In this section we prove our main results. The first one is a generalization of Theorem 2.

### Theorem 3

*Let*
\(\vec{v}=(v_{0}, v_{1})\in {\mathbb {R}}^{2}\) (*or*
\(\in {\mathbb {C}}^{2}\)) *and*
\((s_{n}(\vec{v}))_{n\in\mathbb{N} _{0}}\)
*be the solution to equation* (1) *such that*

$$\begin{aligned} s_{0}=v_{0}\quad\mbox{and}\quad s_{1}=v_{1}. \end{aligned}$$

(30)

*Then all the solutions to equation* (1) *have representation* (29) *if and only if*

$$\begin{aligned} bv_{0}^{2}+av_{0}v_{1}\ne v_{1}^{2}. \end{aligned}$$

(31)

*Further*, *if* (31) *holds*, *then for every solution*
\((x_{n})_{n \in\mathbb{N} _{0}}\)
*to equation* (1) *the following representation holds*:

$$\begin{aligned} x_{n}=\frac{(bv_{0}x_{0}+(av_{0}-v_{1})x_{1})s_{n}(\vec{v})+b(v_{0}x _{1}-v_{1}x_{0})s_{n-1}(\vec{v})}{bv_{0}^{2}+av_{0}v_{1}-v_{1}^{2}} \end{aligned}$$

(32)

*for*
\(n\in\mathbb{N} _{0}\).

### Proof

Since \(s_{n}(\vec{v})\) is a solution to linear equation (1) so is every sequence of the following form:

$$ c_{1}s_{n}(\vec{v})+c_{2}s_{n-1}(\vec{v}), \quad n\in\mathbb{N} _{0}, $$

where \(c_{1}\) and \(c_{2}\) are arbitrary constants.

This means that the set

$$ S:=\bigl\{ c_{1}s_{n}(\vec{v})+c_{2}s_{n-1}( \vec{v}): c_{j}\in {\mathbb {R}}\, ( \mbox{or} \in {\mathbb {C}}), j=1,2\bigr\} $$

is a subset of the set of all solutions to equation (1). Now we prove that these two sets are equal if and only if condition (31) is satisfied.

Since each solution \((x_{n})_{n\in\mathbb{N} _{0}}\) to equation (1) is completely determined by its initial values \(x_{0}\) and \(x_{1}\), we see that (29) holds for every solution \((x_{n})_{n\in\mathbb{N} _{0}}\) to the equation if and only if, for every \((x_{0},x_{1})\in {\mathbb {R}}^{2}\) (or \(\in {\mathbb {C}}^{2}\)),

$$\begin{aligned} &x_{0}=c_{1}s_{0}(\vec{v})+c_{2}s_{-1}( \vec{v}), \\ &x_{1}=c_{1}s_{1}(\vec{v})+c_{2}s_{0}( \vec{v}) \end{aligned}$$

for some constants \(c_{1}\) and \(c_{2}\).

Due to (30), this means that (29) holds for every solution to the equation if and only if the linear system

$$\begin{aligned} \begin{aligned} &v_{0}c_{1}+v_{-1}c_{2}=x_{0}, \\ &v_{1}c_{1}+v_{0}c_{2}=x_{1} \end{aligned} \end{aligned}$$

(33)

in variables \(c_{1}\) and \(c_{2}\) has a solution for every \((x_{0},x _{1})\in {\mathbb {R}}^{2}\) (or \(\in {\mathbb {C}}^{2}\)).

This will happen if and only if the determinant of system (33) is different from zero, that is, if and only if

$$\begin{aligned} \Delta:= \left \vert \textstyle\begin{array}{@{}c@{\quad}c@{}} v_{0} & v_{-1} \\ v_{1} & v_{0} \end{array}\displaystyle \right \vert =v_{0}^{2}-v_{1}v_{-1}= \frac{bv_{0}^{2}+av_{0}v_{1}-v_{1}^{2}}{b}\ne0, \end{aligned}$$

(34)

from which the first part of the theorem follows.

Now assume that (31) holds. Then from (33) it follows that

$$\begin{aligned} c_{1}=\frac{1}{\Delta} \left \vert \textstyle\begin{array}{@{}c@{\quad}c@{}} x_{0} & (v_{1}-av_{0})/b \\ x_{1} & v_{0} \end{array}\displaystyle \right \vert = \frac{bv_{0}x_{0}+(av_{0}-v_{1})x_{1}}{bv_{0}^{2}+av_{0}v _{1}-v_{1}^{2}} \end{aligned}$$

(35)

and

$$\begin{aligned} c_{2}=\frac{1}{\Delta} \left \vert \textstyle\begin{array}{@{}c@{\quad}c@{}} v_{0} & x_{0} \\ v_{1} & x_{1} \end{array}\displaystyle \right \vert = \frac{b(v_{0}x_{1}-v_{1}x_{0})}{bv_{0}^{2}+av_{0}v_{1}-v_{1} ^{2}} . \end{aligned}$$

(36)

Using (35) and (36) in (29), we get (32), as claimed. □

### Remark 1

The set *S̃* consisting of all \(\vec{v}\in {\mathbb {C}}^{2}\) such that (31) does not hold consists of the zero vector \(\vec{0}\in {\mathbb {C}}^{2}\), and of those *v⃗* such that

$$ \biggl(\frac{v_{1}}{v_{0}} \biggr)^{2}-a\frac{v_{1}}{v_{0}}-b=0, $$

that is,

$$ v_{1}=\frac{a+\sqrt{a^{2}+4b}}{2}v_{0} $$

or

$$ v_{1}=\frac{a-\sqrt{a^{2}+4b}}{2}v_{0}, $$

so

$$ \widetilde{S}= \biggl\{ \biggl(v_{0}, \frac{a\pm\sqrt{a^{2}+4b}}{2}v _{0} \biggr): v_{0}\in {\mathbb {C}}\biggr\} . $$

By representation (32) a generalization of the representation of solutions to difference equation (8) given in (28) can be obtained. Namely, the following theorem holds.

