# Characterization of entire solutions of systems of quadratic trinomial
difference and partial differential difference equations in $\mathbb{C}^{n}$

Goutam Haldar  Department of Mathematics, Malda College, Malda, West Bengal
732101, India<EMAIL_ADDRESS>
(Date: Received: xxxxxx; Revised: yyyyyy; Accepted: zzzzzz.
∗ Corresponding author)

###### Abstract.

In this paper we establish some results about the existence and precise forms
of finite order entire solutions of some systems of quadratic trinomial
functional equations one of which in $\mathbb{C}^{n}$, $n\in\mathbb{N}$ and
other two in $\mathbb{C}^{2}$. Our results are the generalizations and
improvements of the previous theorems given by Xu-Cao [42, 43] and Xu-Jiang
[45]. Moreover, we exhibit some examples in support of our claims.

###### Key words and phrases:

entire solutions, Fermat-type, differential difference equations, Nevanlinna
theory.

###### 1991 Mathematics Subject Classification:

39A45, 30D35, 32H30, 39A14, 35A20

##  1\. Introduction

It has a long history on the study of entire and meromorphic solutions of
Fermat type equations by employing Nevanlinna theory as a tool. For detail
study, we insist the readers to go through [32, 11, 12, 19, 5, 6, 7, 50, 48,
40, 41, 32] and the references therein. We assume the readers are familiar
with the notations and basic definitions of Nevanlinna theory (see [21, 17]).
In 1939, Iyer [19] proved that the entire solutions of Fermat type equation

$\displaystyle f(z)^{2}+g(z)^{2}=1$ (1.1)

are $f=\cos a(z)$, $g=\sin a(z)$, where $a(z)$ is an entire function.

Utilizing difference Nevanlinna theory of meromorphic functions developed by
Halburd-Korhonen [13, 14], and Chiang-Feng [4], independently, Liu et al.
[29], in $2012$, proved that any finite order transcendental entire solutions
of

$\displaystyle f(z)^{2}+f(z+c)^{2}=1,$ (1.2)

has the form $f(z)=\sin(az+b)$, where $b$ is a constant and $a=(4k+1)\pi/2c$,
with $k$ is an integer. In the same paper, they also proved that any finite
order transcendental entire solution of

$\displaystyle f^{\prime}(z)^{2}+f(z+c)^{2}=1$ (1.3)

has the form $f(z)=\sin(z\pm Bi)$, where $B$ is a constant and $c=2p\pi$ or
$c=(2p+1)\pi$, $p\in\mathbb{Z}$, whereas any transcendental entire solutions
with finite order of

$\displaystyle f^{\prime}(z)^{2}+[f(z+c)-f(z)]^{2}=1,$ (1.4)

must take the form $f(z)=12\sin(2z+bi)$, where $c=(2p+1)\pi$, $p$ being an
integer, and $b$ is a constant.

Although there are many important and interesting results on the existence of
entire and meromorphic solutions of complex difference as well as
differential-difference equations (see [28, 29, 31, 26, 27, 30, 25, 39]), only
a few number of results are there on the existence and forms of entire and
meromorphic solutions of quadratic trinomial functional equation. To the best
of our knowledge, Saleeby [37] first investigated the existence and forms of
entire and meromorphic solutions of quadratic trinomial functional equation

$\displaystyle f^{2}+2wfg+g^{2}=1,$ (1.5)

where $w$ is a constant. Observe that (1.1) can be seen as the case of $w=0$.
For $w\neq 0$, Saleeby [37] proved the following classical results about the
Fermat type functional equations (1.5).

######  Theorem A.

[37] Let $\alpha\in\mathbb{C}$ with $\alpha^{2}\neq 0,1$. Then any
transcendental entire solution of (1.5) has the form
$f(z)=\dfrac{1}{\sqrt{2}}\left(\dfrac{\cos(h(z))}{\sqrt{1+w}}+\dfrac{\sin(h(z))}{\sqrt{1-w}}\right)$
and
$\dfrac{1}{\sqrt{2}}\left(\dfrac{\cos(h(z))}{\sqrt{1+w}}-\dfrac{\sin(h(z))}{\sqrt{1-w}}\right)$,
where $h$ is entire $\mathbb{C}^{n}$. The meromorphic solutions of (1.5) is of
the form
$f(z)=\dfrac{\alpha_{1}-\alpha_{2}\beta(z)^{2}}{(\alpha_{1}-\alpha_{2})\beta(z)}$
and $g(z)=\dfrac{1-\beta(z)^{2}}{(\alpha_{1}-\alpha_{2})\beta(z)}$, where
$\beta(z)$ is meromorphic in $\mathbb{C}^{n}$ and
$\alpha_{1}=-w+\sqrt{w^{2}-1}$, $\alpha_{2}=-w-\sqrt{w^{2}-1}$.

By utilizing Theorem A, in 2016, Liu-Yang [31] obtained more precise
properties of entire and meromorphic solutions of

$\displaystyle f(z)^{2}+2wf(z)f^{\prime}(z)+f^{\prime}(z)^{2}=1$ (1.6)

and

$\displaystyle f(z)^{2}+2wf(z)f(z+c)+f(z+c)^{2}=1,\;\;w\neq 0,\pm 1.$ (1.7)

To the best of our knowledge, there are only a few number of results on the
system of complex difference and differential-difference equations in the
literature (see [9]). It was Gao [9], who, in $2016$, first discussed the
existence and form of transcendental entire solutions of system of
differential difference equations and obtained the result as follows: If
$(f,g)$ be a pair of finite order transcendental entire solution of the system
of differential-difference equations

$\displaystyle\begin{cases}f^{\prime}(z)^{2}+g(z+c)^{2}=1,\\\
g^{\prime}(z)^{2}+f(z+c)^{2}=1.\end{cases}$ (1.8)

Then $(f,g)$ must be of the form

$\displaystyle(f,g)=(\sin(z-ib),\sin(z-ib_{1})),\;\;\text{or}\;\;(\sin(z+ib),\sin(z+ib_{1})),$

where $b,b_{1}$ are constants and $c=k\pi$, $k$ is a integer.

##  2\. Solutions of Fermat type equations in several complex variables

Recently, the study of entire and meromorphic solutions of Fermat type partial
differential equations in several complex variables has received considerable
attention in the literature (see [24, 23, 35, 36, 22]). The study partial
differential equations which is a generalizations of the well-known eikonal
equation in real variable case has a long history. We insist the readers to go
through [8, 10, 33] and the references therein. In 1995, Khavinson pointed out
that in $\mathbb{C}^{2}$, the entire solution of the Fermat type partial
differential equations $f_{z_{1}}^{2}+f_{z_{2}}^{2}=1$ must necessarily be
linear. After the development of the difference Nevanlinna theory in several
complex variables, specially the difference version of logarithmic derivative
lemma (see [2, 1, 3, 20]), many Researchers started studying the existence and
the precise form of entire and meromorphic solutions of different variants of
Fermat type difference and partial differential difference equations, and
obtained very remarkable and interesting results (see [42, 43, 15, 16, 44, 34,
38, 46]).

Hereinafter, we denote by $z+w=(z_{1}+w_{1},z_{2}+w_{2},\ldots,z_{n}+w_{n})$,
where
$z=(z_{1},z_{2},\ldots,z_{n}),w=(w_{1},w_{2},\ldots,w_{n})\in\mathbb{C}^{n}$,
$n$ being a positive integer.

In 2018, Xu-Cao [42] established the existence of entire solutions for some
Fermat type complex difference and partial differential difference equations
with several variables. We list some of them below.

######  Theorem B.

[42] Let
$c=(c_{1},c_{2},\ldots,c_{n})\in\mathbb{C}^{n}\setminus\\{(0,0,\ldots,0)\\}$.
If $f:\mathbb{C}^{n}\rightarrow\mathbb{P}^{1}(\mathbb{C})$ be an entire
solution with finite order of the Fermat type difference equation

$\displaystyle f(z)^{2}+f(z+c)^{2}=1,$ (2.1)

the $f(z)$must assume the form $f(z)=\cos(L(z)+B)$, where $L$ is a linear
function of the form $L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$ on $\mathbb{C}^{n}$
such that $L(c)=-\pi/2-2k\pi$ $(k\in\mathbb{Z})$, and $B$ is a constant on
$\mathbb{C}$.

######  Theorem C.

[42] Let $c=(c_{1},c_{2})\in\mathbb{C}^{2}$. If $f(z)$ be transcendental
entire solution with finite order of the Fermat type partial differential-
difference equation

$\displaystyle\left(\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+f(z_{1}+c_{1},z_{2}+c_{2})^{2}=1,$ (2.2)

$f$ must be of the form $f(z_{1},z_{2})=\sin(Az_{1}+Bz_{2}+\phi(z_{2}))$,
Where $A,B$ are constant on $\mathbb{C}$ satisfying $A^{2}=1$ and
$Ae^{i(Ac_{1}+Bc_{2})}=1$, and $\phi(z_{2})$ is a polynomial in one variable
$z_{2}$ such that $\phi(z_{2})\equiv\phi(z_{2}+c_{2})$. In the special case
whenever $c_{2}\neq 0$, we have $f(z_{1},z_{2})=\sin(Az_{1}+Bz_{2}+Constant)$.

Very recently, corresponding to Equations (2.5) and (1.6), Xu-Jiang [45]
considered the following system of trinomial difference and differential-
difference equations in $\mathbb{C}^{2}$

$\displaystyle\begin{cases}f(z)^{2}+2wf(z)g(z+c)+g(z+c)^{2}=1\\\
g(z)^{2}+2wg(z)f(z+c)+f(z+c)^{2}=1\end{cases}$ (2.3)

and

$\displaystyle\begin{cases}\left(\dfrac{\partial f}{\partial
z_{1}}\right)^{2}+2w\dfrac{\partial f}{\partial
z_{1}}g(z+c)+g(z+c)^{2}=1\vspace{1mm}\\\ \left(\dfrac{\partial g}{\partial
z_{1}}\right)^{2}+2w\dfrac{\partial g}{\partial
z_{1}}f(z+c)+f(z+c)^{2}=1,\end{cases}$ (2.4)

and obtained the results as follows.

######  Theorem D.

[45] Let $c=(c_{1},c_{2})\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$. Then any pair
of finite order transcendental entire solutions of (2.3) must assume one of
the following forms:

  1. (i)

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left(\frac{\cos(\gamma(z)+b_{1})}{\sqrt{1+w}}+\frac{\sin(\gamma(z)+b_{1})}{\sqrt{1-w}}\right),$
$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left(\frac{\cos(\gamma(z)+b_{2})}{\sqrt{1+w}}+\frac{\sin(\gamma(z)+b_{2})}{\sqrt{1-w}}\right),$

where $\gamma(z)=L(z)+H(s)$, $L(z)=a_{1}z_{1}+a_{2}z_{2}$, $H(s)$ is a
polynomial in $s:=c_{2}z_{1}-c_{1}z_{2}$, $a_{i},b_{i},c_{i}$, $i=1,2$ satisfy
the relations

$\displaystyle
e^{2iL(c)}=\frac{-w+\sqrt{w^{2}-1}}{-w-\sqrt{w^{2}-1}},\;\;e^{2i(b_{1}-b_{2})}=1.$

  2. (ii)

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left(\frac{\cos(\gamma(z)+b_{1})}{\sqrt{1+w}}+\frac{\sin(\gamma(z)+b_{1})}{\sqrt{1-w}}\right),$
$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left(\frac{\cos(\gamma(z)+b_{2})}{\sqrt{1+w}}-\frac{\sin(\gamma(z)+b_{2})}{\sqrt{1-w}}\right),$

where $\gamma(z)$ is the same as in (i), and $a_{i},b_{i},c_{i}$, $i=1,2$
satisfy the relations

$\displaystyle e^{2iL(c)}=1,\;\;e^{2i(b_{1}-b_{2})}=1.$

######  Theorem E.

[45] Let $c=(c_{1},c_{2})\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$ with
$c_{2}\neq 0$. Then any pair of finite order transcendental entire solutions
of (2.4) must assume one of the following forms:

  1. (i)

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left(\frac{\cos(L(z)-L(c)+b_{1})}{\sqrt{1+w}}+\frac{\sin(L(z)-L(c)+b_{1})}{\sqrt{1-w}}\right),$
$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left(\frac{\cos(L(z)-L(c)+b_{2})}{\sqrt{1+w}}+\frac{\sin(L(z)-L(c)+b_{2})}{\sqrt{1-w}}\right),$

where $L(z)=a_{1}z_{1}+a_{2}z_{2}$, $a_{i},b_{i},c_{i}$, $i=1,2$ satisfy the
relations

$\displaystyle
a_{1}^{2}=1,\;\;e^{2iL(c)}=\frac{-w+\sqrt{w^{2}-1}}{-w-\sqrt{w^{2}-1}},\;\;e^{2i(b_{1}-b_{2})}=1.$

  2. (ii)

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left(\frac{\cos(L(z)-L(c)+b_{1})}{\sqrt{1+w}}+\frac{\sin(L(z)-L(c)+b_{1})}{\sqrt{1-w}}\right),$
$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left(\frac{\cos(L(z)-L(c)+b_{2})}{\sqrt{1+w}}-\frac{\sin(L(z)-L(c)+b_{2})}{\sqrt{1-w}}\right),$

where $L(z)$ is the same as in (i), and $a_{i},b_{i},c_{i}$, $i=1,2$ satisfy
the relations

$\displaystyle a_{1}^{2}=1,\;\;e^{2iL(c)}=1,\;\;e^{2i(b_{1}-b_{2})}=1.$

Observing Theorems B and E, it is inevitable to ask the following open
questions.

######  Question 2.1.

What can be said about the entire solutions of (2.3) when one replace the
right hand sides of each equation by $e^{g_{1}(z)}$ and $e^{g_{2}(z)}$,
respectively, where $g_{1}(z),g_{2}(z)$ are any two polynomials in
$\mathbb{C}^{n}$, $n$ being an integer$?$

######  Question 2.2.

What can be said about the existence and precise form of transcendental entire
solutions of (2.4) when one replace (a) first order partial derivative by
$k$-th order partial derivative for any $k\in\mathbb{N}$ and (b) right hand
sides of each of the equations by $e^{g_{1}(z_{1},z_{2})}$ and
$e^{g_{2}(z_{1},z_{2})}$, respectively, where
$g_{1}(z_{1},z_{2}),g_{2}(z_{1},z_{2})$ are any two polynomials in
$\mathbb{C}^{2}$$?$

######  Question 2.3.

What can be said about the existence and precise form of transcendental entire
solutions of

$\displaystyle\begin{cases}\left[\dfrac{\partial^{k}f}{\partial
z_{1}^{k}}\right]^{2}+2w\dfrac{\partial^{k}f}{\partial
z_{1}^{k}}[g(z+c)-g(z)]+[g(z+c)-g(z)]^{2}=1,\vspace{1mm}\\\
\left[\dfrac{\partial^{k}g}{\partial
z_{1}^{k}}\right]^{2}+2w\dfrac{\partial^{k}g}{\partial
z_{1}^{k}}[f(z+c)-f(z)]+[f(z+c)-f(z)]^{2}=1?\end{cases}$ (2.5)

##  3\. Results and Examples

Inspired by the above results and questions, we investigate the existence and
form of transcendental entire solutions of the system of equations

$\displaystyle\begin{cases}f(z)^{2}+2wf(z)g(z+c)+g(z+c)^{2}=e^{g_{1}(z)}\\\
g(z)^{2}+2wg(z)f(z+c)+f(z+c)^{2}=e^{g_{2}(z)},\end{cases}$ (3.1)
$\displaystyle\begin{cases}\left[\dfrac{\partial^{k}f}{\partial
z_{1}^{k}}\right]^{2}+2w\dfrac{\partial^{k}f}{\partial
z_{1}^{k}}g(z+c)+g(z+c)^{2}=e^{g_{1}(z_{1},z_{2})}\vspace{1mm}\\\
\left[\dfrac{\partial^{k}g}{\partial
z_{1}^{k}}\right]^{2}+2w\dfrac{\partial^{k}g}{\partial
z_{1}^{k}}f(z+c)+f(z+c)^{2}=e^{g_{2}(z_{1},z_{2})},\end{cases}$ (3.2)

where $g_{1}(z_{1},z_{2})$, $g_{2}(z_{1},z_{2})$ are any two polynomials and
$k\in\mathbb{N}$ and Eq. (2.5). Our results are the generalizations of
Theorems B–C to a large extent.

