This section considers the problem of solvability of systems (18)–(26). The systems are considered separately, one by one. It is shown that they all are really solvable by presenting some closed-form formulas for their general solutions. As a consequence, some closed-form formulas for general solutions to systems (8)–(16) are obtained. In this way it is shown that each of the systems of difference equations is practically solvable.

### 3.1 System (18)

From the first equation in the system of difference equations (18) we have

$$\begin{aligned} \zeta _{n}=\zeta _{n-1}^{2}, \quad n\in {\mathbb {N}}, \end{aligned}$$

(27)

from which by iteration and a simple inductive argument we obtain

$$\begin{aligned} \zeta _{n}=\zeta _{0}^{2^{n}}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

(28)

By using (28) in the second equation in (18) it follows that

$$\begin{aligned} \eta _{n}=\zeta _{0}^{2^{n}}, \quad n\in {\mathbb {N}}. \end{aligned}$$

(29)

Using relations (28) and (29) in (17), we see that the following theorem holds.

### Theorem 1

*Let* \(a\ne 0\). *Then the following closed*-*form formulas*

$$\begin{aligned} &x_{n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}}-1}, \quad n\in {\mathbb {N}}_{0}, \\ &y_{n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}}-1}, \quad n\in {\mathbb {N}}, \end{aligned}$$

*present a general solution to system* (8).

### 3.2 System (19)

Bearing in mind that the first equation in (19) is the same as in (18), we have that formula (28) also holds in this case. Employing (28) in the second equation in (19), we have

$$\begin{aligned} \eta _{n}=\eta _{n-1}\zeta _{n-1}=\eta _{n-1}\zeta _{0}^{2^{n-1}}, \quad n\in {\mathbb {N}}. \end{aligned}$$

(30)

From (30) and by a simple inductive argument, we obtain

$$ \eta _{n}=\eta _{0}\prod_{j=1}^{n} \zeta _{0}^{2^{j-1}}=\eta _{0} \zeta _{0}^{\sum _{j=1}^{n}2^{j-1}}, $$

from which together with the formula for the sum of a finite geometric progression it follows that

$$\begin{aligned} \eta _{n}=\eta _{0}\zeta _{0}^{2^{n}-1},\quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

(31)

Using relations (28) and (31) in (17), we see that the following theorem holds.

### Theorem 2

*Let* \(a\ne 0\). *Then the following closed*-*form formulas*

$$\begin{aligned} &x_{n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}}-1}, \quad n\in {\mathbb {N}}_{0}, \\ &y_{n}=\sqrt{a} \frac{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} ) (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}-1}+1}{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} ) (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}-1}-1}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

*present a general solution to system* (9).

### 3.3 System (20)

Since the first equation in (20) is the same as in (18), formula (28) also holds in this case. On the other hand, the second equation in (20) is obtained from the first one by interchanging letters *ζ* and *η*, from which along with (28) it follows that

$$\begin{aligned} \eta _{n}=\eta _{0}^{2^{n}}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

(32)

Using relations (28) and (32) in (17), we see that the following theorem holds.

### Theorem 3

*Let* \(a\ne 0\). *Then the following closed*-*form formulas*

$$\begin{aligned} &x_{n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n}}-1}, \quad n\in {\mathbb {N}}_{0}, \\ &y_{n}=\sqrt{a} \frac{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}}+1}{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}}-1}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

*present a general solution to system* (10).

### 3.4 System (21)

From the equations in (21) we have

$$\begin{aligned} \zeta _{n}=\zeta _{n-1}\eta _{n-1}=\zeta _{n-1}\zeta _{n-2}^{2}, \quad n\ge 2. \end{aligned}$$

(33)

If we use the following notations:

$$ a_{1}:=1 \quad \text{and} \quad b_{1}:=2, $$

then equation (33) can be written as follows:

$$\begin{aligned} \zeta _{n}=\zeta _{n-1}^{a_{1}}\zeta _{n-2}^{b_{1}}, \quad n\ge 2. \end{aligned}$$

(34)

