Solvable product-type system of difference equations with two dependent variables

Abstract

It has been recently noticed that there is a finite number of two-dimensional classes of product-type systems of difference equations solvable in closed form. We present a new class of this type. A detailed analysis of the form of its solutions is given. Our results complement the previous ones on such systems and present one of the final steps in describing the forms of their solutions.

1 Introduction

Many types of difference equations and systems have been studied so far. A part of the studies can be found in [1–24]. Some types of the systems essentially obtained by symmetrization of scalar ones were studied in [8–10], which was a motivation for further investigations in the field [6, 7, 11, 12, 14–24]. Historically, perhaps the first main problem of interest in the whole area was finding formulas for their solutions. For known methods for finding the formulas the reader can consult, for example, [1–5]. A note of ours from 2004 has influenced some investigation in this direction since that time (see, for example, [13, 15–24] and the references therein).

In the study of some classes of equations and systems, product-type ones appear as boundary cases. Finding formulas for positive solutions to the equations and systems in the boundary cases is a routine problem, so not of theoretical interest nowadays. It can be of practical interest only if another system or equation is reduced to such one. However, if all solutions are not positive, the problem is very complicated. The boundary cases of equations and systems have motivated us to study them for the case of non-positive initial values. In fact, the equations and systems on the complex domain have attracted our special attention. Our study started in [21], where a system with two dependent variables was investigated. The form of the system in [21] strikingly suggested the study of the solvability of the other systems of related forms (see, e.g., [16, 22]). Since the system, as well as a couple of other ones later studied (see, e.g., [22]), was of the form

$$\begin{aligned} z_{n}=z_{n-m_{1}}^{a_{1}}w_{n-m_{2}}^{a_{2}}, \qquad w_{n}=w_{n-m_{3}}^{a_{3}}z_{n-m_{4}}^{a_{4}}, \quad n\in\mathbb{N}_{0}, \end{aligned}$$

(1)

it naturally suggested the study of the solvability of this, as well as of some related systems. This motivated us to include some coefficients in (1) and study the solvability of such systems, which was for the first time done in [15], where we showed the solvability theoretically and gave some hints on how to deal with more concrete cases, that is, for some special values of parameters a, b, c and d. Later we realized that complete pictures of the form of the solutions of this type of systems could be given by studying all the quantities appearing there in detail. References [18] and [24] were the first ones which gave the complete pictures of the forms of the solutions to the systems studied therein. Later in [17] we devised another method which deals with the solvability problem, although technically somewhat complex. For some quite recent results on product-type systems see [19], [20] and [23].

To finish the project of studying the solvability of product-type systems with two dependent variables (see [15, 17–24] and the related references therein), we have to study a few more. Here we study the system

$$\begin{aligned} z_{n+1}=\alpha z_{n}^{a}w_{n}^{b}, \qquad w_{n+1}=\beta w_{n-2}^{c}z_{n-1}^{d}, \quad n\in\mathbb{N}_{0}, \end{aligned}$$

(2)

where $a,b,c,d\in\mathbb{Z}$, $\alpha , \beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in \mathbb{C}$. In fact, we assume that $\alpha , \beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in\mathbb{C}\setminus\{0\}$, to avoid dealing with non-defined or trivial solutions. We will give a complete picture of the forms of the solutions to system (2) for all the values of the parameters and initial values.

2 Auxiliary results

Some classical auxiliary results that are employed in the section that follows are quoted in this one.

Lemma 1

see, e.g., [3, 25]

Let

$$R_{k}(s)=b_{k}\prod_{j=1}^{k}(s-s_{j}), $$

$s_{j}\ne s_{t}$, $j\ne t$, and $b_{k}\ne0$. Then

$$\sum_{j=1}^{k}\frac{s_{j}^{m}}{R'_{k}(s_{j})}=0, $$

for each $m\in\{0,1,\ldots,k-2\}$, and

$$\sum_{j=1}^{k}\frac{s_{j}^{k-1}}{R'_{k}(s_{j})}= \frac{1}{b_{k}}. $$

Four more or less widely known formulas are listed in the following lemma (see, e.g., [3, 5]). A recurrent relation connecting this type of sums is given in [18].

Lemma 2

Let

$$s_{n}^{(m)}(z)=\sum_{j=1}^{n}j^{m}z^{j-1}, \quad n\in\mathbb{N}, $$

$m\in\mathbb{N}_{0}$ and $z\in\mathbb{C}$.

Then

$$\begin{aligned}& s_{n}^{(0)}(z) =\frac{1-z^{n}}{1-z}, \\& s_{n}^{(1)}(z) =\frac{1-(n+1)z^{n}+nz^{n+1}}{(1-z)^{2}}, \\& s_{n}^{(2)}(z) =\frac {1+z-(n+1)^{2}z^{n}+(2n^{2}+2n-1)z^{n+1}-n^{2}z^{n+2}}{(1-z)^{3}}, \\& s_{n}^{(3)}(z) =\frac {n^{3}z^{n}(z-1)^{3}-3n^{2}z^{n}(z-1)^{2}+3nz^{n}(z^{2}-1)-(z^{n}-1)(z^{2}+4z+1)}{(1-z)^{4}}, \end{aligned}$$

for every $z\in\mathbb{C}\setminus\{1\}$ and $n\in\mathbb{N}$.

The following lemma describes the nature of the zeros of a polynomial of the fourth order in detail (see [26]).

Lemma 3

Let

$$\begin{aligned}& P_{4}(t)=t^{4}+bt^{3}+ct^{2}+dt+e, \\& \Delta _{0}=c^{2}-3bd+12e,\qquad \Delta _{1}=2c^{3}-9bcd+27b^{2}e+27d^{2}-72ce, \\& \Delta =\frac{1}{27}\bigl(4\Delta _{0}^{3}- \Delta _{1}^{2}\bigr),\qquad P=8c-3b^{2}, \\& Q=b^{3}+8d-4bc,\qquad D=64e-16c^{2}+16b^{2}c-16bd-3b^{4}. \end{aligned}$$

• (a) If $\Delta <0$, then two zeros of $P_{4}$ are real and different, and two are complex conjugate.

• (b) If $\Delta >0$, then all the zeros of $P_{4}$ are real or none is. More precisely,

  1^∘ :

  if $P<0$ and $D<0$, then all four zeros of $P_{4}$ are real and different;

  2^∘ :

  if $P>0$ or $D>0$, then there are two pairs of complex conjugate zeros of $P_{4}$.

• (c) If $\Delta =0$, then and only then $P_{4}$ has a multiple zero. The following cases can occur:

  1^∘ :

  if $P<0$, $D<0$ and $\Delta _{0}\ne0$, then two zeros of $P_{4}$ are real and equal and two are real and simple;

  2^∘ :

  if $D>0$ or ($P>0$ and ($D\ne0$ or $Q\ne0$)), then two zeros of $P_{4}$ are real and equal and two are complex conjugate;

  3^∘ :

  if $\Delta _{0}=0$ and $D\ne0$, there is a triple zero of $P_{4}$ and one simple, all real;

  4^∘ :

  if $D=0$, then

  4.1^∘ :

  if $P<0$ there are two double real zeros of $P_{4}$;

  4.2^∘ :

  if $P>0$ and $Q=0$ there are two double complex conjugate zeros of $P_{4}$;

  4.3^∘ :

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

3 Main results

The main results in this paper are proved in this section.

Theorem 1

Assume that $b,c,d\in\mathbb{Z}$, $a=0$, $\alpha, \beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in\mathbb{C}\setminus\{ 0\}$. Then

• (a) if $c+bd\ne1$, the general solution to (2) is given by

  $$\begin{aligned}& z_{3m}= \alpha ^{\frac{1-c-bd(c+bd)^{m-1}}{1-c-bd}}\beta ^{b\frac {1-(c+bd)^{m}}{1-c-bd}}z_{0}^{bd(c+bd)^{m-1}}w_{-1}^{bc(c+bd)^{m-1}}, \end{aligned}$$

  (3)
  $$\begin{aligned}& z_{3m+1}= \alpha ^{\frac{1-c-bd(c+bd)^{m}}{1-c-bd}}\beta ^{b\frac {1-(c+bd)^{m}}{1-c-bd}}w_{0}^{b(c+bd)^{m}}, \end{aligned}$$

  (4)
  $$\begin{aligned}& z_{3m+2}= \alpha ^{\frac{1-c-bd(c+bd)^{m}}{1-c-bd}}\beta ^{b\frac {1-(c+bd)^{m+1}}{1-c-bd}}z_{-1}^{bd(c+bd)^{m}}w_{-2}^{bc(c+bd)^{m}}, \end{aligned}$$

  (5)
  $$\begin{aligned}& w_{3m}= \alpha ^{d\frac{1-(c+bd)^{m}}{1-c-bd}}\beta ^{\frac {1-(c+bd)^{m}}{1-c-bd}}w_{0}^{(c+bd)^{m}}, \end{aligned}$$

  (6)
  $$\begin{aligned}& w_{3m+1}= \alpha ^{d\frac{1-(c+bd)^{m}}{1-c-bd}}\beta ^{\frac {1-(c+bd)^{m+1}}{1-c-bd}}z_{-1}^{d(c+bd)^{m}}w_{-2}^{c(c+bd)^{m}}, \end{aligned}$$

  (7)
  $$\begin{aligned}& w_{3m+2}= \alpha ^{d\frac{1-(c+bd)^{m}}{1-c-bd}}\beta ^{\frac {1-(c+bd)^{m+1}}{1-c-bd}}z_{0}^{d(c+bd)^{m}}w_{-1}^{c(c+bd)^{m}}, \end{aligned}$$

  (8)

• (b) if $c+bd=1$, the general solution to (2) is given by

  $$\begin{aligned}& z_{3m}= \alpha ^{1+bd(m-1)}\beta ^{bm}z_{0}^{bd}w_{-1}^{bc}, \end{aligned}$$

  (9)
  $$\begin{aligned}& z_{3m+1}= \alpha ^{1+bdm}\beta ^{bm}w_{0}^{b}, \end{aligned}$$

