# On Entire solutions of system of Fermat-type 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.

The equation $f^{n}+g^{n}=1$, $n\in\mathbb{N}$ can be regarded as the Fermat
Diophantine equation over the function field. In this paper we study the
characterization of entire solutions of some system of Fermat type functional
equations by taking $e^{g_{1}(z)}$ and $e^{g_{2}(z)}$ in the right hand side
of each equation, where $g_{1}(z)$ and $g_{2}(z)$ are polynomials in
$\mathbb{C}^{n}$. Our results extend and generalize some recent results.
Moreover, some examples have been exhibited to show that our results are
precise to some extent.

###### Key words and phrases:

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

###### 1991 Mathematics Subject Classification:

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

##  1\. Introduction and Results

It is well known that Nevanlinna theory is an important tool to study value
distribution of entire and meromorphic solutions on complex differential
equations (see [14, 18]). The study of finding the entire and meromorphic
solutions with precise form of different variants Fermat type functional
equation is an interesting topic in the field of complex analysis. The Fermat
type functional equation has been originated from the Fermat type equation
$x^{n}+y^{m}=1$, where $m,n\in\mathbb{N}$, which were studied by Wiles and
Taylor [35, 36], around 1995, and they pointed out that it does not admit
nontrivial solution in rational numbers for $m=n\geq 3$, and does exist
nontrivial solution in rational numbers for $m=n=2$. Initially, Fermat-type
functional equations were investigated by Montel [27], Gross [8, 9] and Iyer
[16], independently.

Recently, the difference analogues of Nevanlinna theory, specially the
development of difference analogous lemma of the logarithmic derivative has
been established by Halburd and Korhonen [10, 11], and Chiang and Feng [4],
independently. Since then many researchers have started to study the existence
of entire and meromorphic solutions of complex difference as well as complex
differential-difference equations, and obtained a number of important and
interesting results in the literature (see [23, 25, 26, 34]).

In 2012, Liu et al. [25] proved that the entire solutions of the Fermat type
difference equation $f(z)^{2}+f(z+c)^{2}=1$ must be of the form
$f(z)=\sin(Az+B)$, where $c(\neq 0)$, $B\in\mathbb{C}$ and $A=(4k+1)\pi/2c$,
$k$ is an integer. Later in $2013$, Liu and Yang [24] extended this result by
considering the Fermat-type difference equation
$f(z)^{2}+P(z)^{2}f(z+c)^{2}=Q(z)$, where $P(z)$ and $Q(z)$ are two non-zero
polynomials. In the same paper they have also proved that the finite order
transcendental entire solutions of $f^{\prime}(z)^{2}+f(z+c)^{2}=1$ must
assumes the form $f(z)=\sin(z\pm Bi)$, where $B$ is a constant and $c=2k\pi$
or $c=(2k+1)\pi$, $k$ is an integer.

To the best of our knowledge, Gao [6], in 2016, first studied about the entire
solutions of system of differential difference equations and obtained the
result as follows. If $(f_{1},f_{2})$ be a pair of finite order transcendental
entire solution of the system of differential-difference equations

$\displaystyle\begin{cases}f_{1}^{\prime}(z)^{2}+f_{2}(z+c)^{2}=1\\\
f_{2}^{\prime}(z)^{2}+f_{1}(z+c)^{2}=1,\end{cases}$ (1.1)

then $(f_{1},f_{2})$ must be of the form
$(f_{1},f_{2})=(\sin(z-ib),\sin(z-ib_{1})),\;\;\text{or}\;\;(\sin(z+ib),\sin(z+ib_{1})),$
where $b,b_{1}$ are constants and $c=k\pi$, $k$ is a integer. In 2022, Wu-Hu
[37] characterized transcendental entire solutions for a system of Fermat type
q-differential-difference equations which generalizes the result of Gao [6].

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

It has a long history to study partial differential equations which is a
generalizations of the well-known eikonal equation in real variable case. We
insist the readers to go through [5, 7, 29] and the references therein.
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 [22, 21, 31, 32, 20]). 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 [1, 3, 17]), many mathematicians have
started to study 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 [38, 39, 41, 43, 12, 13, 40]).

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 [38] proved the following results which are the extensions and
generalizations of some previous results given by Liu et al. [25] from single
complex variable to several complex variables. We list some of them here.

######  Theorem A.

[38] 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 B.

[38] 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 $f(z_{1},z_{2})=\sin(Az_{1}+Bz_{2}+Constant)$.

Motivated be Theorems A and B, Xu et al., in 2020 [42], studied about entire
solutions for the system of the Fermat type difference as well as partial
differential difference equations in $\mathbb{C}^{2}$ and proved the following
two results.

######  Theorem C.

[42] Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$. Then any pair of
transcendental entire solutions with finite order for the system of Fermat-
type difference equations

$\displaystyle\begin{cases}f_{1}(z_{1},z_{2})^{2}+f_{2}(z_{1}+c_{1},z_{2}+c_{2})^{2}=1,\\\
f_{2}(z_{1},z_{2})^{2}+f_{1}(z_{1}+c_{1},z_{2}+c_{2})^{2}=1\end{cases}$ (2.3)

have the following forms

$\displaystyle(f_{1}(z),f_{2}(z))=\left(\frac{e^{L(z)+B_{1}}+e^{-(L(z)+B_{1})}}{2},\frac{A_{21}e^{L(z)+B_{1}}+A_{22}e^{-(L(z)+B_{1})}}{2}\right),$

where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$, $B_{1}$ is a constant in
$\mathbb{C}$, and $c,A_{21},A_{22}$ satisfy one of the following cases

  1. (i)

$L(c)=2k\pi i$, $A_{21}=-i$ and $A_{22}=i$, or $L(c)=(2k+1)\pi i$, $A_{21}=i$
and $A_{22}=-i$, here and below $k$ is an integer;

  2. (ii)

$L(c)=(2k+1/2)\pi i$, $A_{21}=-1$ and $A_{22}=-1$, or $L(c)=(2k-1/2)\pi i$,
$A_{21}=1$ and $A_{22}=1$.

######  Theorem D.

[42] Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$. Then any pair of
transcendental entire solutions with finite order for the system of Fermat-
type partial differential-difference equations

$\displaystyle\begin{cases}\left(\frac{\partial f_{1}(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+f_{2}(z_{1}+c_{1},z_{2}+c_{2})^{2}=1,\\\
\left(\frac{\partial f_{2}(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+f_{1}(z_{1}+c_{1},z_{2}+c_{2})^{2}=1\end{cases}$ (2.4)

have the following forms

$\displaystyle(f_{1}(z),f_{2}(z))=\left(\frac{e^{L(z)+B_{1}}+e^{-(L(z)+B_{1})}}{2},\frac{A_{21}e^{L(z)+B_{1}}+A_{22}e^{-(L(z)+B_{1})}}{2}\right),$

where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$, $B_{1}$ is a constant in
$\mathbb{C}$, and $c,A_{21},A_{22}$ satisfy one of the following cases

  1. (i)

$A_{21}=-i$, $A_{22}=i$, and $a_{1}=i$, $L(c)=\left(2k+\frac{1}{2}\right)\pi
i$, or $a_{1}=-i$, $L(c)=\left(2k-\frac{1}{2}\right)\pi i$;

  2. (ii)

$A_{21}=i$, $A_{22}=-i$, and $a_{1}=i$, $L(c)=\left(2k-\frac{1}{2}\right)\pi
i$, or $a_{1}=-i$, $L(c)=\left(2k+\frac{1}{2}\right)\pi i$;

  3. (iii)

$A_{21}=1$, $A_{22}=1$, and $a_{1}=i$, $L(c)=2k\pi i$, or $a_{1}=-i$,
$L(c)=\left(2k+1\right)\pi i$

  4. (iv)

$A_{21}=-1$, $A_{22}=-1$, and $a_{1}=i$, $L(c)=\left(2k+1\right)\pi ii$, or
$a_{1}=-i$, $L(c)=2k\pi i$.

Theorems A–C suggest us the following open questions.

######  Question 2.1.

Can we generalize Theorem C from $\mathbb{C}^{2}$ to $\mathbb{C}^{n}$, where
$n$ is any positive integer?

######  Question 2.2.

Can we characterize the entire solutions of the system of equations (2.4) when
the right hand sides of each of the equations of (2.4) are replaced by
$e^{g_{1}(z)}$ and $e^{g_{2}(z)}$, respectively, where $g_{1}(z)$ and
$g_{2}(z)$ are any two polynomials in $\mathbb{C}^{n}$?

Inspired by the above two questions and utilizing the Nevanlinna theory and
difference Nevanlinna theory for several complex variables (see [2, 17]), we
consider the following system of equations

$\displaystyle\begin{cases}f_{1}(z)^{2}+f_{2}(z+c)^{2}=e^{g_{1}(z)},\\\
f_{2}(z)^{2}+f_{1}(z+c)^{2}=e^{g_{2}(z)},\end{cases}$ (2.5)

and obtain Theorem 2.1 below. Before we state it, let us set the following.

$\displaystyle\Phi(z)$
$\displaystyle=\sum_{i_{1}=1}^{{}^{n}C_{2}}H_{i_{1}}^{2}(s_{i_{1}}^{2})+\sum_{i_{2}=1}^{{}^{n}C_{3}}H_{i_{2}}^{3}(s_{i_{2}}^{3})+\sum_{i_{3}=1}^{{}^{n}C_{4}}H_{i_{3}}^{4}(s_{i_{3}}^{4})+\cdots$
(2.6)
$\displaystyle+\sum_{i_{n-2}=1}^{{}^{n}C_{n-1}}H_{i_{n-2}}^{n-1}(s_{i_{n-2}}^{n-1})+H_{n-1}^{n}(s_{n-1}^{n}),$
$\displaystyle\Psi(z)$
$\displaystyle=\sum_{i_{1}=1}^{{}^{n}C_{2}}L_{i_{1}}^{2}(t_{i_{1}}^{2})+\sum_{i_{2}=1}^{{}^{n}C_{3}}L_{i_{2}}^{3}(t_{i_{2}}^{3})+\sum_{i_{3}=1}^{{}^{n}C_{4}}L_{i_{3}}^{4}(t_{i_{3}}^{4})+\cdots$
(2.7)
$\displaystyle+\sum_{i_{n-2}=1}^{{}^{n}C_{n-1}}L_{i_{n-2}}^{n-1}(t_{i_{n-2}}^{n-1})+L_{n-1}^{n}(t_{n-1}^{n}),$

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 state our first result as follows.