Before we state our main results, let us denote two non-zero complex numbers
$A_{1},A_{2}$ as follows.

$\displaystyle
A_{1}=\frac{1}{2\sqrt{1+w}}+\frac{1}{2i\sqrt{1-w}}\;\;\text{and}\;\;A_{2}=\frac{1}{2\sqrt{1+w}}-\frac{1}{2i\sqrt{1-w}}.$

Before we state our main results, let us set the following.

$\displaystyle\Phi(z)$
$\displaystyle=\sum_{i_{1}=1}^{{}^{n}C_{2}}H_{i_{1}}^{2}(s_{i_{1}}^{2})+\sum_{i_{2}=1}^{{}^{n}C_{3}}H_{i_{2}}^{3}(s_{i_{2}}^{3})+\sum_{i_{3}=1}^{{}^{n}C_{4}}H_{i_{3}}^{4}(s_{i_{3}}^{4})+\cdots$
(3.3)
$\displaystyle+\sum_{i_{n-2}=1}^{{}^{n}C_{n-1}}H_{i_{n-2}}^{n-1}(s_{i_{n-2}}^{n-1})+H_{n-1}^{n}(s_{n-1}^{n}),$
$\displaystyle\Psi(z)$
$\displaystyle=\sum_{i_{1}=1}^{{}^{n}C_{2}}L_{i_{1}}^{2}(t_{i_{1}}^{2})+\sum_{i_{2}=1}^{{}^{n}C_{3}}L_{i_{2}}^{3}(t_{i_{2}}^{3})+\sum_{i_{3}=1}^{{}^{n}C_{4}}L_{i_{3}}^{4}(t_{i_{3}}^{4})+\cdots$
(3.4)
$\displaystyle+\sum_{i_{n-2}=1}^{{}^{n}C_{n-1}}L_{i_{n-2}}^{n-1}(t_{i_{n-2}}^{n-1})+L_{n-1}^{n}(t_{n-1}^{n}),$

where $H_{i_{1}}^{2}$ is a polynomial in
$s_{i_{1}}^{2}:=d_{i_{1}j_{1}}^{2}z_{j_{1}}+d_{i_{1}j_{2}}^{2}z_{j_{2}}$ with
$d_{i_{1}j_{1}}^{2}c_{j_{1}}+d_{i_{1}j_{2}}^{2}c_{j_{2}}=0$, $1\leq
i_{1}\leq\;^{n}C_{2}$, $1\leq j_{1}<j_{2}\leq n$, $H_{i_{2}}^{3}$ is a
polynomial in
$s_{i_{2}}^{3}:=d_{i_{2}j_{1}}^{3}z_{j_{1}}+d_{i_{2}j_{2}}^{3}z_{j_{2}}+d_{i_{2}j_{3}}^{3}z_{j_{3}}$
with
$d_{i_{2}j_{1}}^{3}c_{j_{1}}+d_{i_{2}j_{2}}^{3}c_{j_{2}}+d_{i_{2}j_{3}}^{3}z_{3}=0$,
$1\leq i_{2}\leq\;^{n}C_{3}$, $1\leq j_{1}<j_{2}<j_{3}\leq n\ldots$,
$H_{i_{n-2}}^{n-1}$ is a polynomial in
$s_{i_{n-2}}^{n-1}:=d_{i_{n-2}j_{1}}z_{j_{1}}+d_{i_{n-2}j_{2}}z_{j_{2}}+\cdots+d_{i_{n-2}j_{n-1}}z_{j_{n-1}}$
with
$d_{i_{n-2}j_{1}}c_{j_{1}}+d_{i_{n-2}j_{2}}c_{j_{2}}+\cdots+d_{i_{n-2}j_{n-1}}c_{j_{n-1}}=0$,
$1\leq i_{n-2}\leq\;^{n}C_{n-1}$, $1\leq j_{1}<j_{2}<\cdots<j_{n-1}\leq n$ and
$H_{n-1}^{n}$ is a polynomial in
$s_{n-1}^{n}:=d_{i_{n-1}1}z_{1}+d_{i_{n-1}2}z_{2}+\cdots+d_{i_{n-1}n}z_{n}$
with $d_{i_{n-1}1}c_{1}+d_{i_{n-1}2}c_{2}+\cdots+d_{i_{n-1}n}c_{n}=0$,
$L_{i_{1}}^{2}$ is a polynomial in
$t_{i_{1}}^{2}:=e_{i_{1}j_{1}}^{2}z_{j_{1}}+e_{i_{1}j_{2}}^{2}z_{j_{2}}$ with
$e_{i_{1}j_{1}}^{2}c_{j_{1}}+e_{i_{1}j_{2}}^{2}c_{j_{2}}=0$, $1\leq
i_{1}\leq\;^{n}C_{2}$, $1\leq j_{1}<j_{2}\leq n$, $H_{i_{2}}^{3}$ is a
polynomial in
$t_{i_{2}}^{3}:=e_{i_{2}j_{1}}^{3}z_{j_{1}}+e_{i_{2}j_{2}}^{3}z_{j_{2}}+e_{i_{2}j_{3}}^{3}z_{j_{3}}$
with
$e_{i_{2}j_{1}}^{3}c_{j_{1}}+e_{i_{2}j_{2}}^{3}c_{j_{2}}+e_{i_{2}j_{3}}^{3}z_{3}=0$,
$1\leq i_{2}\leq\;^{n}C_{3}$, $1\leq j_{1}<j_{2}<j_{3}\leq n\ldots$,
$L_{i_{n-2}}^{n-1}$ is a polynomial in
$t_{i_{n-2}}^{n-1}:=e_{i_{n-2}j_{1}}z_{j_{1}}+e_{i_{n-2}j_{2}}z_{j_{2}}+\cdots+e_{i_{n-2}j_{n-1}}z_{j_{n-1}}$
with
$e_{i_{n-2}j_{1}}c_{j_{1}}+e_{i_{n-2}j_{2}}c_{j_{2}}+\cdots+e_{i_{n-2}j_{n-1}}c_{j_{n-1}}=0$,
$1\leq i_{n-2}\leq\;^{n}C_{n-1}$, $1\leq j_{1}<j_{2}<\cdots<j_{n-1}\leq n$ and
$L_{n-1}^{n}$ is a polynomial in
$t_{n-1}^{n}:=e_{i_{n-1}1}z_{1}+e_{i_{n-1}2}z_{2}+\cdots+e_{i_{n-1}n}z_{n}$
with $e_{i_{n-1}1}c_{1}+e_{i_{n-1}2}c_{2}+\cdots+e_{i_{n-1}n}c_{n}=0$, where
for each $k$, the representation of $s_{i}^{k}$ and $t_{i}^{k}$ in terms of
the conditions of $j_{1},j_{2},\ldots,j_{k}$ are unique.

Now, we list our main results as follows.

######  Theorem 3.1.

Let $c=(c_{1},c_{2},\ldots,c_{n})\in\mathbb{C}^{n}$, $w\in\mathbb{C}$ such
that $w^{2}\neq 0,1$ and $g_{1}(z)$, $g_{2}(z)$ be any two polynomials in
$\mathbb{C}^{n}$. If $(f(z),g(z))$ be a pair of transcendental entire solution
with finite order of the system Fermat type difference equation (3.1), then
one of the following must occur.

  1. (i)

$g_{1}(z)=L(z)+\Phi(z)+d_{1}$, $g_{2}(z)=L(z)+\Phi(z)+d_{2}$,

$\displaystyle
f(z)=\frac{A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}}{\sqrt{2}}e^{\frac{1}{2}[L(z)+\Phi(z)+d_{1}]},\;g(z)=\frac{A_{1}\xi_{2}+A_{2}\xi_{2}^{-1}}{\sqrt{2}}e^{\frac{1}{2}[L(z)+\Phi(z)+d_{2}]},$

where $L(z)=\sum_{j=1}^{n}a_{j}z_{j}$, $\Phi(z)$ is defined as in (3.3) and
satisfy the relations

$\displaystyle
e^{L(c)}=\frac{(A_{2}\xi_{2}+A_{1}\xi_{2}^{-1})(A_{2}\xi_{1}+A_{1}\xi_{1}^{-1})}{(A_{1}\xi_{1}+A_{2}\xi_{1}^{-1})(A_{1}\xi_{2}+A_{2}\xi_{2}^{-1})},$
$\displaystyle
e^{d_{1}-d_{2}}=\frac{(A_{2}\xi_{2}+A_{1}\xi_{2}^{-1})(A_{1}\xi_{2}+A_{2}\xi_{2}^{-1})}{(A_{1}\xi_{1}+A_{2}\xi_{1}^{-1})(A_{2}\xi_{1}+A_{1}\xi_{1}^{-1})},$

$a_{j},d_{1},d_{2},\xi_{1}(\neq 0),\xi_{2}(\neq 0)\in\mathbb{C}$ for
$j=1,2,\ldots,n$.

  2. (ii)

$\displaystyle g_{1}(z)=L_{1}(z)+L_{2}(z)+\Phi(z)+\Psi(z)+d_{1}+d_{3},$
$\displaystyle g_{2}(z)=L_{1}(z)+L_{2}(z)+\Phi(z)+\Psi(z)+d_{2}+d_{4},$
$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{L_{1}(z)+\Phi(z)+d_{1}}+A_{2}e^{L_{2}(z)+\Psi(z)+d_{3}}\right],$
$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{L_{2}(z)+\Psi(z)+d_{4}}+A_{2}e^{L_{1}(z)+\Phi(z)+d_{2}}\right],$

where $L_{1}(z)=\sum_{j=1}^{n}a_{j}z_{j}$,
$L_{2}(z)=\sum_{j=1}^{n}b_{j}z_{j}$, $\Phi(z)$, $\Psi(z)$ are defined in (3.3)
and $\eqref{e2.6}$, respectively, $a_{j},b_{j},d_{i}\in\mathbb{C}$ for
$j=1,2,\ldots,n$, $i=1,\ldots,4$ such that $L_{1}(z)+\Phi(z)\neq
L_{2}(z)+\Psi(z)$ and satisfy one of the following relations:

    1. (a)

$e^{L_{1}(c)}=1$, $e^{L_{2}(c)}=1$, $e^{d_{1}-d_{2}}=1$ and
$e^{d_{3}-d_{4}}=1$,

    2. (b)

$e^{L_{1}(c)}=1$, $e^{L_{2}(c)}=-1$, $e^{d_{1}-d_{2}}=1$ and
$e^{d_{3}-d_{4}}=-1$,

    3. (c)

$e^{L_{1}(c)}=-1$, $e^{L_{2}(c)}=1$, $e^{d_{1}-d_{2}}=-1$ and
$e^{d_{3}-d_{4}}=1$,

    4. (d)

$e^{L_{1}(c)}=-1$, $e^{L_{2}(c)}=-1$, $e^{d_{1}-d_{2}}=-1$ and
$e^{d_{3}-d_{4}}=-1$.

  3. (iii)

$g_{1}(z)$ and $g_{2}(z)$ have the same form as in (ii),

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{L_{1}(z)+\Phi(z)+d_{1}}+A_{2}e^{L_{2}(z)+\Psi(z)+d_{3}}\right],$
$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{L_{1}(z)+\Phi(z)+d_{2}}+A_{2}e^{L_{2}(z)+\Psi(z)+d_{4}}\right],$

where $L_{1},L_{2},\Phi,\Psi$ are same as in (ii) with $L_{1}(z)+\Phi(z)\neq
L_{2}(z)+\Psi(z)$, and satisfy one of the following relations:

    1. (a)

$e^{L_{1}(c)}=A_{2}/A_{1}$, $e^{L_{2}(c)}=A_{1}/A_{2}$, $e^{d_{1}-d_{2}}=1$
and $e^{d_{3}-d_{4}}=1$,

    2. (b)

$e^{L_{1}(c)}=A_{2}/A_{1}$, $e^{L_{2}(c)}=-A_{1}/A_{2}$, $e^{d_{1}-d_{2}}=1$
and $e^{d_{3}-d_{4}}=-1$,

    3. (c)

$e^{L_{1}(c)}=-A_{2}/A_{1}$, $e^{L_{2}(c)}=A_{1}/A_{2}$, $e^{d_{1}-d_{2}}=-1$
and $e^{d_{3}-d_{4}}=1$,

    4. (d)

$e^{L_{1}(c)}=-A_{2}/A_{1}$, $e^{L_{2}(c)}=-A_{1}/A_{2}$, $e^{d_{1}-d_{2}}=-1$
and $e^{d_{3}-d_{4}}=-1$.

The following examples show the existence of transcendental entire solutions
with finite order of the system (3.1).

######  Example 3.1.

Let $n=2$, $w=3$, $\xi_{1}=\xi_{2}=1$, $c_{1}=c_{2}=\pi i$, $d_{1}=\pi
i,d_{2}=-\pi i$, $g_{1}(z)=z_{1}+z_{2}+(z_{1}-z_{2})^{3}+d_{1}$,
$g_{2}(z)=z_{1}+z_{2}+(z_{1}-z_{2})^{3}+d_{2}$. Then in view of the conclusion
(i) of Theorem 3.1, we can see that

$\displaystyle(f(z),g(z))=\left(\frac{1}{2\sqrt{2}}e^{\frac{1}{2}[z_{1}+z_{2}+(z_{1}-z_{2})^{3}+d_{1}]},\frac{1}{2\sqrt{2}}e^{\frac{1}{2}[z_{1}+z_{2}+(z_{1}-z_{2})^{3}+d_{2}]}\right)$

is a solution of (3.1).

######  Example 3.2.

Let $w=2$, $n=3$, $L(z)=z_{1}+z_{2}+z_{3}$, $L_{2}(z)=z_{1}+z_{2}-z_{3}$,
$c_{1}=\pi i,c_{2}=2\pi i,c_{3}=-\pi i$,
$\Phi(z)=-\pi^{2}(z_{1}-z_{2})^{2}+i\pi^{3}(z_{1}+z_{2})^{3}+\pi^{4}(z_{2}-z_{3})^{4}$,
$\Psi(z)=\pi^{4}(z_{1}+2z_{2}-z_{3})^{2}$. Choose $d_{j}\in\mathbb{C}$ such
that $d_{1}-d_{2}=2\pi i$ and $d_{3}-d_{4}=4\pi i$. Then, in view of the
conclusion (ii) of Theorem 3.1, $(f(z),g(z))$, where

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left[-\frac{3-\sqrt{3}}{6}e^{L_{1}(z)+\Phi(z)+d_{1}}+\frac{3+\sqrt{3}}{6}e^{L_{2}(z)+\Psi(z)+d_{3}}\right]\;\;\text{and}$
$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left[-\frac{3-\sqrt{3}}{6}e^{L_{2}(z)+\Psi(z)+d_{3}}+\frac{3+\sqrt{3}}{6}e^{L_{1}(z)+\Phi(z)+d_{1}}\right]$

is a solution of (3.1).

######  Theorem 3.2.

Let $k\in\mathbb{N}$, $c=(c_{1},c_{2})\in\mathbb{C}^{2}$, $w\in\mathbb{C}$
such that $w^{2}\neq 0,1$, $g_{1}(z_{1},z_{2}),g_{2}(z_{1},z_{2})$ be any two
polynomials in $\mathbb{C}^{2}$. If $(f,g)$ is a pair of finite order
transcendental entire solution of (3.2), then one of the following cases must
occur.

