
mention here that in some cases the sequence $\theta(ab-d)^n$ (mod
$2\pi$), $n\in\mathbb{N}_0$, can have a set of accumulation points which
is nowhere dense in the interval $[0,2\pi]$, but is some other
cases it can be everywhere dense in the interval.
\end{remark}

\begin{theorem} \label{thm3}
 Consider system \eqref{ms}. Assume that $a,b,c\in\mathbb{Z}$, $d=0$, and initial
values $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$.
Then the following statements hold:
\begin{itemize}

\item[(a)] If $ac=b$, then the solution $(z_n,w_n)_{n\ge -1}$ is
eventually constant.

\item[(b)] If $ac-b=1$, then the solution $(z_n,w_n)_{n\ge -1}$ is
two-periodic, while if $ac-b=-1$, then the solution
$(z_n,w_n)_{n\ge -1}$ is four-periodic.

\item[(c)] If $ac-b\ge 2$ and $0<|w_0^a/z_{-1}^b|<1$, then $z_{2n+1}\to
0$, as $n\to\infty$.

\item[(d)] If $ac-b\ge 2$ and $|w_0^a/z_{-1}^b|>1$, then $z_{2n+1}\to
\infty$, as $n\to\infty$.

\item[(e)] If $w_0^a=z_{-1}^b$, then $z_{2n+1}=1$, $n\in\mathbb{N}_0$.

\item[(f)] If $ac\ne b$ and $w_0^a=z_{-1}^be^{i\theta}$,
$\theta=p\pi/q$, $q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $z_{2n+1}$ is
periodic with period $T\le 2q$.

\item[(g)] If $ac-b\le -2$ and  $0<|w_0^a/z_{-1}^b|<1$, then
$z_{4n+1}\to 0$, as $n\to\infty$, while if $|w_0^a/z_{-1}^b|>1$,
then $z_{4n+1}\to \infty$, as $n\to\infty$.

\item[(h)] If $ac-b\le -2$ and $0<|w_0^a/z_{-1}^b|<1$, then
$z_{4n+3}\to \infty$, as $n\to\infty$, while if
$|w_0^a/z_{-1}^b|>1$, then $z_{4n+3}\to 0$, as $n\to\infty$.

\item[(i)] If $ac-b\ge 2$ and $0<|z_0|<1$, then $z_{2n+2}\to 0$, as
$n\to\infty$.

\item[(j)] If $ac-b\ge 2$ and $|z_0|>1$, then $z_{2n+2}\to \infty$, as
$n\to\infty$.

\item[(k)] If $ac=b$ or $z_0=1$, then $z_{2n+2}=1$, $n\in\mathbb{N}_0$.

\item[(l)] If $ac\ne b$ and $z_0=e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $z_{2n+2}$ is periodic with period
$T\le 2q$.


\item[(m)] If $ac-b\le -2$ and $0<|z_0|<1$, then $z_{4n+2}\to \infty$,
as $n\to\infty$, while if $|z_0|>1$, then $z_{4n+2}\to 0$, as
$n\to\infty$.

\item[(n)] If $ac-b\le -2$ and $0<|z_0|<1$, then $z_{4n}\to 0$, as
$n\to\infty$, while if $|z_0|>1$, then $z_{4n}\to \infty$, as
$n\to\infty$.


\item[(o)] If $ac-b\ge 2$ and $0<|z_0^c|<1$, then $w_{2n+1}\to 0$, as
$n\to\infty$.

\item[(p)] If $ac-b\ge 2$ and $|z_0^c|>1$, then $w_{2n+1}\to \infty$,
as $n\to\infty$.

\item[(q)] If $z_0^c=1$, then $w_{2n+1}=1$, $n\in\mathbb{N}_0$.

\item[(r)] If $ac\ne b$, and $z_0^c=e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $w_{2n+1}$ is periodic with period
$T\le 2q$.


\item[(s)] If $ac-b\le -2$ and $0<|z_0^c|<1$, then $w_{4n+1}\to 0$, as
$n\to\infty$, while if $|z_0^c|>1$, then $w_{4n+1}\to \infty$, as
$n\to\infty$.

\item[(t)] If $ac-b\le -2$ and $0<|z_0^c|<1$, then $w_{4n+3}\to
\infty$, as $n\to\infty$, while if $|z_0^c|>1$, then $w_{4n+3}\to
0$, as $n\to\infty$.



\item[(u)] If $ac-b\ge 2$ and $0<|w_0^{ac}/z_{-1}^{bc}|<1$, then
$w_{2n+2}\to 0$, as $n\to\infty$.

\item[(v)] If $ac-b\ge 2$ and $|w_0^{ac}/z_{-1}^{bc}|>1$, then
$w_{2n+2}\to \infty$, as $n\to\infty$.

\item[(w)] If $w_0^{ac}=z_{-1}^{bc}$, then $w_{2n+2}=1$, $n\in\mathbb{N}_0$.

\item[(x)] If $ac\ne b$ and $w_0^{ac}=z_{-1}^{bc}e^{i\theta}$,
$\theta=p\pi/q$, $q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $w_{2n+2}$ is
periodic with period $T\le 2q$.


\item[(y)] If $ac-b\le -2$ and $0<|w_0^{ac}/z_{-1}^{bc}|<1$, then
$w_{4n+2}\to 0$, as $n\to\infty$, while if
$|w_0^{ac}/z_{-1}^{bc}|>1$, then $w_{4n+2}\to \infty$, as
$n\to\infty$.