######  Theorem 2.1.

Let $c=(c_{1},c_{2},\ldots,c_{n})\in\mathbb{C}^{n}$ and $g_{1}(z)$, $g_{1}(z)$
be any two polynomials in $\mathbb{C}^{n}$. If $(f_{1},f_{2})$ is a pair of
transcendental entire solution with finite order of simultaneous Fermat-type
difference equation (2.5), then one of the following cases must occur.

  1. (i)

$g_{1}(z)=L(z)+\Phi(z)+B_{1}$, $g_{2}(z)=L(z)+\Phi(z)+B_{2}$, where
$L(z)=\sum_{j=1}^{n}a_{j}z_{j}$, $\Phi(z)$ is a polynomial in $\mathbb{C}^{n}$
as defined in (2.6), $a_{j},B_{1},B_{2}\in\mathbb{C}$ for $j=1,2,\ldots,n$ and

$\displaystyle
f_{1}(z)=\frac{\xi_{1}^{2}+1}{2\xi_{1}}e^{\frac{1}{2}[L(z)+\Phi(z)+B_{1}]},\;f_{2}(z)=\frac{\xi_{2}^{2}+1}{2\xi_{2}}e^{\frac{1}{2}[L(z)+\Phi(z)+B_{2}]},$

where $\xi_{j}$ is a non-zero complex number in $\mathbb{C}$ such that
$\xi_{j}^{4}\neq 1$, $j=1,2$, and

$\displaystyle
e^{L(c)}=-\frac{(\xi_{1}^{2}-1)(\xi_{2}^{2}-1)}{(\xi_{1}^{2}+1)(\xi_{2}^{2}+1)}\;\;\text{and}\;\;e^{B_{1}-B_{2}}=\frac{\xi_{1}^{2}(\xi_{2}^{4}-1)}{\xi_{2}^{2}(\xi_{1}^{4}-1)}.$

  2. (ii)

$g_{1}(z)=L(z)+R(z)+D_{1}$, $g_{2}(z)=L(z)+R(z)+D_{2}$, where
$L(z)=L_{1}(z)+L_{2}(z)$, $R(z)=\Phi(z)+\Psi(z)$, $D_{1}=B_{1}+B_{2}$,
$D_{2}=B_{3}+B_{4}$ with $L_{1}(z)+\Phi(z)\neq L_{2}(z)+\Psi(z)$,
$L_{j}(z)=a_{j1}z_{1}+a_{j2}z_{2}$, $\Phi(z),\Psi(z)$ are defined in (2.7) and
(2.7), respectively, $a_{j1},a_{j2},D_{i}\in\mathbb{C}$, $j=1,2,\ldots,n$
$i=1,2$, and

$\displaystyle
f_{1}(z)=\frac{1}{2}\left[e^{L_{1}(z)+\Phi(z)+B_{1}}+e^{L_{2}(z)+\Psi(z)+B_{3}}\right],$
$\displaystyle
f_{2}(z)=\frac{1}{2}\left[e^{L_{1}(z)+\Phi(z)+B_{2}}+e^{L_{2}(z)+\Psi(z),+B_{4}}\right],$

where $L_{1}(z),L_{2}(z)$, $B_{1},B_{2},B_{3},B_{4}$ satisfy one of the
following relations.

    1. (a)

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

    2. (b)

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

    3. (c)

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

    4. (d)

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

  3. (iii)

$g_{1}(z)=L(z)+R(z)+D_{1}$, $g_{2}(z)=L(z)+R(z)+D_{2}$, where
$L(z)=L_{1}(z)+L_{2}(z)$, $R(z)=\Psi(z)+\chi(z)$, $D_{1}=B_{1}+B_{3}$,
$D_{2}=B_{2}+B_{4}$ with $L_{1}(z)+\Phi(z)\neq L_{2}(z)+\Psi(z)$,
$L_{j}(z)=a_{j1}z_{1}+a_{j2}z_{2}$, $\Phi(z),\Psi(z)$ are defined in (2.7) and
(2.7),respectively, $a_{j1},a_{j2},D_{i}\in\mathbb{C}$, $j=1,2,\ldots,n$
$i=1,2$, and

$\displaystyle
f_{1}(z)=\frac{1}{2}\left[e^{L_{1}(z)+\Phi(z)+B_{1}}+e^{L_{2}(z)+\chi(z)+B_{3}}\right],$
$\displaystyle
f_{2}(z)=\frac{1}{2}\left[e^{L_{1}(z)+\Phi(z)+B_{2}}+e^{L_{2}(z)+\chi(z),+B_{4}}\right],$

where $L_{1}(z),L_{2}(z)$, $B_{1},B_{2},B_{3},B_{4}$ satisfy one of the
following relations.

    1. (a)

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

    2. (b)

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

    3. (c)

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

    4. (d)

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

The following examples show the existence of transcendental entire solutions
with finite order of the system (2.5).

######  Example 2.1.

Let $\xi_{1}=\xi_{2}=2$, $L(z)=z_{1}+2z_{2}$. Choose
$c=(c_{1},c_{2})\in\mathbb{C}^{2}$ such that $e^{c_{1}+2c_{2}}=-9/25$ and
$\Phi(z)=(c_{2}z_{1}-c_{1}z_{2})^{5}$. Let
$g_{1}(z_{1},z_{2})=L(z)+\Phi(z)+B_{1}$ and
$g_{2}(z_{1},z_{2})=L(z)+\Phi(z)+B_{2}$, where $B_{1},B_{2}$ are constants in
$\mathbb{C}$ such that $B_{1}-B_{2}=2\pi i$. Then in view of the conclusion
(i) of Theorem 2.1, we can see that

$\displaystyle(f_{1},f_{2})=\left(\frac{5}{4}e^{\frac{1}{2}[z_{1}+2z_{2}+(c_{2}z_{1}-c_{1}z_{2})^{3}+B_{1}]},\frac{5}{4}e^{\frac{1}{2}[z_{1}+2z_{2}+(c_{2}z_{1}-c_{1}z_{2})^{3}+B_{2}]}\right)$

is a solution of (2.5).

######  Example 2.2.

Let $c_{1}=7\pi i/5,c_{2}=-2\pi i/5$, $L_{1}(z)=z_{1}+z_{2}$,
$L_{2}(z)=2z_{1}-3z_{2}$, $\Phi(z)=-\dfrac{\pi^{2}}{25}(2z_{1}+7z_{2})^{2}$,
$\Psi(z)\equiv 0$, $B_{1}=3\pi i/2$, $B_{2}=\pi i$, $B_{3}=2\pi i$, $B_{4}=\pi
i/2$ and

$\displaystyle
f_{1}(z_{1},z_{2})=\frac{1}{2}\left[e^{z_{1}+z_{2}-\frac{\pi^{2}}{25}(2z_{1}+7z_{2})^{2}+\frac{3\pi
i}{2}}+e^{2z_{1}-3z_{2}+2\pi i}\right],$ $\displaystyle
f_{2}(z_{1},z_{2})=\frac{1}{2}\left[e^{z_{1}+z_{2}-\frac{\pi^{2}}{25}(2z_{1}+7z_{2})^{2}+\pi
i}+e^{2z_{1}-3z_{2}+\frac{\pi i}{2}}\right].$

Then in view of the conclusion (ii) of Theorem 2.1, we can see that
$(f_{1},f_{2})$ is a solution of (2.5) with
$g_{1}(z_{1},z_{2})=3z_{1}-2z_{2}-\dfrac{\pi^{2}}{25}(2z_{1}+7z_{2})^{2}+\dfrac{7\pi
i}{2}$ and
$g_{2}(z_{1},z_{2})=3z_{1}-2z_{2}-\dfrac{\pi^{2}}{25}(2z_{1}+7z_{2})^{2}+\dfrac{3\pi
i}{2}$

In order to generalize Theorem D to a large extent, let us consider the
following system of equations.

$\displaystyle\begin{cases}\left(\frac{\partial^{k}f_{1}(z_{1},z_{2})}{\partial
z_{1}^{k}}\right)^{2}+f_{2}(z_{1}+c_{1},z_{2}+c_{2})^{2}=e^{g_{1}(z_{1},z_{2})},\vspace{1mm}\\\
\left(\frac{\partial^{k}f_{2}(z_{1},z_{2})}{\partial
z_{1}^{k}}\right)^{2}+f_{1}(z_{1}+c_{1},z_{2}+c_{2})^{2}=e^{g_{2}(z_{1},z_{2})},\end{cases}$
(2.8)

where $e^{g_{1}(z_{1},z_{2})}$, $e^{g_{2}(z_{1},z_{2})}$ are any two
polynomials in $\mathbb{C}^{2}$ and obtain the following result.

######  Theorem 2.2.

Let $c=(c_{1},c_{2})\in\mathbb{C}^{2}$ and $g_{1}(z_{1},z_{2})$,
$g_{1}(z_{1},z_{2})$ be any two polynomials in $\mathbb{C}^{2}$. If
$(f_{1},f_{2})$ is a pair of transcendental entire solution with finite order
of simultaneous Fermat type difference equation (2.8), then one of the
following cases must occur.