### Theorem 4

*Consider equation* (8) *with*
\(\alpha \delta \ne \beta \gamma \). *Assume that*
\(\vec{v}=(v_{0}, v_{1})\in {\mathbb {R}}^{2}\) (*or*
\(\in {\mathbb {C}}^{2}\)) *satisfies the following condition*:

$$\begin{aligned} (\beta \gamma -\alpha \delta )v_{0}^{2}+(\alpha +\delta )v_{0}v_{1} \ne v_{1}^{2}. \end{aligned}$$

(37)

*Let*
\((s_{n})_{n\in\mathbb{N} _{0}}\)
*be the solution to equation* (26) *such that*
\(s_{0}=v_{0}\)
*and*
\(s_{1}=v_{1}\). *Then every well*-*defined solution*
\((z_{n})_{n\in\mathbb{N} _{0}}\)
*to equation* (8) *has the following representation*:

$$ z_{n}=\frac{(\beta \gamma -\alpha \delta )(\beta v_{0}+(\alpha v_{0}-v_{1})z_{0})s_{n-1}+(\beta ((\alpha +\delta )v _{0}-v_{1})+z_{0}(v_{0}(\beta \gamma +\alpha ^{2})-\alpha v_{1}))s_{n}}{((\beta \gamma -\alpha \delta )v _{0}+\alpha v_{1}-\gamma v_{1}z_{0})s_{n}+(\delta v_{0}-v_{1}+\gamma v_{0}z_{0})s_{n+1}} $$

(38)

*for*
\(n\in\mathbb{N} _{0}\).

### Proof

First assume that \(\gamma \ne0\). Then from (14) we see that for every well-defined solution \((z_{n})_{n\in\mathbb{N} _{0}}\) to equation (8) the following holds:

$$\begin{aligned} z_{n}=\frac{1}{\gamma } \biggl(\frac{x_{n+1}}{x_{n}}-\delta \biggr),\quad n \in\mathbb{N} _{0}, \end{aligned}$$

(39)

where \(x_{n}\) is a solution to equation (26).

On the other hand, since (37) holds, from Theorem 3 we have

$$\begin{aligned} x_{n}=\frac{((\beta \gamma -\alpha \delta )v_{0}x_{0}+((\alpha +\delta )v_{0}-v_{1})x_{1})s _{n}+(\beta \gamma -\alpha \delta )(v_{0}x_{1}-v_{1}x_{0})s_{n-1}}{(\beta \gamma -\alpha \delta )v _{0}^{2}+(\alpha +\delta )v_{0}v_{1}-v_{1}^{2}} \end{aligned}$$

(40)

for every solution to equation (26).

From (39), (40), after some standard but time-consuming calculations and use of (26), we have

$$\begin{aligned} z_{n}&=\frac{1}{\gamma } \biggl(\frac{((\beta \gamma -\alpha \delta )v_{0}x_{0}+((\alpha +\delta )v_{0}-v_{1})x_{1})s_{n+1}+(\beta \gamma -\alpha \delta )(v_{0}x_{1}-v_{1}x_{0})s_{n}}{((\beta \gamma -\alpha \delta )v _{0}x_{0}+((\alpha +\delta )v_{0}-v_{1})x_{1})s_{n}+(\beta \gamma -\alpha \delta )(v_{0}x _{1}-v_{1}x_{0})s_{n-1}}-\delta \biggr) \\ &=\frac{1}{\gamma }\frac{((\beta \gamma -\alpha \delta )v_{0}+((\alpha +\delta )v_{0}-v_{1})(\gamma z _{0}+\delta ))(\alpha s_{n}+(\beta \gamma -\alpha \delta )s_{n-1})}{((\beta \gamma -\alpha \delta )v_{0}+((\alpha +\delta )v _{0}-v_{1})(\gamma z_{0}+\delta ))s_{n}+(v_{0}(\gamma z_{0}+\delta )-v_{1})(s_{n+1}-(\alpha +\delta )s _{n})} \\ &\quad{}+\frac{1}{\gamma }\frac{(\beta \gamma -\alpha \delta )(v_{0}(\gamma z_{0}+\delta )-v_{1})(s _{n}-\delta s_{n-1})}{((\beta \gamma -\alpha \delta )v_{0}+((\alpha +\delta )v_{0}-v_{1})(\gamma z _{0}+\delta ))s_{n}+(v_{0}(\gamma z_{0}+\delta )-v_{1})(s_{n+1}-(\alpha +\delta )s_{n})} \\ &=\frac{(\beta \gamma -\alpha \delta )(\beta v_{0}+(\alpha v_{0}-v_{1})z_{0})s_{n-1}+(\beta ((\alpha +\delta )v _{0}-v_{1})+z_{0}(v_{0}(\beta \gamma +\alpha ^{2})-\alpha v_{1}))s_{n}}{((\beta \gamma -\alpha \delta )v _{0}+\alpha v_{1}-\gamma v_{1}z_{0})s_{n}+(\delta v_{0}-v_{1}+\gamma v_{0}z_{0})s_{n+1}}, \end{aligned}$$

finishing the proof, in this case.

Now assume that \(\gamma =0\). Then equation (8) becomes

$$\begin{aligned} z_{n+1}=\frac{\alpha }{\delta } z_{n}+\frac{\beta }{\delta },\quad n\in \mathbb{N} _{0} \end{aligned}$$

(41)

(note that since \(\alpha \delta \ne \beta \gamma =0\), *δ* cannot be equal to zero so that equation (41) is defined).

From (16) and (17), it follows that

$$\begin{aligned} z_{n} &= \biggl(\frac{\alpha }{\delta } \biggr)^{n} \biggl(z_{0}+ \frac{\beta }{\alpha -\delta } \biggr)-\frac{\beta }{\alpha -\delta },\quad \mbox{when } \alpha \ne\delta , \end{aligned}$$

(42)

$$\begin{aligned} z_{n} &=z_{0}+\frac{\beta }{\delta }n,\quad\mbox{when } \alpha =\delta , \end{aligned}$$

(43)

for \(n\in\mathbb{N} _{0}\).