\item[(z)] If $ac-b\le -2$ and $0<|w_0^{ac}/z_{-1}^{bc}|<1$, then
$w_{4n+4}\to \infty$, as $n\to\infty$, while if
$|w_0^{ac}/z_{-1}^{bc}|>1$, then $w_{4n+4}\to 0$, as $n\to\infty$.

\end{itemize}
\end{theorem}


\begin{proof}
 (a) Using the condition $ac=b$ in \eqref{z1d} and
\eqref{z2d} we obtain
$$z_{2n+1}=z_{2n+2}=w_{2n+1}=w_{2n+2}=1,\quad
n\in\mathbb{N},$$ from which the statement follows.

 (b) Using the condition $ac-b=1$ in \eqref{z1d} and
\eqref{z2d} we obtain
$$
z_{2n+1}=\frac{w_0^a}{z_{-1}^b},\quad z_{2n+2}=z_0,\quad
w_{2n+1}=z_0^c,\quad w_{2n+2}=\frac{w_0^{ac}}{z_{-1}^{bc}},\quad
n\in\mathbb{N}_0,
$$
which means that $(z_n,w_n)_{n\ge -1}$ is
two-periodic.

Using the condition $ac-b=-1$ in \eqref{z1d} and \eqref{z2d} we obtain
$$
z_{2n+1}=\frac{w_0^{a(-1)^n}}{z_{-1}^{b(-1)^n}},\quad
z_{2n+2}=z_0^{(-1)^{n+1}},\quad
w_{2n+1}=z_0^{c(-1)^n},\quad
w_{2n+2}=\frac{w_0^{ac(-1)^n}}{z_{-1}^{bc(-1)^n}},
$$
for $n\in\mathbb{N}_0$. From this we have
\begin{gather}
z_{4n+1}=\frac{w_0^a}{z_{-1}^b},\quad z_{4n+2}=\frac1{z_0},\quad
w_{4n+1}=z_0^c,\quad
w_{4n+2}=\frac{w_0^{ac}}{z_{-1}^{bc}} \\
z_{4n+3}=\frac{z_{-1}^b}{w_0^a},\quad z_{4n+4}=z_0,\quad
w_{4n+3}=\frac1{z_0^c},\quad
w_{4n+4}=\frac{z_{-1}^{bc}}{w_0^{ac}},
\end{gather}
for $n\in\mathbb{N}_0$, which means that $(z_n,w_n)_{n\ge -1}$ is
four-periodic.

 (c), (d) If $ac-b\ge 2$, then $(ac-b)^n\to+\infty$ as
$n\to+\infty$. Hence, if $0<|w_0^a/z_{-1}^b|<1$, by using the
formula
\begin{equation}
z_{2n+1}=\Big(\frac{w_0^a}{z_{-1}^b}\Big)^{(ac-b)^n},\quad
n\in\mathbb{N}_0,\label{q19}
\end{equation}
we obtain that $z_{2n+1}\to 0$, as
$n\to\infty$, while if $|w_0^a/z_{-1}^b|>1$, then we obtain that
$z_{2n+1}\to 0$, as $n\to\infty$.

 (e) The statement directly follows by using the condition
$w_0^a=z_{-1}^b$ in \eqref{q19}.

 (f) Using the conditions $w_0=e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, in \eqref{q19} we have \begin{equation}
z_{2n+1}=e^{i\pi \frac{p(ac-d)^n}q},\quad
n\in\mathbb{N}_0.\label{q20}\end{equation} The rest of the proof is similar to the
one of statement (f) in Theorem \ref{thm2}, so is omitted.

 (g), (h) Since $ac-b\le -2$ from \eqref{q19} we have
\begin{equation}
z_{4n+1}=\Big(\frac{w_0^a}{z_{-1}^b}\Big)^{(ac-b)^{2n}},\quad
z_{4n+3}=\Big(\frac{w_0^a}{z_{-1}^b}\Big)^{-|ac-b|^{2n+1}},\quad
n\in\mathbb{N}_0.\label{q21}
\end{equation}
From \eqref{q21} and the posed
conditions these two statements easily follow.

 (i)-(l) The proofs are similar to those ones of
(c)-(f), when $w_0^a/z_{-1}^b$ is replaced by $z_0$, and
$z_{2n+1}$ is replaced by $z_{2n+2}$. Hence, we omit the detail.

 (m), (n) From \eqref{z1d} and since $ac-b\le -2$, we have
\begin{equation}
z_{4n+2}=z_0^{-|ac-b|^{2n+1}},\quad
z_{4n+4}=z_0^{(ac-b)^{2n+2}},\label{q22}
\end{equation}
for $n\in\mathbb{N}_0$. From \eqref{q22} and the posed conditions these
two statements easily follow.

 (o)-(r) Using the formula
\begin{equation}
w_{2n+1}=(z_0^c)^{(ac-b)^n},\quad n\in\mathbb{N}_0,\label{q23}
\end{equation}
statements  (o)-(q) easily follow, while  (r) is proved
similar to (f).

 (s), (t) From \eqref{q23} and since $ac-d\le -2$, it follows
that
\begin{equation}
w_{4n+1}=(z_0^c)^{(ac-b)^{2n}},\quad
w_{4n+3}=(z_0^c)^{-|ac-b|^{2n+1}},\quad n\in\mathbb{N}_0.\label{q25}
\end{equation}
Using the formulas in \eqref{q25} these two statements easily
follow.