  (10)
  $$\begin{aligned}& z_{3m+2}= \alpha ^{1+bdm}\beta ^{b(m+1)}z_{-1}^{bd}w_{-2}^{bc}, \end{aligned}$$

  (11)
  $$\begin{aligned}& w_{3m}= \alpha ^{dm}\beta ^{m}w_{0}, \end{aligned}$$

  (12)
  $$\begin{aligned}& w_{3m+1}= \alpha ^{dm}\beta ^{m+1}z_{-1}^{d}w_{-2}^{c}, \end{aligned}$$

  (13)
  $$\begin{aligned}& w_{3m+2}= \alpha ^{dm}\beta ^{m+1}z_{0}^{d}w_{-1}^{c}. \end{aligned}$$

  (14)

Proof

Since $a=0$, we have

$$ z_{n+1}=\alpha w_{n}^{b},\qquad w_{n+1}= \beta w_{n-2}^{c}z_{n-1}^{d},\quad n\in \mathbb{N}_{0}. $$

(15)

From (15), we have

$$\begin{aligned} w_{n}=\beta \alpha ^{d} w_{n-3}^{c+bd}, \quad n\ge 3, \end{aligned}$$

(16)

which implies that

$$\begin{aligned} w_{3m+i}=\bigl(\alpha ^{d}\beta \bigr)^{\sum_{j=0}^{m-1}(c+bd)^{j}} w_{i}^{(c+bd)^{m}}, \quad m\in\mathbb{N}, i=0,1,2. \end{aligned}$$

(17)

Hence,

$$\begin{aligned}& w_{3m}= \alpha ^{d\sum_{j=0}^{m-1}(c+bd)^{j}}\beta ^{\sum _{j=0}^{m-1}(c+bd)^{j}}w_{0}^{(c+bd)^{m}}, \end{aligned}$$

(18)

$$\begin{aligned}& w_{3m+1}= \bigl(\alpha ^{d}\beta \bigr)^{\sum_{j=0}^{m-1}(c+bd)^{j}} \bigl( \beta w_{-2}^{c} z_{-1}^{d} \bigr)^{(c+bd)^{m}} \\& \hphantom{w_{3m+1}}= \alpha ^{d\sum_{j=0}^{m-1}(c+bd)^{j}}\beta ^{\sum _{j=0}^{m}(c+bd)^{j}}z_{-1}^{d(c+bd)^{m}}w_{-2}^{c(c+bd)^{m}}, \end{aligned}$$

(19)

$$\begin{aligned}& w_{3m+2}= \bigl(\alpha ^{d}\beta \bigr)^{\sum_{j=0}^{m-1}(c+bd)^{j}} \bigl( \beta w_{-1}^{c} z_{0}^{d} \bigr)^{(c+bd)^{m}} \\& \hphantom{w_{3m+2}}= \alpha ^{d\sum_{j=0}^{m-1}(c+bd)^{j}}\beta ^{\sum _{j=0}^{m}(c+bd)^{j}}z_{0}^{d(c+bd)^{m}}w_{-1}^{c(c+bd)^{m}}. \end{aligned}$$

(20)

Using (18)-(20) in the first equality in (15), we get

$$\begin{aligned}& z_{3m}= \alpha ^{1+bd\sum_{j=0}^{m-2}(c+bd)^{j}}\beta ^{b\sum _{j=0}^{m-1}(c+bd)^{j}}z_{0}^{bd(c+bd)^{m-1}}w_{-1}^{bc(c+bd)^{m-1}}, \end{aligned}$$

(21)

$$\begin{aligned}& z_{3m+1}= \alpha ^{1+bd\sum_{j=0}^{m-1}(c+bd)^{j}}\beta ^{b\sum _{j=0}^{m-1}(c+bd)^{j}}w_{0}^{b(c+bd)^{m}}, \end{aligned}$$

(22)

$$\begin{aligned}& z_{3m+2}= \alpha ^{1+bd\sum_{j=0}^{m-1}(c+bd)^{j}}\beta ^{b\sum _{j=0}^{m}(c+bd)^{j}}z_{-1}^{bd(c+bd)^{m}}w_{-2}^{bc(c+bd)^{m}}. \end{aligned}$$

(23)

From (18)-(23) and some calculations, we easily get (3)-(14), as desired. □

Theorem 2

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

Proof

Since $b=0$ system (2) becomes

$$ z_{n+1}=\alpha z_{n}^{a},\qquad w_{n+1}= \beta w_{n-2}^{c}z_{n-1}^{d}, \quad n\in \mathbb{N}_{0}, $$

(24)

which is system (2.11) in [17]. Hence, if $c\ne0$ the theorem follows from Theorem 2.2 in [17], while the case $c=0$ follows from equations (2.13) and (2.14) in [17], as well as the second equation in (24). □

The case $d=0$ has been recently studied in [20], where, among others, the following theorem was proved.

Theorem 3

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

Theorem 4

Assume that $a,b,c,d\in\mathbb{Z}$, $abcd\ne 0$, $\alpha , \beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in\mathbb {C}\setminus\{0\}$. Then system (2) is solvable in closed form.

Proof

From $\alpha , \beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in\mathbb{C} \setminus\{0\}$ and (2) we get $z_{n}w_{n}\ne0$ for $n\in \mathbb{N} _{0}$. Hence,

$$\begin{aligned}& w_{n}^{b}=\frac{z_{n+1}}{\alpha z_{n}^{a}}, \quad n\in\mathbb{N}_{0}, \end{aligned}$$

(25)

$$\begin{aligned}& w_{n+1}^{b}=\beta ^{b} w_{n-2}^{bc}z_{n-1}^{bd}, \quad n\in\mathbb{N}_{0}, \end{aligned}$$

(26)

and consequently

$$\begin{aligned} z_{n+2}=\alpha ^{1-c}\beta ^{b} z_{n+1}^{a}z_{n-1}^{bd+c}z_{n-2}^{-ac}, \end{aligned}$$

(27)

for $n\ge2$.

Note also that

$$ \begin{aligned} &z_{1}=\alpha z_{0}^{a}w_{0}^{b},\qquad z_{2}=\alpha \bigl(\alpha z_{0}^{a}w_{0}^{b} \bigr)^{a}\bigl(\beta w_{-2}^{c}z_{-1}^{d} \bigr)^{b}=\alpha ^{1+a}\beta ^{b}z_{-1}^{bd}z_{0}^{a^{2}}w_{-2}^{bc}w_{0}^{ab}, \\ &z_{3}=\alpha z_{2}^{a}w_{2}^{b}= \alpha ^{1+a+a^{2}}\beta ^{b(1+a)}z_{-1}^{abd}z_{0}^{a^{3}+bd}w_{-2}^{abc}w_{-1}^{bc}w_{0}^{a^{2}b}. \end{aligned} $$

(28)

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

$$\begin{aligned} a_{1}=a,\qquad b_{1}=0,\qquad c_{1}=bd+c, \qquad d_{1}=-ac,\qquad y_{1}=1, \end{aligned}$$

(29)

then

$$\begin{aligned} z_{n+2}=\delta ^{y_{1}}z_{n+1}^{a_{1}}z_{n}^{b_{1}}z_{n-1}^{c_{1}}z_{n-2}^{d_{1}}, \quad n\ge2, \end{aligned}$$

(30)

and consequently

$$\begin{aligned} z_{n+2}&=\delta^{y_{1}}\bigl(\delta z_{n}^{a_{1}}z_{n-1}^{b_{1}}z_{n-2}^{c_{1}}z_{n-3}^{d_{1}} \bigr)^{a_{1}}z_{n}^{b_{1}}z_{n-1}^{c_{1}}z_{n-2}^{d_{1}} \\ &=\delta ^{y_{1}+a_{1}}z_{n}^{a_{1}a_{1}+b_{1}}z_{n-1}^{b_{1}a_{1}+c_{1}}z_{n-2}^{c_{1}a_{1}+d_{1}}z_{n-3}^{d_{1}a_{1}} \\ &=\delta^{y_{2}}z_{n}^{a_{2}}z_{n-1}^{b_{2}}z_{n-2}^{c_{2}}z_{n-3}^{d_{2}}, \end{aligned}$$

for $n\ge3$, where

$$\begin{aligned}& a_{2}:=a_{1}a_{1}+b_{1},\qquad b_{2}:=b_{1}a_{1}+c_{1},\qquad c_{2}:=c_{1}a_{1}+d_{1}, \\& d_{2}:=d_{1}a_{1},\qquad y_{2}:=y_{1}+a_{1}. \end{aligned}$$

Assume

$$\begin{aligned} z_{n+2}=\delta ^{y_{k}}z_{n+2-k}^{a_{k}}z_{n+1-k}^{b_{k}}z_{n-k}^{c_{k}}z_{n-k-1}^{d_{k}}, \end{aligned}$$

(31)

for a $k\ge2$ and every $n\ge k+1$, and

$$\begin{aligned}& \begin{aligned} &a_{k}=a_{1}a_{k-1}+b_{k-1}, \qquad b_{k}=b_{1}a_{k-1}+c_{k-1}, \\ &c_{k}=c_{1}a_{k-1}+d_{k-1},\qquad d_{k}=d_{1}a_{k-1}, \end{aligned} \end{aligned}$$

(32)

$$\begin{aligned}& y_{k}=y_{k-1}+a_{k-1}. \end{aligned}$$

(33)

Using (30) in (31), we get

$$\begin{aligned} z_{n+2}&=\delta^{y_{k}}\bigl(\delta z_{n+1-k}^{a_{1}}z_{n-k}^{b_{1}}z_{n-k-1}^{c_{1}}z_{n-k-2}^{d_{1}} \bigr)^{a_{k}}z_{n+1-k}^{b_{k}}z_{n-k}^{c_{k}}z_{n-k-1}^{d_{k}} \\ &=\delta ^{y_{k}+a_{k}}z_{n+1-k}^{a_{1}a_{k}+b_{k}}z_{n-k}^{b_{1}a_{k}+c_{k}}z_{n-k-1}^{c_{1}a_{k}+d_{k}}z_{n-k-2}^{d_{1}a_{k}} \\ &=\delta ^{y_{k+1}}z_{n+1-k}^{a_{k+1}}z_{n-k}^{b_{k+1}}z_{n-k-1}^{c_{k+1}}z_{n-k-2}^{d_{k+1}}, \end{aligned}$$

for $n\ge k+2$, where

$$\begin{aligned}& a_{k+1}=a_{1}a_{k}+b_{k},\qquad b_{k+1}=b_{1}a_{k}+c_{k},\qquad c_{k+1}=c_{1}a_{k}+d_{k}, \qquad d_{k+1}:=d_{1}a_{k}, \\& y_{k+1}:=y_{k}+a_{k}. \end{aligned}$$

Hence, by induction we have proved that (31)-(33) hold.