Let the right-hand side in (38) with \(\gamma =0\) be denoted by \(\tilde{z}_{n}\). Then we have

$$\begin{aligned} \tilde{z}_{n} &:=\frac{\alpha \delta (\beta v_{0}+(\alpha v_{0}-v_{1})z_{0})s_{n-1}-(\beta ((\alpha +\delta )v _{0}-v_{1})+\alpha (\alpha v_{0}-v_{1})z_{0})s_{n}}{\alpha (\delta v_{0}-v_{1})s_{n}+(v _{1}-\delta v_{0}) s_{n+1}} \end{aligned}$$

(44)

for \(n\in\mathbb{N} _{0}\).

By de Moivre’s formula (2) we know that

$$\begin{aligned} s_{n}=\frac{(\delta v_{0}-v_{1})\alpha ^{n}+(v_{1}-\alpha v_{0})\delta ^{n}}{\delta -\alpha }, \quad n\in\mathbb{N} _{0}, \end{aligned}$$

(45)

if \(\alpha \ne\delta \), while

$$\begin{aligned} s_{n}=\bigl(v_{1}n+\alpha v_{0}(1-n)\bigr)\alpha ^{n-1},\quad n\in\mathbb{N} _{0}, \end{aligned}$$

(46)

if \(\alpha =\delta \), since *α* and *δ* are the zeros of the characteristic polynomial

$$ P_{2}(\lambda )=\lambda ^{2}-(\alpha +\delta )\lambda +\alpha \delta $$

associated to equation (26) in the case \(\gamma =0\).

Assume that \(\alpha \ne\delta \). Then, by using (45) in (44), after some calculation we obtain

$$\begin{aligned} \tilde{z}_{n}&=\frac{\alpha \delta (\beta v_{0}+(\alpha v_{0}-v_{1})z_{0})((\delta v_{0}-v _{1})\alpha ^{n-1}+(v_{1}-\alpha v_{0})\delta ^{n-1})}{\alpha (\delta v_{0}-v_{1})((\delta v _{0}-v_{1})\alpha ^{n}+(v_{1}-\alpha v_{0})\delta ^{n})+(v_{1}-\delta v_{0}) ((\delta v _{0}-v_{1})\alpha ^{n+1}+(v_{1}-\alpha v_{0})\delta ^{n+1})} \\ &\quad{}-\frac{(\beta ((\alpha +\delta )v_{0}-v_{1})+\alpha (\alpha v_{0}-v_{1})z_{0})((\delta v _{0}-v_{1})\alpha ^{n}+(v_{1}-\alpha v_{0})\delta ^{n})}{\alpha (\delta v_{0}-v_{1})((\delta v _{0}-v_{1})\alpha ^{n}+(v_{1}-\alpha v_{0})\delta ^{n})+(v_{1}-\delta v_{0}) ((\delta v _{0}-v_{1})\alpha ^{n+1}+(v_{1}-\alpha v_{0})\delta ^{n+1})} \\ &=\frac{(v_{1}-\alpha v_{0})(v_{1}-\delta v_{0})(\beta \delta ^{n}+(-\beta +(\delta -\alpha )z _{0})\alpha ^{n})}{(v_{1}-\alpha v_{0})(v_{1}-\delta v_{0})(\delta -\alpha )\delta ^{n}}. \end{aligned}$$

(47)

Because of (37) it must be

$$\begin{aligned} v_{1}\ne \alpha v_{0}\quad\mbox{and}\quad v_{1}\ne\delta v_{0}. \end{aligned}$$

(48)

From (48) and since \(\alpha \ne\delta \), from (47) we obtain

$$\begin{aligned} \tilde{z}_{n}=\frac{\beta }{\delta -\alpha }+ \biggl(-\frac{\beta }{\delta -\alpha }+z_{0} \biggr) \biggl(\frac{\alpha }{\delta } \biggr)^{n}. \end{aligned}$$

(49)

From (42) and (49) we see that

$$\begin{aligned} \tilde{z}_{n}=z_{n},\quad n\in\mathbb{N} _{0}, \end{aligned}$$

(50)

from which the result follows in this case.

If \(\alpha =\delta \), then by using the assumption and (46) in (44), after some calculation we obtain

$$\begin{aligned} \tilde{z}_{n} &=\frac{\alpha \delta (\beta v_{0}+(\alpha v_{0}-v_{1})z_{0})((v_{1}-\alpha v _{0})(n-1)+\alpha v_{0})\alpha ^{n-2}}{\alpha (\delta v_{0}-v_{1})((v_{1}-\alpha v_{0})n+\alpha v _{0})\alpha ^{n-1}+(v_{1}-\delta v_{0}) ((v_{1}-\alpha v_{0})(n+1)+\alpha v_{0})\alpha ^{n}} \\ &\quad-\frac{(\beta ((\alpha +\delta )v_{0}-v_{1})+\alpha (\alpha v_{0}-v_{1})z_{0})((v _{1}-\alpha v_{0})n+\alpha v_{0})\alpha ^{n-1}}{\alpha (\delta v_{0}-v_{1})((v_{1}-\alpha v _{0})n+\alpha v_{0})\alpha ^{n-1}+(v_{1}-\delta v_{0}) ((v_{1}-\alpha v_{0})(n+1)+\alpha v _{0})\alpha ^{n}} \\ &=\frac{(\beta v_{0}+(\alpha v_{0}-v_{1})z_{0})((v_{1}-\alpha v_{0})n+2\alpha v_{0}-v _{1})\alpha ^{n}}{((\alpha v_{0}-v_{1})((v_{1}-\alpha v_{0})n+\alpha v_{0})+(v_{1}-\alpha v _{0}) ((v_{1}-\alpha v_{0})n+v_{1}))\alpha ^{n}} \\ &\quad-\frac{(\beta (2\alpha v_{0}-v_{1})+\alpha (\alpha v_{0}-v_{1})z_{0})((v_{1}-\alpha v _{0})n+\alpha v_{0})\alpha ^{n-1}}{((\alpha v_{0}-v_{1})((v_{1}-\alpha v_{0})n+\alpha v _{0})+(v_{1}-\alpha v_{0}) ((v_{1}-\alpha v_{0})n+v_{1}))\alpha ^{n}} \\ &=\frac{(v_{1}-\alpha v_{0})^{2}(\beta n+\alpha z_{0})\alpha ^{n-1}}{(v_{1}-\alpha v _{0})^{2}\alpha ^{n}} \end{aligned}$$

(51)

for \(n\in\mathbb{N} _{0}\).