(u)-(x) Using the formula
\begin{equation}
w_{2n+2}=\Big(\frac{w_0^{ac}}{z_{-1}^{bc}}\Big)^{(ac-b)^n},\quad
n\in\mathbb{N}_0,\label{q26}
\end{equation}
statements  (u)-(w) easily follow,
while (x) is proved similar to (f).

 (y), (z) From \eqref{q26} and since $ac-b\le -2$, it follows
that
\begin{equation}
w_{4n+2}=\Big(\frac{w_0^{ac}}{z_{-1}^{bc}}\Big)^{(ac-b)^{2n}},\quad
w_{4n+4}=\Big(\frac{w_0^{ac}}{z_{-1}^{bc}}\Big)^{-|ac-b|^{2n+1}},\quad
n\in\mathbb{N}_0.\label{q27}
\end{equation}
Using the formulas in \eqref{q27} these
two statements easily follow.
\end{proof}

\begin{theorem} \label{thm4}
 Consider system \eqref{ms}. Assume
that $a,b,c,d\in\mathbb{Z}$, $(ac-b-d)^2=4bd$ and $ac-b-d=2$, and initial
values $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$. Then the
following statements hold:
\begin{itemize}

\item[(a)] If $0<|w_0^a/z_{-1}^{b(ac-b-1)}|<1$, then $z_{2n+1}\to 0$,
as $n\to\infty$.

\item[(b)] If $|w_0^a/z_{-1}^{b(ac-b-1)}|>1$, then $z_{2n+1}\to
\infty$, as $n\to\infty$.

\item[(c)] If $w_0^a=z_{-1}^{b(ac-b-1)}$, then
$z_{2n+1}=w_0^a/z_{-1}^b$, as $n\to\infty$.

\item[(d)] If $w_0^a=z_{-1}^{b(ac-b-1)}e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $z_{2n+1}$ is periodic with period
$T\le 2q$.

\item[(e)] If $0<|z_0^{ac-b-1}/w_{-1}^{ad}|<1$, then $z_{2n+2}\to 0$,
as $n\to\infty$.

\item[(f)] If $|z_0^{ac-b-1}/w_{-1}^{ad}|>1$, then $z_{2n+2}\to
\infty$, as $n\to\infty$.

\item[(g)] If $z_0^{ac-b-1}=w_{-1}^{ad}$, then
$z_{2n+2}=z_0^{ac-b}/w_{-1}^{ad}$, as $n\to\infty$.

\item[(h)] If $z_0^{ac-b-1}=w_{-1}^{ad}e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $z_{2n+2}$ is periodic with period
$T\le 2q$.

\item[(i)] If $0<|z_0^c/w_{-1}^{d(ac-d-1)}|<1$, then $w_{2n+1}\to 0$,
as $n\to\infty$.

\item[(j)] If $|z_0^c/w_{-1}^{d(ac-d-1)}|>1$, then $w_{2n+1}\to
\infty$, as $n\to\infty$.

\item[(k)] If $z_0^c=w_{-1}^{d(ac-d-1)}$, then
$w_{2n+1}=z_0^c/w_{-1}^d$, as $n\to\infty$.

\item[(l)] If $z_0^c=w_{-1}^{d(ac-d-1)}e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $w_{2n+1}$ is periodic with period
$T\le 2q$.

\item[(m)] If $0<|w_0^{ac-d-1}/z_{-1}^{bc}|<1$, then $w_{2n+2}\to 0$,
as $n\to\infty$.

\item[(n)] If $|w_0^{ac-d-1}/z_{-1}^{bc}|>1$, then $w_{2n+2}\to
\infty$, as $n\to\infty$.

\item[(o)] If $w_0^{ac-d-1}=z_{-1}^{bc}$, then
$w_{2n+2}=w_0^{ac-d}/z_{-1}^{bc}$, as $n\to\infty$.

\item[(p)] If $w_0^{ac-d-1}=z_{-1}^{bc}e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $z_{2n+2}$ is periodic with period
$T\le 2q$.

\end{itemize}
\end{theorem}

\begin{proof}
First note that in this case the characteristic roots
of polynomial \eqref{q28} are $\lambda_1=\lambda_2=1$. Using it into
\eqref{b5}-\eqref{b8} we obtain the following formulas
\begin{gather*}
z_{2n+1}=\frac{w_0^a}{z_{-1}^b}\Big(\frac{w_0^a}{z_{-1}^{b(ac-b-1)}}\Big)^n,\quad
z_{2n+2}=\frac{z_0^{ac-b}}{w_{-1}^{ad}}\Big(\frac{z_0^{ac-b-1}}{w_{-1}^{ad}}\Big)^n,\\
w_{2n+1}=\frac{z_0^c}{w_{-1}^d}\Big(\frac{z_0^c}{w_{-1}^{d(ac-d-1)}}\Big)^n,\quad
w_{2n+2}=\frac{w_0^{ac-d}}{z_{-1}^{bc}}\Big(\frac{w_0^{ac-d-1}}{z_{-1}^{bc}}\Big)^n,
\end{gather*}
for $n\in\mathbb{N}_0$, from which all the statements of the theorem
easily follow.
\end{proof}

\begin{theorem} \label{thm5}
Consider system \eqref{ms}. Assume
that $a,b,c,d\in\mathbb{Z}$, $(ac-b-d)^2=4bd$ and $ac-b-d=-2$, and
initial values $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$.
Then the following statements hold:
\begin{itemize}

\item[(a)] If $0<|w_0^az_{-1}^{b(ac-b+1)}|<1$, then $z_{4n+1}\to 0$ and
$z_{4n+3}\to \infty$, as $n\to\infty$.