From (31)-(33) and (28), we get

$$\begin{aligned} z_{n+2} =&\delta ^{y_{n-1}}z_{3}^{a_{n-1}}z_{2}^{b_{n-1}}z_{1}^{c_{n-1}}z_{0}^{d_{n-1}} \\ =&\bigl(\alpha ^{1-c}\beta ^{b}\bigr)^{y_{n-1}}\bigl( \alpha ^{1+a+a^{2}}\beta ^{b(1+a)}z_{-1}^{abd}z_{0}^{a^{3}+bd}w_{-2}^{abc}w_{-1}^{bc}w_{0}^{a^{2}b} \bigr)^{a_{n-1}} \\ &{}\times\bigl(\alpha ^{1+a}\beta ^{b}z_{-1}^{bd}z_{0}^{a^{2}}w_{-2}^{bc}w_{0}^{ab} \bigr)^{b_{n-1}}\bigl(\alpha z_{0}^{a}w_{0}^{b} \bigr)^{c_{n-1}}z_{0}^{d_{n-1}} \\ =&\alpha ^{(1-c)y_{n-1}+(1+a+a^{2})a_{n-1}+(1+a)b_{n-1}+c_{n-1}}\beta ^{by_{n-1}+b(1+a)a_{n-1}+bb_{n-1}} \\ &{}\times z_{-1}^{abda_{n-1}+bdb_{n-1}}z_{0}^{(a^{3}+bd)a_{n-1}+a^{2}b_{n-1}+ac_{n-1}+d_{n-1}}w_{-2}^{abca_{n-1}+bcb_{n-1}}w_{-1}^{bca_{n-1}} \\ &{}\times w_{0}^{a^{2}ba_{n-1}+abb_{n-1}+bc_{n-1}} \\ =&\alpha ^{y_{n+2}-cy_{n-1}}\beta ^{by_{n+1}}z_{-1}^{bda_{n}}z_{0}^{a_{n+2}-ca_{n-1}}w_{-2}^{bca_{n}}w_{-1}^{bca_{n-1}}w_{0}^{ba_{n+1}}, \end{aligned}$$

(34)

for $n\ge2$.

From (32) one sees that $a_{k}$, $b_{k}$, $c_{k}$ and $d_{k}$ are solutions to

$$\begin{aligned} \hat{x}_{k+4}=a_{1}\hat{x}_{k+3}+b_{1} \hat{x}_{k+2}+c_{1}\hat{x}_{k+1}+d_{1} \hat{x}_{k},\quad k\in\mathbb{N}, \end{aligned}$$

(35)

and, along with (33) (for $k=1, 0, -1, -2$), we also obtain

$$\begin{aligned}& a_{-3}=0,\qquad a_{-2}=0,\qquad a_{-1}=0, \qquad a_{0}=1, \end{aligned}$$

(36)

$$\begin{aligned}& y_{-3}=y_{-2}=y_{-1}=y_{0}=0,\qquad y_{1}=1, \end{aligned}$$

(37)

and

$$\begin{aligned} y_{k}=\sum_{j=0}^{k-1}a_{j}. \end{aligned}$$

(38)

The solvability of (35) is well known, from which, along with (36), a formula for $a_{k}$ is obtained. Using it in (38), a formula for $y_{k}$ is obtained by Lemma 2. Hence, (27) is solvable.

We have

$$\begin{aligned}& z_{n-1}^{d}=\frac{w_{n+1}}{\beta w_{n-2}^{c}},\quad n\in \mathbb{N}_{0}, \end{aligned}$$

(39)

$$\begin{aligned}& z_{n+1}^{d}=\alpha ^{d}z_{n}^{ad}w_{n}^{bd}, \quad n\in\mathbb{N}_{0}, \end{aligned}$$

(40)

so that

$$\begin{aligned} w_{n+3}=\alpha ^{d}\beta ^{1-a}w_{n+2}^{a}w_{n}^{bd+c}w_{n-1}^{-ac}, \quad n\in\mathbb{N}_{0}. \end{aligned}$$

(41)

We also have

$$\begin{aligned} w_{1}=\beta w_{-2}^{c}z_{-1}^{d} \quad \mbox{and}\quad w_{2}=\beta w_{-1}^{c}z_{0}^{d}. \end{aligned}$$

(42)

As above we get

$$\begin{aligned} w_{n+3}=\eta ^{y_{k}}w_{n+3-k}^{a_{k}}w_{n+2-k}^{b_{k}}w_{n+1-k}^{c_{k}}w_{n-k}^{d_{k}}, \quad n\ge k-1, \end{aligned}$$

(43)

where $\eta=\alpha ^{d}\beta ^{1-a}$, $a_{k}$ satisfies (35) and (36), and $y_{k}$ is given by (38).

From (43) with $k=n+1$ and by using (42) we get

$$\begin{aligned} w_{n+3} =&\eta ^{y_{n+1}}w_{2}^{a_{n+1}}w_{1}^{b_{n+1}}w_{0}^{c_{n+1}}w_{-1}^{d_{n+1}} \\ =&\bigl(\alpha ^{d}\beta ^{1-a}\bigr)^{y_{n+1}}\bigl( \beta w_{-1}^{c}z_{0}^{d} \bigr)^{a_{n+1}}\bigl(\beta w_{-2}^{c}z_{-1}^{d} \bigr)^{b_{n+1}}w_{0}^{c_{n+1}}w_{-1}^{d_{n+1}} \\ =&\alpha ^{dy_{n+1}}\beta ^{(1-a)y_{n+1}+a_{n+1}+b_{n+1}}z_{-1}^{db_{n+1}}z_{0}^{da_{n+1}}w_{-2}^{cb_{n+1}}w_{-1}^{ca_{n+1}+d_{n+1}}w_{0}^{c_{n+1}} \\ =&\alpha ^{dy_{n+1}}\beta ^{y_{n+3}-ay_{n+2}}z_{-1}^{d(a_{n+2}-aa_{n+1})}z_{0}^{da_{n+1}}w_{-2}^{c(a_{n+2}-aa_{n+1})} \\ &{}\times w_{-1}^{c(a_{n+1}-aa_{n})}w_{0}^{a_{n+3}-aa_{n+2}}, \end{aligned}$$

(44)

for $n\in\mathbb{N}_{0}$.

As we have already seen, formulas for $a_{k}$ and $y_{k}$ can be found. Using them in (44) we show the solvability of (41). Some calculations show that (34) and (44) present a solution to (2), from which the result follows. □

Corollary 1

Assume that $a,b,c,d\in\mathbb{Z}$, $abcd\ne0$, $\alpha ,\beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in\mathbb{C}\setminus\{0\}$. Then the general solution to (2) is given by (34) and (44), where $a_{k}$ satisfies (35) and (36), and $y_{k}$ is given by (37) and (38).

Theorem 4 gives a general form of solutions to system (2) when $abcd\ne0$, but does not present explicit formulas for sequences $a_{n}$ and $y_{n}$ involved in the solutions. Now we give some explicit formulas for them in more concrete cases, following some arguments related to the system in [19]. Since $ac\ne0$, we can find the zeros of the characteristic polynomial associated to (35)

$$\begin{aligned} p_{4}(\lambda )=\lambda ^{4}-a\lambda ^{3}-(bd+c) \lambda +ac. \end{aligned}$$

(45)

To do this, we consider the following equivalent equation with a parameter [25]:

$$\begin{aligned} \biggl(\lambda ^{2}-\frac{a}{2} \lambda +\frac {s}{2} \biggr)^{2}- \biggl( \biggl(\frac{a^{2}}{4}+s \biggr)\lambda ^{2}- \biggl(\frac{as}{2}-(bd+c) \biggr)\lambda + \frac {s^{2}}{4}-ac \biggr)=0. \end{aligned}$$

(46)

The parameter is chosen so that $(as-2(bd+c))^{2}=(a^{2}+4s)(s^{2}-4ac)$, that is,

$$\begin{aligned} s^{3}+a(bd-3c)s-a^{3}c-(bd+c)^{2}=0. \end{aligned}$$

(47)

We have

$$\begin{aligned} \biggl(\lambda ^{2}-\frac{a}{2} \lambda +\frac {s}{2} \biggr)^{2}- \biggl(\frac {\sqrt{a^{2}+4s}}{2} \lambda -\frac{as-2(bd+c)}{2\sqrt{a^{2}+4s}} \biggr)^{2}=0, \end{aligned}$$

(48)

or equivalently

$$\begin{aligned} &\lambda ^{2}- \biggl(\frac{a}{2}+\frac{\sqrt{a^{2}+4s}}{2} \biggr) \lambda +\frac {s}{2}+\frac{as-2(bd+c)}{2\sqrt{a^{2}+4s}}=0, \end{aligned}$$

(49)

$$\begin{aligned} &\lambda ^{2}- \biggl(\frac{a}{2}-\frac{\sqrt{a^{2}+4s}}{2} \biggr) \lambda +\frac {s}{2}-\frac{as-2(bd+c)}{2\sqrt{a^{2}+4s}}=0. \end{aligned}$$

(50)

Let $p=a(bd-3c)$, $q=-a^{3}c-(bd+c)^{2}$, and $s=u+v$. Assuming that $uv=-p/3$, from (47) we get $u^{3}+v^{3}=-q$. Hence, $u^{3}$ and $v^{3}$ are solutions to $z^{2}+qz-p^{3}/27$. Thus

$$\begin{aligned} s=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac {p^{3}}{27}}}+ \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac {p^{3}}{27}}}, \end{aligned}$$

(51)

or

$$\begin{aligned} s=\frac{1}{3\sqrt[3]{2}} \Bigl(\sqrt[3]{\Delta _{1}-\sqrt { \Delta_{1}^{2}-4\Delta_{0}^{3}}}+ \sqrt[3]{\Delta_{1}+\sqrt{\Delta _{1}^{2}-4 \Delta_{0}^{3}}} \Bigr), \end{aligned}$$

(52)

by using the change of variables $p=-\Delta_{0}/3$ and $q=-\Delta_{1}/27$ in (51).