Now note that due to (37) it must be

$$\begin{aligned} v_{1}\ne \alpha v_{0}. \end{aligned}$$

(52)

From (52) and since \(\alpha \ne0\), from (51) we obtain

$$\begin{aligned} \tilde{z}_{n}=\frac{\beta n+\alpha z_{0}}{\alpha } \end{aligned}$$

(53)

for \(\in\mathbb{N} _{0}\).

From (43) and (53) we see that (50) holds also in this case, finishing the proof of the theorem. □

### On a bilinear system of difference equations

It is a natural problem to see if there is a representation similar to the one in Theorem 4 for the case of bilinear systems of difference equations.

The corresponding close-to-symmetric system (a recently introduced terminology, see, e.g, [58]) to equation (8) is

$$\begin{aligned} z_{n+1}=\frac{\alpha w_{n}+\beta }{\gamma w_{n}+\delta }, \qquad w_{n+1}=\frac {az_{n}+b}{cz_{n}+d}, \quad n\in\mathbb{N} _{0}, \end{aligned}$$

(54)

where \(\alpha ,\beta ,\gamma ,\delta , a, b, c, d, z_{0}, w_{0}\in {\mathbb {R}}\) (or \(\in {\mathbb {C}}\)).

From (54) we easily obtain

$$\begin{aligned} z_{n+1} &= \frac{(a\alpha +\beta c) z_{n-1}+\alpha b+\beta d}{(a\gamma +c\delta )z_{n-1}+b\gamma +d\delta }, \\ w_{n+1} &=\frac{(a\alpha +b\gamma ) w_{n-1}+a\beta +b \delta }{(\alpha c+\gamma d)w_{n-1}+\beta c+d\delta } \end{aligned}$$

for \(n\in\mathbb{N} \).

Hence, the sequences

$$ z_{n}^{(j)}:=z_{2n+j},\quad n\in\mathbb{N} _{0}, j=0,1, $$

are two solutions to the equation

$$\begin{aligned} \widetilde{z}_{n+1}=\frac{\widetilde{\alpha }\widetilde{z}_{n}+ \widetilde{\beta }}{\widetilde{\gamma }\widetilde{z}_{n}+\widetilde {\delta }}, \quad n\in\mathbb{N} _{0}, \end{aligned}$$

(55)

where

$$\begin{aligned} \widetilde{\alpha }=a\alpha +\beta c,\qquad\widetilde{\beta }=\alpha b+\beta d,\qquad \widetilde{\gamma }=a\gamma +c\delta ,\qquad\widetilde{\delta }=b\gamma +d\delta , \end{aligned}$$

(56)

while the sequences

$$ w_{n}^{(j)}:=w_{2n+j},\quad n\in\mathbb{N} _{0}, j=0,1, $$

are two solutions to the equation

$$\begin{aligned} \widetilde{w}_{n+1}=\frac{\widehat{\alpha }\widetilde{w}_{n}+ \widehat{\beta }}{\widehat{\gamma }\widetilde{w}_{n}+\widehat{\delta }},\quad n \in\mathbb{N} _{0}, \end{aligned}$$

(57)

where

$$\begin{aligned} \widehat{\alpha }=a\alpha +b\gamma ,\qquad\widehat{\beta }=a\beta +b\delta ,\qquad \widehat{\gamma }=\alpha c+ \gamma d,\qquad\widehat{\delta }=\beta c+d\delta . \end{aligned}$$

(58)

For the case of bilinear equations (55) and (57), the corresponding equations in (26) are the same and are given by

$$\begin{aligned} s_{n+1}-(a\alpha +b\gamma +c\beta +d\delta )s_{n}+(ad-bc) (\alpha \delta -\beta \gamma )s_{n-1}=0 \end{aligned}$$

(59)

for \(n\in\mathbb{N} _{0}\).

It is not difficult to see that

$$\begin{aligned} a\alpha +b\gamma +c\beta +d\delta =\widetilde{\alpha }+\widetilde{\delta }=\widehat{\alpha }+ \widehat{\delta } \end{aligned}$$

(60)

and

$$\begin{aligned} (ad-bc) (\alpha \delta -\beta \gamma )=\widetilde{\alpha }\widetilde{\delta }-\widetilde{\beta } \widetilde{\gamma }=\widehat{\alpha }\widehat{\delta }-\widehat{\beta }\widehat{\gamma }. \end{aligned}$$

(61)

From (60) and (61), we see that (59) can be written in the following two forms:

$$\begin{aligned} s_{n+1}-(\widetilde{\alpha }+\widetilde{\delta })s_{n}+(\widetilde{\alpha } \widetilde{\delta }-\widetilde{\beta }\widetilde{\gamma })s_{n-1}=0,\quad n \in \mathbb{N} _{0}, \end{aligned}$$

(62)

and

$$\begin{aligned} s_{n+1}-(\widehat{\alpha }+\widehat{\delta })s_{n}+(\widehat{\alpha } \widehat{\delta }-\widehat{\beta }\widehat{\gamma })s_{n-1}=0,\quad n\in \mathbb{N} _{0}, \end{aligned}$$

(63)

which are the corresponding equations in (26) associated to equations (55) and (57), respectively.

By using Theorem 4, equations (62) and (63), as well as the following two equalities:

$$ z_{1}=\frac{\alpha w_{0}+\beta }{\gamma w_{0}+\delta }\quad\mbox{and}\quad w_{1}=\frac{az_{0}+b}{cz _{0}+d}, $$

we obtain the following theorem.