Employing relation (33), where index *n* is replaced by \(n-1\) in (34), we have

$$\begin{aligned} \zeta _{n}&=\zeta _{n-1}^{a_{1}}\zeta _{n-2}^{b_{1}}=\bigl(\zeta _{n-2} \zeta _{n-3}^{2}\bigr)^{a_{1}}\zeta _{n-2}^{b_{1}}=\zeta _{n-2}^{a_{1}+b_{1}} \zeta _{n-3}^{2a_{1}} \\ &=\zeta _{n-2}^{a_{2}}\zeta _{n-3}^{b_{2}}, \end{aligned}$$

(35)

where \(a_{2}\) and \(b_{2}\) are clearly defined by

$$\begin{aligned} a_{2}:=a_{1}+b_{1} \quad \text{and}\quad b_{2}:=2a_{1}. \end{aligned}$$

(36)

Relations (35) and (36) suggest that the following ones hold:

$$\begin{aligned} \zeta _{n}=\zeta _{n-k}^{a_{k}}\zeta _{n-k-1}^{b_{k}} \end{aligned}$$

(37)

and

$$\begin{aligned} a_{k}=a_{k-1}+b_{k-1},\qquad b_{k}=2a_{k-1} \end{aligned}$$

(38)

for \(2\le k\le n-1\).

Indeed, assume that (37) and (38) hold for *k* such that \(2\le k\le n-1\). Then, by using relation (33) where index *n* is replaced by \(n-k\) in (37), we have

$$\begin{aligned} \zeta _{n}&=\zeta _{n-k}^{a_{k}}\zeta _{n-k-1}^{b_{k}} \\ &=\bigl(\zeta _{n-k-1} \zeta _{n-k-2}^{2}\bigr)^{a_{k}}\zeta _{n-k-1}^{b_{k}} \\ &=\zeta _{n-k-1}^{a_{k}+b_{k}} \zeta _{n-k-2}^{2a_{k}} \\ &=\zeta _{n-k-1}^{a_{k+1}}\zeta _{n-k-2}^{b_{k+1}}, \end{aligned}$$

where \(a_{k+1}\) and \(b_{k+1}\) are clearly defined by

$$ a_{k+1}:=a_{k}+b_{k} \quad \text{and}\quad b_{k+1}:=2a_{k} $$

for \(2\le k\le n-2\). This inductive step along with (35) and (36) shows that (37) and (38) hold, as claimed.

Let \(k=n-1\). Then from (37) we have

$$ \zeta _{n}=\zeta _{1}^{a_{n-1}}\zeta _{0}^{b_{n-1}}=(\zeta _{0}\eta _{0})^{a_{n-1}} \zeta _{0}^{b_{n-1}}= \zeta _{0}^{a_{n-1}+b_{n-1}}\eta _{0}^{a_{n-1}}. $$

From this and since \(a_{n-1}+b_{n-1}=a_{n}\), we have

$$\begin{aligned} \zeta _{n}=\zeta _{0}^{a_{n}}\eta _{0}^{a_{n-1}}. \end{aligned}$$

(39)

Now note that from (38) we have

$$\begin{aligned} a_{n}=a_{n-1}+2a_{n-2},\quad n\ge 3. \end{aligned}$$

(40)

Moreover, relation (40) can be used to calculate \(a_{n}\) also for \(n\le 0\) by using the following obvious consequence of it:

$$\begin{aligned} a_{n-2}=\frac{a_{n}-a_{n-1}}{2}. \end{aligned}$$

(41)

Since \(a_{1}=1\) and \(a_{2}=3\), from (41) it follows that

$$\begin{aligned} a_{0}=1 \quad \text{and}\quad a_{-1}=0. \end{aligned}$$

(42)

By using (42) it is easy to see that formula (39) holds also for \(n=0\).