  1. (i)

$g_{1}(z_{1},z_{2})=L(z)+H(s)+B_{1}$, $g_{2}(z_{1},z_{2})=L(z)+H(s)+B_{2}$,
where $L(z)=a_{1}z_{1}+a_{2}z_{2}$, $H(s)$ is a polynomial in $s:=d_{2}z_{2}$
with $d_{2}c_{2}=0$, $a_{1},a_{2},B_{1},B_{2}$ constants in $\mathbb{C}$, and

$\displaystyle
f_{1}(z_{1},z_{2})=\frac{\xi_{2}^{2}-1}{2i\xi_{2}}e^{\frac{1}{2}[L(z)+H(s)+B_{2}-L(c)]},\;f_{2}(z)=\frac{\xi_{1}^{2}-1}{2i\xi_{1}}e^{\frac{1}{2}[L(z)+H(s)+B_{1}-L(c)]},$

where $\xi_{j}$ is a non-zero complex number in $\mathbb{C}$ such that
$\xi_{j}^{4}\neq 0,1$, $j=1,2$, and

$\displaystyle
e^{L(c)}=-\left(\frac{a_{1}}{2}\right)^{k}\frac{(\xi_{1}^{2}-1)(\xi_{2}^{2}-1)}{(\xi_{1}^{2}+1)(\xi_{2}^{2}+1)}\;\;\text{and}\;\;e^{B_{1}-B_{2}}=\frac{\xi_{1}^{2}(\xi_{2}^{4}-1)}{\xi_{2}^{2}(\xi_{1}^{4}-1)}.$

  2. (ii)

$g_{1}(z_{1},z_{2})=L_{1}(z)+L_{2}(z)+H_{1}(s)+H_{2}(s)+B_{1}+B_{3}$,
$g_{2}(z_{1},z_{2})=L_{1}(z)+L_{2}(z)+H_{1}(s)+H_{2}(s)+B_{2}+B_{4}$, where
$L_{1}(z)+H_{1}(s)\neq L_{2}(z)+H_{2}(s)$, $L_{j}(z)=a_{j1}z_{1}+a_{j2}z_{2}$,
$H_{j}(s)$ is a polynomial in $s:=d_{2}z_{2}$ with $d_{2}c_{2}=0$,
$a_{11},a_{12},a_{21},a_{22},B_{1},B_{2},B_{3},B_{4}\in\mathbb{C}$, and

$\displaystyle
f_{1}(z_{1},z_{2})=\frac{1}{2i}\left[e^{L_{1}(z)+H_{1}(s)-L_{1}(c)+B_{2}}-e^{L_{2}(z)+H_{2}(s)-L_{2}(c)+B_{4}}\right],$
$\displaystyle
f_{2}(z_{1},z_{2})=\frac{1}{2i}\left[e^{L_{1}(z)+H_{1}(s)-L_{1}(c)+B_{1}}-e^{L_{2}(z)+H_{2}(s)-L_{2}(c)+B_{3}}\right],$

where $L_{1}(z),L_{2}(z)$, $B_{1},B_{2},B_{3},B_{4}$ satisfies one of the
following relations

    1. (a)

$e^{L_{1}(c)}=ia_{11}^{k}$, $e^{L_{2}(c)}=ia_{21}^{k}$, $e^{B_{1}-B_{2}}=-1$,
$e^{B_{3}-B_{4}}=1$.

    2. (b)

$e^{L_{1}(c)}=ia_{11}^{k}$, $e^{L_{2}(c)}=-ia_{21}^{k}$, $e^{B_{1}-B_{2}}=-1$,
$e^{B_{3}-B_{4}}=-1$.

    3. (c)

$e^{L_{1}(c)}=-ia_{11}^{k}$, $e^{L_{2}(c)}=ia_{21}^{k}$, $e^{B_{1}-B_{2}}=1$,
$e^{B_{3}-B_{4}}=1$.

    4. (d)

$e^{L_{1}(c)}=-ia_{11}^{k}$, $e^{L_{2}(c)}=-ia_{21}^{k}$, $e^{B_{1}-B_{2}}=1$,
$e^{B_{3}-B_{4}}=-1$.

  3. (iii)

$g_{1}(z_{1},z_{2})=L_{1}(z)+L_{2}(z)+H_{1}(s)+H_{2}(s)+B_{1}+B_{3}$,
$g_{2}(z_{1},z_{2})=L_{1}(z)+L_{2}(z)+H_{1}(s)+H_{2}(s)+B_{2}+B_{4}$, where
$L_{1}(z)+H_{1}(s)\neq L_{2}(z)+H_{2}(s)$, $L_{j}(z)=a_{j1}z_{1}+a_{j2}z_{2}$,
$H_{j}(s)$ is a polynomial in $s:=d_{2}z_{2}$ with $d_{2}c_{2}=0$,
$a_{11},a_{12},a_{21},a_{22},B_{1},B_{2},B_{3},B_{4}\in\mathbb{C}$, and

$\displaystyle
f_{1}(z_{1},z_{2})=-\frac{1}{2i}\left[e^{L_{1}(z)+H_{1}(s)-L_{1}(c)+B_{2}}-e^{L_{2}(z)+H_{2}(s)-L_{2}(c)+B_{4}}\right],$
$\displaystyle
f_{2}(z_{1},z_{2})=\frac{1}{2i}\left[e^{L_{1}(z)+H_{1}(s)-L_{1}(c)+B_{1}}-e^{L_{2}(z)+H_{2}(s)-L_{2}(c)+B_{3}}\right],$

where $L_{1}(z),L_{2}(z)$, $B_{1},B_{2},B_{3},B_{4}$ satisfies one of the
following relations

    1. (a)

$e^{L_{1}(c)}=a_{11}^{k}$, $e^{L_{2}(c)}=a_{21}^{k}$, $e^{B_{1}-B_{2}}=i$,
$e^{B_{3}-B_{4}}=-i$.

    2. (b)

$e^{L_{1}(c)}=a_{11}^{k}$, $e^{L_{2}(c)}=-a_{21}^{k}$, $e^{B_{1}-B_{2}}=i$,
$e^{B_{3}-B_{4}}=i$.

    3. (c)

$e^{L_{1}(c)}=-a_{11}^{k}$, $e^{L_{2}(c)}=a_{21}^{k}$, $e^{B_{1}-B_{2}}=-i$,
$e^{B_{3}-B_{4}}=-i$.

    4. (d)

$e^{L_{1}(c)}=-a_{11}^{k}$, $e^{L_{2}(c)}=-a_{21}^{k}$, $e^{B_{1}-B_{2}}=-i$,
$e^{B_{3}-B_{4}}=i$.

The following examples show the existence of transcendental entire solutions
with finite order of the system (2.8).

######  Example 2.3.

Let $\xi_{1}=\xi_{2}=2$, $L(z)=2z_{1}+z_{2}$,
$g_{1}(z_{1},z_{2})=g_{2}(z_{1},z_{2})=L(z)+\pi i$. Choose
$c=(c_{1},c_{2})\in\mathbb{C}^{2}$ such that $e^{L(c)}=-9/25$. Then in view of
the conclusion $(i)$ of Theorem 2.2, it follows that $(f_{1}(z),f_{2}(z))$ is
a solution of

$\displaystyle\begin{cases}\left(\frac{\partial f_{1}(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+f_{2}(z_{1}+c_{1},z_{2}+c_{2})^{2}=e^{g_{1}(z_{1},z_{2})},\vspace{1mm}\\\
\left(\frac{\partial f_{2}(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+f_{1}(z_{1}+c_{1},z_{2}+c_{2})^{2}=e^{g_{2}(z_{1},z_{2})},\end{cases},$

where

$\displaystyle f_{1}(z)=f_{2}(z)=-\frac{5i}{4}e^{\frac{1}{2}[2z_{1}+z_{2}]}.$

######  Example 2.4.

Let $k=2$, $L_{1}(z)=z_{1}+z_{2}$, $L_{2}(z)=2z_{1}+z_{2}$,
$(c_{1},c_{2})=(2\log 2,\pi i/2-2\log 2)$, $B_{1}=2\pi i$, $B_{2}=\pi i$,
$B_{3}=B_{4}=\pi i$. Then in view of the conclusion $(ii)(a)$ of Theorem 2.2,
we can easily see that

$\displaystyle(f_{1}(z),f_{2}(z))=\left(\frac{1}{8}\left(4e^{z_{1}+z_{2}}-e^{2z_{1}+z_{2}}\right),-\frac{1}{8}\left(4e^{z_{1}+z_{2}}+e^{2z_{1}+z_{2}}\right)\right)$

is a solution of

$\displaystyle\begin{cases}\left(\frac{\partial^{2}f_{1}(z_{1},z_{2})}{\partial
z_{1}^{2}}\right)^{2}+f_{2}(z_{1}+c_{1},z_{2}+c_{2})^{2}=e^{g_{1}(z_{1},z_{2})},\vspace{1mm}\\\
\left(\frac{\partial^{2}f_{2}(z_{1},z_{2})}{\partial
z_{1}^{2}}\right)^{2}+f_{1}(z_{1}+c_{1},z_{2}+c_{2})^{2}=e^{g_{2}(z_{1},z_{2})},\end{cases},$

where $g_{1}(z_{1},z_{2})=3z_{1}+2z_{2}+3\pi i$ and
$g_{1}(z_{1},z_{2})=3z_{1}+2z_{2}+2\pi i$

######  Example 2.5.

Let $k=1$, $L_{1}(z)=2z_{1}+3z_{2}$, $L_{2}(z)=2z_{1}-z_{2}$,
$(c_{1},c_{2})=(\frac{1}{2}\log 2,0)$, $B_{1}=\pi i$, $B_{2}=\pi i/2$,
$B_{3}=\pi i/2$, $B_{4}=\pi i$. Then in view of the conclusion $(iii)(a)$ of
Theorem 2.2, we can easily see that

$\displaystyle(f_{1}(z),f_{2}(z))=\left(-\frac{1}{4i}\left(ie^{2z_{1}+3z_{2}+z_{2}^{3}}+e^{2z_{1}-z_{2}+z_{2}^{5}}\right),-\frac{1}{4i}\left(e^{2z_{1}+3z_{2}+z_{2}^{3}}+ie^{2z_{1}-z_{2}+z_{2}^{5}}\right)\right)$

is a solution of

$\displaystyle\begin{cases}\left(\frac{\partial f_{1}(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+f_{2}(z_{1}+c_{1},z_{2}+c_{2})^{2}=e^{g_{1}(z_{1},z_{2})},\vspace{1mm}\\\
\left(\frac{\partial f_{2}(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+f_{1}(z_{1}+c_{1},z_{2}+c_{2})^{2}=e^{g_{2}(z_{1},z_{2})},\end{cases},$

where
$g_{1}(z_{1},z_{2})=g_{1}(z_{1},z_{2})=4z_{1}+2z_{2}+z_{2}^{3}+z_{2}^{5}+3\pi
i/2$.