For s given in (52) we solve equations (49) and (50). So, the zeros of polynomial (45) are

$$\begin{aligned}& \lambda _{1}= \frac{a}{4}+\frac{\sqrt{a^{2}+4s}}{4} + \frac{1}{2}\sqrt{\frac{a^{2}}{2}-s-\frac{Q}{2\sqrt{a^{2}+4s}}}, \end{aligned}$$

(53)

$$\begin{aligned}& \lambda _{2}= \frac{a}{4}+\frac{\sqrt{a^{2}+4s}}{4} - \frac{1}{2}\sqrt{\frac{a^{2}}{2}-s-\frac{Q}{2\sqrt{a^{2}+4s}}}, \end{aligned}$$

(54)

$$\begin{aligned}& \lambda _{3}= \frac{a}{4}-\frac{\sqrt{a^{2}+4s}}{4} + \frac{1}{2}\sqrt{\frac{a^{2}}{2}-s+\frac{Q}{2\sqrt{a^{2}+4s}}}, \end{aligned}$$

(55)

$$\begin{aligned}& \lambda _{4}= \frac{a}{4}-\frac{\sqrt{a^{2}+4s}}{4} - \frac{1}{2}\sqrt{\frac{a^{2}}{2}-s+\frac{Q}{2\sqrt{a^{2}+4s}}}, \end{aligned}$$

(56)

where

$$\begin{aligned}& \Delta _{0}:= 3a(3c-bd), \end{aligned}$$

(57)

$$\begin{aligned}& \Delta _{1}:= 27\bigl(a^{3}c+(bd+c)^{2} \bigr), \end{aligned}$$

(58)

$$\begin{aligned}& Q:= -a^{3}-8bd-8c. \end{aligned}$$

(59)

By Lemma 3, the nature of $\lambda _{j}$, $j=\overline {1,4}$, depends also on

$$\begin{aligned}& \Delta = \frac{1}{27}\bigl(4\Delta _{0}^{3}-\Delta _{1}^{2}\bigr), \end{aligned}$$

(60)

$$\begin{aligned}& P= -3a^{2} , \end{aligned}$$

(61)

$$\begin{aligned}& D= a\bigl(48c-16bd-3a^{3}\bigr). \end{aligned}$$

(62)

Zeros of $p_{4}$ are mutually different and different from 1. If $a=1$, $c=2$ and $bd=3$, polynomial (45) becomes

$$p_{4}(\lambda )=\lambda ^{4}-\lambda ^{3}-5 \lambda +2=(\lambda -2) \bigl(\lambda ^{3}+\lambda ^{2}+2 \lambda -1\bigr). $$

Since in this case $\Delta<0$, all the zeros of the polynomial are different. Since $p_{4}(1)\ne0$, 1 is not a zero of the polynomial. In fact, there are many polynomials of the form in (45) such that $\Delta<0$. For example, they are those for which holds $3ac< abd$, that is, $\Delta _{0}<0$.

Since $\lambda _{j}\ne\lambda _{i}$, $i\ne j$,

$$\begin{aligned} a_{n}=\gamma _{1}\lambda _{1}^{n}+ \gamma _{2}\lambda _{2}^{n}+\gamma _{3} \lambda _{3}^{n}+\gamma _{4}\lambda _{4}^{n},\quad n\in\mathbb{N}, \end{aligned}$$

(63)

where $\gamma _{i}$, $i=\overline {1,4}$ are constants, is the general solution to (35).

Equalities (36), along with Lemma 1 applied to polynomial (45), yield

$$\begin{aligned} a_{n} =&\sum_{j=1}^{4} \frac{\lambda _{j}^{n+3}}{p_{4}'(\lambda_{j})} \\ =&\frac{\lambda _{1}^{n+3}}{(\lambda _{1}-\lambda _{2})(\lambda _{1}-\lambda _{3})(\lambda _{1}-\lambda _{4})}+\frac{\lambda _{2}^{n+3}}{(\lambda _{2}-\lambda _{1})(\lambda _{2}-\lambda _{3})(\lambda _{2}-\lambda _{4})} \\ &{} +\frac{\lambda _{3}^{n+3}}{(\lambda _{3}-\lambda _{1})(\lambda _{3}-\lambda _{2})(\lambda _{3}-\lambda _{4})}+\frac{\lambda _{4}^{n+3}}{(\lambda _{4}-\lambda _{1})(\lambda _{4}-\lambda _{2})(\lambda _{4}-\lambda _{3})}, \end{aligned}$$

(64)

for $n\ge-3$, from which, along with (38) and the fact that $\lambda _{i}\ne1$, $i=\overline {1,4}$, is obtained:

$$\begin{aligned} y_{n}=\sum_{j=0}^{n-1}\sum _{i=1}^{4}\frac{\lambda _{i}^{j+3}}{p_{4}'(\lambda_{i})}=\sum _{i=1}^{4}\frac{\lambda _{i}^{3}(\lambda _{i}^{n}-1)}{p_{4}'(\lambda_{i})(\lambda _{i}-1)}, \quad n\in \mathbb{N}. \end{aligned}$$

(65)

Moreover, (65) holds for $n\ge-3$.

Zeros of $p_{4}$ are different and one of them is 1. In this case it must be $p_{4}(1)=1-a-bd-c+ac=0$. Hence,

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

(66)

which implies

$$\begin{aligned} p_{4}(\lambda )&=\lambda ^{4}-a\lambda ^{3}-(ac-a+1)\lambda +ac \\ &=(\lambda -1) \bigl(\lambda ^{3}-(a-1)\lambda ^{2}-(a-1) \lambda -ac\bigr). \end{aligned}$$

(67)

Let $\lambda _{1}=1$. To find the other zeros of $p_{4}$, we have to solve the equation

$$\lambda ^{3}-(a-1)\lambda ^{2}-(a-1)\lambda -ac=0. $$

By using the change of variables $\lambda =t+\frac{a-1}{3}$ and some simple calculations, we get

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

where

$$\tilde{p}=\frac{(1-a)(a+2)}{3} \quad \mbox{and} \quad \tilde{q}=- \biggl( \frac {2(a-1)^{3}}{27}+\frac{(a-1)^{2}}{3}+ac \biggr). $$

Using the standard arguments, as those in getting (51), we obtain

$$\begin{aligned} \lambda _{j}=\frac{a-1}{3}+\varepsilon ^{j-2}\sqrt [3]{-\frac{\tilde{q}}{2}-\sqrt {\frac{\tilde{q}^{2}}{4}+\frac{\tilde{p}^{3}}{27}}}+\bar{\varepsilon}^{j-2}\sqrt [3]{-\frac{\tilde{q}}{2}+\sqrt{ \frac{\tilde{q}^{2}}{4}+\frac{\tilde{p}^{3}}{27}}}, \quad j=\overline {2,4}, \end{aligned}$$

(68)

where ε is such that $\varepsilon ^{3}=1$, $\varepsilon \ne1$.

For example, if $a=3$ and $c=2$, then $bd=2\ne0$, $\Delta \ne0$ and

$$p_{4}(\lambda )=\lambda ^{4}-3\lambda ^{3}-4 \lambda +6=(\lambda -1) \bigl(\lambda ^{3}-2\lambda ^{2}-2 \lambda -6\bigr), $$

so by Lemma 3, the polynomial has four different zeros, and one of them is 1.

Equality (64) holds with, say, $\lambda _{1}=1$. Further, we have

$$ y_{n}= \sum_{j=0}^{n-1} \frac{1}{p_{4}'(1)}+\sum_{j=0}^{n-1}\sum _{i=2}^{4}\frac{\lambda _{i}^{j+3}}{p_{4}'(\lambda_{i})} = \frac{n}{3-2a-ac}+\sum_{i=2}^{4} \frac{\lambda _{i}^{3}(\lambda _{i}^{n}-1)}{p_{4}'(\lambda_{i})(\lambda _{i}-1)}, $$

(69)

for $n\in\mathbb{N}$. It is easily shown that (69) also holds for $n=-j$, $j=\overline {0,3}$.

This analysis, along with Corollary 1, implies the following result.

Corollary 2

Assume that $a,b,c,d\in\mathbb{Z}$, $abcd\ne0$, $\alpha , \beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in\mathbb{C}\setminus\{0\}$ and $\Delta\ne0$. Then the following statements are true:

• (a) If $(a-1)(c-1)\ne bd$, then the general solution to (2) is given by (34) and (44), where $(a_{n})_{n\ge-3}$ is given by (64), $(y_{n})_{n\ge-3}$ is given by (65), while $\lambda _{j}$, $j=\overline {1,4}$, are given by (53)-(56).

• (b) If $(a-1)(c-1)=bd$ and $3-2a\ne ac$, then the general solution to (2) is given by (34) and (44), where $(a_{n})_{n\ge-3}$ is given by (64) with $\lambda _{1}=1$, $(y_{n})_{n\ge -3}$ is given by (69), $\lambda _{1}=1$, while $\lambda _{j}$, $j=\overline {2,4}$, are given by (68).

1 is the only double zero of $p_{4}$. Polynomial $p_{4}$ has a double zero equal to 1 if (66) holds and

$$\begin{aligned} p_{4}'(1)=3-2a-ac=0, \end{aligned}$$

(70)

that is, if and only if

$$\begin{aligned} c=\frac{3}{a}-2. \end{aligned}$$

(71)

Then we have

$$p_{4}(\lambda )=\lambda ^{4}-a\lambda ^{3}+(3a-4) \lambda +3-2a=(\lambda -1)^{2}\bigl(\lambda ^{2}+(2-a)\lambda +3-2a\bigr), $$

and consequently

$$\begin{aligned} \lambda _{1,2}=1,\qquad \lambda _{3,4}=\frac{a-2\pm \sqrt {a^{2}+4a-8}}{2}. \end{aligned}$$

(72)

From (71) we must have $a=3$ and $c=-1$, or $a=1$ and $c=1$, or $a=-1$ and $c=-5$, or $a=-3$ and $c=-3$.