### Theorem 5

*Consider system* (54) *with*
\((\alpha \delta -\beta \gamma )(ad-bc)\ne0\). *Assume that*
\(\vec{v}=(v_{0}, v_{1}) \in {\mathbb {R}}^{2}\) (*or*
\(\in {\mathbb {C}}^{2}\)) *satisfies the condition*

$$ (\beta \gamma -\alpha \delta ) (ad-bc)v_{0}^{2}+(a\alpha +b\gamma +c\beta +d\delta)v_{0}v_{1}\ne v _{1}^{2}. $$

(64)

*Let*
\((s_{n})_{n\in\mathbb{N} _{0}}\)
*be the solution to equation* (59) *such that*
\(s_{0}=v_{0}\)
*and*
\(s_{1}=v_{1}\). *Then every well*-*defined solution*
\((z_{n}, w_{n})_{n\in\mathbb{N} _{0}}\)
*to system* (54) *has the following representation*:

$$\begin{aligned}& z_{2n} =\frac{(\widetilde{\beta }\widetilde{\gamma }-\widetilde{\alpha } \widetilde{\delta })(\widetilde{\beta }v_{0}+(\widetilde{\alpha }v_{0}-v_{1})z _{0})s_{n-1}+(\widetilde{\beta }((\widetilde{\alpha }+\widetilde{\delta })v_{0}-v _{1})+z_{0}(v_{0}(\widetilde{\beta }\widetilde{\gamma }+\widetilde{\alpha }^{2})- \widetilde{\alpha }v_{1}))s_{n}}{((\widetilde{\beta }\widetilde{\gamma }- \widetilde{\alpha }\widetilde{\delta })v_{0}+\widetilde{\alpha }v_{1}- \widetilde{\gamma }v_{1}z_{0})s_{n}+(\widetilde{\delta }v_{0}-v_{1}+ \widetilde{\gamma }v_{0}z_{0})s_{n+1}}, \\& z_{2n+1} =\frac{(\widetilde{\beta }\widetilde{\gamma }-\widetilde{\alpha } \widetilde{\delta })(\widetilde{\beta }v_{0}(\gamma w_{0}+\delta )+(\widetilde{\alpha }v _{0}-v_{1})(\alpha w_{0}+\beta ))s_{n-1}}{(((\widetilde{\beta }\widetilde{\gamma }- \widetilde{\alpha }\widetilde{\delta })v_{0}+\widetilde{\alpha }v_{1})(\gamma w_{0}+\delta )- \widetilde{\gamma }v_{1}(\alpha w_{0}+\beta ))s_{n}+((\widetilde{\delta }v_{0}-v_{1})(\gamma w _{0}+\delta )+\widetilde{\gamma }v_{0}(\alpha w_{0}+\beta ))s_{n+1}} \\& \hphantom{z_{2n+1} =} +\frac{(\widetilde{\beta }((\widetilde{\alpha }+\widetilde{\delta })v_{0}-v_{1})(\gamma w _{0}+\delta )+(v_{0}(\widetilde{\beta }\widetilde{\gamma }+\widetilde{\alpha }^{2})- \widetilde{\alpha }v_{1})(\alpha w_{0}+\beta ))s_{n}}{(((\widetilde{\beta } \widetilde{\gamma }-\widetilde{\alpha }\widetilde{\delta })v_{0}+\widetilde {\alpha }v _{1})(\gamma w_{0}+\delta )-\widetilde{\gamma }v_{1}(\alpha w_{0}+\beta ))s_{n}+(( \widetilde{\delta }v_{0}-v_{1})(\gamma w_{0}+\delta )+\widetilde{\gamma }v_{0}(\alpha w _{0}+\beta ))s_{n+1}}, \\& w_{2n} =\frac{(\widehat{\beta }\widehat{\gamma }-\widehat{\alpha }\widehat {\delta })( \widehat{\beta }v_{0}+(\widehat{\alpha }v_{0}-v_{1})w_{0})s_{n-1}+( \widehat{\beta }((\widehat{\alpha }+\widehat{\delta })v_{0}-v_{1})+w_{0}(v_{0}( \widehat{\beta }\widehat{\gamma }+\widehat{\alpha }^{2})-\widehat{\alpha }v_{1}))s _{n}}{((\widehat{\beta }\widehat{\gamma }-\widehat{\alpha }\widehat{\delta })v_{0}+ \widehat{\alpha }v_{1}-\widehat{\gamma }v_{1}w_{0})s_{n}+(\widehat{\delta }v_{0}-v _{1}+\widehat{\gamma }v_{0}w_{0})s_{n+1}}, \\& w_{2n+1} =\frac{(\widehat{\beta }\widehat{\gamma }-\widehat{\alpha } \widehat{\delta })(\widehat{\beta }v_{0}(cz_{0}+d)+(\widehat{\alpha }v_{0}-v_{1})(az _{0}+b))s_{n-1}}{(((\widehat{\beta }\widehat{\gamma }-\widehat{\alpha } \widehat{\delta })v_{0}+\widehat{\alpha }v_{1})(cz_{0}+d)-\widehat{\gamma }v_{1}(az _{0}+b))s_{n}+((\widehat{\delta }v_{0}-v_{1})(cz_{0}+d)+\widehat{\gamma }v _{0}(az_{0}+b))s_{n+1}} \\& \hphantom{w_{2n+1} =} +\frac{(\widehat{\beta }((\widehat{\alpha }+\widehat{\delta })v_{0}-v_{1})(cz _{0}+d)+(v_{0}(\widehat{\beta }\widehat{\gamma }+\widehat{\alpha }^{2})- \widehat{\alpha }v_{1})(az_{0}+b))s_{n}}{(((\widehat{\beta }\widehat{\gamma }- \widehat{\alpha }\widehat{\delta })v_{0}+\widehat{\alpha }v_{1})(cz_{0}+d)- \widehat{\gamma }v_{1}(az_{0}+b))s_{n}+((\widehat{\delta }v_{0}-v_{1})(cz_{0}+d)+ \widehat{\gamma }v_{0}(az_{0}+b))s_{n+1}}, \end{aligned}$$

\(n\in\mathbb{N} _{0}\), *where*
*α̃*, *β̃*, *γ̃*, *δ̃*, *α̂*, *β̂*, *γ̂*, *δ̂*
*are defined in* (56) *and* (58).

### On a three-dimensional generalization of equation (8)

Here we show that it is also possible to generalize Theorem 4 for the case of a three-dimensional close-to-cyclic system of bilinear difference equations which extends difference equation (8).