\item[(b)] If $|w_0^az_{-1}^{b(ac-b+1)}|>1$, then $z_{4n+1}\to \infty$
and $z_{4n+3}\to 0$, as $n\to\infty$.

\item[(c)] If $w_0^az_{-1}^{b(ac-b+1)}=1$, then
$z_{4n+1}=w_0^a/z_{-1}^b=1/z_{4n+3}$, as $n\in\mathbb{N}_0$.

\item[(d)] If $w_0^az_{-1}^{b(ac-b+1)}=e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $z_{4n+1}$ and $z_{4n+3}$ are
periodic with period $T\le 2q$.


\item[(e)] If $0<|z_0^{ac-b+1}/w_{-1}^{ad}|<1$, then $z_{4n+2}\to 0$
and $z_{4n+4}\to \infty$, as $n\to\infty$.

\item[(f)] If $|z_0^{ac-b+1}/w_{-1}^{ad}|>1$, then $z_{4n+2}\to \infty$
and $z_{4n+4}\to 0$, as $n\to\infty$.

\item[(g)] If $z_0^{ac-b+1}=w_{-1}^{ad}$, then
$z_{4n+2}=z_0^{ac-b}/w_{-1}^{ad}=1/z_{4n+4}$, as $n\in\mathbb{N}_0$.

\item[(h)] If $z_0^{ac-b+1}=w_{-1}^{ad}e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then  $z_{4n+2}$ and $z_{4n+4}$ are
periodic with period $T\le 2q$.

\item[(i)] If $0<|z_0^cw_{-1}^{d(ac-d+1)}|<1$, then $w_{4n+1}\to 0$ and
$w_{4n+3}\to \infty$, as $n\to\infty$.

\item[(j)] If $|z_0^cw_{-1}^{d(ac-d+1)}|>1$, then $w_{4n+1}\to \infty$
and $w_{4n+3}\to 0$, as $n\to\infty$.

\item[(k)] If $z_0^cw_{-1}^{d(ac-d+1)}=1$, then
$w_{4n+1}=z_0^{c}/w_{-1}^{d}=1/w_{4n+3}$, as $n\in\mathbb{N}_0$.

\item[(l)] If $z_0^cw_{-1}^{d(ac-d+1)}=e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $w_{4n+1}$ and $w_{4n+3}$ are
periodic with period $T\le 2q$.

\item[(m)] If $0<|w_0^{ac-d+1}/z_{-1}^{bc}|<1$, then $w_{4n+2}\to 0$
and $w_{4n+4}\to\infty$, as $n\to\infty$.

\item[(n)] If $|w_0^{ac-d+1}/z_{-1}^{bc}|>1$, then $w_{4n+2}\to \infty$
and $w_{4n+4}\to 0$, as $n\to\infty$.

\item[(o)] If $w_0^{ac-d+1}=z_{-1}^{bc}$, then
$w_{4n+2}=w_0^{ac-d}/z_{-1}^{bc}=1/w_{4n+4}$, as $n\in\mathbb{N}_0$.

\item[(p)] If $w_0^{ac-d+1}=z_{-1}^{bc}e^{i\theta}$, $\theta=p\pi/q$,
$q\in\mathbb{N}$ and $p\in\mathbb{Z}$, then $w_{4n+2}$ and $w_{4n+4}$ are
periodic with period $T\le 2q$.

\end{itemize}
\end{theorem}


\begin{proof}
First note that in this case the characteristic roots
of polynomial \eqref{q28} are $\lambda_1=\lambda_2=-1$. Using it into
\eqref{b5}-\eqref{b8} we obtain the following formulas
\begin{gather*}
z_{2n+1}=\frac{w_0^{a(-1)^n}}{z_{-1}^{b(-1)^n}}
 \Big(\frac{w_0^a}{z_{-1}^{-b(ac-b+1)}}\Big)^{n(-1)^n},\quad
z_{2n+2}=\frac{z_0^{(ac-b)(-1)^n}}{w_{-1}^{ad(-1)^n}}
 \Big(\frac{z_0^{ac-b+1}}{w_{-1}^{ad}}\Big)^{n(-1)^n},\\
w_{2n+1}=\frac{z_0^{c(-1)^n}}{w_{-1}^{d(-1)^n}}
 \Big(\frac{z_0^c}{w_{-1}^{-d(ac-d+1)}}\Big)^{n(-1)^n},\quad
w_{2n+2}=\frac{w_0^{(ac-d)(-1)^n}}{z_{-1}^{bc(-1)^n}}
 \Big(\frac{w_0^{ac-d+1}}{z_{-1}^{bc}}\Big)^{n(-1)^n}\!,
\end{gather*}
for $n\in\mathbb{N}_0$, from which it follows that
\begin{gather}
z_{4n+1}=\frac{w_0^a}{z_{-1}^b}\Big(\frac{w_0^a}{z_{-1}^{-b(ac-b+1)}}\Big)^{2n},\quad
z_{4n+3}=\frac{z_{-1}^b}{w_0^a}\Big(\frac{z_{-1}^{-b(ac-b+1)}}{w_0^a}\Big)^{2n+1},
\label{q29a}\\
z_{4n+2}=\frac{z_0^{ac-b}}{w_{-1}^{ad}}\Big(\frac{z_0^{ac-b+1}}{w_{-1}^{ad}}
\Big)^{2n},\quad
z_{4n+4}=\frac{w_{-1}^{ad}}{z_0^{ac-b}}\Big(\frac{w_{-1}^{ad}}{z_0^{ac-b+1}}
\Big)^{2n+1},\label{q30a}\\
w_{4n+1}=\frac{z_0^{c}}{w_{-1}^{d}}\Big(\frac{z_0^c}{w_{-1}^{-d(ac-d+1)}}
\Big)^{2n},\quad
w_{4n+3}=\frac{w_{-1}^{d}}{z_0^{c}}\Big(\frac{w_{-1}^{-d(ac-d+1)}}{z_0^c}
\Big)^{2n+1},\label{q31a}\\
w_{4n+2}=\frac{w_0^{ac-d}}{z_{-1}^{bc}}\Big(\frac{w_0^{ac-d+1}}{z_{-1}^{bc}}
\Big)^{2n},\quad
w_{4n+4}=\frac{z_{-1}^{bc}}{w_0^{ac-d}}\Big(\frac{z_{-1}^{bc}}{w_0^{ac-d+1}}
\Big)^{2n+1},\label{q32a}
\end{gather}
for $n\in\mathbb{N}_0$. Using formulas \eqref{q29a}-\eqref{q32a} all the
statements of the theorem easily follow.
\end{proof}