  1. (i)

$g_{1}(z_{1},z_{2})=\mathcal{L}(z)+\chi(z_{2})+d_{1}$,
$g_{2}(z_{1},z_{2})=\mathcal{L}(z)+\chi(z_{2})+d_{2}$,

$\displaystyle
f(z)=\frac{A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}}{\sqrt{2}}e^{\frac{1}{2}[\mathcal{L}(z)+\chi(z_{2})-\mathcal{L}(c)+d_{2}]},\;g(z)=\frac{A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}}{\sqrt{2}}e^{\frac{1}{2}[\mathcal{L}(z)+\chi(z_{2})-\mathcal{L}(c)+d_{1}]},$

where $\mathcal{L}(z)=a_{1}z_{1}+a_{2}z_{2}$, $\chi(z_{2})$ is a polynomial in
$z_{2}$, only such that $\chi(z_{2}+c_{2})=\chi(z_{2})$, $\xi_{j}\neq
0,a_{j},d_{j}\in\mathbb{C}$, $j=1,2$, $a_{1}\neq 0$, and satisfy the
following:

$\displaystyle\begin{cases}e^{\mathcal{L}(c)}=\left(\dfrac{a_{1}}{2}\right)^{2k}\dfrac{(A_{2}\xi_{2}+A_{1}\xi_{2}^{-1})(A_{2}\xi_{1}+A_{1}\xi_{1}^{-1})}{(A_{1}\xi_{1}+A_{2}\xi_{1}^{-1})(A_{1}\xi_{2}+A_{2}\xi_{2}^{-1})},\vspace{1mm}\\\
e^{\frac{1}{2}[\mathcal{L}(c)+d_{1}-d_{2}]}=\left(\dfrac{a_{1}}{2}\right)^{k}\dfrac{A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}}{A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}}.\end{cases}$

In particular, if $c_{2}\neq 0$, then $\chi(z_{2})$ becomes linear in $z_{2}$.

  2. (ii)

$\displaystyle
g_{1}(z_{1},z_{2})=L_{1}(z)+L_{2}(z)+\Phi_{1}(z_{2})+\Psi_{1}(z_{2})+d_{1}+d_{3},$
$\displaystyle
g_{2}(z_{1},z_{2})=L_{1}(z)+L_{2}(z)+\Phi_{1}(z_{2})+\Psi_{1}(z_{2})+d_{2}+d_{4},$
$\displaystyle
f(z_{1},z_{2})=\frac{A_{2}e^{L_{2}(z)+\Psi_{1}(z_{2})-L_{2}(c)+d_{4}}+A_{1}e^{L_{1}(z)+\Phi_{1}(z_{2})-L_{1}(c)+d_{2}}}{\sqrt{2}},$
$\displaystyle
g(z_{1},z_{2})=\frac{A_{2}e^{L_{1}(z)+\Phi_{1}(z_{2})-L_{1}(c)+d_{1}}+A_{1}e^{L_{2}(z)+\Psi_{1}(z_{2})-L_{2}(c)+d_{3}}}{\sqrt{2}},$

where $L_{1}(z)=a_{1}z_{1}+a_{2}z_{2}$, $L_{2}(z)=b_{1}z_{1}+b_{2}z_{2}$,
$\Phi_{1},\Psi_{1}$ are two polynomials in $z_{2}$, only with
$\Phi_{1}(z_{2}+c_{2})=\Phi_{1}(z_{2})$,
$\Psi_{1}(z_{2}+c_{2})=\Psi_{1}(z_{2})$ such that
$L_{1}(z)+\Phi_{1}(z_{2})\neq L_{2}(z)+\Psi_{1}(z_{2})$,
$a_{i},b_{1},d_{j}\in\mathbb{C}$ for $i=1,2$, $j=1,2.3.4$, and satisfy one of
the following relations:

    1. (a)

$e^{L_{1}(c)}=a_{1}^{k}$, $e^{L_{2}(c)}=b_{1}^{k}$, $e^{d_{1}-d_{2}}=1$ and
$e^{d_{3}-d_{4}}=1$,

    2. (b)

$e^{L_{1}(c)}=a_{1}^{k}$, $e^{L_{2}(c)}=-b_{1}^{k}$, $e^{d_{1}-d_{2}}=1$ and
$e^{d_{3}-d_{4}}=-1$,

    3. (c)

$e^{L_{1}(c)}=-a_{1}^{k}$, $e^{L_{2}(c)}=b_{1}^{k}$, $e^{d_{1}-d_{2}}=-1$ and
$e^{d_{3}-d_{4}}=1$,

    4. (d)

$e^{L_{1}(c)}=-a_{1}^{k}$, $e^{L_{2}(c)}=-b_{1}^{k}$, $e^{d_{1}-d_{2}}=-1$ and
$e^{d_{3}-d_{4}}=-1$.

  3. (iii)

The form of $g_{1}(z_{1},z_{2})$ and $g_{2}(z_{1},z_{2})$ are the same as in
(ii),

$\displaystyle
f(z_{1},z_{2})=\frac{A_{2}e^{L_{1}(z)+\Phi_{1}(z_{2})-L_{1}(c)+d_{2}}+A_{1}e^{L_{2}(z)+\Psi_{1}(z_{2})-L_{2}(c)+d_{4}}}{\sqrt{2}},$
$\displaystyle
g(z_{1},z_{2})=\frac{A_{2}e^{L_{1}(z)+\Phi_{1}(z_{2})-L_{1}(c)+d_{1}}+A_{1}e^{L_{2}(z)+\Psi_{1}(z_{2})-L_{2}(c)+d_{3}}}{\sqrt{2}},$

where $L_{1},L_{2},\Phi_{1},\Psi_{1}$ are the same as in (ii), and satisfy one
of the following conditions:

    1. (a)

$e^{L_{1}(c)}=A_{2}a_{1}^{k}/A_{1}$, $e^{L_{2}(c)}=A_{1}b_{1}^{k}/A_{2}$,
$e^{d_{1}-d_{2}}=1$ and $e^{d_{3}-d_{4}}=1$,

    2. (b)

$e^{L_{1}(c)}=A_{2}a_{1}^{k}/A_{1}$, $e^{L_{2}(c)}=-A_{1}b_{1}^{k}/A_{2}$,
$e^{d_{1}-d_{2}}=1$ and $e^{d_{3}-d_{4}}=-1$,

    3. (c)

$e^{L_{1}(c)}=-A_{2}a_{1}^{k}/A_{1}$, $e^{L_{2}(c)}=A_{1}b_{1}^{k}/A_{2}$,
$e^{d_{1}-d_{2}}=-1$ and $e^{d_{3}-d_{4}}=1$,

    4. (d)

$e^{L_{1}(c)}=-A_{2}a_{1}^{k}/A_{1}$, $e^{L_{2}(c)}=-A_{1}b_{1}^{k}/A_{2}$,
$e^{d_{1}-d_{2}}=-1$ and $e^{d_{3}-d_{4}}=-1$.

The following examples show the existence of transcendental entire solutions
with finite order of the system (3.2).

######  Example 3.3.

Let $k=1$, $L(z)=z_{1}+z_{2}$, $\xi_{1}=1$, $\xi_{2}=-1$ and $w$ be a complex
number such that $w\neq 0,\pm 1$. Choose $c=(c_{1},c_{2})\in\mathbb{C}$ such
that $e^{L(c)/2}=1/2$ and $d_{1},d_{2}\in\mathbb{C}$ such that
$e^{(d_{1}-d_{2})/2}=-1$. Then, in view of the conclusion (i) of Theorem 3.2,
we easily see that $(f,g)$, where

$\displaystyle
f(z_{1},z_{2})=-\frac{2}{\sqrt{2(1+w)}}e^{\frac{1}{2}[z_{1}+z_{2}+d_{2}]}\;\;\text{and}\;\;g(z_{1},z_{2})=\frac{2}{\sqrt{2(1+w)}}e^{\frac{1}{2}[z_{1}+z_{2}+d_{1}]}$

is a solution of (3.2) with $g_{1}(z_{1},z_{2})=z_{1}+z_{2}+d_{1}$ and
$g_{2}(z_{1},z_{2})=z_{1}+z_{2}+d_{2}$.

######  Example 3.4.

Let $k=1$, $w=2$, $L_{1}(z)=z_{1}+z_{2}$, $L_{2}(z)=z_{1}-2z_{2}$, $c_{1}=8\pi
i/3,c_{2}=-2\pi i/3$. Choose $d_{j}\in\mathbb{C}$ for $j=1,\ldots,4$ such that
$d_{1}-d_{2}=2m\pi i$ and $d_{3}-d_{4}=2p\pi i$, $m,p\in\mathbb{Z}$. Then, in
view of the conclusion (ii) of Theorem 3.2, we easily see that $(f(z),g(z))$,
where

$\displaystyle
f(z_{1},z_{2})=\frac{1}{6\sqrt{2}}\left[-(3-\sqrt{3})e^{z_{1}+z_{2}+d_{2}}+(3+\sqrt{3})e^{z_{1}-2z_{2}+d_{4}}\right]\;\;\text{and}$
$\displaystyle
g(z_{1},z_{2})=\frac{1}{6\sqrt{2}}\left[(3+\sqrt{3})e^{z_{1}+z_{2}+d_{1}}-(3-\sqrt{3})e^{z_{1}-2z_{2}+d_{3}}\right]$

is a solution of (3.2) with $g_{1}(z_{1},z_{2})=2z_{1}-z_{2}+d_{1}+d_{3}$ and
$g_{2}(z_{1},z_{2})=2z_{1}-z_{2}+d_{2}+d_{4}$.

######  Theorem 3.3.

Let $k$ be a positive integer, $c=(c_{1},c_{2})\in\mathbb{C}^{2}$,
$w\in\mathbb{C}$ such that $w^{2}\neq 0,1$ and $f(z),g(z)$ are transcendental
entire functions of finite order not of $c$-periodic. If $(f(z),g(z))$ be a
solution of (2.5), then one of the following conclusions must hold.

  1. (i)

$\displaystyle
f(z)=\frac{A_{1}e^{L(z)+\Phi_{1}(z_{2})+\gamma}+A_{2}e^{-(L(z)+\Phi_{1}(z_{2})+\gamma)}}{\sqrt{2}\alpha^{k}}\;\;\text{and}$
$\displaystyle
g(z)=\frac{A_{1}e^{L(z)+\Phi_{1}(z_{2})+\gamma+\eta}+A_{2}e^{-(L(z)+\Phi_{1}(z_{2})+\gamma+\eta)}}{\sqrt{2}\alpha^{k}}.$

where $L(z)=\alpha z_{1}+\beta z_{2}+\gamma$, $\Phi_{1}(z_{2})$ is a
polynomial in $z_{2}$ with $\Phi_{1}(z_{2}+c_{2})=\Phi_{1}(z_{2})$,
$\alpha,\beta,\gamma,\eta\in\mathbb{C}$, and satisfy the following relations:

$\displaystyle
e^{L(c)}=\frac{A_{2}e^{\eta}}{(-1)^{k}A_{1}\alpha^{k}+A_{2}e^{\eta}},\;\;e^{2\eta}=1\;\;\text{and}\;\;\alpha^{k}=-\frac{(-1)^{k}A_{1}^{2}+A_{2}^{2}}{(-1)^{k}A_{1}A_{2}e^{\eta}}.$

In particular, if $c_{2}\neq 0$, then the polynomial $\Phi_{1}(z_{2})$ becomes
linear in $z_{2}$.

  2. (ii)

$\displaystyle
f(z)=\frac{A_{1}e^{L(z)+\Phi_{1}(z_{2})+\gamma}+A_{2}e^{-(L(z)+\Phi_{1}(z_{2})+\gamma)}}{\sqrt{2}\alpha^{k}},$
$\displaystyle
g(z)=\frac{A_{1}e^{-(L(z)+\Phi_{1}(z_{2})+\gamma)+\eta}+A_{2}e^{L(z)+\Phi_{1}(z_{2})+\gamma-\eta}}{\sqrt{2}\alpha^{k}},$

where $k$ is an even positive integer, $L(z),\Phi_{1}(z_{2})$ are the same as
in (i), and satisfy the following relations:

$\displaystyle
e^{L(c)}=1+e^{-\eta}\alpha^{k},\;\;e^{2\eta}=1\;\;\text{and}\;\;\alpha^{k}=-2e^{\eta}.$

The following example shows that the existence of transcendental entire
solutions of (2.5) is precise.

######  Example 3.5.

Let $k=2,w=2$ and choose $\alpha,\eta,c_{1},c_{2}\in\mathbb{C}$ such that
$\alpha^{2}=-2,\eta=2p\pi i$ and $c_{1}+c_{2}=(2m+1)\pi i$,
$p,m\in\mathbb{Z}$. Then in view of the conclusion (ii) of Theorem 3.3, we see
that $(f,g)$, where

$\displaystyle f(z_{1},z_{2})=\frac{1}{12\sqrt{2}}\left[(3-\sqrt{3})e^{\alpha
z_{1}+z_{2}+1}-(3+\sqrt{3})e^{-(\alpha z_{1}+z_{2}+1)}\right]\;\;\text{and}$
$\displaystyle
g(z_{1},z_{2})=\frac{1}{12\sqrt{2}}\left[(3-\sqrt{3})e^{-(\alpha
z_{1}+z_{2}+1)}-(3+\sqrt{3})e^{\alpha z_{1}+z_{2}+1}\right]\;\;\text{and}$

is a solution of (2.5).

##  4\. Proof of the main results

Before we starting the proof of the main results, we present here some
necessary lemmas which will play key role to prove the main results of this
paper.

######  Lemma 4.1.

[49] Let $f_{j}\not\equiv 0$ $(j=1,2\ldots,m;\;m\geq 3)$ be meromorphic
functions such that $f_{1},\ldots,f_{m-1}$ are not constants,
$f_{1}+f_{2}+\cdots+f_{m}=1$ and such that

$\displaystyle\sum_{j=1}^{m}\left\\{N_{n-1}\left(r,\frac{1}{f_{j}}\right)+(m-1)\overline{N}(r,f_{j})\right\\}<\lambda
T(r,f_{j})+O(\log^{+}T(r,f_{j}))$

holds for $j=1,\ldots,m-1$ and all $r$ outside possibly a set with finite
logarithmic measure, where $\lambda<1$ is a positive number. Then $f_{m}=1$.

######  Lemma 4.2.

[18] Let $f_{j}\not\equiv 0$ $(j=1,2,3)$ be meromorphic functions on such that
$f_{1}$ are not constant, $f_{1}+f_{2}+f_{3}=1$, and such that

$\displaystyle\sum_{j=1}^{3}\left\\{N_{2}\left(r,\frac{1}{f_{j}}\right)+2\overline{N}(r,f_{j})\right\\}<\lambda
T(r,f_{j})+O(\log^{+}T(r,f_{j}))$

holds for all $r$ outside possibly a set with finite logarithmic measure,
where $\lambda<1$ is a positive number. Then, either $f_{2}=1$ or $f_{3}=1$.

######  Lemma 4.3.

[49] Let $a_{0}(z),a_{1}(z),\ldots,a_{n}(z)$ $(n\geq 1)$ be meromorphic
functions on and $g_{0}(z),g_{1}(z),\ldots,g_{n}(z)$ are entire functions such
that $g_{j}(z)-g_{k}(z)$ are not constants for $0\leq j<k\leq n$. If
$\sum_{j=0}^{n}a_{j}(z)e^{g_{j}(z)}\equiv 0$, and $T(r,a_{j})=o(T(r))$, where
$T(r)=\text{min}_{0\leq j<k\leq n}T(r,e^{g_{j}-g_{k}})$ for $j=0,1,\ldots,n$,
then $a_{j}(z)\equiv 0$ for each $j=0,1,\ldots,n$.