The characteristic polynomial associated with equation (40) is \(P_{2}(s)=s^{2}-s-2\), and clearly their roots are

$$\begin{aligned} s_{1}=2 \quad \text{and} \quad s_{2}=-1. \end{aligned}$$

(43)

Hence, by using the de Moivre formula (3), with \(s_{1}\) and \(s_{2}\) given in (43), and \(c_{1}=2\) and \(c_{2}=1\), we obtain

$$\begin{aligned} a_{n}=\frac{2^{n+1}-(-1)^{n+1}}{3}, \quad n\in {\mathbb {Z}}. \end{aligned}$$

(44)

Using (44) in (39), we get

$$\begin{aligned} \zeta _{n}=\zeta _{0}^{\frac{2^{n+1}-(-1)^{n+1}}{3}}\eta _{0}^{ \frac{2^{n}-(-1)^{n}}{3}},\quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

(45)

By using (45) in the second equation in (21), we obtain

$$\begin{aligned} \eta _{n}=\zeta _{0}^{\frac{2^{n+1}-2(-1)^{n}}{3}}\eta _{0}^{ \frac{2^{n}-2(-1)^{n-1}}{3}}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

(46)

Using relations (45) and (46) in (17), we see that the following theorem holds.

### Theorem 4

*Let* \(a\ne 0\). *Then the following closed*-*form formulas*

$$\begin{aligned} &x_{n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{\frac{2^{n+1}-(-1)^{n+1}}{3}} (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{\frac{2^{n}-(-1)^{n}}{3}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{\frac{2^{n+1}-(-1)^{n+1}}{3}} (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{\frac{2^{n}-(-1)^{n}}{3}}-1}, \quad n\in {\mathbb {N}}_{0}, \\ &y_{n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{\frac{2^{n+1}-2(-1)^{n}}{3}} (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{\frac{2^{n}-2(-1)^{n-1}}{3}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{\frac{2^{n+1}-2(-1)^{n}}{3}} (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{\frac{2^{n}-2(-1)^{n-1}}{3}}-1}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

*present a general solution to system* (11).

### 3.5 System (22)

First note that we have

$$\begin{aligned} \zeta _{n}=\eta _{n},\quad n\in {\mathbb {N}}. \end{aligned}$$

(47)

Using (47) in the first equation in (22), we have that (27) holds, but this time for \(n\ge 2\), from which it follows that

$$ \zeta _{n}=\zeta _{1}^{2^{n-1}},\quad n\in {\mathbb {N}}, $$

and consequently,

$$\begin{aligned} \zeta _{n}=\zeta _{0}^{2^{n-1}}\eta _{0}^{2^{n-1}}, \quad n\in {\mathbb {N}}. \end{aligned}$$

(48)

From (47) and (48) we have

$$\begin{aligned} \eta _{n}=\zeta _{0}^{2^{n-1}}\eta _{0}^{2^{n-1}}, \quad n\in {\mathbb {N}}. \end{aligned}$$

(49)

Using relations (48) and (49) in (17), we see that the following theorem holds.

### Theorem 5

*Let* \(a\ne 0\). *Then the following closed*-*form formulas*

$$\begin{aligned} &x_{n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n-1}} (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n-1}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n-1}} (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n-1}}-1},\quad n\in {\mathbb {N}}, \\ &y_{n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n-1}} (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n-1}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{n-1}} (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n-1}}-1}, \quad n\in {\mathbb {N}}, \end{aligned}$$

*present a general solution to system* (12).

### 3.6 System (23)

This system is obtained from the system in (19) by interchanging letters *ζ* and *η*. Hence, we have that the following formulas hold:

$$\begin{aligned} \zeta _{n}=\zeta _{0}\eta _{0}^{2^{n}-1},\quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

(50)

and

$$\begin{aligned} \eta _{n}=\eta _{0}^{2^{n}}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

(51)

Using relations (50) and (51) in (17), we see that the following theorem holds.

### Theorem 6

*Let* \(a\ne 0\). *Then the following closed*-*form formulas*

$$\begin{aligned} &x_{n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} ) (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}-1}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} ) (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}-1}-1}, \quad n\in {\mathbb {N}}_{0}, \\ &y_{n}=\sqrt{a} \frac{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}}+1}{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}}-1}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

*present a general solution to system* (13).