The following three-dimensional close-to-cyclic system of bilinear difference equations is a natural generalization of difference equation (8), as well as of system of equations (54)

$$\begin{aligned} z_{n+1}=\frac{a w_{n}+b}{c w_{n}+d},\qquad w_{n+1}=\frac{eu_{n}+f}{gu _{n}+h}, \qquad u_{n+1}=\frac{pz_{n}+q}{rz_{n}+s},\quad n\in\mathbb{N} _{0}, \end{aligned}$$

(65)

where \(a, b, c, d, e, f, g, h, p, q, r, s, z_{0}, w_{0}, u_{0}\in {\mathbb {R}}\) (or \(\in {\mathbb {C}}\)).

By using the method applied to system (54), it is easy to see that from the equations in (65) it follows that

$$\begin{aligned}& z_{n+1} =\frac{(aep+bgp+afr+bhr) z_{n-2}+aeq+bgq+afs+bhs}{(cep+dgp+cfr+dhr) z_{n-2}+ceq+dgq+cfs+dhs}, \\& w_{n+1} =\frac{(aep+afr+ceq+cfs) w_{n-2}+bep+bfr+deq+dfs}{(agp+ahr+cgq+chs) w_{n-2}+bgp+bhr+dgq+dhs}, \\& u_{n+1} =\frac{(pae+qce+pbg+qdg) w_{n-2}+paf+qcf+pbh+qdh}{(rae+sce+rbg+sdg) w_{n-2}+raf+scf+rbh+sdh} \end{aligned}$$

for \(n\ge2\).

Hence, the sequences

$$ z_{n}^{(j)}:=z_{3n+j},\quad n\in\mathbb{N} _{0}, j=0,1,2, $$

are three solutions to the equation

$$\begin{aligned} \widetilde{z}_{n+1}=\frac{\widetilde{a}\widetilde{z}_{n}+ \widetilde{b}}{\widetilde{c}\widetilde{z}_{n}+\widetilde{d}},\quad n \in\mathbb{N} _{0}, \end{aligned}$$

(66)

where

$$\begin{aligned}& \widetilde{a} =aep+bgp+afr+bhr, \end{aligned}$$

(67)

$$\begin{aligned}& \widetilde{b} =aeq+bgq+afs+bhs, \end{aligned}$$

(68)

$$\begin{aligned}& \widetilde{c} =cep+dgp+cfr+dhr, \end{aligned}$$

(69)

$$\begin{aligned}& \widetilde{d} =ceq+dgq+cfs+dhs, \end{aligned}$$

(70)

the sequences

$$ w_{n}^{(j)}:=w_{3n+j},\quad n\in\mathbb{N} _{0}, j=0,1,2, $$

are three solutions to the equation

$$\begin{aligned} \widetilde{w}_{n+1}=\frac{\widetilde{e}\widetilde{w}_{n}+ \widetilde{f}}{\widetilde{g}\widetilde{w}_{n}+\widetilde{h}},\quad n \in\mathbb{N} _{0}, \end{aligned}$$

(71)

where

$$\begin{aligned}& \widetilde{e} =aep+afr+ceq+cfs, \end{aligned}$$

(72)

$$\begin{aligned}& \widetilde{f} =bep+bfr+deq+dfs, \end{aligned}$$

(73)

$$\begin{aligned}& \widetilde{g} =agp+ahr+cgq+chs, \end{aligned}$$

(74)

$$\begin{aligned}& \widetilde{h} =bgp+bhr+dgq+dhs, \end{aligned}$$

(75)

while the sequences

$$ u_{n}^{(j)}:=u_{3n+j},\quad n\in\mathbb{N} _{0}, j=0,1,2, $$

are three solutions to the equation

$$\begin{aligned} \widetilde{u}_{n+1}=\frac{\widetilde{p}\widetilde{u}_{n}+ \widetilde{q}}{\widetilde{r}\widetilde{u}_{n}+\widetilde{s}},\quad n \in\mathbb{N} _{0}, \end{aligned}$$

(76)

where

$$\begin{aligned} \widetilde{p} &=pae+qce+pbg+qdg, \end{aligned}$$

(77)

$$\begin{aligned} \widetilde{q} &=paf+qcf+pbh+qdh, \end{aligned}$$

(78)

$$\begin{aligned} \widetilde{r} &=rae+sce+rbg+sdg, \end{aligned}$$

(79)

$$\begin{aligned} \widetilde{s} &=raf+scf+rbh+sdh. \end{aligned}$$

(80)

For the case of bilinear equations (66), (71), and (76), the corresponding equations in (26) are the same and are given by

$$\begin{aligned} &s_{n+1} -(aep+bgp+afr+bhr+ceq+dgq+cfs+dhs)s_{n} \\ &\quad {}+(ad-bc) (eh-fg) (ps-qr)s_{n-1}=0 \end{aligned}$$

(81)

for \(n\in\mathbb{N} _{0}\).

It is not difficult to see that

$$\begin{aligned} aep+bgp+afr+bhr+ceq+dgq+cfs+dhs=\widetilde{a}+\widetilde{d}= \widetilde{e}+ \widetilde{h}=\widetilde{p}+\widetilde{s} \end{aligned}$$

(82)

and

$$\begin{aligned} (ad-bc) (eh-fg) (ps-qr)=\widetilde{a}\widetilde{d}-\widetilde{b} \widetilde{c}= \widetilde{e}\widetilde{h}-\widetilde{f}\widetilde{g}= \widetilde {p}\widetilde{s}- \widetilde{q}\widetilde{r}. \end{aligned}$$

(83)

From (82) and (83), we see that (81) can be written in the following three forms:

$$\begin{aligned}& s_{n+1}-(\widetilde{a}+\widetilde{d})s_{n}+(\widetilde{a} \widetilde{d}-\widetilde{b}\widetilde{c})s_{n-1}=0, \end{aligned}$$

(84)

$$\begin{aligned}& s_{n+1}-(\widetilde{e}+\widetilde{h})s_{n}+(\widetilde{e} \widetilde{h}-\widetilde{f}\widetilde{g})s_{n-1}=0, \end{aligned}$$

(85)

and

$$\begin{aligned} s_{n+1}-(\widetilde{p}+\widetilde{s})s_{n}+(\widetilde{p} \widetilde{s}-\widetilde{q}\widetilde{r})s_{n-1}=0 \end{aligned}$$

(86)

for \(n\in\mathbb{N} _{0}\), which are the corresponding equations in (26) associated to equations (66), (71), and (76), respectively.