Beside these, we are interested to characterize the entire solutions of the
following system of $k$-th partial differential difference equations

$\displaystyle\begin{cases}\left(\frac{\partial^{k}f_{1}(z_{1},z_{2})}{\partial
z_{1}^{k}}\right)^{2}+(f_{2}(z_{1}+c_{1},z_{2}+c_{2})-f_{2}(z_{1},z_{2}))^{2}=1,\vspace{1mm}\\\
\left(\frac{\partial^{k}f_{2}(z_{1},z_{2})}{\partial
z_{1}^{k}}\right)^{2}+(f_{1}(z_{1}+c_{1},z_{2}+c_{2})-f_{1}(z_{1},z_{2}))^{2}=1,\end{cases}$
(2.9)

which has not been discussed earlier in the literature. Regarding this, we
obtain the following result.

######  Theorem 2.3.

Let $c$ be a non-zero constant in $\mathbb{C}^{2}$, $k$ be a positive integer
and $f_{1}(z),f_{2}(z)$ are transcendental entire functions of finite order
not of $c$-periodic. If $(f_{1}(z),f_{2}(z))$ be a solution with finite order
of (2.9), then one of the following conclusions must hold.

  1. (i)

$\displaystyle
f_{1}(z_{1},z_{2})=\frac{e^{L(z)+H(z_{2})+\beta}-e^{-(L(z)+H(z_{2})+\beta)}}{2\alpha_{1}^{k}},$
$\displaystyle
f_{2}(z_{1},z_{2})=\frac{e^{L(z)+H(z_{2})+\beta+\eta}-e^{-(L(z)+H(z_{2})+\beta+\eta)}}{2\alpha_{1}^{k}}.$

where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$, $H(z_{2})$ is a polynomial in
$z_{2}$, only, $k$ is an odd positive integer, $\alpha_{1}(\neq
0),\alpha_{2},\beta,\eta\in\mathbb{C}$ and satisfy the following relations:

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

  2. (ii)

$\displaystyle
f_{1}(z_{1},z_{2})=\frac{e^{L(z)+H(z_{2})+\beta}+e^{-(L(z)+H(z_{2})+\beta)}}{2\alpha_{1}^{k}},$
$\displaystyle
f_{2}(z_{1},z_{2})=\frac{e^{-(L(z)+H(z_{2})+\beta)+\eta}+e^{L(z)+H(z_{2})+\beta-\eta}}{2\alpha_{1}^{k}}.$

where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$, $H(z_{2})$ is a polynomial in
$z_{2}$, only $k$ is an even positive integer, $\alpha_{1}(\neq
0),\alpha_{2},\beta,\eta\in\mathbb{C}$ and satisfy the following relations:

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

The following examples show that transcendental entire solutions of (2.9) are
precise.

######  Example 2.6.

Let $k=1$, $\eta=4\pi i$ and $\alpha_{1}=-2i$. Choose
$c=(c_{1},c_{2})\in\mathbb{C}^{2}$ such that $-2ic_{1}+c_{2}=(2m+\pi i)$, $m$
being an integer. Then, it can be easily verified that

$\displaystyle(f_{1}(z),f_{2}(z))=\left(\frac{e^{2iz_{1}-z_{2}-z_{2}^{10}}-e^{-2iz_{1}+z_{2}+z_{2}^{10}}}{4i},\frac{e^{2iz_{1}-z_{2}-z_{2}^{10}}-e^{-2iz_{1}+z_{2}+z_{2}^{10}}}{4i}\right)$

is a solution of

$\displaystyle\begin{cases}\left(\frac{\partial f_{1}}{\partial
z_{1}}\right)^{2}+[f_{2}(z+c)-f_{2}(z)]^{2}=1,\\\ \left(\frac{\partial
f_{2}}{\partial z_{1}}\right)^{2}+[f_{1}(z+c)-f_{1}(z)]^{2}=1.\end{cases}$

######  Example 2.7.

Let $k=2$, $\eta=5\pi i/2$. Choose $\alpha_{1}\in\mathbb{C}$ such that
$\alpha_{1}^{2}=-2$ and $c=(c_{1},c_{2})\in\mathbb{C}^{2}$ such that
$\alpha_{1}c_{1}+c_{2}+1=(2m+1)\pi i$, $m\in\mathbb{Z}$. Then one can see that

$\displaystyle(f_{1}(z),f_{2}(z))=\left(\frac{e^{\alpha_{1}z_{1}+z_{2}+1}+e^{(-\alpha_{1}z_{1}+z_{2}+1)}}{-4},\frac{e^{\alpha_{1}z_{1}+z_{2}+1}-e^{-(\alpha_{1}z_{1}+z_{2}+1)}}{-4i}\right)$

is a solution of

$\displaystyle\begin{cases}\left(\frac{\partial^{2}f_{1}}{\partial
z_{1}^{2}}\right)^{2}+[f_{2}(z+c)-f_{2}(z)]^{2}=1,\vspace{.5mm}\\\
\left(\frac{\partial^{2}f_{2}}{\partial
z_{1}^{2}}\right)^{2}+[f_{1}(z+c)-f_{1}(z)]^{2}=1.\end{cases}$

##  3\. Proof of the main results

Before we start the proof of the main results, we present some important
lemmas which will play key role to prove the main results of the paper.

######  Lemma 3.1.

[15] Let $f_{j}\not\equiv 0$ $(j=1,2\ldots,m;\;m\geq 3)$ be meromorphic
functions on $\mathbb{C}^{n}$ 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 3.2.

[19, 30, 33] For an entire function $F$ on $\mathbb{C}^{n}$, $F(0)\not\equiv
0$ and put $\rho(n_{F})=\rho<\infty$. Then there exist a canonical function
$f_{F}$ and a function $g_{F}\in\mathbb{C}^{n}$ such that
$F(z)=f_{F}(z)e^{g_{F}(z)}$. For the special case $n=1$, $f_{F}$ is the
canonical product of Weierstrass.

######  Lemma 3.3.

[28] If $g$ and $h$ are entire functions on the complex plane $\mathbb{C}$ and
$g(h)$ is an entire function of finite order, then there are only two possible
cases: either

  1. (i)

the internal function $h$ is a polynomial and the external function $g$ is of
finite order; or else

  2. (ii)

the internal function $h$ is not a polynomial but a function of finite order,
and the external function $g$ is of zero order.

######  Lemma 3.4.

[15] Let $f_{j}\not\equiv 0$ $(j=1,2,3)$ be meromorphic functions on
$\mathbb{C}^{n}$ 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 3.5.

[15] Let $a_{0}(z),a_{1}(z),\ldots,a_{n}(z)$ $(n\geq 1)$ be meromorphic
functions on $\mathbb{C}^{m}$ and $g_{0}(z),g_{1}(z),\ldots,g_{n}(z)$ are
entire functions on $\mathbb{C}^{m}$ 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 2.1.

Suppose that $(f_{1},f_{2})$ is a pair of transcendental entire functions of
finite order satisfying system (2.5). First we write (2.5) as follows.

$\displaystyle\begin{cases}\left[\frac{f_{1}(z)}{e^{\frac{g_{1}(z)}{2}}}+i\frac{f_{2}(z+c)}{e^{\frac{g_{1}(z)}{2}}}\right]\left[\frac{f_{1}(z)}{e^{\frac{g_{1}(z)}{2}}}-i\frac{f_{2}(z+c)}{e^{\frac{g_{1}(z)}{2}}}\right]=1,\vspace{1mm}\\\
\left[\frac{f_{2}(z)}{e^{\frac{g_{2}(z)}{2}}}+i\frac{f_{1}(z+c)}{e^{\frac{g_{2}(z)}{2}}}\right]\left[\frac{f_{2}(z)}{e^{\frac{g_{2}(z)}{2}}}-i\frac{f_{1}(z+c)}{e^{\frac{g_{2}(z)}{2}}}\right]=1.\end{cases}$
(3.1)

Since $f_{1}$, $f_{2}$ are transcendental entire functions with finite order,
in view of Lemmas 3.2 and 3.3, there exist polynomials $p_{1}(z)$, $p_{2}(z)$
in $\mathbb{C}^{n}$ such that

$\displaystyle\begin{cases}\frac{f_{1}(z)}{e^{\frac{g_{1}(z)}{2}}}+i\frac{f_{2}(z+c)}{e^{\frac{g_{1}(z)}{2}}}=e^{p_{1}(z)},\vspace{1mm}\\\
\frac{f_{1}(z)}{e^{\frac{g_{1}(z)}{2}}}-i\frac{f_{2}(z+c)}{e^{\frac{g_{1}(z)}{2}}}=e^{-p_{1}(z)},\vspace{1mm}\\\
\frac{f_{2}(z)}{e^{\frac{g_{2}(z)}{2}}}+i\frac{f_{1}(z+c)}{e^{\frac{g_{2}(z)}{2}}}=e^{p_{2}(z)},\vspace{1mm}\\\
\frac{f_{2}(z)}{e^{\frac{g_{2}(z)}{2}}}-i\frac{f_{1}(z+c)}{e^{\frac{g_{2}(z)}{2}}}=e^{-p_{2}(z)}.\end{cases}$
(3.2)

Set

$\displaystyle\begin{cases}\alpha_{1}=\frac{1}{2}g_{1}(z)+p_{1}(z),\;\;\alpha_{2}=\frac{1}{2}g_{1}(z)-p_{1}(z),\\\
\alpha_{3}=\frac{1}{2}g_{2}(z)+p_{2}(z),\;\;\alpha_{4}=\frac{1}{2}g_{2}(z)-p_{2}(z).\end{cases}$
(3.3)

In view of (3.2) and (3.3), it follows that

$\displaystyle\begin{cases}f_{1}(z)=\displaystyle\frac{1}{2}\left[e^{\alpha_{1}(z)}+e^{\alpha_{2}(z)}\right]\\\
f_{2}(z+c)=\displaystyle\frac{1}{2i}\left[e^{\alpha_{1}(z)}-e^{\alpha_{2}(z)}\right]\\\
f_{2}(z)=\displaystyle\frac{1}{2}\left[e^{\alpha_{3}(z)}+e^{\alpha_{4}(z)}\right]\\\
f_{1}(z+c)=\displaystyle\frac{1}{2i}\left[e^{\alpha_{3}(z)}-e^{\alpha_{4}(z)}\right].\end{cases}$
(3.4)

After simple calculation, it follows from (3.4) that

$\displaystyle\begin{cases}ie^{\alpha_{1}(z+c)-\alpha_{3}(z)}+ie^{\alpha_{2}(z+c)-\alpha_{3}(z)}+e^{\alpha_{4}(z)-\alpha_{3}(z)}=1,\\\
ie^{\alpha_{3}(z+c)-\alpha_{1}(z)}+ie^{\alpha_{4}(z+c)-\alpha_{1}(z)}+e^{\alpha_{2}(z)-\alpha_{1}(z)}=1.\end{cases}$
(3.5)

Now, we consider the following four possible cases.