### 3.7 System (24)

By using the second equation in (24) in the first one, we obtain

$$\begin{aligned} \zeta _{n}=\eta _{n-1}^{2}=\zeta _{n-2}^{4}, \quad n\ge 2. \end{aligned}$$

(52)

From (52) we have

$$ \zeta _{2n}=\zeta _{2(n-1)}^{4}, \quad n\in {\mathbb {N}}, $$

from which by iteration and a simple inductive argument we obtain

$$\begin{aligned} \zeta _{2n}=\zeta _{0}^{4^{n}}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

(53)

and

$$ \zeta _{2n+1}=\zeta _{2(n-1)+1}^{4}, \quad n\in {\mathbb {N}}, $$

from which by iteration and a simple inductive argument we obtain

$$\begin{aligned} \zeta _{2n+1}=\zeta _{1}^{4^{n}}=\eta _{0}^{2^{2n+1}}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

(54)

Since this system is symmetric, we have

$$\begin{aligned} \eta _{2n}=\eta _{0}^{4^{n}}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

(55)

and

$$\begin{aligned} \eta _{2n+1}=\zeta _{0}^{2^{2n+1}},\quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

(56)

Using relations (53)–(56) in (17), we see that the following theorem holds.

### Theorem 7

*Let* \(a\ne 0\). *Then the following closed*-*form formulas*

$$\begin{aligned} &x_{2n}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{4^{n}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{4^{n}}-1}, \quad n\in {\mathbb {N}}_{0}, \\ &x_{2n+1}=\sqrt{a} \frac{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{2n+1}}+1}{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{2n+1}}-1}, \quad n\in {\mathbb {N}}_{0}, \\ &y_{2n}=\sqrt{a} \frac{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{4^{n}}+1}{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{4^{n}}-1}, \quad n\in {\mathbb {N}}_{0}, \\ &y_{2n+1}=\sqrt{a} \frac{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{2n+1}}+1}{ (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{2^{2n+1}}-1},\quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

*present a general solution to system* (14).

### 3.8 System (25)

This system is obtained from the system in (21) by interchanging letters *ζ* and *η*. Hence, we have that the following formulas hold:

$$\begin{aligned} \zeta _{n}=\eta _{0}^{\frac{2^{n+1}-2(-1)^{n}}{3}}\zeta _{0}^{ \frac{2^{n}-2(-1)^{n-1}}{3}} \end{aligned}$$

(57)

and

$$\begin{aligned} \eta _{n}=\eta _{0}^{\frac{2^{n+1}-(-1)^{n+1}}{3}}\zeta _{0}^{ \frac{2^{n}-(-1)^{n}}{3}}. \end{aligned}$$

(58)

Using relations (57) and (58) in (17), we see that the following theorem holds.

### Theorem 8

*Let* \(a\ne 0\). *Then the following closed*-*form formulas*

$$\begin{aligned} &x_{n}=\sqrt{a} \frac{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{\frac{2^{n+1}-2(-1)^{n}}{3}} (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{\frac{2^{n}-2(-1)^{n-1}}{3}}+1}{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{\frac{2^{n+1}-2(-1)^{n}}{3}} (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{\frac{2^{n}-2(-1)^{n-1}}{3}}-1}, \quad n\in {\mathbb {N}}_{0}, \\ &y_{n}=\sqrt{a} \frac{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{\frac{2^{n+1}-(-1)^{n+1}}{3}} (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{\frac{2^{n}-(-1)^{n}}{3}}+1}{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{\frac{2^{n+1}-(-1)^{n+1}}{3}} (\frac{x_{0}+\sqrt{a}}{x_{0}-\sqrt{a}} )^{\frac{2^{n}-(-1)^{n}}{3}}-1}, \quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

*present a general solution to system* (15).

### 3.9 System (26)

This system is obtained from the system in (18) by interchanging letters *ζ* and *η*. Hence, we have that the following formulas hold:

$$\begin{aligned} \zeta _{n}=\eta _{0}^{2^{n}},\quad n\in {\mathbb {N}}, \quad \text{and} \quad \eta _{n}=\eta _{0}^{2^{n}}, \quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

(59)

Using (59) in (17), we see that the following theorem holds.

### Theorem 9

*Let* \(a\ne 0\). *Then the following closed*-*form formulas*

$$\begin{aligned} &x_{n}=\sqrt{a} \frac{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}}+1}{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}}-1},\quad n\in {\mathbb {N}}, \\ &y_{n}=\sqrt{a} \frac{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}}+1}{ (\frac{y_{0}+\sqrt{a}}{y_{0}-\sqrt{a}} )^{2^{n}}-1},\quad n\in {\mathbb {N}}_{0}, \end{aligned}$$

*present a general solution to system* (16).