###### Proof of Theorem 3.1.

Let $(f(z),g(z))$ be a pair of finite order transcendental entire solution of
(3.1). Let us first make a transformation

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}(u+v)\;\;\text{and}\;\;g(z+c)=\frac{1}{\sqrt{2}}(u-v).$
(4.1)

Then first equation of (3.1) yields

$\displaystyle(1+w)u^{2}+(1-w)v^{2}=e^{g_{1}(z)},$

which can be rewritten as

$\displaystyle\left(\frac{\sqrt{1+w}u}{e^{g_{1}(z)/2}}+i\frac{\sqrt{1-w}v}{e^{g_{1}(z)/2}}\right)\left(\frac{\sqrt{1+w}u}{e^{g_{1}(z)/2}}-i\frac{\sqrt{1-w}v}{e^{g_{1}(z)/2}}\right)=1.$

In view of the above equation and assumption, it follows that there exists a
polynomial $p_{1}(z)$ in $\mathbb{C}^{n}$ such that

$\displaystyle\frac{u\sqrt{1+w}}{e^{g_{1}(z)/2}}+i\frac{v\sqrt{1-w}}{e^{g_{1}(z)/2}}=e^{p_{1}(z)}\;\;\text{and}\;\;\frac{u\sqrt{1+w}}{e^{g_{1}(z)/2}}-i\frac{v\sqrt{1-w}}{e^{g_{1}(z)/2}}=e^{-p_{1}(z)}.$
(4.2)

Let

$\displaystyle\gamma_{1}(z)=p_{1}(z)+\frac{1}{2}g_{1}(z),\;\;\gamma_{2}(z)=-p_{1}(z)+\frac{1}{2}g_{1}(z).$
(4.3)

Then from (4.2) and (4.3), we obtain

$\displaystyle
u=\frac{1}{2\sqrt{1+w}}\left[e^{\gamma_{1}}+e^{\gamma_{2}}\right],\;\;v=\frac{1}{2i\sqrt{1-w}}\left[e^{\gamma_{1}}-e^{\gamma_{2}}\right].$
(4.4)

Therefore, in view of (4.1) and (4.4), we obtain

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{\gamma_{1}}+A_{2}e^{\gamma_{2}}\right],\;\;g(z+c)=\frac{1}{\sqrt{2}}\left[A_{2}e^{\gamma_{1}}+A_{1}e^{\gamma_{2}}\right].$
(4.5)

Similarly, in view of the second equation of (3.1), there exists a polynomial
$p_{2}(z)$ such that

$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{\gamma_{3}}+A_{2}e^{\gamma_{4}}\right],\;\;f(z+c)=\frac{1}{\sqrt{2}}\left[A_{2}e^{\gamma_{3}}+A_{1}e^{\gamma_{4}}\right],$
(4.6)

where

$\displaystyle\gamma_{3}(z)=p_{2}(z)+\frac{1}{2}g_{2}(z),\;\;\gamma_{4}(z)=-p_{2}(z)+\frac{1}{2}g_{2}(z).$
(4.7)

From (4.5) and (4.6), we easily get

$\displaystyle\begin{cases}A_{1}e^{\gamma_{1}(z+c)-\gamma_{4}(z)}+A_{2}e^{\gamma_{2}(z+c)-\gamma_{4}(z)}-A_{2}e^{\gamma_{3}(z)-\gamma_{4}(z)}=A_{1},\\\
A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}+A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}-A_{2}e^{\gamma_{1}(z)-\gamma_{1}(z)}=A_{1}.\end{cases}$
(4.8)

Now we discuss the following four possible cases.

Case 1: Let $\gamma_{1}(z)-\gamma_{2}(z)=\zeta_{1}$ and
$\gamma_{3}(z)-\gamma_{4}(z)=\zeta_{2}$, where $\zeta_{1},\zeta_{2}$ are two
complex constants.

Then, it clearly follows from (4.3) and (4.7) that $p_{1}(z)$ and $p_{2}(z)$
both are constants in $\mathbb{C}$. Let $e^{p_{1}}=\xi_{1}$ and
$e^{p_{2}}=\xi_{2}$.

Then, from (4.5) and (4.6), it follows that

$\displaystyle\begin{cases}f(z)=\frac{A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}}{\sqrt{2}}e^{\frac{1}{2}g_{1}(z)},\\\
g(z+c)=\frac{A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}}{\sqrt{2}}e^{\frac{1}{2}g_{1}(z)},\\\
g(z)=\frac{A_{1}\xi_{2}+A_{2}\xi_{2}^{-1}}{\sqrt{2}}e^{\frac{1}{2}g_{2}(z)},\\\
f(z+c)=\frac{A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}}{\sqrt{2}}e^{\frac{1}{2}g_{2}(z)}.\end{cases}$
(4.9)

Since $f,g$ are transcendental entire, it follows from (4.9) that
$A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}$, $A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}$,
$A_{1}\xi_{2}+A_{2}\xi_{2}^{-1}$ and $A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}$ are all
non zero constants. Since $p_{1}$ and $p_{2}$ are constants and $f,g$ are
transcendental entire, it follows from (4.9) that $g_{1}$ and $g_{2}$ are non-
constants.

After simple calculations, we have from (4.9) that

$\displaystyle\begin{cases}(A_{1}\xi_{1}+A_{2}\xi_{1}^{-1})e^{\frac{1}{2}[g_{1}(z+c)-g_{2}(z)]}=A_{2}\xi_{2}+A_{1}\xi_{2}^{-1},\\\
(A_{1}\xi_{2}+A_{2}\xi_{2}^{-1})e^{\frac{1}{2}[g_{2}(z+c)-g_{1}(z)]}=A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}.\end{cases}$
(4.10)

Since $g_{1},g_{2}$ are non-constant polynomials in $\mathbb{C}^{n}$, from
(4.10), we see that $g_{1}(z+c)-g_{2}(z)=\eta_{1}$ and
$g_{2}(z+c)-g_{1}(z)=\eta_{2}$, where $\eta_{1},\eta_{2}\in\mathbb{C}$.
Otherwise, left hand sides of both the equations are transcendental, whereas
right hand sides will be polynomial, a contradiction. Therefore,
$g_{1}(z+2c)-g_{2}(z)=g_{2}(z+2c)-g_{1}(z)=\zeta_{1}+\zeta_{2}$. Thus,

$\displaystyle
g_{1}(z)=L(z)+\Phi(z)+d_{1}\;\;\text{and}\;\;g_{2}(z)=L(z)+\Phi(z)+d_{2},$

where $L(z)=\sum_{j=1}^{n}a_{j}z_{j}$ and $\Phi(z)$ is a polynomial in
$\mathbb{C}^{n}$ defined as in (3.3), $a_{j},d_{1},d_{2}\in\mathbb{C}$,
$j=1,2,\ldots,n$.

Thus, from (4.10), we get

$\displaystyle
e^{\frac{1}{2}[L(c)+d_{1}-d_{2}]}=\frac{A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}}{A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}}\;\;\text{and}\;\;e^{\frac{1}{2}[L(c)+d_{2}-d_{1}]}=\frac{A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}}{A_{1}\xi_{2}+A_{2}\xi_{2}^{-1}},$
(4.11)

which imply

$\displaystyle
e^{L(c)}=\frac{(A_{2}\xi_{2}+A_{1}\xi_{2}^{-1})(A_{2}\xi_{1}+A_{1}\xi_{1}^{-1})}{(A_{1}\xi_{1}+A_{2}\xi_{1}^{-1})(A_{1}\xi_{2}+A_{2}\xi_{2}^{-1})}\;\;\text{and}\;\;e^{d_{1}-d_{2}}=\frac{(A_{2}\xi_{2}+A_{1}\xi_{2}^{-1})(A_{1}\xi_{2}+A_{2}\xi_{2}^{-1})}{(A_{1}\xi_{1}+A_{2}\xi_{1}^{-1})(A_{2}\xi_{1}+A_{1}\xi_{1}^{-1})}.$

Therefore, from (4.9), we obtain

$\displaystyle
f(z)=\frac{A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}}{\sqrt{2}}e^{\frac{1}{2}[L(z)+\Phi(z)+d_{1}]}\;\;\text{and}\;\;g(z)=\frac{A_{1}\xi_{2}+A_{2}\xi_{2}^{-1}}{\sqrt{2}}e^{\frac{1}{2}[L(z)+\Phi(z)+d_{2}]}.$

Case 2. Let $\gamma_{1}(z)-\gamma_{2}(z)$ and $\gamma_{3}(z)-\gamma_{4}(z)$
both are non-constants. Then, by Lemma 4.2, it follows from (4.8) that

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)-\gamma_{4}(z)}=A_{1}\;\;\text{or}\;\;A_{2}e^{\gamma_{2}(z+c)-\gamma_{4}(z)}=A_{1}$

and

$\displaystyle
A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}=A_{1}\;\;\text{or}\;\;A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}=A_{1}.$

Now, we consider the following four possible subcases.

Subcase 2.1. Let

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)-\gamma_{4}(z)}=A_{1}\;\;\text{and}\;\;A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}=A_{1}.$
(4.12)

Then, from (4.8) and (4.12), we obtain

$\displaystyle
A_{2}e^{\gamma_{2}(z+c)-\gamma_{3}(z)}=A_{2}\;\;\text{and}\;\;A_{2}e^{\gamma_{4}(z+c)-\gamma_{1}(z)}=A_{2}.$
(4.13)

Since $\gamma_{j}$’s, $j=1,2,3,4$ are all polynomials, it follows from (4.9)
and (4.10) that $\gamma_{1}(z+c)-\gamma_{4}(z)=\zeta_{1}$,
$\gamma_{3}(z+c)-\gamma_{2}(z)=\zeta_{2}$,
$\gamma_{2}(z+c)-\gamma_{3}(z)=\zeta_{3}$ and
$\gamma_{4}(z+c)-\gamma_{1}(z)=\zeta_{4}$, $\zeta_{j}\in\mathbb{C}$ for
$j=1,2,3,4$. This implies that

$\displaystyle\gamma_{1}(z+2c)-\gamma_{1}(z)=\gamma_{4}(z+2c)-\gamma_{4}(z)=\zeta_{1}+\zeta_{2}\;\;\text{and}$
$\displaystyle\gamma_{2}(z+2c)-\gamma_{2}(z)=\gamma_{3}(z+2c)-\gamma_{3}(z)=\zeta_{3}+\zeta_{4}.$

Therefore, we must get

$\displaystyle\gamma_{1}(z)=L_{1}(z)+\Phi(z)+d_{1},\;\;\gamma_{4}(z)=L_{1}(z)+\Phi(z)+d_{2},$
$\displaystyle\gamma_{2}(z)=L_{2}(z)+\Psi(z)+d_{3},\;\;\text{and}\;\;\gamma_{3}(z)=L_{2}(z)+\Psi(z)+d_{4},$

where $L_{1}(z)=\sum_{j=1}^{n}a_{j}z_{j}$,
$L_{2}(z)=\sum_{j=1}^{n}b_{j}z_{j}$, $\Phi(z)$ and $\Psi(z)$ are defined in
(3.3) and (3.4), respectively, $a_{j},,b_{j},d_{i}\in\mathbb{C}$ for $i=1,2$
and $j=1,\ldots,n$. As $\gamma_{4}(z)-\gamma_{3}(z)$ and
$\gamma_{2}(z)-\gamma_{1}(z)$ both are non-constants, we must have

$L_{1}(z)+\Phi(z)\neq L_{2}(z)+\Psi(z).$

Therefore, in view of (4.3) and (4.7), we have

$\displaystyle
g_{1}(z)=L_{1}(z)+L_{2}(z)+\Phi(z)+\Psi(z)+d_{1}+d_{3}\;\;\text{and}$
$\displaystyle g_{2}(z)=L_{1}(z)+L_{2}(z)+\Phi(z)+\Psi(z)+d_{2}+d_{4}.$

Therefore, from (4.12) and (4.13), we obtain

$\displaystyle\begin{cases}e^{L_{1}(c)+d_{1}-d_{2}}=1,\;\;e^{L_{1}(c)+d_{2}-d_{1}}=1,\\\
e^{L_{2}(c)+d_{3}-d_{4}}=1,\;\;e^{L_{2}(c)+d_{4}-d_{3}}=1.\end{cases}$ (4.14)

After simple computation, we obtain from (4.14) that

$\displaystyle
e^{2L_{1}(c)}=1,\;\;e^{2L_{2}(c)}=1,\;\;e^{2(d_{1}-d_{2})}=1\;\;\text{and}\;\;e^{2(d_{3}-d_{4})}=1,$

which yield

$\displaystyle e^{L_{1}(c)}=\pm 1,\;\;e^{d_{1}-d_{2}}=\pm
1,\;\;e^{L_{2}(c)}=\pm 1\;\;\text{and}\;\;e^{d_{3}-d_{4}}=\pm 1.$

If $e^{L_{1}(c)}=1$, then from (4.14), we obtain $e^{d_{1}-d_{2}}=1$. If
$e^{L_{1}(c)}=-1$, then from (4.14), we obtain $e^{d_{1}-d_{2}}=-1$.
Similarly, If $e^{L_{2}(c)}=1$, then from (4.14), it follows that
$e^{d_{3}-d_{4}}=1$. If $e^{L_{2}(c)}=-1$, then it follows from (4.14) that
$e^{d_{3}-d_{4}}=-1$.

Hence, from (4.5) and (4.6), we get

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{L_{1}(z)+\Phi(z)+d_{1}}+A_{2}e^{L_{2}(z)+\Psi(z)+d_{3}}\right]\;\;\text{and}$
$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{L_{2}(z)+\Psi(z)+d_{4}}+A_{2}e^{L_{1}(z)+\Phi(z)+d_{2}}\right].$

Subcase 2.2. Let

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)-\gamma_{4}(z)}=A_{1}\;\;\text{and}\;\;A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}=A_{1}.$
(4.15)

Therefore, from (4.3) and (4.15), we obtain

$\displaystyle
A_{2}e^{\gamma_{2}(z+c)-\gamma_{3}(z)}=A_{2}\;\;\text{and}\;\;A_{1}e^{\gamma_{3}(z+c)-\gamma_{1}(z)}=A_{2}.$
(4.16)

Since $\gamma_{j}$’s are polynomials, $j=1,2,3,4$, it follows from (4.15) and
(4.16) that $\gamma_{1}(z+c)-\gamma_{4}(z)=\zeta_{1}$,
$\gamma_{4}(z+c)-\gamma_{2}(z)=\zeta_{2}$,
$\gamma_{2}(z+c)-\gamma_{3}(z)=\zeta_{3}$ and
$\gamma_{3}(z+c)-\gamma_{1}(z)=\zeta_{4}$, $\zeta_{j}\in\mathbb{C}$ for
$j=1,2,3,4$. This implies that
$\gamma_{1}(z+4c)-\gamma_{1}(z)=\gamma_{2}(z+4c)-\gamma_{2}(z)=\sum_{j=1}^{4}\zeta_{j}$.
Hence, we can have $\gamma_{1}(z)=L(z)+\Phi(z)+k_{1}$ and
$\gamma_{2}(z)=L(z)+\Phi(z)+k_{2}$, where $L(z)$ and $\Phi(z)$ are the same as
defined in Case 1. But, then we must have
$\gamma_{2}(z)-\gamma_{1}(z)=k_{2}-k_{1}\in\mathbb{C}$, a contradiction.

Subcase 2.3. Let

$\displaystyle
A_{2}e^{\gamma_{2}(z+c)-\gamma_{4}(z)}=A_{1}\;\;\text{and}\;\;A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}=A_{1}.$

By similar argument as used in Subcase 2.2, we can easily get a contradiction.

Subcase 2.4. Let

$\displaystyle
A_{2}e^{\gamma_{2}(z+c)-\gamma_{4}(z)}=A_{1}\;\;\text{and}\;\;A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}=A_{1}.$
(4.17)

In view of (4.3) and (4.17), it follows that

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)-\gamma_{3}(z)}=A_{2}\;\;\text{and}\;\;A_{1}e^{\gamma_{3}(z+c)-\gamma_{1}(z)}=A_{2}.$
(4.18)

Since $\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4}$ are all polynomials, from
(4.17) and (4.18), we get $\gamma_{2}(z+c)-\gamma_{4}(z)=l_{1}$,
$\gamma_{4}(z+c)-\gamma_{2}(z)=l_{2}$, $\gamma_{1}(z+c)-\gamma_{3}(z)=l_{3}$
and $\gamma_{3}(z+c)-\gamma_{1}(z)=l_{4}$, $l_{j}\in\mathbb{C}$ for
$j=1,2,3,4$. This implies that
$\gamma_{1}(z+2c)-\gamma_{1}(z)=\gamma_{3}(z+2c)-\gamma_{3}(z)=l_{3}+l_{4}$
and
$\gamma_{2}(z+2c)-\gamma_{2}(z)=\gamma_{4}(z+2c)-\gamma_{4}(z)=l_{1}+l_{2}$.