\begin{theorem} \label{thm6}
Consider system \eqref{ms}. Assume
that $a,b,c,d\in\mathbb{Z}$, $bd\ne 0$, $(ac-b-d)^2=4bd$ and $ac-b-d>2$,
and initial values $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$.
Then the following statements hold:
\begin{itemize}

\item[(a)] If $0<|w_0^{a\lambda_1}/z_{-1}^{b(ac-b-\lambda_1)}|<1$, then
$z_{2n+1}\to 0$, as $n\to\infty$.

\item[(b)] If $|w_0^{a\lambda_1}/z_{-1}^{b(ac-b-\lambda_1)}|>1$, then
$z_{2n+1}\to \infty$, as $n\to\infty$.

\item[(c)] If $w_0^{a\lambda_1}=z_{-1}^{b(ac-b-\lambda_1)}$ and
$0<|w_0^a/z_{-1}^b|<1$, then $z_{2n+1}\to 0$, as $n\to\infty$.

\item[(d)] If $w_0^{a\lambda_1}=z_{-1}^{b(ac-b-\lambda_1)}$ and
$|w_0^a/z_{-1}^b|>1$, then $z_{2n+1}\to \infty$, as $n\to\infty$.

\item[(e)] If $w_0^{a\lambda_1}=z_{-1}^{b(ac-b-\lambda_1)}$ and $w_0^a=z_{-1}^b$,
then $z_{2n+1}=1$, as $n\in\mathbb{N}_0$.

\item[(f)] If $w_0^{a\lambda_1}=z_{-1}^{b(ac-b-\lambda_1)}$ and
$w_0^a=z_{-1}^be^{i\theta}$, $\theta=p\pi/q$, $q\in\mathbb{N}$ and
$p\in\mathbb{Z}$, then $z_{2n+1}$ is periodic with period $T\le 2q$.

\item[(g)] If $0<|z_0^{ac-b-\lambda_1}/w_{-1}^{ad}|<1$, then $z_{2n+2}\to
0$, as $n\to\infty$.

\item[(h)] If $|z_0^{ac-b-\lambda_1}/w_{-1}^{ad}|>1$, then $z_{2n+2}\to
\infty$, as $n\to\infty$.

\item[(i)] If $z_0^{ac-b-\lambda_1}=w_{-1}^{ad}$ and
$0<|z_0^{ac-b}/w_{-1}^{ad}|<1$, then $z_{2n+2}\to 0$, as
$n\to\infty$.

\item[(j)] If $z_0^{ac-b-\lambda_1}=w_{-1}^{ad}$ and
$|z_0^{ac-b}/w_{-1}^{ad}|>1$, then $z_{2n+2}\to \infty$, as
$n\to\infty$.

\item[(k)] If $z_0^{ac-b-\lambda_1}=w_{-1}^{ad}$ and
$z_0^{ac-b}=w_{-1}^{ad}$, then $z_{2n+2}=1$, as $n\in\mathbb{N}_0$.

\item[(l)] If $z_0^{ac-b-\lambda_1}=w_{-1}^{ad}$ and
$z_0^{ac-b}=w_{-1}^{ad}e^{i\theta}$, $\theta=p\pi/q$, $q\in\mathbb{N}$
and $p\in\mathbb{Z}$, then $z_{2n+2}$ is periodic with period $T\le 2q$.

\item[(m)] If $0<|z_0^{c\lambda_1}/w_{-1}^{d(ac-d-\lambda_1)}|<1$, then
$w_{2n+1}\to 0$, as $n\to\infty$.

\item[(n)] If $|z_0^{c\lambda_1}/w_{-1}^{d(ac-d-\lambda_1)}|>1$, then
$w_{2n+1}\to \infty$, as $n\to\infty$.

\item[(o)] If $z_0^{c\lambda_1}=w_{-1}^{d(ac-d-\lambda_1)}$ and
$0<|z_0^c/w_{-1}^d|<1$, then $w_{2n+1}\to 0$, as $n\to\infty$.

\item[(p)] If $z_0^{c\lambda_1}=w_{-1}^{d(ac-d-\lambda_1)}$ and
$|z_0^c/w_{-1}^d|>1$, then $w_{2n+1}\to \infty$, as $n\to\infty$.