Case 1: Let $\alpha_{4}(z)-\alpha_{3}(z)=\eta_{1}$ and
$\alpha_{2}(z)-\alpha_{1}(z)=\eta_{2}$, where $\eta_{1},\eta_{2}$ are
constants in $\mathbb{C}$.

Then, it clearly follows from (3.4) 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, (3.4) can be rewritten as

$\displaystyle\begin{cases}f_{1}(z)=\frac{\xi_{1}+\xi_{1}^{-1}}{2}e^{\frac{1}{2}g_{1}(z)},\\\
f_{2}(z+c)=\frac{\xi_{1}-\xi_{1}^{-1}}{2i}e^{\frac{1}{2}g_{1}(z)},\\\
f_{2}(z)=\frac{\xi_{2}+\xi_{2}^{-1}}{2}e^{\frac{1}{2}g_{2}(z)},\\\
f_{1}(z+c)=\frac{\xi_{2}-\xi_{2}^{-1}}{2i}e^{\frac{1}{2}g_{2}(z)}.\end{cases}$
(3.6)

After simple computations, it follows from (3.6) that

$\displaystyle\begin{cases}i(\xi_{1}+\xi_{1}^{-1})e^{\frac{1}{2}[g_{1}(z+c)-g_{2}(z)]}=\xi_{2}-\xi_{2}^{-1}\\\
i(\xi_{2}+\xi_{2}^{-1})e^{\frac{1}{2}[g_{2}(z+c)-g_{1}(z)]}=\xi_{1}-\xi_{1}^{-1}.\end{cases}$
(3.7)

Since $\xi_{j}^{4}\neq 0,1$ and $g_{j}(z_{1},z_{2})$ are polynomials in
$\mathbb{C}^{n}$ for $=1,2$, it follows from (3.7) that
$g_{1}(z+c)-g_{2}(z)=\zeta_{1}$ and $g_{2}(z+c)-g_{1}(z)=\zeta_{2}$, where
$\zeta_{1},\zeta_{2}\in\mathbb{C}$. This implies that
$g_{1}(z+2c)-g_{2}(z)=g_{2}(z+2c)-g_{1}(z)=\zeta_{1}+\zeta_{2}$. Thus,
$g_{1}(z)=L(z)+\Phi(z)+B_{1}$ and $g_{2}(z)=L(z)+\Phi(z)+B_{2}$, where
$L(z)=\sum_{j=1}^{n}a_{j}z_{j}$, $\Phi(z)$ is a polynomial in $\mathbb{C}^{n}$
defined in (2.6). Thus, from (3.7), we get

$\displaystyle\begin{cases}e^{\frac{1}{2}[L(c)+B_{1}-B_{2}]}=\frac{\xi_{2}-\xi_{2}^{-1}}{i(\xi_{1}+\xi_{1}^{-1})}\\\
e^{\frac{1}{2}[L(c)+B_{2}-B_{1}]}=\frac{\xi_{1}-\xi_{1}^{-1}}{i(\xi_{2}+\xi_{2}^{-1})}.\end{cases}$
(3.8)

Therefore, from (3.8), we obtain

$\displaystyle
e^{L(c)}=-\frac{(\xi_{1}^{2}-1)(\xi_{2}^{2}-1)}{(\xi_{1}^{2}+1)(\xi_{2}^{2}+1)}\;\;\text{and}\;\;e^{B_{1}-B_{2}}=\frac{\xi_{1}^{2}(\xi_{2}^{4}-1)}{\xi_{2}^{2}(\xi_{1}^{2}+1)}.$

Case 2. Let $\alpha_{4}(z)-\alpha_{3}(z)$ and $\alpha_{2}(z)-\alpha_{1}(z)$
both are non-constants. Then, by Lemma 3.4, it follows from (3.5) that

$\displaystyle
ie^{\alpha_{1}(z+c)-\alpha_{3}(z)}=1,\;\;\text{or}\;\;ie^{\alpha_{2}(z+c)-\alpha_{3}(z)}=1$

and

$\displaystyle
ie^{\alpha_{3}(z+c)-\alpha_{1}(z)}=1,\;\;\text{or}\;\;ie^{\alpha_{4}(z+c)-\alpha_{1}(z)}=1.$

Now, we consider the following subcases.

Subcase 2.1. Let

$\displaystyle
ie^{\alpha_{1}(z+c)-\alpha_{3}(z)}=1\;\;\text{and}\;\;ie^{\alpha_{3}(z+c)-\alpha_{1}(z)}=1.$
(3.9)

Then, from (3.5) and (3.9), we obtain

$\displaystyle-
ie^{\alpha_{2}(z+c)-\alpha_{4}(z)}=1\;\;\text{and}\;\;-ie^{\alpha_{4}(z+c)-\alpha_{1}(z)}=1.$
(3.10)

Since $\alpha_{j}$’s, $j=1,2,3,4$ are all polynomials in $\mathbb{C}^{n}$, it
follows from (3.9) and (3.10) that $\alpha_{1}(z+c)-\alpha_{3}(z)=\eta_{1}$,
$\alpha_{3}(z+c)-\alpha_{1}(z)=\eta_{2}$,
$\alpha_{2}(z+c)-\alpha_{4}(z)=\eta_{3}$ and
$\alpha_{4}(z+c)-\alpha_{2}(z)=\eta_{4}$, $\eta_{j}\in\mathbb{C}$ for
$j=1,2,3,4$. This implies that
$\alpha_{1}(z+2c)-\alpha_{1}(z)=\eta_{1}+\eta_{2}=\alpha_{3}(z+2c)-\alpha_{3}(z)$
and
$\alpha_{2}(z+2c)-\alpha_{2}(z)=\eta_{3}+\eta_{4}=\alpha_{4}(z+2c)-\alpha_{4}(z)$.
Thus, $\alpha_{1}(z)=L_{1}(z)+\Phi(z)+B_{1}$,
$\alpha_{3}(z)=L_{1}(z)+\Phi(z)+B_{2}$, $\alpha_{2}(z)=L_{2}(z)+\Psi(z)+B_{3}$
and $\alpha_{4}(z)=L_{2}(z)+\Psi(z)+B_{4}$, where
$L_{i}(z)=\sum_{j=1}^{n}a_{ij}z_{j}$, $i=1,2$ and $\Phi(z),\Psi(z)$ are
polynomials in $\mathbb{C}^{n}$ defined as in (2.6) and (2.7), respectively,
$a_{1j},a_{2j},B_{i}\in\mathbb{C}$ for $j=1,2,\ldots,n$ and $i=1,\ldots,4$.
Thus, $g_{1}(z)=L(z)+R(z)+D_{1}$ and $g_{2}(z)=L(z)+R(z)+D_{2}$, where
$L(z)=L_{1}(z)+L_{2}(z)$, $R(z)=\Phi(z)+\Psi(z)$, $D_{1}=B_{1}+B_{2}$ and
$D_{2}=B_{3}+B_{4}$.

As $\alpha_{4}-\alpha_{3}$ and $\alpha_{2}-\alpha_{1}$ both are non-constants,
it follows that $L_{1}(z)+\Phi(z)\neq L_{2}(z)+\Psi(z)$. Therefore, from (3.9)
and (3.10), we obtain that

$\displaystyle\begin{cases}ie^{L_{1}(c)+B_{1}-B_{2}}=1,\;\;ie^{L_{1}(c)+B_{2}-B_{1}}=1,\\\
ie^{L_{2}(c)+B_{3}-B_{4}}=-1,\;\;ie^{L_{2}(c)+B_{4}-B_{2}}=-1.\end{cases}$
(3.11)

After simple calculation, it follows from (3.11) that

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

From the above relations, we have $e^{L_{1}(c)}=\pm i$, $e^{L_{2}(c)}=\pm i$,
$e^{B_{1}-B_{2}}=\pm 1$ and $e^{B_{3}-B_{4}}=\pm 1$. Therefore, in view of
(3.11), we obtain the following four possible relations.

  1. (a)

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

  2. (b)

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

  3. (c)

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

  4. (d)

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

Hence, it follows from (3.4) that

$\displaystyle\begin{cases}f_{1}(z)=\frac{1}{2}\left[e^{L_{1}(z)+\Phi(z)+B_{1}}+e^{L_{2}(z)+\Psi(z)+B_{3}}\right],\vspace{1mm}\\\
f_{2}(z)=\frac{1}{2}\left[e^{L_{1}(z)+\Phi(z)+B_{2}}+e^{L_{2}(z)+\Psi(z)+B_{4}}\right].\end{cases}$

Subcase 2.2. Let

$\displaystyle
ie^{\alpha_{1}(z+c)-\alpha_{3}(z)}=1\;\;\text{and}\;\;ie^{\alpha_{4}(z+c)-\alpha_{3}(z)}=1.$
(3.12)

Therefore, it follows from (3.5) and (3.12) that

$\displaystyle
ie^{\alpha_{2}(z+c)-\alpha_{4}(z)}=-1\;\;\text{and}\;\;ie^{\alpha_{3}(z+c)-\alpha_{2}(z)}=-1.$
(3.13)

As all $\alpha_{j}$’s are polynomials in $\mathbb{C}^{n}$, it follows from
(3.12) and (3.13) that $\alpha_{1}(z+c)-\alpha_{3}(z)=\zeta_{1}$,
$\alpha_{4}(z+c)-\alpha_{3}(z)=\zeta_{2}$,
$\alpha_{2}(z+c)-\alpha_{4}(z)=\zeta_{3}$ and
$\alpha_{3}(z+c)-\alpha_{2}(z)=\zeta_{4}$, $\zeta_{j}\in\mathbb{C}$ for
$j=1,2,3,4$. This implies that
$\alpha_{1}(z+2c)-\alpha_{2}(z)=\zeta_{1}+\zeta_{4}$ and
$\alpha_{2}(z+2c)-\alpha_{1}(z)=\zeta_{2}+\zeta_{3}$. Thus,
$\alpha_{1}(z+4c)-\alpha_{1}(z)=\alpha_{2}(z+4c)-\alpha_{2}(z)=\sum_{j=1}^{4}\zeta_{j}$.
Hence, we can get that $\alpha_{1}(z)=L(z)+\Phi(z)+k_{1}$ and
$\alpha_{2}(z)=L(z)+\Phi(z)+k_{2}$, where $L(z)=\sum_{j=1}^{n}a_{j}z_{j}$,
$\Phi(z)$ is defined as in (2.6), $a_{j},k_{1},k_{2}\in\mathbb{C}$ for
$j=1,2,\ldots,n$. But, the we must obtain that
$\alpha_{2}(z)-\alpha_{1}(z)=k_{2}-k_{1}\in\mathbb{C}$, a contradiction.