Therefore, we must have

$\displaystyle\gamma_{1}(z)=L_{1}(z)+\Phi(z)+d_{1},\;\;\gamma_{3}(z)=L_{1}(z)+\Phi(z)+d_{2},$
$\displaystyle\gamma_{2}(z)=L_{2}(z)+\Psi(z)+d_{3},\;\;\text{and}\;\;\gamma_{4}(z)=L_{2}(z)+\Psi(z)+d_{4},$

where $L_{1}(z),L_{2}(z),\Phi(z),\Psi(z)$ are defined in Subcase 2.1. As
$\gamma_{2}-\gamma_{1}$ and $\gamma_{4}-\gamma_{3}$ are non-constants, we have
$L_{1}(z)+\Phi(z)\neq L_{2}(z)+\Psi(z)$.

Therefore, in view of (4.3) and (4.7), we obtain

$\displaystyle
g_{1}(z)=L_{1}(z)+L_{2}(z)+\Phi(z)+\Psi(z)+d_{1}+d_{3}\;\;\text{and}$
$\displaystyle g_{2}(z)=L_{1}(z)+L_{2}(z)+\Phi(z)+\Psi(z)+d_{2}+d_{4}.$

Hence, from (4.17) and (4.18), we have

$\displaystyle\begin{cases}A_{2}e^{L_{2}(c)+d_{3}-d_{4}}=A_{1},\;\;A_{2}e^{L_{2}(c)+d_{4}-d_{3}}=A_{1}\\\
A_{1}e^{L_{1}(c)+d_{1}-d_{2}}=A_{2},\;\;A_{1}e^{L_{1}(c)+d_{2}-d_{1}}=A_{2}.\end{cases}$
(4.19)

From (4.19), it follows that

$\displaystyle
e^{2L_{1}(c)}=A_{2}^{2}/A_{1}^{2},\;\;e^{2L_{2}(c)}=A_{1}^{2}/A_{2}^{2},\;\;e^{2(d_{1}-d_{4})}=1\;\;\text{and}\;\;e^{2(d_{3}-d_{4})}=1,$

which yield

$\displaystyle e^{L_{1}(c)}=\pm A_{2}/A_{1},\;\;e^{L_{2}(c)}=\pm
A_{1}/A_{2},\;\;e^{d_{1}-d_{4}}=\pm 1\;\;\text{and}\;\;e^{2(d_{3}-d_{4})}=\pm
1.$

If $e^{L_{1}(c)}=A_{2}/A_{1}$, then from (4.19), we get $e^{d_{1}-d_{2}}=1$.
If $e^{L_{1}(c)}=-A_{2}/A_{1}$, then it follows from (4.19) that
$e^{d_{1}-d_{2}}=-1$. Similarly, If $e^{L_{2}(c)}=A_{1}/A_{2}$, then from
(4.19), we get $e^{d_{3}-d_{4}}=1$. If $e^{L_{2}(c)}=-A_{1}/A_{2}$, then from
(4.19), we get $e^{d_{3}-d_{4}}=-1$.

Hence, from (4.5) and (4.6), we get

$\displaystyle
f(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{L_{1}(z)+\Phi(z)+d_{1}}+A_{2}e^{L_{2}(z)+\Psi(z)+d_{3}}\right]\;\;\text{and}$
$\displaystyle
g(z)=\frac{1}{\sqrt{2}}\left[A_{1}e^{L_{1}(z)+\Phi(z)+d_{2}}+A_{2}e^{L_{2}(z)+\Psi(z)+d_{4}}\right].$

Case 3. Let $\gamma_{3}(z)-\gamma_{4}(z)=\eta\in\mathbb{C}$ and
$\gamma_{1}(z)-\gamma_{2}(z)$ be non-constant. Then, in view of (4.7), we see
that $p_{2}(z)$ is a constant.

Since $\gamma_{1}-\gamma_{2}$ is non-constant, in view of Lemma 4.2, it
follows from the second equation of (4.8) that either
$A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}=A_{1}$ or
$A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}=A_{1}$.

First suppose that $A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}=A_{1}$. Then, from
the second equation of (4.8), we have
$A_{2}e^{\gamma_{4}(z+c)-\gamma_{1}(z)}=A_{2}$. As all $\gamma_{j}$,
$j=1,2,3,4$ are polynomials, it follows that
$\gamma_{3}(z+c)-\gamma_{2}(z)=m_{1}$ and
$\gamma_{4}(z+c)-\gamma_{1}(z)=m_{2}$, where $m_{1},m_{2}\in\mathbb{C}$.
Therefore, $\gamma_{1}(z)-\gamma_{2}(z)=m_{1}-m_{2}-\eta$, a constant in
$\mathbb{C}$, which contradicts to our assumption.

Similarly, we can get a contradiction for the case
$A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}=A_{1}$.

Case 4. Let $\gamma_{2}-\gamma_{1}=\eta_{1}\in\mathbb{C}$ and
$\gamma_{4}-\gamma_{3}$ be non-constant. Then, by similar argument as used in
Subcase 3, we easily get a contradiction.∎

###### Proof of Theorem 3.2.

Let $(f,g)$ is a pair of finite transcendental entire solutions of the system
(3.2).

Then, in a similar manner as in Theorem 3.1, we obtain that

$\displaystyle\begin{cases}\dfrac{\partial^{k}f}{\partial
z_{1}^{k}}=\displaystyle\frac{1}{\sqrt{2}}\left[A_{1}e^{\gamma_{1}(z)}+A_{2}e^{\gamma_{2}(z)}\right]\\\
g(z+c)=\displaystyle\frac{1}{\sqrt{2}}\left[A_{2}e^{\gamma_{1}(z)}+A_{1}e^{\gamma_{2}(z)}\right]\\\
\dfrac{\partial^{k}g}{\partial
z_{1}^{k}}=\displaystyle\frac{1}{\sqrt{2}}\left[A_{1}e^{\gamma_{3}(z)}+A_{2}e^{\gamma_{4}(z)}\right]\\\
f(z+c)=\displaystyle\frac{1}{\sqrt{2}}\left[A_{2}e^{\gamma_{3}(z)}+A_{1}e^{\gamma_{4}(z)}\right],\end{cases}$
(4.20)

where $\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4}$ are defined in (4.3) and
(4.7).

Differentiating partially $k$ times with respect to $z_{1}$, second and fourth
equations of (4.20), we obtain

$\displaystyle\begin{cases}\dfrac{\partial^{k}f(z+c)}{\partial
z_{1}^{k}}=\frac{1}{\sqrt{2}}\left[A_{2}h_{1}(z)e^{\gamma_{3}(z)}+A_{1}h_{2}(z)e^{\gamma_{4}(z)}\right],\vspace{1mm}\\\
\dfrac{\partial^{k}g(z+c)}{\partial
z_{1}^{k}}=\frac{1}{\sqrt{2}}\left[A_{2}h_{3}(z)e^{\gamma_{1}(z)}+A_{1}h_{4}(z)e^{\gamma_{2}(z)}\right],\end{cases}$
(4.21)

where $h_{1},h_{2},h_{3},h_{4}$ are given by

$\displaystyle\begin{cases}h_{1}(z)=\left(\dfrac{\partial\gamma_{3}}{\partial
z_{1}}\right)^{k}+M_{k}\left(\dfrac{\partial^{k}\gamma_{3}}{\partial
z_{1}^{k}},\ldots,\dfrac{\partial\gamma_{3}}{\partial z_{1}}\right),\\\
h_{2}(z)=\left(\dfrac{\partial\gamma_{4}}{\partial
z_{1}}\right)^{k}+N_{k}\left(\dfrac{\partial^{k}\gamma_{4}}{\partial
z_{1}^{k}},\ldots,\dfrac{\partial\gamma_{4}}{\partial z_{1}}\right)\\\
h_{3}(z)=\left(\dfrac{\partial\gamma_{1}}{\partial
z_{1}}\right)^{k}+O_{k}\left(\dfrac{\partial^{k}\gamma_{1}}{\partial
z_{1}^{k}},\ldots,\dfrac{\partial\gamma_{1}}{\partial z_{1}}\right),\\\
h_{4}(z)=\left(\dfrac{\partial\gamma_{2}}{\partial
z_{1}}\right)^{k}+R_{k}\left(\dfrac{\partial^{k}\gamma_{2}}{\partial
z_{1}^{k}},\ldots,\dfrac{\partial\gamma_{2}}{\partial
z_{1}}\right),\end{cases}$

where $M_{k}$ is the partial differential polynomial in
$\frac{\partial\gamma_{3}}{\partial z_{1}}$ of order less than $k$ in which
$\frac{\partial\gamma_{3}}{\partial z_{1}}$ appears in product with at least
one more partial derivative of $\gamma_{3}(z)$ of order greater than or equal
to $2$. Similar definitions for $N_{k},O_{k}$ and $R_{k}$.

Now, from the first and third equations of (4.20) and (4.21), we obtain

$\displaystyle\begin{cases}A_{1}e^{\gamma_{1}(z+c)-\gamma_{4}(z)}+A_{2}e^{\gamma_{2}(z+c)-\gamma_{4}(z)}-A_{2}h_{1}(z)e^{\gamma_{3}(z)-\gamma_{4}(z)}=A_{1}h_{2}(z),\\\
A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}+A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}-A_{2}h_{3}(z)e^{\gamma_{1}(z)-\gamma_{2}(z)}=A_{1}h_{4}(z).\end{cases}$
(4.22)

Now, we consider the following four possible cases.

Case 1. Let $\gamma_{2}(z)-\gamma_{1}(z)=\eta_{1}$ and
$\gamma_{4}(z)-\gamma_{3}(z)=\eta_{2}$, where
$\eta_{1},\eta_{2}\in\mathbb{C}$. Then, in view of (4.3) and (4.7), it follows
that $h_{1}(z)$ and $h_{2}(z)$ are both constants. Set $e^{h_{1}}=\xi_{1}$ and
$e^{h_{2}}=\xi_{2}$, $\xi_{1},\xi_{2}$ are non-zero constants in $\mathbb{C}$.
Therefore, in view of (4.20), we obtain

$\displaystyle\begin{cases}\dfrac{\partial^{k}f}{\partial
z_{1}^{k}}=\frac{A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}}{\sqrt{2}}e^{\frac{1}{2}g_{1}(z)},\vspace{.5mm}\\\
g(z+c)=\frac{A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}}{\sqrt{2}}e^{\frac{1}{2}g_{1}(z)},\vspace{.5mm}\\\
\dfrac{\partial^{k}g}{\partial
z_{1}^{k}}=\frac{A_{1}\xi_{2}+A_{2}\xi_{2}^{-1}}{\sqrt{2}}e^{\frac{1}{2}g_{2}(z)},\vspace{.5mm}\\\
f(z+c)=\frac{A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}}{\sqrt{2}}e^{\frac{1}{2}g_{2}(z)}.\end{cases}$
(4.23)

As $f,g$ are transcendental entire, in view of (4.23), we see that
$g_{1}(z),g_{2}(z)$ both are non-constant polynomials and
$A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}$, $A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}$,
$A_{1}\xi_{2}+A_{2}\xi_{2}^{-1}$, $A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}$ are all
non-zero constants.

Also, after simple computations, we obtain from (4.23) that

$\displaystyle\begin{cases}e^{\frac{1}{2}[g_{1}(z+c)-g_{2}(z)]}=\dfrac{A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}}{A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}}r_{1}(z),\\\
e^{\frac{1}{2}[g_{2}(z+c)-g_{1}(z)]}=\dfrac{A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}}{A_{1}\xi_{2}+A_{2}\xi_{2}^{-1}}r_{2}(z)\end{cases}$
(4.24)

with

$\displaystyle r_{1}(z)=\left(\frac{1}{2}\dfrac{\partial g_{2}(z)}{\partial
z_{1}}\right)^{k}+G_{1}\left(\dfrac{\partial^{k}g_{2}(z)}{\partial
z_{1}^{k}},\ldots,\dfrac{\partial g_{2}(z)}{\partial
z_{1}}\right)\;\;\text{and}$ $\displaystyle
r_{2}(z)=\left(\frac{1}{2}\dfrac{\partial g_{1}(z)}{\partial
z_{1}}\right)^{k}+G_{2}\left(\dfrac{\partial^{k}g_{1}(z)}{\partial
z_{1}^{k}},\ldots,\dfrac{\partial g_{1}(z)}{\partial z_{1}}\right),$

where $G_{1}$ is a differential polynomial in $g_{2}(z)$ of order less than
$k$ in which $g_{2}^{\prime}(z)$ appears in the product with at least one more
partial derivative of order greater than or equal to $2$, and similar
definition for $G_{2}$.

Since $g_{1}(z),g_{2}(z)$ are non-constant polynomials, from (4.24), conclude
that $r_{1},r_{2},g_{1}(z+c)-g_{2}(z)$ and $g_{2}(z+c)-g_{1}(z)$ are all
constants. Let $g_{1}(z+c)-g_{2}(z)=\eta_{3}$ and
$g_{2}(z+c)-g_{1}(z)=\eta_{4}$, $\eta_{3},\eta_{4}\in\mathbb{C}$. This implies
that $g_{1}(z+2c)-g_{1}(z)=g_{2}(z+2c)-g_{2}(z)=\eta_{3}+\eta_{4}$. Hence, we
can get

$\displaystyle
g_{1}(z_{1},z_{2})=\mathcal{L}(z)+\chi(z_{2})+d_{1},\;\;g_{2}(z_{1},z_{2})=\mathcal{L}(z)+\chi(z_{2})+d_{2},$

where $\mathcal{L}(z)=a_{1}z_{1}+a_{2}z_{2}$, $\chi(z_{2})$ is a polynomial in
$z_{2}$ such that $\chi(z_{2}+c_{2})=\chi(z_{2})$,
$a_{1},a_{2},d_{1},d_{2}\in\mathbb{C}$. Since, $r_{1},r_{2}$ both are
constants, it follows that

$r_{1}=r_{2}=\left(\frac{a_{1}}{2}\right)^{k}.$

Therefore, from (4.24), we see that

$\displaystyle\begin{cases}e^{\frac{1}{2}[\mathcal{L}(c)+d_{1}-d_{2}]}=\left(\dfrac{a_{1}}{2}\right)^{k}\dfrac{A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}}{A_{1}\xi_{1}+A_{2}\xi_{1}^{-1}},\vspace{1mm}\\\
e^{\frac{1}{2}[\mathcal{L}(c)+d_{2}-d_{1}]}=\left(\dfrac{a_{1}}{2}\right)^{k}\dfrac{A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}}{A_{1}\xi_{2}+A_{2}\xi_{2}^{-1}}.\end{cases}$
(4.25)

Hence, from (4.25), it follows that

$\displaystyle
e^{\mathcal{L}(c)}=\left(\dfrac{a_{1}}{2}\right)^{2k}\dfrac{(A_{2}\xi_{2}+A_{1}\xi_{2}^{-1})(A_{2}\xi_{1}+A_{1}\xi_{1}^{-1})}{(A_{1}\xi_{1}+A_{2}\xi_{1}^{-1})(A_{1}\xi_{2}+A_{2}\xi_{2}^{-1})}.$

Thus, from (4.23), we get

$\displaystyle
f(z_{1},z_{2})=\dfrac{A_{2}\xi_{2}+A_{1}\xi_{2}^{-1}}{\sqrt{2}}e^{\frac{1}{2}[\mathcal{L}(z)+\chi(z_{2})-\mathcal{L}(c)+d_{2}]}\;\;\text{and}$
$\displaystyle
g(z_{1},z_{2})=\dfrac{A_{2}\xi_{1}+A_{1}\xi_{1}^{-1}}{\sqrt{2}}e^{\frac{1}{2}[\mathcal{L}(z)+\chi(z_{2})-\mathcal{L}(c)+d_{1}]}.$

Case 2. Let $\gamma_{1}(z)-\gamma_{2}(z)$ and $\gamma_{3}(z)-\gamma_{4}(z)$
both are non-constant.