If $a=c=1$, then

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

and consequently

$$\lambda _{1,2}=1,\qquad \lambda _{3,4}=\frac{-1\pm i\sqrt{3}}{2}. $$

Since (66) holds we see that this case is not possible when $abcd\ne0$.

If $a=3$, $c=-1$, then

$$p_{4}(\lambda )=\lambda ^{4}-3\lambda ^{3}+5 \lambda -3=(\lambda -1)^{2}\bigl(\lambda ^{2}-\lambda -3 \bigr), $$

and consequently

$$\begin{aligned} \lambda _{1,2}=1,\qquad \lambda _{3,4}=\frac{1\pm \sqrt{13}}{2}. \end{aligned}$$

(73)

If $a=c=-3$, then

$$p_{4}(\lambda )=\lambda ^{4}+3\lambda ^{3}-13 \lambda +9=(\lambda -1)^{2}\bigl(\lambda ^{2}+5\lambda +9 \bigr), $$

and consequently

$$\begin{aligned} \lambda _{1,2}=1, \qquad \lambda _{3,4}= \frac{-5\pm i\sqrt{11}}{2}. \end{aligned}$$

(74)

If $a=-1$ and $c=-5$, then

$$p_{4}(\lambda )=\lambda ^{4}+\lambda ^{3}-7 \lambda +5=(\lambda -1)^{2}\bigl(\lambda ^{2}+3\lambda +5 \bigr), $$

and consequently

$$\begin{aligned} \lambda _{1,2}=1, \qquad \lambda _{3,4}= \frac{-3\pm i\sqrt{11}}{2}. \end{aligned}$$

(75)

In these four cases, we have [19]

$$\begin{aligned} a_{n}={}&\frac{n(1-\lambda _{3})(1-\lambda _{4})+3\lambda _{3}\lambda _{4}-2\lambda _{3}-2\lambda _{4}+1}{(1-\lambda _{3})^{2}(1-\lambda _{4})^{2}} \\ &{}+\frac{\lambda _{3}^{n+3}}{(\lambda _{3}-1)^{2}(\lambda _{3}-\lambda _{4})}+\frac{\lambda _{4}^{n+3}}{(\lambda _{4}-1)^{2}(\lambda _{4}-\lambda _{3})} \end{aligned}$$

(76)

and

$$\begin{aligned} y_{n}={}&\sum_{j=0}^{n-1} \biggl(\frac{j(1-\lambda _{3})(1-\lambda _{4})+3\lambda _{3}\lambda _{4}-2\lambda _{3}-2\lambda _{4}+1}{(1-\lambda _{3})^{2}(1-\lambda _{4})^{2}} \\ &{} +\frac{\lambda _{3}^{j+3}}{(\lambda _{3}-1)^{2}(\lambda _{3}-\lambda _{4})}+\frac{\lambda _{4}^{j+3}}{(\lambda _{4}-1)^{2}(\lambda _{4}-\lambda _{3})} \biggr) \\ ={}&\frac{(n-1)n}{2(1-\lambda _{3})(1-\lambda _{4})} +\frac{n(3\lambda _{3}\lambda _{4}-2\lambda _{3}-2\lambda _{4}+1)}{(1-\lambda _{3})^{2}(1-\lambda _{4})^{2}} \\ &{}+\frac{\lambda _{3}^{3}(\lambda _{3}^{n}-1)}{(\lambda _{3}-1)^{3}(\lambda _{3}-\lambda _{4})}+\frac{\lambda _{4}^{3}(\lambda _{4}^{n}-1)}{(\lambda _{4}-1)^{3}(\lambda _{4}-\lambda _{3})}. \end{aligned}$$

(77)

$p_{4}$ has only one double zero different from 1. Assume that $\lambda =m\notin\{0,1\}$ is a double zero of $p_{4}$. Then we have

$$\begin{aligned} m^{4}-am^{3}-(bd+c)m+ac=0 \quad \mbox{and}\quad 4m^{3}-3am^{2}-bd-c=0. \end{aligned}$$

(78)

If m is not a triple zero, then it must be $12m^{2}-6am\ne0$, that is, $a\ne2m$.

From (78), we get

$$\begin{aligned} p_{4}(\lambda )&=\lambda ^{4}-a\lambda ^{3}+ \bigl(3am^{2}-4m^{3}\bigr)\lambda +3m^{4}-2am^{3} \\ &=(\lambda -m)^{2}\bigl(\lambda ^{2}+(2m-a)\lambda +m(3m-2a) \bigr), \end{aligned}$$

(79)

and consequently

$$\begin{aligned} \lambda _{1,2}=m, \qquad \lambda _{3,4}= \frac{a-2m\pm \sqrt {-8m^{2}+4am+a^{2}}}{2}. \end{aligned}$$

(80)

Hence, if we additionally assume that $2a\ne3m$, $3am^{2}-4m^{3}\in \mathbb{Z}$, $3m^{4}-2am^{3}\in\mathbb{Z}$, we get a family of polynomials of the form in (45) which have double zeros different from 1. For example, if $a=m\in\mathbb{Z}\setminus\{0,1\}$, then from (79) it follows that

$$p_{4}(\lambda )=(\lambda -a)^{2}\bigl(\lambda ^{2}+a\lambda +a^{2}\bigr). $$

Since, in the case $\lambda _{1}=\lambda _{2}$, $\lambda _{i}\ne\lambda _{j}$, $2\le i,j\le4$, we have

$$\begin{aligned} a_{n}=(\gamma _{1}+\gamma _{2}n)\lambda _{2}^{n}+\gamma _{3}\lambda _{3}^{n}+ \gamma _{4}\lambda _{4}^{n},\quad n\in\mathbb{N} , \end{aligned}$$

(81)

where $\gamma _{i}$ and $i=\overline {1,4}$ are constants, and the solution satisfying (36) is

$$\begin{aligned} a_{n}={}&\frac{\lambda _{2}^{n+2}((n+3)(\lambda _{2}-\lambda _{3})(\lambda _{2}-\lambda _{4})-\lambda _{2}(2\lambda _{2}-\lambda _{3}-\lambda _{4})) }{(\lambda _{2}-\lambda _{3})^{2}(\lambda _{2}-\lambda _{4})^{2}} \\ &{} +\frac{\lambda _{3}^{n+3}}{(\lambda _{3}-\lambda _{2})^{2}(\lambda _{3}-\lambda _{4})}+\frac {\lambda _{4}^{n+3}}{(\lambda _{4}-\lambda _{2})^{2}(\lambda _{4}-\lambda _{3})}. \end{aligned}$$

(82)

From (38) and (82) and by Lemma 2, we get

$$\begin{aligned} y_{n}={}&\sum_{j=0}^{n-1} \biggl(\frac{\lambda _{2}^{j+2}((j+3)(\lambda _{2}-\lambda _{3})(\lambda _{2}-\lambda _{4})-\lambda _{2}(2\lambda _{2}-\lambda _{3}-\lambda _{4})) }{(\lambda _{2}-\lambda _{3})^{2}(\lambda _{2}-\lambda _{4})^{2}} \\ &{} +\frac{\lambda _{3}^{j+3}}{(\lambda _{3}-\lambda _{2})^{2}(\lambda _{3}-\lambda _{4})}+\frac{\lambda _{4}^{j+3}}{(\lambda _{4}-\lambda _{2})^{2}(\lambda _{4}-\lambda _{3})} \biggr) \\ ={}&\frac{\lambda _{2}^{3}-n\lambda _{2}^{n+2}+(n-1)\lambda _{2}^{n+3}}{(\lambda _{2}-\lambda _{3})(\lambda _{2}-\lambda _{4})(1-\lambda _{2})^{2}} +\frac{(\lambda _{2}^{4}-2\lambda _{2}^{3}\lambda _{3}-2\lambda _{2}^{3}\lambda _{4}+3\lambda _{2}^{2}\lambda _{3}\lambda _{4})(\lambda _{2}^{n}-1)}{(\lambda _{2}-\lambda _{3})^{2}(\lambda _{2}-\lambda _{4})^{2}(\lambda _{2}-1)} \\ &{}+\frac{\lambda _{3}^{3}(\lambda _{3}^{n}-1)}{(\lambda _{3}-\lambda _{2})^{2}(\lambda _{3}-\lambda _{4})(\lambda _{3}-1)}+\frac{\lambda _{4}^{3}(\lambda _{4}^{n}-1)}{(\lambda _{4}-\lambda _{2})^{2}(\lambda _{4}-\lambda _{3})(\lambda _{4}-1)}. \end{aligned}$$

(83)

Corollary 3

Assume that $a,b,c,d\in\mathbb{Z}$, $abcd\ne0$ and $\alpha , \beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in\mathbb{C}\setminus\{0\}$. Then the following statements are true:

• (a) If only one of the zeros of $p_{4}$ is double and different from 1, say $\lambda _{1}$ and $\lambda _{2}$, then the general solution to (2) is given by (34) and (44), where $(a_{n})_{n\ge-3}$ is given by (82), $(y_{n})_{n\ge-3}$ is given by (83), while $\lambda _{j}$, $j=\overline {1,4}$, are given by (80), where $m\notin\{0,1\}$, $2m\ne a\ne3m$ and $3am^{2}-4m^{3}, 3m^{4}-2am^{3}\in\mathbb{Z}$.

• (b) If 1 is a unique double zero of polynomial $p_{4}$, say $\lambda _{1}=\lambda _{2}=1$, then the general solution to (2) is given by (34) and (44), where $(a_{n})_{n\ge-3}$ is given by (76), $(y_{n})_{n\ge-3}$ is given by (77), while $\lambda _{j}$, $j=\overline {1,4}$, are given by (73) when $a=3$, $c=-1$, by (74) when $a=c=-3$, and by (75) when $a=-1$, $c=-5$.