Subcase 2.3. Let

$\displaystyle
ie^{\alpha_{2}(z+c)-\alpha_{3}(z)}=1\;\;\text{and}\;\;ie^{\alpha_{3}(z+c)-\alpha_{3}(z)}=1.$

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

Subcase 2.4. Let

$\displaystyle
ie^{\alpha_{2}(z+c)-\alpha_{3}(z)}=1\;\;ie^{\alpha_{4}(z+c)-\alpha_{1}(z)}=1.$
(3.14)

In view of (3.5) and (3.14), it follows that

$\displaystyle
ie^{\alpha_{1}(z+c)-\alpha_{4}(z)}=-1\;\;\text{and}\;\;ie^{\alpha_{3}(z+c)-\alpha_{2}(z)}=-1.$
(3.15)

Since $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$ are all polynomials in
$\mathbb{C}^{n}$, it follows from (3.14) and (3.15) that
$\alpha_{2}(z+c)-\alpha_{3}(z)=l_{1}$, $\alpha_{4}(z+c)-\alpha_{1}(z)=l_{2}$,
$\alpha_{1}(z+c)-\alpha_{4}(z)=l_{3}$ and
$\alpha_{3}(z+c)-\alpha_{2}(z)=l_{4}$, $l_{j}\in\mathbb{C}$ for $j=1,2,3,4$.
This implies that
$\alpha_{2}(z+2c)-\alpha_{2}(z)=\alpha_{3}(z+2c)-\alpha_{3}(z)=l_{1}+l_{4}$
and
$\alpha_{4}(z+2c)-\alpha_{4}(z)=\alpha_{1}(z+2c)-\alpha_{1}(z)=l_{2}+l_{3}$.
Therefore, we must have $\alpha_{2}(z)=L_{1}(z)+\Phi(z)+B_{1}$,
$\alpha_{3}(z)=L_{1}(z)+\Phi(z)+B_{2}$, $\alpha_{1}(z)=L_{2}(z)+\Psi(z)+B_{3}$
and $\alpha_{4}(z)=L_{2}(z)+\Psi(z)+B_{4}$, where
$L_{i}(z)=\sum_{j=1}^{n}a_{ij}z_{j}$, $i=1,2$ and $\Phi(z),\Psi(z)$ are
polynomials in $\mathbb{C}^{n}$ defined as in (2.6) and (2.7), respectively,
$a_{1j},a_{2j},B_{i}\in\mathbb{C}$ for $j=1,2,\ldots,n$ and $i=1,\ldots,4$.
Thus, $g_{1}(z)=L(z)+R(z)+D_{1}$ and $g_{2}(z)=L(z)+R(z)+D_{2}$, where
$L(z)=L_{1}(z)+L_{2}(z)$, $R(z)=\Phi(z)+\Psi(z)$, $D_{1}=B_{1}+B_{3}$ and
$D_{2}=B_{2}+B_{4}$.

Hence, from (3.14) and (3.15), we get

$\displaystyle\begin{cases}ie^{L_{1}(c)+B_{1}-B_{2}}=1,\;\;ie^{L_{2}(c)+B_{4}-B_{3}}=1,\\\
ie^{L_{2}(c)+B_{3}-B_{4}}=-1\;\;ie^{L_{1}(c)+B_{2}-B_{1}}=-1.\end{cases}$
(3.16)

After simple computation, it follows from (3.16) that

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

Therefore, form the above relation, we get $e^{L_{1}(c)}=\pm 1$,
$e^{L_{2}(c)}=\pm 1$, $e^{B_{1}-B_{2}}=\pm i$ and $e^{B_{3}-B_{4}}=\pm i$.
Therefore, in view of (3.11), we obtain the following four possible relations.

  1. (a)

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

  2. (b)

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

  3. (c)

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

  4. (d)

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

Hence, it follows from (3.4) that

$\displaystyle\begin{cases}f_{1}(z)=\frac{1}{2}\left[e^{L_{1}(z)+\Psi(z)+B_{3}}+e^{L_{2}(z)+\Phi(z)+B_{1}}\right],\vspace{1mm}\\\
f_{2}(z)=\frac{1}{2}\left[e^{L_{1}(z)+\Phi(z)+B_{2}}+e^{L_{2}(z)+\Psi(z)+B_{4}}\right].\end{cases}$

Case 3. Let $\alpha_{4}-\alpha_{3}=\eta\in\mathbb{C}$ and
$\alpha_{2}-\alpha_{1}$ be non-constant. Then from (3.3), we see that
$p_{2}(z)$ is a constant in $\mathbb{C}$.

Since $\alpha_{2}-\alpha_{1}$ is non-constant, by Lemma 3.4 and from second
equation of (3.5), we have either $ie^{\alpha_{3}(z+c)-\alpha_{1}(z)}=1$ or
$ie^{\alpha_{4}(z+c)-\alpha_{1}(z)}=1$.

First suppose that $ie^{\alpha_{3}(z+c)-\alpha_{1}(z)}=1$. Then, from the
second equation of (3.5), we have $ie^{\alpha_{4}(z+c)-\alpha_{2}(z)}=-1$. As
all $\alpha_{j}$, $j=1,2,3,4$ are polynomials in $\mathbb{C}^{n}$, it follows
that $\alpha_{3}(z+c)-\alpha_{1}(z)=m_{1}$ and
$\alpha_{4}(z+c)-\alpha_{2}(z)=m_{2}$, where $m_{1},m_{2}\in\mathbb{C}$.
Therefore, from (3.3), it follows that
$\alpha_{2}(z)-\alpha_{1}(z)=-2h_{1}(z)=m_{1}-m_{2}-2h_{2}$, a constant in
$\mathbb{C}$, which contradicts to our assumption.

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

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

###### Proof of Theorem 2.2.

Let $(f_{1},f_{2})$ is a pair of finite transcendental entire solutions of the
system (2.8).

Then, in a similar manner as in Theorem 2.1, we obtain that

$\displaystyle\begin{cases}\dfrac{\partial^{k}f_{1}}{\partial
z_{1}^{k}}=\displaystyle\frac{1}{2}\left[e^{\alpha_{1}(z)}+e^{\alpha_{2}(z)}\right]\\\
f_{2}(z+c)=\displaystyle\frac{1}{2i}\left[e^{\alpha_{1}(z)}-e^{\alpha_{2}(z)}\right]\\\
\dfrac{\partial^{k}f_{2}}{\partial
z_{1}^{k}}=\displaystyle\frac{1}{2}\left[e^{\alpha_{3}(z)}+e^{\alpha_{4}(z)}\right]\\\
f_{1}(z+c)=\displaystyle\frac{1}{2i}\left[e^{\alpha_{3}(z)}-e^{\alpha_{4}(z)}\right],\end{cases}$
(3.17)

where $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$ are defined in (3.3).

After simple computations, it follows from (3.17) that

$\displaystyle\begin{cases}ie^{\alpha_{1}(z+c)-\alpha_{3}(z)}+ie^{\alpha_{2}(z+c)-\alpha_{3}(z)}+p_{2}(z)e^{\alpha_{4}(z)-\alpha_{3}(z)}=p_{1}(z),\\\
ie^{\alpha_{3}(z+c)-\alpha_{1}(z)}+ie^{\alpha_{4}(z+c)-\alpha_{1}(z)}+p_{4}(z)e^{\alpha_{2}(z)-\alpha_{1}(z)}=p_{3}(z),\end{cases}$
(3.18)

with

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

where $M_{k}$ is the partial differential polynomial in
$\frac{\partial\alpha_{3}}{\partial z_{1}}$ of order less than $k$ in which
$\frac{\partial\alpha_{3}}{\partial z_{1}}$ appears in product with at least
one higher order partial derivatives with respect to $z_{1}$ of $\alpha_{3}$.
Similar definitions for $N_{k},O_{k}$ and $R_{k}$.

Now, we consider the following four possible cases.

Case 1. Let $\alpha_{2}(z)-\alpha_{1}(z)=\eta_{1}$ and
$\alpha_{4}(z)-\alpha_{3}(z)=\eta_{2}$, where
$\eta_{1},\eta_{2}\in\mathbb{C}$. Then, in view of (3.3), it follows that
$h_{1}(z)$ and $h_{2}(z)$ are both constants in $\mathbb{C}$. 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 (3.17), we obtain that

$\displaystyle\begin{cases}\dfrac{\partial^{k}f_{1}}{\partial
z_{1}^{k}}=\dfrac{\xi_{1}+\xi_{1}^{-1}}{2}e^{\frac{1}{2}g_{1}(z)},\\\
f_{2}(z+c)=\dfrac{\xi_{1}-\xi_{1}^{-1}}{2i}e^{\frac{1}{2}g_{1}(z)},\\\
\dfrac{\partial^{k}f_{2}}{\partial
z_{1}^{k}}=\dfrac{\xi_{2}+\xi_{2}^{-1}}{2}e^{\frac{1}{2}g_{2}(z)},\\\
f_{1}(z+c)=\dfrac{\xi_{2}-\xi_{2}^{-1}}{2i}e^{\frac{1}{2}g_{2}(z)}.\end{cases}$
(3.19)

It is clear from (3.19) that $\xi_{1}^{4},\xi_{2}^{4}\neq 1$. Also, after
simple computations, we obtain from (3.19) that

$\displaystyle\begin{cases}e^{\frac{1}{2}[g_{1}(z+c)-g_{2}(z)]}=\dfrac{\xi_{2}-\xi_{2}^{-1}}{2i(\xi_{1}+\xi_{1}^{-1})}q_{1}(z),\\\
e^{\frac{1}{2}[g_{2}(z+c)-g_{1}(z)]}=\dfrac{\xi_{1}-\xi_{1}^{-1}}{2i(\xi_{2}+\xi_{2}^{-1})}q_{2}(z),\end{cases}$
(3.20)

with

$\displaystyle q_{1}(z)=\left(\frac{1}{2}\frac{\partial g_{2}}{\partial
z_{1}}\right)^{k}+T_{1}\left(\frac{\partial^{k}g_{2}}{\partial
z_{1}^{k}},\ldots,\frac{\partial g_{2}}{\partial
z_{1}}\right),\;q_{2}(z)=\left(\frac{1}{2}\frac{\partial g_{1}}{\partial
z_{1}}\right)^{k}+T_{2}\left(\frac{\partial^{k}g_{1}}{\partial
z_{1}^{k}},\ldots,\frac{\partial g_{1}}{\partial z_{1}}\right),$

where $T_{1}$ is a partial differential polynomial in $\frac{\partial
g_{2}}{\partial z_{1}}$ of order less than $k$ where $\frac{\partial
g_{2}}{\partial z_{1}}$ appears in the product with at least one more partial
derivative of higher dimension, and similar definition for $T_{2}$.