Hence, by using Theorem 4, (84)–(86), and the following equalities:

$$ z_{1}=\frac{a w_{0}+b}{c w_{0}+d},\qquad w_{1}= \frac{eu_{0}+f}{gu_{0}+h}, \qquad u_{1}=\frac{pz_{0}+q}{rz_{0}+s}, $$

we get the following result.

### Theorem 6

*Consider system* (65). *Assume that*

$$ (ad-bc) (eh-fg) (ps-qr)\ne0, $$

\(\vec{v}=(v_{0}, v_{1})\in {\mathbb {R}}^{2}\) (*or*
\(\in {\mathbb {C}}^{2}\)) *satisfies the condition*

$$\begin{aligned} (bc-ad) (eh-fg) (ps-qr)v_{0}^{2}+(aep+bgp+afr+bhr+ceq+dgq+cfs+dhs)v_{0}v _{1}\ne v_{1}^{2}. \end{aligned}$$

(87)

*Let*
\((s_{n})_{n\in\mathbb{N} _{0}}\)
*be the solution to equation* (81) *such that*
\(s_{0}=v_{0}\)
*and*
\(s_{1}=v_{1}\). *Then every well*-*defined solution*
\((z_{n}, w_{n}, u_{n})_{n\in\mathbb{N} _{0}}\)
*to system* (65) *has the following representation*:

$$\begin{aligned} \begin{aligned} &z_{3n} =\frac{(\widetilde{b}\widetilde{c}-\widetilde{a}\widetilde{d})( \widetilde{b} v_{0}+(\widetilde{a} v_{0}-v_{1})z_{0})s_{n-1}+( \widetilde{b}((\widetilde{a}+\widetilde{d})v_{0}-v_{1})+z_{0}(v_{0}( \widetilde{b}\widetilde{c}+\widetilde{a}^{2})-\widetilde{a} v_{1}))s _{n}}{((\widetilde{b}\widetilde{c}-\widetilde{a}\widetilde{d})v_{0}+ \widetilde{a} v_{1}-\widetilde{c} v_{1}z_{0})s_{n}+(\widetilde{d} v _{0}-v_{1}+\widetilde{c} v_{0}z_{0})s_{n+1}}, \\ &w_{3n} =\frac{(\widetilde{f}\widetilde{g}-\widetilde{e}\widetilde{h})( \widetilde{f} v_{0}+(\widetilde{e} v_{0}-v_{1})w_{0})s_{n-1}+( \widetilde{f}((\widetilde{e}+\widetilde{h})v_{0}-v_{1})+w_{0}(v_{0}( \widetilde{f}\widetilde{g}+\widetilde{e}^{2})-\widetilde{e} v_{1}))s _{n}}{((\widetilde{f}\widetilde{g}-\widetilde{e}\widetilde{h})v_{0}+ \widetilde{e} v_{1}-\widetilde{g} v_{1}w_{0})s_{n}+(\widetilde{h} v _{0}-v_{1}+\widetilde{g} v_{0}w_{0})s_{n+1}}, \\ &u_{3n} =\frac{(\widetilde{q}\widetilde{r}-\widetilde{p}\widetilde{s})( \widetilde{q} v_{0}+(\widetilde{p} v_{0}-v_{1})u_{0})s_{n-1}+( \widetilde{q}((\widetilde{p}+\widetilde{s})v_{0}-v_{1})+u_{0}(v_{0}( \widetilde{q}\widetilde{r}+\widetilde{p}^{2})-\widetilde{p} v_{1}))s _{n}}{((\widetilde{q}\widetilde{r}-\widetilde{p}\widetilde{s})v_{0}+ \widetilde{p} v_{1}-\widetilde{r} v_{1}u_{0})s_{n}+(\widetilde{s} v _{0}-v_{1}+\widetilde{r} v_{0}u_{0})s_{n+1}}, \\ &z_{3n+1} =\frac{(\widetilde{b}\widetilde{c}-\widetilde{a} \widetilde{d})(\widetilde{b} v_{0}(c w_{0}+d)+(\widetilde{a} v_{0}-v _{1})(a w_{0}+b))s_{n-1}}{(((\widetilde{b}\widetilde{c}-\widetilde{a} \widetilde{d})v_{0}+\widetilde{a} v_{1})(c w_{0}+d)-\widetilde{c} v _{1}(a w_{0}+b))s_{n}+((\widetilde{d} v_{0}-v_{1})(c w_{0}+d)+ \widetilde{c} v_{0}(a w_{0}+b))s_{n+1}} \\ & \hphantom{z_{3n+1} =}{} +\frac{(\widetilde{b}((\widetilde{a}+\widetilde {d})v_{0}-v_{1})(c w _{0}+d)+(v_{0}(\widetilde{b}\widetilde{c}+\widetilde{a}^{2})- \widetilde{a} v_{1})(a w_{0}+b))s_{n}}{(((\widetilde{b}\widetilde{c}- \widetilde{a}\widetilde{d})v_{0}+\widetilde{a} v_{1})(c w_{0}+d)- \widetilde{c} v_{1}(a w_{0}+b))s_{n}+((\widetilde{d} v_{0}-v_{1})(c w _{0}+d)+\widetilde{c} v_{0}(a w_{0}+b))s_{n+1}}, \\ &w_{3n+1} =\frac{(\widetilde{f}\widetilde{g}-\widetilde{e} \widetilde{h})(\widetilde{f} v_{0}(gu_{0}+h)+(\widetilde{e} v_{0}-v _{1})(eu_{0}+f))s_{n-1}}{(((\widetilde{f}\widetilde{g}-\widetilde{e} \widetilde{h})v_{0}+\widetilde{e} v_{1})(gu_{0}+h)-\widetilde{g} v _{1}(eu_{0}+f))s_{n}+((\widetilde{h} v_{0}-v_{1})(gu_{0}+h)+ \widetilde{g} v_{0}(eu_{0}+f))s_{n+1}} \\ &\hphantom{w_{3n+1} =}{} +\frac{(\widetilde{f}((\widetilde{e}+\widetilde{h})v_{0}-v_{1})(gu _{0}+h)+(v_{0}(\widetilde{f}\widetilde{g}+\widetilde{e}^{2})- \widetilde{e} v_{1})(eu_{0}+f))s_{n}}{(((\widetilde{f}\widetilde{g}- \widetilde{e}\widetilde{h})v_{0}+\widetilde{e} v_{1})(gu_{0}+h)- \widetilde{g} v_{1}(eu_{0}+f))s_{n}+((\widetilde{h} v_{0}-v_{1})(gu _{0}+h)+\widetilde{g} v_{0}(eu_{0}+f))s_{n+1}}, \\ &u_{3n+1} =\frac{(\widetilde{q}\widetilde{r}-\widetilde{p} \widetilde{s})(\widetilde{q} v_{0}(rz_{0}+s)+(\widetilde{p} v_{0}-v _{1})(pz_{0}+q))s_{n-1}}{(((\widetilde{q}\widetilde{r}-\widetilde{p} \widetilde{s})v_{0}+\widetilde{p} v_{1})(rz_{0}+s)-\widetilde{r} v _{1}(pz_{0}+q))s_{n}+((\widetilde{s} v_{0}-v_{1})(rz_{0}+s)+ \widetilde{r} v_{0}(pz_{0}+q))s_{n+1}} \\ &\hphantom{u_{3n+1} =}{}+\frac{(\widetilde{q}((\widetilde{p}+\widetilde{s})v_{0}-v_{1})(rz _{0}+s)+(v_{0}(\widetilde{q}\widetilde{r}+\widetilde{p}^{2})- \widetilde{p} v_{1})(pz_{0}+q))s_{n}}{(((\widetilde{q}\widetilde{r}- \widetilde{p}\widetilde{s})v_{0}+\widetilde{p} v_{1})(rz_{0}+s)- \widetilde{r} v_{1}(pz_{0}+q))s_{n}+((\widetilde{s} v_{0}-v_{1})(rz _{0}+s)+\widetilde{r} v_{0}(pz_{0}+q))s_{n+1}}, \\ &z_{3n+2} =\frac{(\widetilde{b}\widetilde{c}-\widetilde{a} \widetilde{d})(\widetilde{b} v_{0}+(\widetilde{a} v_{0}-v_{1})z_{2})s _{n-1}+(\widetilde{b}((\widetilde{a}+\widetilde{d})v_{0}-v_{1})+z_{2}(v _{0}(\widetilde{b}\widetilde{c}+\widetilde{a}^{2})-\widetilde{a} v _{1}))s_{n}}{((\widetilde{b}\widetilde{c}-\widetilde{a}\widetilde{d})v _{0}+\widetilde{a} v_{1}-\widetilde{c} v_{1}z_{2})s_{n}+( \widetilde{d} v_{0}-v_{1}+\widetilde{c} v_{0}z_{2})s_{n+1}}, \\ &w_{3n+2} =\frac{(\widetilde{f}\widetilde{g}-\widetilde{e} \widetilde{h})(\widetilde{f} v_{0}+(\widetilde{e} v_{0}-v_{1})w_{2})s _{n-1}+(\widetilde{f}((\widetilde{e}+\widetilde{h})v_{0}-v_{1})+w_{2}(v _{0}(\widetilde{f}\widetilde{g}+\widetilde{e}^{2})-\widetilde{e} v _{1}))s_{n}}{((\widetilde{f}\widetilde{g}-\widetilde{e}\widetilde{h})v _{0}+\widetilde{e} v_{1}-\widetilde{g} v_{1}w_{2})s_{n}+( \widetilde{h} v_{0}-v_{1}+\widetilde{g} v_{0}w_{2})s_{n+1}}, \\ &u_{3n+2} =\frac{(\widetilde{q}\widetilde{r}-\widetilde{p} \widetilde{s})(\widetilde{q} v_{0}+(\widetilde{p} v_{0}-v_{1})u_{2})s _{n-1}+(\widetilde{q}((\widetilde{p}+\widetilde{s})v_{0}-v_{1})+u_{2}(v _{0}(\widetilde{q}\widetilde{r}+\widetilde{p}^{2})-\widetilde{p} v _{1}))s_{n}}{((\widetilde{q}\widetilde{r}-\widetilde{p}\widetilde{s})v _{0}+\widetilde{p} v_{1}-\widetilde{r} v_{1}u_{2})s_{n}+( \widetilde{s} v_{0}-v_{1}+\widetilde{r} v_{0}u_{2})s_{n+1}} \end{aligned} \end{aligned}$$