$p_{4}$ has two pairs of different double zeros. In this case it must be $D=0$, which implies that $a=0$ or $16bd=48c-3a^{3}$. The case $a=0$ is impossible due to the condition $abcd\ne0$. In the other case we have $\Delta =0$ if and only if

$$4 \biggl(3a \biggl(3c+\frac{3a^{3}-48c}{16} \biggr) \biggr)^{3}= \biggl(27 \biggl(a^{3}c+ \biggl(4c-\frac{3a^{3}}{16} \biggr)^{2} \biggr) \biggr)^{2}, $$

that is,

$$\begin{aligned} 2^{3}a^{6}=\pm\bigl(3^{2}a^{6}-2^{7}a^{3}c+2^{12}c^{2} \bigr). \end{aligned}$$

(84)

From (84) we have

$$17a^{6}-2^{7}a^{3}c+2^{12}c^{2}=0, $$

from which it follows that $a^{3}/c=\frac{2^{6}}{17}(1\pm4i)$, which is impossible due to the rationality of $a^{3}/c$, or is

$$a^{6}-2^{7}a^{3}c+2^{12}c^{2}= \bigl(a^{3}-2^{6}c\bigr)^{2}=0, $$

which implies $c=a^{3}/2^{6}$.

Assume $c=a^{3}/2^{6}$. Then

$$ p_{4}(\lambda )=\lambda ^{4}-a\lambda ^{3}+ \frac{a^{3}}{8}\lambda +\frac {a^{4}}{2^{6}}= \biggl(\lambda ^{2}- \frac{a\lambda }{2}-\frac{a^{2}}{8} \biggr)^{2} $$

(for more details see [19], p.14).

Hence

$$\begin{aligned} \lambda _{1,2}=\frac{a}{4}(1+\sqrt{3}), \qquad \lambda _{3,4}=\frac {a}{4}(1-\sqrt{3}), \end{aligned}$$

(85)

are two double zeros of $p_{4}$, for each $a\ne0$.

Since

$$a(1\pm\sqrt{3})/4\ne1, $$

for every $a\in\mathbb{Z}$, $p_{4}$ cannot have two pairs of double zeros such that one of them is equal to 1.

The general solution to (35) in this case is of the following form:

$$\begin{aligned} a_{n}=(\gamma _{1}+\gamma _{2}n)\lambda _{1}^{n}+(\gamma _{3}+\gamma _{4}n) \lambda _{3}^{n}, \quad n\in\mathbb{N}, \end{aligned}$$

(86)

for some constants $\gamma _{i}$, $i=\overline {1,4}$. The solution with initial conditions (36) is

$$\begin{aligned} \begin{aligned}[b] a_{n}={}&\frac{\lambda _{2}^{n+2}(n(\lambda _{2}-\lambda _{4})^{2}+\lambda _{2}^{2}-4\lambda _{2}\lambda _{4}+3\lambda _{4}^{2}) }{(\lambda _{2}-\lambda _{4})^{4}} \\ &{}+\frac{\lambda _{4}^{n+2}(n(\lambda _{4}-\lambda _{2})^{2}+\lambda _{4}^{2}-4\lambda _{2}\lambda _{4}+3\lambda _{2}^{2})}{(\lambda _{4}-\lambda _{2})^{4}}. \end{aligned} \end{aligned}$$

(87)

From (38), (87) and Lemma 2, we get

$$\begin{aligned} y_{n} ={}&\sum_{j=0}^{n-1} \biggl(\frac{\lambda _{2}^{j+2}(j(\lambda _{2}-\lambda _{4})^{2}+\lambda _{2}^{2}-4\lambda _{2}\lambda _{4}+3\lambda _{4}^{2})}{(\lambda _{2}-\lambda _{4})^{4}} \\ &{} +\frac{\lambda _{4}^{j+2}(j(\lambda _{4}-\lambda _{2})^{2}+\lambda _{4}^{2}-4\lambda _{2}\lambda _{4}+3\lambda _{2}^{2})}{(\lambda _{4}-\lambda _{2})^{4}} \biggr) \\ ={}&\frac{\lambda _{2}^{3}-n\lambda _{2}^{n+2}+(n-1)\lambda _{2}^{n+3}}{(\lambda _{2}-\lambda _{4})^{2}(1-\lambda _{2})^{2}} +\frac{(\lambda _{2}^{4}-4\lambda _{2}^{3}\lambda _{4}+3\lambda _{2}^{2}\lambda _{4}^{2})(\lambda _{2}^{n}-1)}{(\lambda _{2}-\lambda _{4})^{4}(\lambda _{2}-1)} \\ &{}+\frac{\lambda _{4}^{3}-n\lambda _{4}^{n+2}+(n-1)\lambda _{4}^{n+3}}{(\lambda _{4}-\lambda _{2})^{2}(1-\lambda _{4})^{2}} +\frac{(\lambda _{4}^{4}-4\lambda _{2}\lambda _{4}^{3}+3\lambda _{2}^{2}\lambda _{4}^{2})(\lambda _{4}^{n}-1)}{(\lambda _{4}-\lambda _{2})^{4}(\lambda _{4}-1)}. \end{aligned}$$

(88)

Corollary 4

Assume that $a,b,c,d\in\mathbb{Z}$, $abcd\ne0$ and $\alpha , \beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in\mathbb{C}\setminus\{0\}$. Then the following statements are true:

• (a) If polynomial $p_{4}$ has two pairs of double zeros both different from 1, then the general solution to (2) is given by (34) and (44), where $(a_{n})_{n\ge-3}$ is given by (87), $(y_{n})_{n\ge-3}$ is given by (88), while $\lambda _{j}$, $j=\overline {1,4}$, are given by (85).

• (b) The characteristic polynomial (45) cannot have two pairs of double zeros such that one of them is equal to 1.

Triple zero case. In this case we must have $\Delta =\Delta _{0}=0$ or equivalently $\Delta _{0}=\Delta _{1}=0$, that is,

$$a=0 \quad \mbox{or}\quad bd=3c. $$

Since $abcd\ne0$, the case $a=0$ is not possible. If $c=bd/3$, then

$$\Delta _{1}=27c\bigl(a^{3}+16c\bigr). $$

Since the case $c=0$ is excluded, we must have $c=-a^{3}/16$.

We have

$$ p_{4}(\lambda )=\lambda ^{4}-a\lambda ^{3}+ \frac{a^{3}}{4}\lambda -\frac {a^{4}}{16}= \biggl(\lambda -\frac {a}{2} \biggr)^{3} \biggl(\lambda +\frac{a}{2} \biggr) $$

(see, for example, [19], p.15), and consequently

$$\begin{aligned} \lambda _{j}=\frac{a}{2}, \quad j=\overline {1,3},\qquad \lambda _{4}=-\frac{a}{2}. \end{aligned}$$

(89)

Thus, for every $a\ne0$, $a/2$ is a triple zero of $p_{4}$, and $p_{4}$ cannot have a zero of the fourth order.

Hence

$$\begin{aligned} a_{n}=\bigl(\gamma _{1}+\gamma _{2}n+\gamma _{3}n^{2}\bigr)\lambda _{1}^{n}+\gamma _{4}\lambda _{4}^{n}, \quad n\in \mathbb{N}, \end{aligned}$$

(90)

where $\gamma _{i}$ and $i=\overline {1,4}$ are constants, is the general solution to (35) in this case.

Further, by using the initial conditions in (36), we obtain

$$\begin{aligned}& \gamma _{1}=1-\frac{\lambda _{4}^{3}}{(\lambda _{4}-\lambda _{1})^{3}},\qquad \gamma _{2}= \frac{\lambda _{1}(3\lambda _{1}-5\lambda _{4})}{2(\lambda _{4}-\lambda _{1})^{2}}, \\& \gamma _{3}=\frac {\lambda _{1}}{2(\lambda _{1}-\lambda _{4})},\qquad \gamma _{4}=\frac{\lambda ^{3}_{4}}{(\lambda _{4}-\lambda _{1})^{3}}. \end{aligned}$$

Thus

$$\begin{aligned} a_{n}=1-\frac{\lambda _{4}^{3}}{(\lambda _{4}-1)^{3}}+\frac {3-5\lambda _{4}}{2(\lambda _{4}-1)^{2}}n+\frac{1}{2(1-\lambda _{4})}n^{2}+ \frac{\lambda ^{n+3}_{4}}{(\lambda _{4}-1)^{3}},\quad n\ge-3, \end{aligned}$$

(91)

when $\lambda _{1}=1$, while if $\lambda _{1}\ne1$, then

$$\begin{aligned} a_{n}= \biggl(1-\frac{\lambda _{4}^{3}}{(\lambda _{4}-\lambda _{1})^{3}}+\frac{\lambda _{1}(3\lambda _{1}-5\lambda _{4})}{2(\lambda _{4}-\lambda _{1})^{2}}n+ \frac{\lambda _{1}}{2(\lambda _{1}-\lambda _{4})}n^{2} \biggr)\lambda _{1}^{n}+ \frac{\lambda ^{n+3}_{4}}{(\lambda _{4}-\lambda _{1})^{3}}, \end{aligned}$$

(92)

for $n\ge-3$.