Since $g_{1}(z),g_{2}(z)$ are polynomials in $\mathbb{C}^{2}$, it follows from
(3.20) that $q_{1},q_{2},g_{2}(z+c)-g_{1}(z)$ and $g_{2}(z+c)-g_{1}(z)$ are
all constants in $\mathbb{C}$. 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 $g_{1}(z+2c)-g_{1}(z)=g_{2}(z+2c)-g_{2}(z)=\eta_{3}+\eta_{4}$.

Thus, we have

$\displaystyle
g_{1}(z)=L(z)+H(s)+B_{1},\;\;\text{and}\;\;g_{2}(z)=L(z)+H(s)+B_{2},$

where $L(z)=a_{1}z_{1}+a_{2}z_{2}$, $H(s)$ is a polynomial in
$s:=d_{1}z_{1}+d_{2}z_{2}$ with $d_{1}c_{1}+d_{2}c_{2}=0$,
$a_{1},a_{2},B_{1},B_{2},d_{1},d_{2}\in\mathbb{C}$. Also from the form of
$q_{1}(z)$ and $q_{2}(z)$, we conclude that $\frac{\partial g_{1}}{\partial
z_{1}}$ and $\frac{\partial g_{2}}{\partial z_{1}}$ both are constants.
Therefore, we must have $s:=d_{2}z_{2}$ with $d_{2}c_{2}=0$.

Therefore, it follows from (3.20) that

$\displaystyle\begin{cases}e^{\frac{1}{2}[L(c)+B_{1}-B_{2}]}=\left(\dfrac{a_{1}}{2}\right)^{k}\dfrac{\xi_{2}-\xi_{2}^{-1}}{i(\xi_{1}+\xi_{1}^{-1})},\\\
e^{\frac{1}{2}[L(c)+B_{2}-B_{1}]}=\left(\dfrac{a_{1}}{2}\right)^{k}\dfrac{\xi_{1}-\xi_{1}^{-1}}{i(\xi_{2}+\xi_{2}^{-1})}.\end{cases}$
(3.21)

Hence, from we get (3.21) that

$\displaystyle
e^{L(c)}=-\left(\dfrac{a_{1}}{2}\right)^{2k}\dfrac{(\xi_{1}-\xi_{1}^{-1})(\xi_{2}-\xi_{2}^{-1})}{(\xi_{1}+\xi_{1}^{-1})(\xi_{2}+\xi_{2}^{-1})}\;\;\text{and}\;\;e^{B_{1}-B_{2}}=\dfrac{\xi_{2}^{2}-\xi_{2}^{-2}}{\xi_{1}^{2}-\xi_{1}^{-2}}.$

Thus, from (3.19), we get

$\displaystyle\begin{cases}f_{1}(z_{1},z_{2})=\dfrac{\xi_{2}^{2}-1}{2i\xi_{2}}e^{\frac{1}{2}[L(z)+H(s)-L(c)+B_{2}]},\\\
f_{2}(z_{1},z_{2})=\dfrac{\xi_{1}^{2}-1}{2i\xi_{1}}e^{\frac{1}{2}[L(z)+H(s)-L(c)+B_{1}]}.\end{cases}$

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

Observe that $p_{1}(z)$ and $p_{2}(z)$ both can not be simultaneously zero.
Otherwise, from the first equation of (3.18), we obtain
$e^{\alpha_{1}(z+c)-\alpha_{2}(z+c)}=-1$, which is a contradiction as
$\alpha_{2}(z)-\alpha_{1}(z)$ is non-constant.

Similarly, in view of the second equation of (3.18), we conclude that
$p_{3}(z)$ and $p_{4}(z)$ can not be simultaneously zero.

Now, let $p_{1}(z)\not\equiv 0$ and $p_{2}(z)\equiv 0$. Then, first equation
of (3.18) reduces to

$\displaystyle
e^{\alpha_{1}(z+c)-\alpha_{3}(z)}+e^{\alpha_{2}(z+c)-\alpha_{3}(z)}=-ip_{1}(z).$
(3.22)

As $\alpha_{2}(z)-\alpha_{1}(z)$ is a non-constant polynomial in
$\mathbb{C}^{2}$, it follows from (3.22) that $\alpha_{1}(z+c)-\alpha_{3}(z)$
and $\alpha_{2}(z+c)-\alpha_{3}(z)$ both are non-constants.

Now, rewrite (3.22) as

$\displaystyle
e^{\alpha_{1}(z+c)}+e^{\alpha_{2}(z+c)}+ip_{1}(z)e^{\alpha_{3}(z)}\equiv 0.$
(3.23)

Therefore, in view of Lemma 3.5, we can easily get a contradiction from
(3.23).

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

Therefore, by (3.18) and Lemma 3.4, we obtain that

$\displaystyle\text{either}\;\;ie^{\alpha_{1}(z+c)-\alpha_{3}(z)}=p_{1}(z)\;\;\text{or}\;\;ie^{\alpha_{2}(z+c)-\alpha_{3}(z)}=p_{1}(z)$

and

$\displaystyle\text{either}\;\;ie^{\alpha_{3}(z+c)-\alpha_{1}(z)}=p_{3}(z)\;\;\text{or}\;\;ie^{\alpha_{4}(z+c)-\alpha_{1}(z)}=p_{3}(z).$

Now, we consider the four possible subcases.

Subcase 2.1. Let

$\displaystyle
ie^{\alpha_{1}(z+c)-\alpha_{3}(z)}=p_{1}(z)\;\;\text{and}\;\;ie^{\alpha_{3}(z+c)-\alpha_{1}(z)}=p_{3}(z).$
(3.24)

Then, by (3.18) and (3.24), we get

$\displaystyle-
ie^{\alpha_{2}(z+c)-\alpha_{4}(z)}=p_{2}(z)\;\;\text{and}\;\;-ie^{\alpha_{4}(z+c)-\alpha_{2}(z)}=p_{4}(z).$
(3.25)

Since all $\alpha_{j}$’s are polynomials in $\mathbb{C}^{2}$, $j=1,\ldots,4$,
from (3.24) and (3.25), we conclude that $p_{1},p_{2},p_{3},p_{4}$,
$\alpha_{1}(z+c)-\alpha_{3}(z),\alpha_{3}(z+c)-\alpha_{1}(z),\alpha_{2}(z+c)-\alpha_{4}(z)=k_{3}$
and $\alpha_{4}(z+c)-\alpha_{2}(z)=k_{4}$ are all constants in $\mathbb{C}$.
Let $\alpha_{1}(z+c)-\alpha_{3}(z)=k_{1}$,
$\alpha_{3}(z+c)-\alpha_{1}(z)=k_{2}$, $\alpha_{2}(z+c)-\alpha_{4}(z)=k_{3}$
and $\alpha_{4}(z+c)-\alpha_{2}(z)=k_{4}$, where $k_{j}\in\mathbb{C}$,
$j=1,\ldots,4$. These imply
$\alpha_{1}(z+2c)-\alpha_{1}(z)=\alpha_{3}(z+2c)-\alpha_{3}(z)=k_{1}+k_{2}$
and
$\alpha_{2}(z+2c)-\alpha_{2}(z)=\alpha_{4}(z+2c)-\alpha_{4}(z)=k_{3}+k_{4}$.
Thus,

$\displaystyle\alpha_{1}(z)=L_{1}(z)+H_{1}(s)+B_{1},\;\alpha_{3}(z)=L_{1}(z)+H_{1}(s)+B_{2},$
$\displaystyle\alpha_{2}(z)=L_{2}(z)+H_{2}(s)+B_{3},\;\alpha_{4}(z)=L_{2}(z)+H_{2}(s)+B_{4},$

where $L_{i}(z)=a_{i1}z_{1}+a_{i2}z_{2}$, $H_{j}(s)$ is a polynomial in
$s:=d_{1}z_{1}+d_{2}z_{2}$ with $d_{1}c_{1}+d_{2}c_{2}=0$, $i=1,2$, $j=1,2$,
$a_{i1},a_{i2},B_{1},B_{2},B_{3},B_{4}\in\mathbb{C}$. As all $p_{j}$’s are
constants, it follows that $s:=d_{2}z_{2}$ with $d_{2}c_{2}=0$.

Since $\alpha_{2}(z)-\alpha_{1}(z)$ and $\alpha_{4}(z)-\alpha_{3}(z)$ are non-
constants, we conclude that $L_{1}(z)+H_{1}(s)\neq L_{2}(z)+H_{2}(s)$. Thus,
in view of (3.3), we have

$\displaystyle
g_{1}(z)=L_{1}(z)+L_{2}(z)+H_{1}(s)+H_{2}(s)+B_{1}+B_{3}\;\;\text{and}$
$\displaystyle g_{2}(z)=L_{1}(z)+L_{2}(z)+H_{1}(s)+H_{2}(s)+B_{2}+B_{4}.$

Therefore, in view of (3.24) and (3.25), we obtain

$\displaystyle\begin{cases}ie^{L_{1}(c)+B_{1}-B_{2}}=a_{11}^{k},\;\;ie^{L_{1}(c)+B_{2}-B_{1}}=a_{11}^{k},\\\
ie^{L_{2}(c)+B_{3}-B_{4}}=-a_{21}^{k},\;\;ie^{L_{2}(c)+B_{4}-B_{3}}=-a_{21}^{k}.\end{cases}$
(3.26)

From first two equations of (3.26), we obtain
$e^{2L_{1}(c)}=-a_{11}^{2k}\;\;\text{and}\;\;e^{2(B_{1}-B_{2})}=1$ and from
last two equations of (3.26), we get
$e^{2L_{2}(c)}=-a_{21}^{2k}\;\;\text{and}\;\;e^{2(B_{3}-B_{4})}=1$. These
imply $e^{L_{1}(c)}=\pm ia_{11}^{k}$, $e^{L_{2}(c)}=\pm ia_{21}^{k}$,
$e^{(B_{1}-B_{2})}=\pm 1$ and $e^{(B_{3}-B_{4})}=\pm 1$.