Observe that $h_{1}(z)$ and $h_{2}(z)$ both can not be simultaneously zero.
Otherwise, from the first equation of (4.22), we obtain
$e^{\gamma_{1}(z+c)-\gamma_{2}(z+c)}=-1$, which is a contradiction as
$\gamma_{2}(z)-\gamma_{1}(z)$ is non-constant.

Now, let $h_{1}(z)\not\equiv 0$ and $h_{2}(z)\equiv 0$. Then, first equation
of (4.22) reduces to

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)-\gamma_{3}(z)}+A_{2}e^{\gamma_{2}(z+c)-\gamma_{3}(z)}=A_{2}h_{1}(z).$
(4.26)

As $\gamma_{2}(z)-\gamma_{1}(z)$ is a non-constant polynomial, in view of
(4.26), it follows that $\gamma_{1}(z+c)-\gamma_{3}(z)$ and
$\gamma_{2}(z+c)-\gamma_{3}(z)$ both are non-constants.

Rewrite (4.26) as

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)}+A_{2}e^{\gamma_{2}(z+c)}-A_{2}h_{1}e^{\gamma_{3}(z)}\equiv
0.$ (4.27)

Therefore, in view of Lemma 4.3, we can easily get a contradiction from
(4.27).

In a similar manner, we can get a contradiction when $h_{1}(z)\equiv 0$ and
$h_{2}(z)\not\equiv 0$. Hence, $h_{1}(z)\not\equiv 0$ and $h_{2}(z)\not\equiv
0$. By similar arguments, we also get $h_{3}(z)\not\equiv 0$ and
$h_{4}(z)\not\equiv 0$.

Therefore, by (4.22) and Lemma 4.2, we obtain

$\displaystyle\text{either}\;\;A_{1}e^{\gamma_{1}(z+c)-\gamma_{4}(z)}=A_{1}h_{2}(z),\;\;\text{or}\;\;A_{2}e^{\gamma_{2}(z+c)-\gamma_{4}(z)}=A_{1}h_{2}(z)$

and

$\displaystyle\text{either}\;\;A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}=A_{1}h_{4}(z),\;\;\text{or}\;\;A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}=A_{1}h_{4}(z).$

Now, we discuss the four possible subcases.

Subcase 2.1. Let

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)-\gamma_{4}(z)}=A_{1}h_{2}(z)\;\;\text{and}\;\;A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}=A_{1}h_{4}(z).$
(4.28)

Then, by (4.22) and (4.28), we get

$\displaystyle
A_{2}e^{\gamma_{2}(z+c)-\gamma_{3}(z)}=A_{2}h_{1}(z)\;\;\text{and}\;\;A_{2}e^{\gamma_{4}(z+c)-\gamma_{1}(z)}=A_{2}h_{3}(z).$
(4.29)

Since $\gamma_{j}$’s are polynomials, $j=1,\ldots,4$, (4.28) and (4.29) yield
that $h_{1},h_{2},h_{3},h_{4}$,
$\gamma_{1}(z+c)-\gamma_{4}(z),\gamma_{3}(z+c)-\gamma_{2}(z),\gamma_{2}(z+c)-\gamma_{3}(z),\gamma_{4}(z+c)-\gamma_{1}(z)$
are all constants. Let $\gamma_{1}(z+c)-\gamma_{4}(z)=k_{1}$,
$\gamma_{3}(z+c)-\gamma_{2}(z)=k_{2}$, $\gamma_{2}(z+c)-\gamma_{3}(z)=k_{3}$
and $\gamma_{4}(z+c)-\gamma_{1}(z)=k_{4}$, where $k_{j}\in\mathbb{C}$,
$j=1,\ldots,4$. These imply
$\gamma_{1}(z+2c)-\gamma_{1}(z)=\gamma_{4}(z+2c)-\gamma_{4}(z)=k_{1}+k_{4}$
and
$\gamma_{2}(z+2c)-\gamma_{2}(z)=\gamma_{3}(z+2c)-\gamma_{3}(z)=k_{2}+k_{3}$.
Thus, we have

$\displaystyle\gamma_{1}(z)=L_{1}(z)+\Phi_{1}(z_{2})+d_{1},\;\;\gamma_{4}(z)=L_{1}(z)+\Phi_{1}(z_{2})+d_{2},$
$\displaystyle\gamma_{2}(z)=L_{2}(z)+\Psi_{1}(z_{2})+d_{3},\;\;\text{and}\;\;\gamma_{3}(z)=L_{2}(z)+\Psi_{1}(z_{2})+d_{4},$

where $L_{1}(z)=a_{1}z_{1}+a_{2}z_{2}$, $L_{2}(z)=b_{1}z_{1}+b_{2}z_{2}$,
$\Phi_{1}(z_{2})$ and $\Psi_{1}(z_{2})$ are polynomials in $z_{2}$ such that
$\Phi_{1}(z_{2}+c_{2})=\Phi_{1}(z_{2})$ and
$\Psi_{1}(z_{2}+c_{2})=\Psi_{1}(z_{2})$,
$a_{j},,b_{j},e_{j},t_{j}\in\mathbb{C}$ for $j=1,2$.

Since $\gamma_{1}(z)-\gamma_{2}(z)$ and $\gamma_{3}(z)-\gamma_{4}(z)$ are non-
constants, we conclude that

$L_{1}(z)+\Phi_{1}(z_{2})\neq L_{2}(z)+\Psi_{1}(z_{2}).$

Therefore, in view of (4.3) and (4.7), we have

$\displaystyle\begin{cases}g_{1}(z_{1},z_{2})=L_{1}(z)+L_{2}(z)+\Phi_{1}(z_{2})+\Psi_{1}(z_{2})+d_{1}+d_{3},\\\
g_{2}(z_{1},z_{2})=L_{1}(z)+L_{2}(z)+\Phi_{1}(z_{2})+\Psi_{1}(z_{2})+d_{2}+d_{4}.\end{cases}$
(4.30)

Hence, from (4.28) and (4.29), we obtain

$\displaystyle\begin{cases}e^{L_{1}(c)+d_{1}-d_{2}}=a_{1}^{k},\;\;e^{L_{1}(c)+d_{2}-d_{1}}=a_{1}^{k},\\\
e^{L_{2}(c)+d_{3}-d_{4}}=b_{1}^{k},\;\;e^{L_{2}(c)+d_{4}-d_{3}}=b_{1}^{k}.\end{cases}$
(4.31)

From (4.31), we obtain

$\displaystyle
e^{2L_{1}(c)}=a_{1}^{2k},\;\;e^{2(d_{1}-d_{2})}=1,\;\;e^{2L_{2}(c)}=b_{1}^{2k}\;\;\text{and}\;\;e^{2(d_{3}-d_{4})}=1.$

Therefore, we have $e^{L_{1}(c)}=\pm a_{1}^{k},e^{L_{2}(c)}=\pm b_{1}^{k},$
$e^{d_{1}-d_{2}}=\pm 1$ and $e^{d_{3}-d_{4}}=\pm 1$.

If $e^{L_{1}(c)}=a_{1}^{k}$, then from (4.31), we get $e^{d_{1}-d_{2}}=1$. If
$e^{L_{1}(c)}=-a_{1}^{k}$, then from (4.31), we get $e^{d_{1}-d_{2}}=-1$. If
$e^{L_{2}(c)}=b_{1}^{k}$, then from (4.31), we get $e^{d_{3}-d_{4}}=1$ and if
$e^{L_{2}(c)}=-b_{1}^{k}$, then we get $e^{d_{3}-d_{4}}=-1$.

Thus, from fourth and second equations of (4.20), we obtain

$\displaystyle\begin{cases}f(z)=\frac{1}{\sqrt{2}}\left[A_{2}e^{L_{2}(z)+\Psi_{1}(z_{2})-L_{2}(c)+d_{4}}+A_{1}e^{L_{1}(z)+\Phi_{1}(z_{2})-L_{1}(c)+d_{2}}\right]\vspace{1mm}\\\
g(z)=\frac{1}{\sqrt{2}}\left[A_{2}e^{L_{1}(z)+\Phi_{1}(z_{2})-L_{1}(c)+d_{1}}+A_{1}e^{L_{2}(z)+\Psi_{1}(z_{2})-L_{2}(c)+d_{3}}\right].\end{cases}$

Subcase 2.2. Let

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)-\gamma_{4}(z)}=A_{1}h_{2}(z)\;\;\text{and}\;\;A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}=A_{1}h_{4}(z).$
(4.32)

Then, in view of (4.22) and (4.32), we obtain

$\displaystyle
A_{2}e^{\gamma_{2}(z+c)-\gamma_{3}(z)}=A_{2}h_{1}(z)\;\;\text{and}\;\;A_{1}e^{\gamma_{3}(z+c)-\gamma_{1}(z)}=A_{2}h_{3}(z).$
(4.33)

Since $\gamma_{1}(z),\gamma_{2}(z),\gamma_{3}(z),\gamma_{4}(z)$ are
polynomials, we conclude from (4.32) and (4.33) that
$\gamma_{1}(z+c)-\gamma_{3}(z)=m_{1}$, $\gamma_{4}(z+c)-\gamma_{1}(z)=m_{2}$,
$\gamma_{2}(z+c)-\gamma_{4}(z)=m_{3}$ and
$\gamma_{3}(z+c)-\gamma_{2}(z)=m_{4}$, where $m_{j}\in\mathbb{C}$,
$j=1,\ldots,4$. These imply
$\gamma_{1}(z+2c)-\gamma_{1}(z)=\gamma_{2}(z+4c)-\gamma_{2}(z)=\sum_{j=1}^{4}m_{j}$.
Therefore, $\gamma_{1}(z)=L_{1}(z)+\Phi_{1}(s_{1})+d_{1}$ and
$\gamma_{2}(z)=L_{1}(z)+\Phi_{1}(s_{1})+d_{2}$, where $L_{1}(z)$ and
$\Phi_{1}(s_{1})$ are defined in Subcase 2.1, $d_{1},d_{2}\in\mathbb{C}$.
Hence, $\gamma_{1}(z)-\gamma_{2}(z)=d_{1}-d_{2}\in\mathbb{C}$, a
contradiction.

Subcase 2.3. Let

$\displaystyle
A_{2}e^{\gamma_{2}(z+c)-\gamma_{4}(z)}=A_{1}h_{2}(z)\;\;\text{and}\;\;A_{1}e^{\gamma_{3}(z+c)-\gamma_{2}(z)}=A_{1}h_{4}(z).$

Then, by similar arguments as in Subcase 2.2, we easily get a contradiction.
Subcase 2.4. Let

$\displaystyle
A_{2}e^{\gamma_{2}(z+c)-\gamma_{4}(z)}=A_{1}h_{2}(z)\;\;\text{and}\;\;A_{2}e^{\gamma_{4}(z+c)-\gamma_{2}(z)}=A_{1}h_{4}(z).$
(4.34)

Then, by (4.22) and (4.34), we get

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)-\gamma_{3}(z)}=A_{2}h_{1}(z)\;\;\text{and}\;\;A_{1}e^{\gamma_{3}(z+c)-\gamma_{1}(z)}=A_{2}h_{3}(z).$
(4.35)

As $\gamma_{1}(z),\gamma_{2}(z),\gamma_{3}(z),\gamma_{4}(z)$ are polynomials,
from (4.34) and (4.35), we conclude that $h_{1},h_{2},h_{3},h_{4}$,
$\gamma_{2}(z+c)-\gamma_{4}(z),\gamma_{4}(z+c)-\gamma_{2}(z),\gamma_{1}(z+c)-\gamma_{3}(z),\gamma_{3}(z+c)-\gamma_{1}(z)$
are all constants. Let $\gamma_{2}(z+c)-\gamma_{4}(z)=n_{1}$,
$\gamma_{4}(z+c)-\gamma_{2}(z)=n_{2}$, $\gamma_{1}(z+c)-\gamma_{3}(z)=n_{3}$
and $\gamma_{3}(z+c)-\gamma_{1}(z)=n_{4}$, where $n_{j}\in\mathbb{C}$,
$j=1,\ldots,4$. Thus, we get
$\gamma_{1}(z+2c)-\gamma_{1}(z)=\gamma_{3}(z+2c)-\gamma_{3}(z)=n_{1}+n_{2}$
and
$\gamma_{2}(z+2c)-\gamma_{2}(z)=\gamma_{4}(z+2c)-\gamma_{4}(z)=n_{3}+n_{4}$.
Hence, by similar arguments as in Subcase 2.1, we obtain

$\displaystyle\gamma_{1}(z)=L_{1}(z)+\Phi_{1}(z_{2})+d_{1},\;\;\gamma_{3}(z)=L_{1}(z)+\Phi_{1}(z_{2})+d_{2},$
$\displaystyle\gamma_{2}(z)=L_{2}(z)+\Psi_{1}(z_{2})+d_{3},\;\;\text{and}\;\;\gamma_{4}(z)=L_{2}(z)+\Psi_{1}(z_{2})+d_{4},$

where $L_{(}z),L_{2}(z),\Phi_{1}(z_{2}),\Psi_{1}(z_{2})$ are same as in
Subcase 2.1 with $L_{1}(z)+\Phi_{1}(z_{2})\neq L_{2}(z)+\Psi_{1}(z_{2})$.
Thus, in view of (4.3) and (4.7), we obtain the same form of
$g_{1}(z_{1},z_{2})$ and $g_{2}(z_{1},z_{2})$ as in (4.30).

Therefore, in view of (4.34) and (4.35), we have

$\displaystyle\begin{cases}A_{2}e^{L_{2}(c)+d_{3}-d_{4}}=A_{1}b_{1}^{k},\;\;A_{2}e^{L_{2}(c)+d_{4}-d_{3}}=A_{1}b_{1}^{k},\\\
A_{1}e^{L_{1}(c)+d_{1}-d_{2}}=A_{2}a_{1}^{k},\;\;A_{1}e^{L_{1}(c)+d_{2}-d_{1}}=A_{2}a_{1}^{k}.\end{cases}$
(4.36)

From (4.36), we obtain

$\displaystyle
e^{2L_{1}(c)}=\frac{A_{2}^{2}}{A_{1}^{2}}a_{1}^{2k}\,\;\;e^{2(d_{1}-d_{2})}=1,\;\;e^{2L_{2}(c)}=\frac{A_{1}^{2}}{A_{2}^{2}}b_{1}^{2k}\;\;\text{and}\;\;e^{2(d_{3}-d_{4})}=1,$

which imply

$\displaystyle
e^{L_{1}(c)}=\pm\frac{A_{2}}{A_{1}}a_{1}^{k}\,\;\;e^{d_{1}-d_{2}}=\pm
1,\;\;e^{L_{2}(c)}=\pm\frac{A_{1}}{A_{2}}b_{1}^{k}\;\;\text{and}\;\;e^{d_{3}-d_{4}}=\pm
1.$

If $e^{L_{1}(c)}=A_{2}a_{1}^{k}/A_{1}$, then from (4.36), we get
$e^{d_{1}-d_{2}}=1$. If $e^{L_{1}(c)}=-A_{2}a_{1}^{k}/A_{1}$, then we have
$e^{d_{1}-d_{2}}=-1$. If $e^{L_{2}(c)}=A_{1}b_{1}^{k}/A_{2}$, then from
(4.36), we get $e^{d_{3}-d_{4}}=1$. If $e^{L_{2}(c)}=-A_{1}b_{1}^{k}/A_{2}$,
then we have $e^{d_{3}-d_{4}}=-1$.