From (38), (91) and Lemma 2, it follows that

$$ y_{n}= \biggl(1-\frac{\lambda _{4}^{3}}{(\lambda _{4}-1)^{3}} \biggr)n+\frac {(3-5\lambda _{4})(n-1)n}{4(\lambda _{4}-1)^{2}}+ \frac {(n-1)n(2n-1)}{12(1-\lambda _{4})}+\frac{\lambda ^{3}_{4}(\lambda ^{n}_{4}-1)}{(\lambda _{4}-1)^{4}}. $$

(93)

From (38), (92) and Lemma 2, it follows that

$$\begin{aligned} y_{n}={}& \biggl(1-\frac{\lambda _{4}^{3}}{(\lambda _{4}-\lambda _{1})^{3}} \biggr)\frac {\lambda _{1}^{n}-1}{\lambda _{1}-1}+ \frac{\lambda _{1}^{2}(3\lambda _{1}-5\lambda _{4})(1-n\lambda _{1}^{n-1}+(n-1)\lambda _{1}^{n})}{2(\lambda _{4}-\lambda _{1})^{2}(1-\lambda _{1})^{2}} \\ &{}+\frac{\lambda _{1}^{2}(1+\lambda _{1}-n^{2}\lambda _{1}^{n-1}+(2n^{2}-2n-1)\lambda _{1}^{n}-(n-1)^{2}\lambda _{1}^{n+1})}{2(\lambda _{1}-\lambda _{4})(1-\lambda _{1})^{3}} \\ &{}+\frac{\lambda ^{3}_{4}(\lambda ^{n}_{4}-1)}{(\lambda _{4}-\lambda _{1})^{3}(\lambda _{4}-1)}, \end{aligned}$$

(94)

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

Corollary 5

Assume that $a,b,c,d\in\mathbb{Z}$, $abcd\ne0$ and $\alpha , \beta , z_{-1}, z_{0}, w_{-2}, w_{-1}, w_{0}\in\mathbb{C}\setminus\{0\}$. Then the following statements are true:

• (a) If polynomial (45) has a triple zero different from 1, then the general solution to system (2) is given by (34) and (44), where $(a_{n})_{n\ge-3}$ is given by (92), $(y_{n})_{n\ge-3}$ is given by (94), while $\lambda _{j}$, $j=\overline {1,4}$, are given by (89).

• (b) If polynomial (45) has a triple zero equal to 1, say $\lambda _{1}$, $\lambda _{2}$ and $\lambda _{3}$, then the general solution to system (2) is given by (34) and (44), where $(a_{n})_{n\ge-3}$ is given by (91), $(y_{n})_{n\ge-3}$ is given by (93), while $\lambda _{j}$, $j=\overline {1,4}$, are given by (89) with $a=2$.

Theorem 5

Assume that $a,b,d\in\mathbb{Z}$, $abd\ne 0$, $c=0$, $\alpha , \beta , z_{-1}, z_{0}, w_{0}\in\mathbb{C}\setminus \{0\}$. Then system (2) is solvable in closed form.

Proof

We modify our method in [18, 24]. We have

$$\begin{aligned} z_{n+1}=\alpha z_{n}^{a}w_{n}^{b}, \qquad w_{n+1}=\beta z_{n-1}^{d}, \quad n\in \mathbb{N}_{0}, \end{aligned}$$

(95)

and consequently

$$\begin{aligned} z_{n+1}=\alpha \beta ^{b} z_{n}^{a}z_{n-2}^{bd}, \quad n\in \mathbb{N}. \end{aligned}$$

(96)

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

$$\begin{aligned} a_{1}=a,\qquad b_{1}=0,\qquad c_{1}=bd, \qquad y_{1}=1. \end{aligned}$$

(97)

Then clearly

$$\begin{aligned} z_{n+1}=\delta^{y_{1}}z_{n}^{a_{1}}z_{n-1}^{b_{1}}z_{n-2}^{c_{1}}, \quad n\in\mathbb{N}. \end{aligned}$$

(98)

Hence,

$$\begin{aligned} z_{n+1}&=\delta^{y_{1}}\bigl(\delta z_{n-1}^{a_{1}}z_{n-2}^{b_{1}}z_{n-3}^{c_{1}} \bigr)^{a_{1}}z_{n-1}^{b_{1}}z_{n-2}^{c_{1}} \\ &=\delta ^{y_{1}+a_{1}}z_{n-1}^{a_{1}a_{1}+b_{1}}z_{n-2}^{b_{1}a_{1}+c_{1}}z_{n-3}^{c_{1}a_{1}} \\ &=\delta^{y_{2}}z_{n-1}^{a_{2}}z_{n-2}^{b_{2}}z_{n-3}^{c_{2}}, \end{aligned}$$

for $n\ge2$, where

$$a_{2}:=a_{1}a_{1}+b_{1}, \qquad b_{2}:=b_{1}a_{1}+c_{1},\qquad c_{2}:=c_{1}a_{1}, \qquad y_{2}:=y_{1}+a_{1}. $$

Assume

$$\begin{aligned} z_{n+1}=\delta ^{y_{k}}z_{n+1-k}^{a_{k}}z_{n-k}^{b_{k}}z_{n-k-1}^{c_{k}}, \end{aligned}$$

(99)

for a $k\ge2$ and every $n\ge k$, and

$$\begin{aligned}& a_{k}=a_{1}a_{k-1}+b_{k-1},\qquad b_{k}=b_{1}a_{k-1}+c_{k-1},\qquad c_{k}=c_{1}a_{k-1}, \end{aligned}$$

(100)

$$\begin{aligned}& y_{k}=y_{k-1}+a_{k-1}. \end{aligned}$$

(101)

Further, by (98), it follows that

$$\begin{aligned} z_{n+1}&=\delta^{y_{k}}\bigl(\delta z_{n-k}^{a_{1}}z_{n-k-1}^{b_{1}}z_{n-k-2}^{c_{1}} \bigr)^{a_{k}}z_{n-k}^{b_{k}}z_{n-k-1}^{c_{k}} \\ &=\delta ^{y_{k}+a_{k}}z_{n-k}^{a_{1}a_{k}+b_{k}}z_{n-k-1}^{b_{1}a_{k}+c_{k}}z_{n-k-2}^{c_{1}a_{k}} \\ &=\delta ^{y_{k+1}}z_{n-k}^{a_{k+1}}z_{n-k-1}^{b_{k+1}}z_{n-k-2}^{c_{k+1}}, \end{aligned}$$

for $n\ge k+1$, where

$$\begin{aligned}& a_{k+1}:=a_{1}a_{k}+b_{k}, \qquad b_{k+1}:=b_{1}a_{k}+c_{k}, \\& c_{k+1}:=c_{1}a_{k},\qquad y_{k+1}:=y_{k}+a_{k}. \end{aligned}$$

Hence, by induction we see that (99)-(101) hold.

Setting $k=n$ in (99), and employing (28), we get

$$\begin{aligned} \begin{aligned}[b] z_{n+1} &=\delta^{y_{n}}z_{1}^{a_{n}}z_{0}^{b_{n}}z_{-1}^{c_{n}} \\ &=\bigl(\alpha \beta ^{b}\bigr)^{y_{n}}\bigl(\alpha z_{0}^{a}w_{0}^{b} \bigr)^{a_{n}}z_{0}^{b_{n}}z_{-1}^{c_{n}} \\ &=\alpha ^{y_{n}+a_{n}}\beta ^{by_{n}}z_{-1}^{c_{n}}z_{0}^{aa_{n}+b_{n}}w_{0}^{ba_{n}} \\ &=\alpha ^{y_{n+1}}\beta ^{by_{n}}z_{-1}^{bda_{n-1}}z_{0}^{a_{n+1}}w_{0}^{ba_{n}}, \end{aligned} \end{aligned}$$

(102)

for $n\ge2$.

From (100) we see that $a_{k}$, $b_{k}$ and $c_{k}$ are solutions to

$$\begin{aligned} \tilde{x}_{k+3}=a_{1}\tilde{x}_{k+2}+b_{1} \tilde{x}_{k+1}+c_{1}\tilde{x}_{k},\quad k\in \mathbb{N}, \end{aligned}$$

(103)

and that along with (100) and (101) (for $k=0, -1, -2$), we obtain

$$\begin{aligned}& a_{-2}=0, \qquad a_{-1}=0,\qquad a_{0}=1, \end{aligned}$$

(104)

$$\begin{aligned}& y_{-2}=y_{-1}=y_{0}=0,\qquad y_{1}=1, \end{aligned}$$

(105)

and

$$\begin{aligned} y_{k}=\sum_{j=0}^{k-1}a_{j}. \end{aligned}$$

(106)

The solvability of (103) is well known, from which along with (104) is obtained a formula for $a_{k}$, which along with (106) and Lemma 2 yields a formula for $y_{k}$. Hence, (96) is solvable.

Using (102), in the second equation in (95), is obtained:

$$\begin{aligned} w_{n}=\alpha ^{dy_{n-2}}\beta ^{bdy_{n-3}+1}z_{-1}^{bd^{2}a_{n-4}}z_{0}^{da_{n-2}}w_{0}^{bda_{n-3}}, \quad n\ge5. \end{aligned}$$

(107)

It is shown that equations (102) and (107) are solutions to system (2), so it is solvable, as claimed. □

Theorem 5 gives a general form of solutions to system (2) in the case $c=0$, $abd\ne0$, but it does not present explicit formulas for $a_{n}$ and $y_{n}$ involved in the solutions. We give some explicit formulas for them, following also some arguments in [19]. Since $bd\ne0$, we find the zeros of the characteristic polynomial associated to (103)

$$\begin{aligned} p_{3}(\lambda )=\lambda ^{3}-a\lambda ^{2}-bd. \end{aligned}$$

(108)

For $\lambda =s+\frac{a}{3}$, the equation $p_{3}(\lambda )=0$ becomes

$$\begin{aligned} s^{3}-\frac{a^{2}}{3}s-\frac{2a^{3}+27bd}{27}=0. \end{aligned}$$

(109)

We know that

$$\begin{aligned} s_{j}=\frac{1}{3\sqrt[3]{2}} \Bigl(\varepsilon ^{j}\sqrt [3]{ \Delta _{1}-\sqrt{\Delta_{1}^{2}-4 \Delta_{0}^{3}}}+\overline {\varepsilon }^{j} \sqrt [3]{\Delta _{1}+\sqrt{\Delta_{1}^{2}-4 \Delta_{0}^{3}}} \Bigr), \end{aligned}$$

(110)

$j=\overline {0,2}$, where

$$\begin{aligned} \Delta_{0}=a^{2}=:-3p,\qquad \Delta _{1}=2a^{3}+27bd=:-27q, \end{aligned}$$

(111)

$\varepsilon ^{3}=1$ and $\varepsilon \ne1$, are the zeros of (109).