\item[(q)] If $z_0^{c\lambda_1}=w_{-1}^{d(ac-d-\lambda_1)}$ and $z_0^c=w_{-1}^d$,
then $w_{2n+1}=1$, as $n\in\mathbb{N}_0$.

\item[(r)] If $z_0^{c\lambda_1}=w_{-1}^{d(ac-d-\lambda_1)}$ and
$z_0^c=w_{-1}^de^{i\theta}$, $\theta=p\pi/q$, $q\in\mathbb{N}$ and
$p\in\mathbb{Z}$, then $w_{2n+1}$ is periodic with period $T\le 2q$.

\item[(s)] If $0<|w_0^{ac-d-\lambda_1}/z_{-1}^{bc}|<1$, then $w_{2n+2}\to
0$, as $n\to\infty$.

\item[(t)] If $|w_0^{ac-d-\lambda_1}/z_{-1}^{bc}|>1$, then $w_{2n+2}\to
\infty$, as $n\to\infty$.

\item[(u)] If $w_0^{ac-d-\lambda_1}=z_{-1}^{bc}$ and
$0<|w_0^{ac-d}/z_{-1}^{bc}|<1$, then $w_{2n+2}\to 0$, as
$n\to\infty$.

\item[(v)] If $w_0^{ac-d-\lambda_1}=z_{-1}^{bc}$ and
$|w_0^{ac-d}/z_{-1}^{bc}|>1$, then $w_{2n+2}\to \infty$, as
$n\to\infty$.

\item[(w)] If $w_0^{ac-d-\lambda_1}=z_{-1}^{bc}$ and
$w_0^{ac-d}=z_{-1}^{bc}$, then $w_{2n+2}=1$, as $n\in\mathbb{N}_0$.

\item[(x)] If $w_0^{ac-d-\lambda_1}=z_{-1}^{bc}$ and
$w_0^{ac-d}=z_{-1}^{bc}e^{i\theta}$, $\theta=p\pi/q$, $q\in\mathbb{N}$
and $p\in\mathbb{Z}$, then $w_{2n+2}$ is periodic with period $T\le 2q$.

\end{itemize}
\end{theorem}


\begin{proof}
First note that in this case the characteristic roots
of polynomial \eqref{q28} are such that
$$
\lambda_1=\lambda_2=\sqrt{|bd|}>1.
$$
Note that they are natural numbers,
since $2\sqrt{|bd|}=|ac-b-d|\in\mathbb{N}$. Equations
\eqref{b5}-\eqref{b8} can be written in the form
\begin{gather}
z_{2n+1}=\Big(\frac{w_0^a}{z_{-1}^b}\Big)^{\lambda_1^n} \Big(\frac{w_0^{a\lambda_1}}{z_{-1}^{b(ac-b-\lambda_1)}}\Big)^{n\lambda_1^{n-1}},\label{b5a}\\
z_{2n+2}=\Big(\frac{z_0^{ac-b}}{w_{-1}^{ad}}\Big)^{\lambda_1^n}
\Big(\frac{z_0^{ac-b-\lambda_1}}{w_{-1}^{ad}}\Big)^{n\lambda_1^n},\label{b6a}\\
w_{2n+1}=\Big(\frac{z_0^c}{w_{-1}^d}\Big)^{\lambda_1^n}
\Big(\frac{z_0^{c\lambda_1}}{w_{-1}^{d(ac-d-\lambda_1)}}\Big)^{n\lambda_1^{n-1}},\label{b7a}\\
w_{2n+2}=\Big(\frac{w_0^{ac-d}}{z_{-1}^{bc}}\Big)^{\lambda_1^n}
\Big(\frac{w_0^{ac-d-\lambda_1}}{z_{-1}^{bc}}\Big)^{n\lambda_1^n},\label{b8a}
\end{gather}
for $n\in\mathbb{N}_0$. Using formulas \eqref{b5a}-\eqref{b8a} all the
statement easily follow.
\end{proof}

\begin{theorem} \label{thm7}
 Consider system \eqref{ms}. Assume
that $a,b,c,d\in\mathbb{Z}$, $bd\ne 0$, $(ac-b-d)^2=4bd$ and $ac-b-d<-2$,
and initial values $z_{-1}, z_0, w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$.
Then the following statements hold:
\begin{itemize}

\item[(a)] If $0<|w_0^{a\lambda_1}/z_{-1}^{b(ac-b-\lambda_1)}|<1$, then
$z_{4n+1}\to \infty$ and $z_{4n+3}\to 0$, as $n\to\infty$.

\item[(b)] If $|w_0^{a\lambda_1}/z_{-1}^{b(ac-b-\lambda_1)}|>1$, then
$z_{4n+1}\to 0$ and $z_{4n+3}\to \infty$, as $n\to\infty$.


\item[(c)] If $w_0^{a\lambda_1}=z_{-1}^{b(ac-b-\lambda_1)}$ and
$0<|w_0^a/z_{-1}^b|<1$, $z_{4n+1}\to 0$ and $z_{4n+3}\to \infty$,
as $n\to\infty$.

\item[(d)] If $w_0^{a\lambda_1}=z_{-1}^{b(ac-b-\lambda_1)}$ and
$|w_0^a/z_{-1}^b|>1$, $z_{4n+1}\to \infty$ and $z_{4n+3}\to 0$, as
$n\to\infty$.

\item[(e)] If $w_0^{a\lambda_1}=z_{-1}^{b(ac-b-\lambda_1)}$ and $w_0^a=z_{-1}^b$,
then $z_{2n+1}=1$, as $n\in\mathbb{N}_0$.