(88)

*for every*
\(n\in\mathbb{N} _{0}\)
*and each*
\(j=0,1,2\), *where*
*ã*, *b̃*, *c̃*, *d̃*, *ẽ*, *f̃*, *g̃*, *h̃*, *p̃*, *q̃*, *r̃*, *s̃*
*are defined by* (67)*–*(70), (72)*–*(75), (77)*–*(80), *respectively*.

### Remark 2

In order to get all the formulas in Theorem 6 in terms of initial values \(z_{0}\), \(w_{0}\), and \(u_{0}\), in the formulas for \(z_{3n+2}\), \(w_{3n+2}\), and \(u_{3n+2}\), the following equalities should be used:

$$\begin{aligned}& z_{2} =\frac{(ae+bg)u_{0}+af+bh}{(ce+dg)u_{0}+cf+dh}, \\& w_{2} =\frac{(ep+fr)z_{0}+eq+fs}{(gp+hr)z_{0}+gq+hs}, \\& u_{2} =\frac{(pa+qc)w_{0}+pb+qd}{(ra+sc) w_{0}+rb+sd}, \end{aligned}$$

which follows from the equations in (65). We have not done this since such obtained formulas are so long that cannot be written in a line without a drastic reduction of letters. Hence, we leave the simple task to the interested reader.

### Remark 3

It is time consuming but not difficult to check that the above procedure employed to the two-dimensional close-to-symmetric and three-dimensional close-to-cyclic bilinear systems of difference equations can be also applied to close-to-cyclic bilinear systems of difference equations of fourth order. However, since the procedure leads to some long calculations and quite long formulas, we also leave getting the corresponding formulas for solutions to the system to the interested reader.