Hence, the zeros of $p_{3}$ are

$$\begin{aligned} \lambda _{j}=\frac{a}{3}+\frac{1}{3\sqrt[3]{2}} \Bigl(\varepsilon ^{j}\sqrt [3]{\Delta_{1}-\sqrt{ \Delta_{1}^{2}-4\Delta_{0}^{3}}}+ \overline {\varepsilon }^{j}\sqrt [3]{\Delta_{1}+\sqrt {\Delta_{1}^{2}-4\Delta_{0}^{3}}} \Bigr),\quad j=\overline {0,2}. \end{aligned}$$

(112)

Zeros of $p_{3}$ are mutually different and different from 1. In the case $\Delta_{1}^{2}-4\Delta_{0}^{3}\ne0$, we get $bd(4a^{3}+27bd)\ne0$. If $0\ne bd\ne-4a^{3}/27$, then the zeros of (108) are mutually different. If also $a+bd\ne1$, then they are different from 1. The case $a=bd=k\in\mathbb{N}$ is such one.

Zeros of $p_{3}$ are different and one of them is 1. In this case we have $a+bd=1$. Hence

$$p_{3}(\lambda )=\lambda ^{3}-a\lambda ^{2}+a-1=( \lambda -1) \bigl(\lambda ^{2}-(a-1)\lambda -(a-1)\bigr), $$

and consequently

$$\begin{aligned} \lambda _{1}=1,\qquad \lambda _{2,3}=\frac{a-1\pm\sqrt {a^{2}+2a-3}}{2}. \end{aligned}$$

(113)

Since $p_{3}'(1)=3-2a\ne0$, when $a\in\mathbb{Z}$, the polynomial in (108) cannot have the unity as a double zero.

It is well known that the general solution to (103) in this case is

$$\begin{aligned} a_{n}=\alpha _{1}\lambda _{1}^{n}+ \alpha _{2}\lambda _{2}^{n}+\alpha _{3} \lambda _{3}^{n},\quad n\in\mathbb{N}, \end{aligned}$$

(114)

where $\alpha _{j}$, $j=\overline {1,3}$, are constants, which due to $c_{1}=bd\ne0$ can be prolonged for every non-positive index.

From (114) and by Lemma 1 with $R_{3}(s)=\prod_{j=1}^{3}(s-\lambda _{j})$, we get

$$ a_{n}=\frac{\lambda _{1}^{n+2}}{(\lambda _{1}-\lambda _{2})(\lambda _{1}-\lambda _{3})}+\frac{\lambda _{2}^{n+2}}{(\lambda _{2}-\lambda _{1})(\lambda _{2}-\lambda _{3})}+\frac {\lambda _{3}^{n+2}}{(\lambda _{3}-\lambda _{1})(\lambda _{3}-\lambda _{2})}, $$

(115)

for $n\ge-2$ (see, for example, [18]).

From (106) and (115), it follows that

$$ y_{n}=\sum_{i=0}^{n-1}\sum _{j=1}^{3}\frac{\lambda _{j}^{i+2}}{p_{3}'(\lambda _{j})}, $$

(116)

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

Equation (116) shows that

$$\begin{aligned} y_{n}={}&\frac{\lambda _{1}^{2}(\lambda _{1}^{n}-1)}{(\lambda _{1}-\lambda _{2})(\lambda _{1}-\lambda _{3})(\lambda _{1}-1)} +\frac{\lambda _{2}^{2}(\lambda _{2}^{n}-1)}{(\lambda _{2}-\lambda _{1})(\lambda _{2}-\lambda _{3})(\lambda _{2}-1)} \\ &{}+\frac{\lambda _{3}^{2}(\lambda _{3}^{n}-1)}{(\lambda _{3}-\lambda _{1})(\lambda _{3}-\lambda _{2})(\lambda _{3}-1)},\quad n\in\mathbb{N}, \end{aligned}$$

(117)

when $\lambda _{j}\ne1$, $j=\overline {1,3}$.

If one of the zeros of $p_{3}$ is 1, say, $\lambda _{3}$, then

$$ y_{n}=\frac{\lambda _{1}^{2}(\lambda _{1}^{n}-1)}{(\lambda _{1}-\lambda _{2})(\lambda _{1}-1)^{2}} +\frac{\lambda _{2}^{2}(\lambda _{2}^{n}-1)}{(\lambda _{2}-\lambda _{1})(\lambda _{2}-1)^{2}}+\frac{n}{(\lambda _{1}-1)(\lambda _{2}-1)}, $$

(118)

for $n\in\mathbb{N}$. Moreover, equations (117) and (118) hold for $n\ge-2$.

Corollary 6

Assume that $a,b,c,d\in\mathbb{Z}$, $abd\ne0$, $c=0$, $\alpha , \beta , z_{-1}, z_{0}, w_{0}\in\mathbb{C}\setminus\{0\}$ and $\Delta_{1}^{2}\ne4\Delta_{0}^{3}$. Then the following statements are true:

• (a) If $a+bd\ne1$, then the general solution to (2) is given by (102) and (107), where $(a_{n})_{n\ge-2}$ is given by (115), $(y_{n})_{n\ge-2}$ is given by (117), while $\lambda _{j}$, $j=\overline {1,3}$, are given by (112).

• (b) If $a+bd=1$, then $p_{3}$ has a unique zero equal to 1, say $\lambda _{3}$, and the general solution to (2) is given by formulas (102) and (107), where $(a_{n})_{n\ge-2}$ is given by (115) with $\lambda _{3}=1$, $(y_{n})_{n\ge-2}$ is given by (118), while $\lambda _{j}$, $j=\overline {1,3}$, are given by (113).

$p_{3}$ has a double zero. Since it must be $\Delta_{1}^{2}=4\Delta_{0}^{3}$, we have $bd=-4a^{3}/27$, so that

$$p_{3}(\lambda )=\lambda ^{3}-a\lambda ^{2}+ \frac{4}{27}a^{3}. $$

The following condition must also be satisfied: $p_{3}'(\lambda )=0$. Hence, $\lambda _{1}=-a/3$, $\lambda _{2,3}=2a/3$, and consequently

$$p_{3}(\lambda )= \biggl(\lambda -\frac{2a}{3} \biggr)^{2} \biggl(\lambda +\frac {a}{3} \biggr). $$

Due to $bd\in\mathbb{Z}$, we get $a=3\hat{a}$, for some $\hat{a}\in \mathbb{Z}$. Now note that $2a/3\ne1$, for every $a\in\mathbb{Z}$, so that 1 cannot be a double zero of $p_{3}$.

Hence

$$\begin{aligned} a_{n}=\hat{\alpha}_{1}\lambda _{1}^{n}+( \hat{\alpha}_{2}+\hat{\alpha}_{3}n)\lambda _{2}^{n}, \quad n\in \mathbb{N}, \end{aligned}$$

(119)

where $\hat{\alpha}_{i}$, $i=\overline {1,3}$, are constants. Using initial conditions (104) we obtain

$$\begin{aligned} a_{n}=\frac{\lambda _{1}^{n+2}+ (\lambda _{2}-2\lambda _{1}+n(\lambda _{2}-\lambda _{1}) )\lambda _{2}^{n+1}}{(\lambda _{2}-\lambda _{1})^{2}}, \end{aligned}$$

(120)

for $n\ge-2$.

From (106) and (120), it follows that

$$ y_{n}=\sum_{j=0}^{n-1} \frac{\lambda _{1}^{j+2}+ (\lambda _{2}-2\lambda _{1}+j(\lambda _{2}-\lambda _{1}) )\lambda _{2}^{j+1}}{(\lambda _{2}-\lambda _{1})^{2}}, $$

(121)

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

Equation (121) along with Lemma 2 yields

$$ y_{n}=\frac{\lambda _{1}^{2}(\lambda _{1}^{n}-1)}{(\lambda _{2}-\lambda _{1})^{2}(\lambda _{1}-1)}+ \frac{(\lambda _{2}-2\lambda _{1})\lambda _{2}(\lambda _{2}^{n}-1)}{(\lambda _{2}-\lambda _{1})^{2}(\lambda _{2}-1)} +\frac{\lambda _{2}^{2}(1-n\lambda _{2}^{n-1}+(n-1)\lambda _{2}^{n})}{(\lambda _{2}-\lambda _{1})(\lambda _{2}-1)^{2}}, $$

(122)

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

On the other hand, if $\lambda _{1}=1\ne\lambda _{2}=\lambda _{3}$ ($a=-3$), then we get

$$ y_{n}=\frac{n}{(\lambda _{2}-1)^{2}}+ \frac{(\lambda _{2}-2)\lambda _{2}(\lambda _{2}^{n}-1)}{(\lambda _{2}-1)^{3}}+\frac{\lambda _{2}^{2}(1-n\lambda _{2}^{n-1}+(n-1)\lambda _{2}^{n})}{(\lambda _{2}-1)^{3}}, $$

(123)

for $n\in\mathbb{N}$. Moreover, (122) and (123) hold for $n\ge-2$.

Corollary 7

Assume that $a,b,c,d\in\mathbb{Z}$, $abd\ne0$, $c=0$, $\alpha , \beta , z_{-1}, z_{0}, w_{0}\in\mathbb{C}\setminus\{0\}$ and $\Delta_{1}^{2}=4\Delta_{0}^{3}$. Then the following statements are true:

• (a) If $a+bd\ne1$, then the general solution to (2) is given by (102) and (107), where $(a_{n})_{n\ge-2}$ is given by (120), $(y_{n})_{n\ge-2}$ is given by (122), where $\lambda _{1}=-a/3$ and $\lambda _{2,3}=2a/3$.

• (b) If only one of the zeros of the polynomial (108) is equal to 1, say, $\lambda _{1}$, then the general solution to system (2) is given by (102) and (107), where $(a_{n})_{n\ge-2}$ is given by (120) with $\lambda _{1}=1$, while $(y_{n})_{n\ge-2}$ is given by (123).

• (c) It is not possible that two zeros of polynomial (108) are equal to one.

Case when all the zeros of $p_{3}$ are equal. We have $p_{3}(\lambda )=p_{3}'(\lambda )=p_{3}''(\lambda )=0$. So, $p_{3}''(\lambda )=0$ would imply $\lambda =a/3$. From $p_{3}'(\lambda )=3\lambda ^{2}-2a\lambda$, we see that $a/3$ is a unique zero of $p_{3}$ if $a=0$, which contradicts the assumption $abd\ne0$. Hence, the case is not possible.