\item[(f)] If $w_0^{a\lambda_1}=z_{-1}^{b(ac-b-\lambda_1)}$ and
$w_0^a=z_{-1}^be^{i\theta}$, $\theta=p\pi/q$, $q\in\mathbb{N}$ and
$p\in\mathbb{Z}$, then $z_{4n+1}$ and $z_{4n+3}$ are periodic with period
$T\le 2q$.

\item[(g)] If $0<|z_0^{ac-b-\lambda_1}/w_{-1}^{ad}|<1$, then $z_{4n+2}\to 0$
and $z_{4n+4}\to \infty$, as $n\to\infty$.

\item[(h)] If $|z_0^{ac-b-\lambda_1}/w_{-1}^{ad}|>1$, then $z_{4n+2}\to
\infty$ and $z_{4n+4}\to 0$, as $n\to\infty$.

\item[(i)] If $z_0^{ac-b-\lambda_1}=w_{-1}^{ad}$ and
$0<|z_0^{ac-b}/w_{-1}^{ad}|<1$, then $z_{4n+2}\to 0$ and
$z_{4n+4}\to \infty$, as $n\to\infty$.

\item[(j)] If $z_0^{ac-b-\lambda_1}=w_{-1}^{ad}$ and
$|z_0^{ac-b}/w_{-1}^{ad}|>1$, then $z_{4n+2}\to \infty$ and
$z_{4n+4}\to 0$, as $n\to\infty$.

\item[(k)] If $z_0^{ac-b-\lambda_1}=w_{-1}^{ad}$ and
$z_0^{ac-b}=w_{-1}^{ad}$, then $z_{2n+2}=1$, as $n\in\mathbb{N}_0$.

\item[(l)] If $z_0^{ac-b-\lambda_1}=w_{-1}^{ad}$ and
$z_0^{ac-b}=w_{-1}^{ad}e^{i\theta}$, $\theta=p\pi/q$, $q\in\mathbb{N}$
and $p\in\mathbb{Z}$, then $z_{4n+2}$ and $z_{4n+4}$ are  periodic with
period $T\le 2q$.

\item[(m)] If $0<|z_0^{c\lambda_1}/w_{-1}^{d(ac-d-\lambda_1)}|<1$, $w_{4n+1}\to
\infty$ and $w_{4n+3}\to 0$, as $n\to\infty$.

\item[(n)] If $|z_0^{c\lambda_1}/w_{-1}^{d(ac-d-\lambda_1)}|>1$, $w_{4n+1}\to 0$
and $w_{4n+3}\to \infty$, as $n\to\infty$.

\item[(o)] If $z_0^{c\lambda_1}=w_{-1}^{d(ac-d-\lambda_1)}$ and
$0<|z_0^c/w_{-1}^d|<1$, then $w_{4n+1}\to 0$ and $w_{4n+3}\to
\infty$, as $n\to\infty$.

\item[(p)] If $z_0^{c\lambda_1}=w_{-1}^{d(ac-d-\lambda_1)}$ and
$|z_0^c/w_{-1}^d|>1$, $w_{4n+1}\to \infty$ and $w_{4n+3}\to 0$, as
$n\to\infty$.

\item[(q)] If $z_0^{c\lambda_1}=w_{-1}^{d(ac-d-\lambda_1)}$ and $z_0^c=w_{-1}^d$,
then $w_{2n+1}=1$, as $n\in\mathbb{N}_0$.

\item[(r)] If $z_0^{c\lambda_1}=w_{-1}^{d(ac-d-\lambda_1)}$ and
$z_0^c=w_{-1}^de^{i\theta}$, $\theta=p\pi/q$, $q\in\mathbb{N}$ and
$p\in\mathbb{Z}$, then $w_{4n+1}$ and $w_{4n+3}$ are  periodic with
period $T\le 2q$.

\item[(s)] If $0<|w_0^{ac-d-\lambda_1}/z_{-1}^{bc}|<1$, then $w_{4n+2}\to 0$
and $w_{4n+4}\to\infty$, as $n\to\infty$.

\item[(t)] If $|w_0^{ac-d-\lambda_1}/z_{-1}^{bc}|>1$, then $w_{4n+2}\to
\infty$ and $w_{4n+4}\to 0$, as $n\to\infty$.

\item[(u)] If $w_0^{ac-d-\lambda_1}=z_{-1}^{bc}$ and
$0<|w_0^{ac-d}/z_{-1}^{bc}|<1$, then $w_{4n+2}\to 0$ and
$w_{4n+4}\to\infty$, as $n\to\infty$.

\item[(v)] If $w_0^{ac-d-\lambda_1}=z_{-1}^{bc}$ and
$|w_0^{ac-d}/z_{-1}^{bc}|>1$, then $w_{4n+2}\to \infty$ and
$w_{4n+4}\to 0$, as $n\to\infty$.

\item[(w)] If $w_0^{ac-d-\lambda_1}=z_{-1}^{bc}$ and
$w_0^{ac-d}=z_{-1}^{bc}$, then $w_{2n+2}=1$, as $n\in\mathbb{N}_0$.

\item[(x)] If $w_0^{ac-d-\lambda_1}=z_{-1}^{bc}$ and
$w_0^{ac-d}=z_{-1}^{bc}e^{i\theta}$, $\theta=p\pi/q$, $q\in\mathbb{N}$
and $p\in\mathbb{Z}$, then $w_{4n+2}$ and $w_{4n+4}$ are periodic with
period $T\le 2q$.