Hence, from fourth and second equations of (4.20), we get

$\displaystyle\begin{cases}f(z)=\frac{1}{\sqrt{2}}\left[A_{2}e^{L_{1}(z)+\Phi_{1}(z_{2})-L_{1}(c)+d_{2}}+A_{1}e^{L_{2}(z)+\Psi_{1}(z_{2})-L_{2}(c)+d_{4}}\right],\vspace{1mm}\\\
g(z)=\frac{1}{\sqrt{2}}\left[A_{2}e^{L_{1}(z)+\Phi_{1}(z_{2})-L_{1}(c)+d_{1}}+A_{1}e^{L_{2}(z)+\Psi_{1}(z_{2})-L_{2}(c)+d_{3}}\right].\end{cases}$

Case 3. Let $\gamma_{3}(z)-\gamma_{4}(z)=\eta\in\mathbb{C}$ and
$\gamma_{1}(z)-\gamma_{2}(z)$ be non-constant. Then the first equation of
(4.22) reduces to

$\displaystyle
A_{1}e^{\gamma_{1}(z+c)-\gamma_{4}(z)}+A_{2}e^{\gamma_{2}(z+c)-\gamma_{4}(z)}=A_{1}h_{2}(z)+A_{2}h_{1}(z)e^{\eta}.$
(4.37)

Note that $A_{1}h_{2}(z)+A_{2}h_{1}(z)e^{\eta}\not\equiv 0$. Otherwise from
(4.37), we have

$\displaystyle A_{1}e^{\gamma_{1}(z+c)-\gamma_{2}(z+c)}=-A_{2},$

which implies that $\gamma_{1}(z+c)-\gamma_{2}(z+c)$ and hence
$\gamma_{1}(z)-\gamma_{2}(z)$ becomes constant. This is a contradiction.

Now, in view of (4.37), it follows that

$\displaystyle
T\left(r,e^{\gamma_{1}(z+c)-\gamma_{4}(z)}\right)=T\left(r,e^{\gamma_{2}(z+c)-\gamma_{4}(z)}\right)+O(1).$
(4.38)

We note that

$\displaystyle
N\left(r,e^{\gamma_{1}(z+c)-\gamma_{4}(z)}\right)=N\left(r,\frac{1}{e^{\gamma_{1}(z+c)-\gamma_{4}(z)}}\right)=S\left(r,e^{\gamma_{1}(z+c)-\gamma_{4}(z)}\right)$

and

$\displaystyle
N\left(r,\frac{1}{e^{\gamma_{1}(z+c)-\gamma_{4}(z)}-w}\right)=N\left(r,\frac{1}{e^{\gamma_{2}(z+c)-\gamma_{4}(z)}}\right)=S\left(r,e^{\gamma_{2}(z+c)-\gamma_{4}(z)}\right),$

where $w=h_{2}+A_{2}h_{1}e^{\eta}/A_{1}$.

Now by the second fundamental theorem of Nevanlinna for several complex
variables, we get

$\displaystyle T\left(r,e^{\gamma_{1}(z+c)-\gamma_{4}(z)}\right)$
$\displaystyle\leq
N\left(r,e^{\gamma_{1}(z+c)-\gamma_{4}(z)}\right)+N\left(r,\frac{1}{e^{\gamma_{1}(z+c)-\gamma_{4}(z)}}\right)+N\left(r,\frac{1}{e^{\gamma_{1}(z+c)-\gamma_{4}(z)}-w}\right)$
$\displaystyle+S\left(r,e^{\gamma_{1}(z+c)-\gamma_{4}(z)}\right)$
$\displaystyle\leq
S\left(r,e^{\gamma_{1}(z+c)-\gamma_{4}(z)}\right)+S\left(r,e^{\gamma_{2}(z+c)-\gamma_{4}(z)}\right).$

This implies that $\gamma_{1}(z+c)-\gamma_{4}(z)$ is constant. In view of
(4.38), we see that $\gamma_{2}(z+c)-\gamma_{4}(z)$ is also constant. These
imply $\gamma_{1}(z+c)-\gamma_{2}(z+c)$ and hence
$\gamma_{1}(z)-\gamma_{2}(z)$ is a constant. This is a contradiction.

Case 4. Let $\gamma_{1}-\gamma_{2}=\eta_{1}\in\mathbb{C}$ and
$\gamma_{3}-\gamma_{4}$ be non-constant. Then, by similar arguments as used in
Subcase 3, we can get a contradiction. So, we omit the details.∎

###### Proof of Theorem 3.3.

Let $(f(z),g(z))$ be a pair of finite order transcendental entire solution of
(2.5).

Then, by similar argument as in Theorem 3.1, we obtain

$\displaystyle\begin{cases}\dfrac{\partial^{k}f}{\partial
z_{1}^{k}}=\dfrac{1}{\sqrt{2}}\left[A_{1}e^{p_{1}(z)}+A_{2}e^{-p_{1}(z)}\right],\\\
g(z+c)-g(z)=\displaystyle\frac{1}{\sqrt{2}}\left[A_{2}e^{p_{1}(z)}+A_{1}e^{-p_{1}(z)}\right],\\\
\dfrac{\partial^{k}g}{\partial
z_{1}^{k}}=\displaystyle\frac{1}{\sqrt{2}}\left[A_{1}e^{p_{2}(z)}+A_{2}e^{-p_{2}(z)}\right],\\\
f(z+c)-f(z)=\displaystyle\frac{1}{\sqrt{2}}\left[A_{2}e^{p_{2}(z)}+A_{1}e^{-p_{2}(z)}\right],\end{cases}$
(4.39)

where $p_{1}(z)$ and $p_{2}(z)$ are two non-constant polynomials.

Differentiating fourth equation of (4.39) $k$ times, we get

$\displaystyle\dfrac{\partial^{k}f(z+c)}{\partial
z_{1}^{k}}-\dfrac{\partial^{k}f}{\partial
z_{1}^{k}}=\frac{1}{\sqrt{2}}\left[A_{2}H_{1}e^{p_{2}(z)}+A_{1}H_{2}e^{-p_{2}(z)}\right],$
(4.40)

where $H_{1}=\left(\frac{\partial p_{2}}{\partial
z_{1}}\right)^{k}+M_{k}\left(\frac{\partial^{k}p_{2}}{\partial
z_{1}^{k}},\ldots,\frac{\partial p_{2}}{\partial z_{1}}\right)$ and
$H_{2}=\left(-\frac{\partial p_{2}}{\partial
z_{1}}\right)^{k}+N_{k}\left(\frac{\partial^{k}p_{2}}{\partial
z_{1}^{k}},\ldots,\frac{\partial p_{2}}{\partial z_{1}}\right)$, where $M_{k}$
is a partial differential polynomial in $p_{2}(z)$ of degree less than $k$ in
which $\frac{\partial p_{2}}{\partial z_{1}}$ appears in the product with at
least one more partial derivative of $p_{2}$ of order greater than $1$.
Similar definition is for $N_{k}$.

Now in view of the first equation of (4.39) and (4.40), we obtain

$\displaystyle
A_{1}e^{p_{1}(z+c)+p_{2}(z)}+A_{2}e^{-p_{1}(z+c)+p_{2}(z)}-A_{1}e^{p_{1}(z)+p_{2}(z)}-A_{2}e^{-p_{1}(z)+p_{2}(z)}$
$\displaystyle-A_{2}H_{1}e^{2p_{2}(z)}=A_{1}H_{2}.$ (4.41)

Similarly in view of the second and third equations of (4.39), we obtain

$\displaystyle
A_{1}e^{p_{2}(z+c)+p_{1}(z)}+A_{2}e^{-p_{2}(z+c)+p_{1}(z)}-A_{1}e^{p_{2}(z)+p_{1}(z)}-A_{2}e^{-p_{2}(z)+p_{1}(z)}$
$\displaystyle-A_{2}H_{3}e^{2p_{1}(z)}=A_{1}H_{4},$ (4.42)

where $H_{3}=\left(\frac{\partial p_{1}}{\partial
z_{1}}\right)^{k}+O_{k}\left(\frac{\partial^{k}p_{1}}{\partial
z_{1}^{k}},\ldots,\frac{\partial p_{1}}{\partial z_{1}}\right)$ and
$H_{4}=\left(-\frac{\partial p_{1}}{\partial
z_{1}}\right)^{k}+R_{k}\left(\frac{\partial^{k}p_{1}}{\partial
z_{1}^{k}},\ldots,\frac{\partial p_{1}}{\partial z_{1}}\right)$, where $O_{k}$
is a differential polynomial in $p_{1}(z)$ of degree less than $k$ in which
$\frac{\partial p_{1}}{\partial z_{1}}$ appears in the product with at least
one higher order derivative of $p_{1}$. Similar definition is for $R_{k}$.

Now, we consider the following two possible cases.

Case 1. Let $p_{2}(z)-p_{1}(z)=\eta$, $\eta\in\mathbb{C}$. Then (4) and (4)
yield

$\displaystyle\begin{cases}A_{1}e^{\eta}e^{p_{1}(z+c)+p_{1}(z)}+A_{2}e^{\eta}e^{p_{1}(z)-p_{1}(z+c)}-e^{\eta}(A_{1}+A_{2}H_{1}e^{\eta})e^{2p_{1}(z)}\\\
=A_{1}H_{2}+A_{2}e^{\eta},\\\
A_{1}e^{\eta}e^{p_{1}(z+c)+p_{1}(z)}+A_{2}e^{-\eta}e^{p_{1}(z)-p_{1}(z+c)}-(A_{1}e^{\eta}+A_{2}H_{3})e^{2p_{1}(z)}\\\
=A_{1}H_{4}+A_{2}e^{-\eta}.\end{cases}$ (4.43)

Note that $A_{1}+A_{2}H_{1}e^{\eta}$ and $A_{1}H_{2}+A_{2}e^{\eta}$ can not be
zero simultaneously. Otherwise, from the first equation of (4.43), we get
$p_{1}(z+c)$, and hence $p_{1}(z)$ is constant, a contradiction.

Next, suppose that $A_{1}+A_{2}H_{1}e^{\eta}\not\equiv 0$ and
$A_{1}H_{2}+A_{2}e^{\eta}\equiv 0$. The the first equation of (4.43) reduces
to

$\displaystyle
A_{1}e^{p_{1}(z+c)-p_{1}(z)}+A_{2}e^{-(p_{1}(z)+p_{1}(z+c))}=A_{1}+A_{2}H_{1}e^{\eta}.$
(4.44)

Now, observe that

$\displaystyle
N\left(r,e^{-(p_{1}(z+c)+p_{1}(z))}\right)=N\left(r,\frac{1}{e^{-(p_{1}(z+c)+p_{1}(z))}}\right)=S\left(r,e^{-(p_{1}(z+c)+p_{1}(z))}\right)$

and in view of (4.44), we have

$\displaystyle
N\left(r,\frac{1}{e^{-(p_{1}(z+c)+p_{1}(z))}-w}\right)=N\left(r,\frac{1}{e^{p_{1}(z+c)-p_{1}(z)}}\right)=S\left(r,e^{-(p_{1}(z+c)+p_{1}(z))}\right),$

where $w=(A_{1}+A_{2}H_{1}e^{\eta})/A_{2}$.

By the second fundamental theorem of Nevanlinna, we obtain

$\displaystyle T\left(r,e^{-(p_{1}(z+c)+p_{1}(z))}\right)$
$\displaystyle\leq\overline{N}\left(r,e^{-(p_{1}(z+c)+p_{1}(z))}\right)+\overline{N}\left(r,\frac{1}{e^{-(p_{1}(z+c)+p_{1}(z))}}\right)$
$\displaystyle+\overline{N}\left(r,\frac{1}{e^{-(p_{1}(z+c)+p_{1}(z))}-w}\right)+S\left(r,e^{-(p_{1}(z+c)+p_{1}(z))}\right)$
$\displaystyle\leq
S\left(r,e^{-(p_{1}(z+c)+p_{1}(z))}\right)+S\left(r,e^{p_{1}(z+c)-p_{1}(z)}\right).$

This implies that $p_{1}(z+c)+p_{1}(z)$ is constant, and hence $p_{1}(z)$ is
constant, which is a contradiction.

Similarly, we can get a contradiction for the case
$A_{1}+A_{2}H_{1}e^{\eta}\equiv 0$ and $A_{1}H_{2}+A_{2}e^{\eta}\not\equiv 0$.
Hence, $A_{1}+A_{2}H_{1}e^{\eta}\not\equiv 0$ and
$A_{1}H_{2}+A_{2}e^{\eta}\not\equiv 0$.

In a similar manner, we can show that $A_{1}e^{\eta}+A_{2}H_{3}\not\equiv 0$
and $A_{1}H_{4}+A_{2}e^{-\eta}\not\equiv 0$.

Therefore, in view of Lemma 4.1 and (4.43), we obtain

$\displaystyle\begin{cases}A_{2}e^{\eta}e^{p_{1}(z)-p_{1}(z+c)}=A_{1}H_{2}+A_{2}e^{\eta},\\\
A_{2}e^{-\eta}e^{p_{1}(z)-p_{1}(z+c)}=A_{1}H_{4}+A_{2}e^{-\eta}.\end{cases}$
(4.45)

By (4.43) and (4.45), we get

$\displaystyle\begin{cases}A_{1}e^{p_{1}(z+c)-p_{1}(z)}=A_{1}+A_{2}H_{1}e^{\eta},\\\
A_{1}e^{\eta}e^{p_{1}(z+c)-p_{1}(z)}=A_{2}H_{3}+A_{1}e^{\eta}.\end{cases}$
(4.46)

Since $p_{1}(z)$ is a non-constant polynomial, it follows from (4.45) and
(4.46) that $p_{1}(z+c)-p_{1}(z)$, $H_{1},H_{2},H_{3},H_{4}$ are all
constants. Otherwise, L.H.S. of both the equations of (4.45) and (4.46) are
transcendental entire, whereas R.H.S’s. are polynomials. Therefore, we must
have $p_{1}(z)=L(z)+\Phi_{1}(z_{2})+\gamma$, where $L(z)=\alpha z_{1}+\beta
z_{2}$, $\Phi_{1}(z_{2})$ is a polynomial in $z_{2}$ with
$\Phi_{1}(z_{2}+c_{2})=\Phi_{1}(z_{2})$, $\alpha,\beta,\gamma\in\mathbb{C}$.
Thus, $p_{2}(z)=L(z)+\Phi_{1}(z_{2})+\gamma+\eta$.

Therefore, in view of (4.45), (4.46) and the form of $H_{1}$, $H_{2}$, $H_{3}$
and $H_{4}$, we obtain

$\displaystyle\begin{cases}A_{2}e^{\eta}e^{-\alpha
c}=A_{1}(-\alpha)^{k}+A_{2}e^{\eta},\;\;A_{2}e^{-\eta}e^{-\alpha
c}=A_{1}(-\alpha)^{k}+A_{2}e^{-\eta},\\\ A_{1}e^{\alpha
c}=A_{1}+A_{2}\alpha^{k}e^{\eta},\;\;A_{1}e^{\eta}e^{\alpha
c}=A_{2}\alpha^{k}+A_{1}e^{\eta}.\end{cases}$ (4.47)

From (4.47), we obtain

$\displaystyle
e^{L(c)}=\frac{A_{2}e^{\eta}}{(-1)^{k}A_{1}\alpha^{k}+A_{2}e^{\eta}},\;\;e^{2\eta}=1\;\;\text{and}\;\;\alpha^{k}=-\frac{(-1)^{k}A_{1}^{2}+A_{2}^{2}}{(-1)^{k}A_{1}A_{2}e^{\eta}}$

Thus, from (4.39), we obtain

$\displaystyle
f(z)=\frac{A_{1}e^{L(z)+\Phi_{1}(z_{2})+\gamma}+A_{2}e^{-(L(z)+\Phi_{1}(z_{2})+\gamma)}}{\sqrt{2}\alpha^{k}}\;\;\text{and}$
$\displaystyle
g(z)=\frac{A_{1}e^{L(z)+\Phi_{1}(z_{2})+\gamma+\eta}+A_{2}e^{-(L(z)+\Phi_{1}(z_{2})+\gamma+\eta)}}{\sqrt{2}\alpha^{k}}.$

Case 2. Let $p_{2}(z)-p_{1}(z)$ be non-constant.

Now we consider two possible subcases.

Subcase 2.1. Let $p_{2}(z)+p_{1}(z)=\eta$, where $\eta$ is a constant in
$\mathbb{C}$.

Then, (4) and (4) yield

$\displaystyle\begin{cases}A_{1}e^{\eta}e^{p_{1}(z+c)-p_{1}(z)}+A_{2}e^{\eta}e^{-(p_{1}(z+c)+p_{1}(z))}-A_{2}e^{\eta}(1+H_{1}e^{\eta})e^{-2p_{1}(z)}\\\
=A_{1}(H_{2}+e^{\eta}),\\\
A_{1}e^{\eta}e^{p_{1}(z)-p_{1}(z+c)}+A_{2}e^{-\eta}e^{p_{1}(z+c)+p_{1}(z)}-A_{2}(H_{3}+e^{-\eta})e^{2p_{1}(z)}\\\
=A_{1}(H_{4}+e^{\eta}).\end{cases}$ (4.48)

Now, by similar argument as in Case 1, we can prove that $1+H_{1}e^{\eta}$,
$H_{2}+e^{\eta}$, $H_{3}+e^{-\eta}$ and $H_{4}+e^{\eta}$ are all non-zero.

Therefore, by Lemma 4.1 and (4.48), we have

$\displaystyle
e^{p_{1}(z+c)-p_{1}(z)}=H_{2}e^{-\eta}+1\;\;\text{and}\;\;e^{p_{1}(z)-p_{1}(z+c)}=H_{4}e^{-\eta}+1.$
(4.49)

From (4.48) and (4.49), we obtain

$\displaystyle
e^{-p_{1}(z+c)+p_{1}(z)}=1+H_{1}e^{\eta}\;\;\text{and}\;\;e^{p_{1}(z+c)-p_{1}(z)}=1+e^{\eta}H_{3}.$
(4.50)

Then by similar arguments as in the Case 1, we obtain that

$\displaystyle
p_{1}(z)=L(z)+\Phi_{1}(z_{2})+\gamma\;\;\text{and}\;\;p_{2}(z)=-(L(z)+\Phi_{1}(z_{2})+\gamma)+\eta.$

Therefore, from (4.49) and (4.50), we obtain

$\displaystyle\begin{cases}e^{L(c)}=1+e^{-\eta}\alpha^{k},\;\;e^{-L(c)}=e^{-\eta}(-\alpha)^{k}+1,\\\
e^{-L(c)}=1+e^{\eta}(-\alpha)^{k},\;\;e^{L(c)}=1+e^{\eta}\alpha^{k}.\end{cases}$
(4.51)

From first and fourth equations of (4.51), we get

$\displaystyle e^{2\eta}=1.$

From the above two equations of (4.51), we obtain

$\displaystyle(-1)^{k}e^{-\eta}\alpha^{k}+1+(-1)^{k}=0.$ (4.52)

If $k$ is odd, then in view of (4.52), we can get a contradiction. If $k$ is
even, then from (4.52), we get

$\displaystyle\alpha^{k}=-2e^{\eta}.$

Therefore, from (4.39), we have

$\displaystyle
f(z)=\frac{A_{1}e^{L(z)+\Phi_{1}(z_{2})+\gamma}+A_{2}e^{-(L(z)+\Phi_{1}(z_{2})+\gamma)}}{\sqrt{2}\alpha^{k}},$
$\displaystyle
g(z)=\frac{A_{1}e^{-(L(z)+\Phi_{1}(z_{2})+\gamma)+\eta}+A_{2}e^{L(z)+\Phi_{1}(z_{2})+\gamma-\eta}}{\sqrt{2}\alpha^{k}}.$

Subcase 2.2. Let $p_{2}(z)+p_{1}(z)$ be a non-constant polynomial.

Now we consider four possible cases below.

Subcase 2.2.1. Let $p_{1}(z+c)+p_{2}(z)=\zeta_{1}$ and
$p_{2}(z+c)+p_{1}(z)=\zeta_{2}$, where $\zeta_{1},\zeta_{2}\in\mathbb{C}$.
Then, we easily see that $p_{1}(z+2c)-p_{1}(z)=\zeta_{1}-\zeta_{2}$ and
$p_{2}(z+2c)-p_{2}(z)=\zeta_{2}-\zeta_{1}$. This implies that $p_{1}(z)=\alpha
z+\beta_{1}$ and $p_{2}(z)=-\alpha z+\beta_{2}$, where
$\alpha,\beta_{1},\beta_{2}\in\mathbb{C}$. But, then
$p_{1}(z)+p_{2}(z)=\beta_{1}+\beta_{2}=$ constant, a contradiction.

Subcase 2.2.2. Let $p_{1}(z+c)+p_{2}(z)=\zeta_{1}\in\mathbb{C}$ and
$p_{2}(z+c)+p_{1}(z)$ is non-constant.

Then, by Lemma 4.1 and (4), we get $A_{2}e^{-p_{2}(z+c)+p_{1}(z)}=A_{1}H_{4}$.
This implies that $-p_{2}(z+c)+p_{1}(z)=\zeta_{2}\in\mathbb{C}$. Therefore, we
must have $p_{1}(z+2c)+p_{1}(z)=\zeta_{1}+\zeta_{2}$, and hence $p_{1}(z)$
must be a constant, a contradiction.

Subcase 2.2.3. Let $p_{2}(z+c)+p_{1}(z)=\zeta_{1}\in\mathbb{C}$ and
$p_{1}(z+c)+p_{2}(z)$ is non-constant.

Then, by similar argument as in subcase 2.2.2, we obtain a contradiction.

Subcase 2.2.4. Let $p_{2}(z+c)+p_{1}(z)$ and $p_{1}(z+c)+p_{2}(z)$ both are
non-constant.

Then, using Lemma 4.1, we obtain from (4) and (4) that

$\displaystyle
A_{2}e^{-p_{1}(z+c)+p_{2}(z)}=A_{1}H_{2}\;\;\text{and}\;\;A_{2}e^{-p_{2}(z+c)+p_{1}(z)}=A_{1}H_{4}.$

As $p_{1}(z)$ and $p_{2}(z)$ are non-constant polynomials, we must have
$-p_{1}(z+c)+p_{2}(z)=\xi_{1}$ and $-p_{2}(z+c)+p_{1}(z)=\xi_{2}$, where
$\xi_{1},\xi_{2}\in\mathbb{C}$. This implies that
$-p_{1}(z+c)+p_{1}(z)=-p_{2}(z+c)+p_{2}(z)=\xi_{1}+\xi_{2}$. Therefore, we
must get $p_{1}(z)=\alpha z+\beta_{1}$ and $p_{2}(z)=\alpha z+\beta_{2}$,
where $\alpha,\beta_{1},\beta_{2}\in\mathbb{C}$. But, then we get
$p_{2}(z)-p_{1}(z)=\beta_{2}-\beta_{1}=$constant, which is a contradiction.∎

##  5\. Statements and Declarations

Conflict of interest The authors declare that there are no conflicts of
interest regarding the publication of this paper.

Funding There is no funding received from any organizations for this research
work.

Data availability statement Data sharing is not applicable to this article as
no database were generated or analyzed during the current study.

Acknowledgment: The authors would like to thank the referee(s) for the helpful
suggestions and comments to improve the exposition of the paper.

## References

  * [1] Biancofiore, A, Stoll, W.: Another proof of the lemma of the logarithmic derivative in several complex variables, Recent developments in several complex variables, Princeton University Press, Princeton, 29–45 ( 1981)
  * [2] Cao, T.B., Korhonen, R.J.: A new version of the second main theorem for meromorphic mappings intersecting hyperplanes in several complex variables, J. Math. Anal. Appl. 444(2), 1114–1132 (2016)
  * [3] T. B. Cao and L. Xu, Logarithmic difference lemma in several complex variables and partial difference equations, Ann. Math. Pure Appl. 199, 767–794 (2020)
  * [4] Chiang, Y.M., Feng, S.J.: On the Nevanlinna characteristic of $f(z+\eta)$ and difference equations in the complex plane, Ramanujan J. 16(1), 105–129 (2008).
  * [5] Chiang,Y.M., Feng, S.J.: On the growth of logarithmic differences, difference quotients and logarithmic derivatives of meromorphic functions, Trans. Am. math. Soc. 361, 3767–3791 (2009).
  * [6] Chen, Z.X.: Growth and zeros of meromorphic solution of some linear difference equations, J. Math. Anal. Appl. 373, 235–241 (2011).
  * [7] Chen, Z.X.: Zeros of entire solutions to complex linear difference euations. Acta Math. Sci. Ser. B Engl. Ed. 32, 1141–1148 (2012).
  * [8] Courant, R., Hilbert, D.: Methods of Mathematical Physics, vol. II: Partial Differential Equations, Interscience (New York, 1962)
  * [9] L. Y. Gao: Entire solutions of two types of systems of complex differential-difference equations, Acta Math. Sinica (Chin. Ser.) 59 (2016), 677–685.
  * [10] Garabedian, P.R.: Partial Differential Equations, Wiley (New York, 1964).
  * [11] Gross, F.: On the equation $f^{n}(z)+g^{n}(z)=1$, Bull. Amer. Math. Soc. 72, 86–88 (1966).
  * [12] Gross, F.: On the equation $f^{n}(z)+g^{n}(z)=h^{n}(z)$, Amer. Math. Monthly. 73, 1093–1096 (1966).
  * [13] Halburd, R.G., Korhonen, R.J.: Difference analogue of the lemma on the logarithmic derivative with applications to difference equations, J. Math. Anal. Appl. 314, 477–487 (2006).
  * [14] Halburd, R.G., Korhonen, R.J.: Nevanlinna theory for the difference operator, Ann. Acad. Sci. Fenn. Math. 31, 463–478 (2006).
  * [15] Haldar, G.: Solutions of Fermat-type partial differential difference equations in $\mathbb{C}^{2}$, Mediterr. J. Math., 20 (2023), 50, DOI: https://doi.org/10.1007/s00009-022-02180-6
  * [16] Haldar, G., Ahamed, M.B.: Entire solutions of several quadratic binomial and trinomial partial differential-difference equations in $\mathbb{C}^{2}$, Anal. Math. Phys. 12 (2022), Article number: 113, DOI: https:10.1007/s13324-022-00722-5
  * [17] Hayman, W.K.: Meromorphic Functions, The Clarendon Press, Oxford, 1964.
  * [18] Hu, P.C., Li, P., Yang, C. C.: Unicity of Meromorphic Mappings, Advances in Complex Analysis and Its Applications, vol. 1, Kluwer Academic Publishers, Dordrecht, Boston, London (2003).
  * [19] Iyer, G.: On certain functional equations. J. Indian. Math. Soc. 3, 312–315 (1939).
  * [20] Korhonen, R.J.: A difference Picard theorem for meromorphic functions of several variables, Comput. Methods Funct. Theory 12(1), 343–361 (2012)
  * [21] Laine, I.: Nevanlinna Theory and Complex Differential Equations, Walter de Gruyter, Berlin, (1993).
  * [22] Li, B.Q.: On reduction of functional-differential equations, Complex Var. 31, 311–324 (1996)
  * [23] Li, B.Q.: On entire solutions of Fermat type partial differential equations, Int. J. Math. 15, 473–485 (2004)
  * [24] B. Q. Li, On certain non-linear differential equations in complex domains, Arch. Math. 91 (2008), 344–353.
  * [25] Liu, M.L., Gao, L.Y.: Transcendental solutions of systems of complex differential-difference equations, Sci. Sin. Math. 49, 1–22 (2019).
  * [26] Liu, K.: Meromorphic functions sharing a set with applications to difference equations. J. Math. Anal. Appl. 359, 384–393 (2009).
  * [27] Liu, K., Yang, L.Z.: On Entire Solutions of Some Differential-Difference Equations. Comput. Methods Funct. Theory 13, 433–447 (2012).
  * [28] Liu, K., Cao, T.B.: Entire solutions of Fermat type q-difference-differential equations. Electron. J. Diff. Equ. 59, 1–10 (2013).
  * [29] Liu, K., Cao, T.B., Cao, H.Z.: Entire solutions of Fermat-type differential-difference equations. Arch. Math. 99, 147–155 (2012).
  * [30] Liu, K., Dong, X.: Fermat-type differential and difference equations, Electron. J. Differ.Equ. 2015, 1–10 (2015).
  * [31] Liu, K., Yang, L.Z.: A note on meromorphic solutions of Fermat types equations, An. Stiint. Univ. Al. I. Cuza Lasi Mat. (N. S.). 1, 317–325 (2016).
  * [32] Montel, P.: Lecons sur les familles de nomales fonctions analytiques et leurs applications. Gauthier-Viuars Paris, 135–136 (1927).
  * [33] Rauch, J.: Partial Differential Equations, Springer-Verlag (New York, 1991)
  * [34] Ronkin, L.I.: Introduction to the Theory of Entire Functions of Several Variables, Moscow: Nauka 1971 (Russian), American Mathematical Society, Providence (1974)
  * [35] Saleeby, E.G.: Entire and meromorphic solutions of Fermat-type partial differential equations, Analysis (Munich) 19, 369–376 (1999)
  * [36] Saleeby, E.G.: On entire and meromorphic solutions of $\lambda u^{k}+\sum_{i=1}^{n}u_{z_{i}}^{m}=1$, Complex Var. Theory Appl. 49, 101–107 (2004)
  * [37] E. G. Saleeby, On complex analytic solutions of certain trinomial functional and partial differential equations, Aequationes Math. 85 (2013), 553–562.
  * [38] Stoll, W.: Holomorphic Functions of Finite Order in Several Complex Variables, American Mathematical Society, Providence, 1974
  * [39] Tang, J.F., Liao, L.W.: The transcendental meromorphic solutions of a certain type of non-linear differential equations, J. Math. Anal. Appl. 334, 517–527 (2007).
  * [40] Taylor, R., Wiles, A.: Ring-theoretic properties of certain Hecke algebra, Ann. Math. 141, 553–572 (1995)
  * [41] Wiles, A.: Modular elliptic curves and Fermat last theorem, Ann. Math. 141, 443–551 (1995)
  * [42] Xu, L., Cao, T.B.: Solutions of complex Fermat-type partial difference and differential-difference equations, Mediterr. J.Math. 15, 1–14 (2018)
  * [43] Xu, L., Cao, T.B.: Correction to: Solutions of Complex Fermat-Type Partial Difference and Differential-Difference Equations, Mediterr. J.Math. 17, 1–4 (2020)
  * [44] Xu, H.Y., Haldar, G.: Solutions of complex nonlinear functional equations including second order partial differential and difference equations in $\mathbb{C}^{2}$, Electron. J. Differ. Equ., 43(2023), 1–18
  * [45] Xu, H.Y., Jiang, Y.Y.: Results on entire and meromorphic solutions for several systems of quadratic trinomial functional equations with two complex variables, RACSAM (2022). https://doi.org/10.1007/s13398-021-01154-9
  * [46] Xu, X.Y., Liu, S.Y., Li, Q.P.: Entire solutions for several systems of nonlinear difference and partial differential-difference equations of Fermat-type, J. Math. Anal. Appl. 483, 123–641 (2020)
  * [47] Yang, C.C.: A generalization of a theorem of P. Montel on entire functions, Proc. Amer. Math. Soc. 26, 332–334 (1970)
  * [48] Yang, C.C., Li, P.: On the transcendental solutions of a certain type of non-linear differential equations, Arch. Math. 82, 442–448 (2004)
  * [49] Yang, C.C., Yi, H.X.: Uniqueness Theory of Meromorphic Functions, Kluwer Acad. Publ. (2003)
  * [50] Zheng, X.M., Tu, J.: Growth of meromorphic solutions of linear difference equations, J. Math. Anal. Appl. 384, 349–356 (2011)