\end{itemize}
\end{theorem}


\begin{proof} First note that in this case the characteristic roots
of polynomial \eqref{q28} are such that
$$
\lambda_1=\lambda_2=-\sqrt{|bd|}<1.
$$
Note that they are negative integers,
since $2\sqrt{|bd|}=|ac-b-d|\in\mathbb{N}$. Equations
\eqref{b5a}-\eqref{b8a} can be written in the form
\begin{gather}
z_{2n+1}=\Big(\frac{w_0^a}{z_{-1}^b}\Big)^{(-|\lambda_1|)^n}
 \Big(\frac{w_0^{a\lambda_1}}{z_{-1}^{b(ac-b-\lambda_1)}}\Big)^{n(-|\lambda_1|)^{n-1}},\label{b5b}\\
z_{2n+2}=\Big(\frac{z_0^{ac-b}}{w_{-1}^{ad}}\Big)^{(-|\lambda_1|)^n}
\Big(\frac{z_0^{ac-b-\lambda_1}}{w_{-1}^{ad}}\Big)^{n(-|\lambda_1|)^n},\label{b6b}\\
w_{2n+1}=\Big(\frac{z_0^c}{w_{-1}^d}\Big)^{(-|\lambda_1|)^n}
\Big(\frac{z_0^{c\lambda_1}}{w_{-1}^{d(ac-d-\lambda_1)}}\Big)^{n(-|\lambda_1|)^{n-1}},\label{b7b}\\
w_{2n+2}=\Big(\frac{w_0^{ac-d}}{z_{-1}^{bc}}\Big)^{(-|\lambda_1|)^n}
\Big(\frac{w_0^{ac-d-\lambda_1}}{z_{-1}^{bc}}\Big)^{n(-|\lambda_1|)^n},\label{b8b}
\end{gather}
for $n\in\mathbb{N}_0$.

From \eqref{b5b}-\eqref{b8b} it follows that
\begin{gather}
z_{4n+1}=\Big(\frac{w_0^a}{z_{-1}^b}\Big)^{\lambda_1^{2n}} \Big(\frac{w_0^{a\lambda_1}}{z_{-1}^{b(ac-b-\lambda_1)}}\Big)^{-2n|\lambda_1|^{2n-1}},\label{b5c}\\
z_{4n+3}=\Big(\frac{w_0^a}{z_{-1}^b}\Big)^{-|\lambda_1|^{2n+1}} \Big(\frac{w_0^{a\lambda_1}}{z_{-1}^{b(ac-b-\lambda_1)}}\Big)^{(2n+1)\lambda_1^{2n}},\label{b5c1}\\
z_{4n+2}=\Big(\frac{z_0^{ac-b}}{w_{-1}^{ad}}\Big)^{\lambda_1^{2n}}
\Big(\frac{z_0^{ac-b-\lambda_1}}{w_{-1}^{ad}}\Big)^{2n\lambda_1^{2n}},\label{b6c}\\
z_{4n+4}=\Big(\frac{z_0^{ac-b}}{w_{-1}^{ad}}\Big)^{-|\lambda_1|^{2n+1}}
\Big(\frac{z_0^{ac-b-\lambda_1}}{w_{-1}^{ad}}\Big)^{-(2n+1)|\lambda_1|^{2n+1}},\label{b6c1}\\
w_{4n+1}=\Big(\frac{z_0^c}{w_{-1}^d}\Big)^{\lambda_1^{2n}}
\Big(\frac{z_0^{c\lambda_1}}{w_{-1}^{d(ac-d-\lambda_1)}}\Big)^{-2n|\lambda_1|^{2n-1}},\label{b7c}\\
w_{4n+3}=\Big(\frac{z_0^c}{w_{-1}^d}\Big)^{-|\lambda_1|^{2n+1}}
\Big(\frac{z_0^{c\lambda_1}}{w_{-1}^{d(ac-d-\lambda_1)}}\Big)^{(2n+1)|\lambda_1|^{2n}},\label{b7c1}\\
w_{4n+2}=\Big(\frac{w_0^{ac-d}}{z_{-1}^{bc}}\Big)^{\lambda_1^{2n}}
\Big(\frac{w_0^{ac-d-\lambda_1}}{z_{-1}^{bc}}\Big)^{2n\lambda_1^{2n}},\label{b8c}\\
w_{4n+4}=\Big(\frac{w_0^{ac-d}}{z_{-1}^{bc}}\Big)^{-|\lambda_1|^{2n+1}}
\Big(\frac{w_0^{ac-d-\lambda_1}}{z_{-1}^{bc}}\Big)^{-(2n+1)|\lambda_1|^{2n+1}},\label{b8c1}
\end{gather}
for $n\in\mathbb{N}_0$.

Using formulas \eqref{b5c}-\eqref{b8c1} all the statements easily
follow.
\end{proof}

Formulations and the proofs of the results on the long-term
behavior of solutions to system \eqref{ms} for the case
$a,b,c,d\in\mathbb{Z}$, $bd\ne 0$, $(ac-b-d)^2\ne 4bd$, $z_{-1}, z_0,
w_{-1}, w_0\in\mathbb{C}\setminus\{0\}$, we leave them for the reader as an
exercise.


\subsection*{Acknowledgements}
This project was funded by the Deanship of
Scientific Research (DSR), King Abdulaziz University, Jeddah,
under grant no. (36-130-33-HiCi). The authors, therefore,
acknowledge technical and financial support of KAU.


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\end{document}