If $e^{L_{1}(c)}=ia_{11}^{k}$, then from (3.26), we get $e^{B_{1}-B_{2}}=-1$
and if $e^{L_{1}(c)}=-ia_{11}^{k}$, then from (3.26), we obtain
$e^{B_{1}-B_{2}}=1$. Similarly, if $e^{L_{2}(c)}=ia_{21}^{k}$, then from
(3.26), we get $e^{B_{3}-B_{4}}=1$ and if $e^{L_{1}(c)}=-ia_{11}^{k}$, then
from (3.26), we obtain $e^{B_{3}-B_{4}}=-1$.

Thus, from fourth and second equations of (3.17), we obtain

$\displaystyle
f_{1}(z)=\frac{1}{2i}\left[e^{L_{1}(z)+H_{1}(s)-L_{1}(c)+B_{2}}-e^{L_{2}(z)+H_{2}(s)-L_{2}(c)+B_{4}}\right]$

and

$\displaystyle
f_{2}(z)=\frac{1}{2i}\left[e^{L_{1}(z)+H_{1}(s)-L_{1}(c)+B_{1}}-e^{L_{2}(z)+H_{2}(s)-L_{2}(c)+B_{3}}\right].$

Subcase 2.2. Let

$\displaystyle
ie^{\alpha_{1}(z+c)-\alpha_{3}(z)}=p_{1}(z)\;\;\text{and}\;\;ie^{\alpha_{4}(z+c)-\alpha_{1}(z)}=p_{3}(z).$
(3.27)

Then, in view of (3.18) and (3.27), we obtain

$\displaystyle-
ie^{\alpha_{2}(z+c)-\alpha_{4}(z)}=p_{2}(z)\;\;\text{and}\;\;-ie^{\alpha_{3}(z+c)-\alpha_{2}(z)}=p_{4}(z).$
(3.28)

Since $\alpha_{1}(z),\alpha_{2}(z),\alpha_{3}(z),\alpha_{4}(z)$ are
polynomials in $\mathbb{C}^{2}$, we conclude from (3.27) and (3.28) that
$\alpha_{1}(z+c)-\alpha_{3}(z)=m_{1}$, $\alpha_{4}(z+c)-\alpha_{1}(z)=m_{2}$,
$\alpha_{2}(z+c)-\alpha_{4}(z)=m_{3}$ and
$\alpha_{3}(z+c)-\alpha_{2}(z)=m_{4}$, where $m_{j}\in\mathbb{C}$,
$j=1,\ldots,4$. These imply
$\alpha_{1}(z+2c)-\alpha_{1}(z)=\alpha_{2}(z+4c)-\alpha_{2}(z)=\sum_{j=1}^{4}m_{j}$.
Therefore, $\alpha_{1}(z)=L(z)+H(s)+B_{1}$ and
$\alpha_{2}(z)=L(z)+H(s)+B_{2}$, where $L(z)=a_{1}z_{1}+a_{2}z_{2}$ and $H(s)$
is a polynomial in $s:=d_{1}z_{1}+d_{2}z_{2}$ with $d_{1}c_{1}+d_{2}c_{2}=0$,
$a_{1},a_{2},B_{1},B_{2}\in\mathbb{C}$. Hence
$\alpha_{2}(z)-\alpha_{1}(z)=B_{2}-B_{1}\in\mathbb{C}$, a contradiction.

Subcase 2.3. Let

$\displaystyle
ie^{\alpha_{2}(z+c)-\alpha_{3}(z)}=p_{1}(z)\;\;\text{and}\;\;ie^{\alpha_{3}(z+c)-\alpha_{1}(z)}=p_{3}(z).$

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

Subcase 2.4. Let

$\displaystyle
ie^{\alpha_{2}(z+c)-\alpha_{3}(z)}=p_{1}(z)\;\;\text{and}\;\;ie^{\alpha_{4}(z+c)-\alpha_{1}(z)}=p_{3}(z).$
(3.29)

Then, by (3.18) and (3.24), we get

$\displaystyle-
ie^{\alpha_{1}(z+c)-\alpha_{4}(z)}=p_{2}(z)\;\;\text{and}\;\;-ie^{\alpha_{3}(z+c)-\alpha_{2}(z)}=p_{4}(z).$
(3.30)

As $\alpha_{1}(z),\alpha_{2}(z),\alpha_{3}(z),\alpha_{4}(z)$ are polynomials
in $\mathbb{C}^{2}$, from (3.29) and (3.30), we conclude that
$p_{1},p_{2},p_{3},p_{4}$, $\alpha_{2}(z+c)-\alpha_{3}(z)$,
$\alpha_{4}(z+c)-\alpha_{1}(z)$, $\alpha_{1}(z+c)-\alpha_{4}(z)$ and
$\alpha_{3}(z+c)-\alpha_{2}(z)$ are all constants in $\mathbb{C}$.

Let $\alpha_{2}(z+c)-\alpha_{3}(z)=n_{1}$,
$\alpha_{4}(z+c)-\alpha_{1}(z)=n_{2}$, $\alpha_{1}(z+c)-\alpha_{4}(z)=n_{3}$
and $\alpha_{3}(z+c)-\alpha_{2}(z)=n_{4}$, where $n_{j}\in\mathbb{C}$,
$j=1,\ldots,4$. Thus, we get
$\alpha_{1}(z+2c)-\alpha_{1}(z)=\alpha_{4}(z+2c)-\alpha_{4}(z)=n_{2}+n_{3}$
and
$\alpha_{2}(z+2c)-\alpha_{2}(z)=\alpha_{3}(z+2c)-\alpha_{3}(z)=n_{1}+n_{4}$.
Hence, we assume

$\displaystyle\alpha_{1}(z)=L_{1}(z)+H_{1}(s)+B_{1},\;\;\alpha_{4}(z)=L_{1}(z)+H_{1}(s)+B_{2},$
$\displaystyle\alpha_{2}(z)=L_{2}(z)+H_{2}(s)+B_{3}\;\;\text{and}\;\;\alpha_{3}(z)=L_{2}(z)+H_{2}(s)+B_{4},$

where $L_{i}(z)=a_{i1}z_{1}+a_{i2}z_{2}$, $H_{j}(s)$ is a polynomial in
$s:=d_{1}z_{1}+d_{2}z_{2}$ with $d_{1}c_{1}+d_{2}c_{2}=0$ for $i=1,2$,
$j=1,2$, $a_{i1},a_{i2},B_{1},B_{2},B_{3},B_{4}\in\mathbb{C}$. Since
$\alpha_{2}(z)-\alpha_{1}(z)$ and $\alpha_{4}(z)-\alpha_{3}(z)$ are non-
constants, it follows that $L_{1}(z)+H_{1}(s)\neq L_{2}(z)+H_{2}(s)$. Also
since $p_{1},p_{2},p_{3},p_{4}$ are all constants, it follows that
$\frac{\partial\alpha_{j}}{\partial z_{1}}$ are constants for all j=1,2,3,4.
Hence, $H_{1}$ and $H_{2}$ are the polynomials in $s:=d_{2}z_{2}$ with
$d_{2}c_{2}=0$.

Now, in view of (3.3), we get

$\displaystyle\begin{cases}g_{1}(z)=L_{1}(z)+L_{2}(z)+H_{1}(s)+H_{2}(s)+B_{1}+B_{3},\\\
g_{2}(z)=L_{1}(z)+L_{2}(z)+H_{1}(s)+H_{2}(s)+B_{2}+B_{4}.\end{cases}$

Therefore, in view of (3.29) and (3.30), we obtain

$\displaystyle\begin{cases}ie^{L_{2}(c)+B_{3}-B_{4}}=a_{21}^{k},\;\;ie^{L_{1}(c)+B_{2}-B_{1}}=a_{11}^{k},\\\
ie^{L_{1}(c)+B_{1}-B_{2}}=-a_{11}^{k},\;\;ie^{L_{2}(c)+B_{4}-B_{3}}=-a_{21}^{k}.\end{cases}$
(3.31)

Thus, from (3.31), we obtain

$\displaystyle
e^{2L_{1}(c)}=a_{11}^{2k},\;\;e^{2L_{2}(c)}=a_{21}^{2k},\;\;e^{2(B_{1}-B_{2})}=-1,\;\;\text{and}\;\;e^{2(B_{3}-B_{4})}=-1,$

which imply

$\displaystyle e^{L_{1}(c)}=\pm a_{11}^{k},\;\;e^{L_{2}(c)}=\pm
a_{21}^{k},\;\;e^{B_{1}-B_{2}}=\pm i,\;\;\text{and}\;\;e^{B_{3}-B_{4}}=\pm i.$

If $e^{L_{1}(c)}=a_{11}^{k}$, then from (3.31), we have $e^{B_{1}-B_{2}}=i$
and if $e^{L_{1}(c)}=-a_{11}^{k}$, then we have $e^{B_{1}-B_{2}}=-i$.
Similarly, if $e^{L_{2}(c)}=a_{21}^{k}$, the from (3.31), we have
$e^{B_{3}-B_{4}}=-i$ and if $e^{L_{2}(c)}=-a_{21}^{k}$, the from (3.31), we
have $e^{B_{3}-B_{4}}=i$.

Hence, from fourth and second equations of (3.17), we get

$\displaystyle\begin{cases}f_{1}(z)=-\frac{1}{2i}\left[e^{L_{1}(z)+H_{1}(z)-L_{1}(c)+B_{2}}-e^{L_{2}(z)+H_{2}(z)-L_{2}(c)+B_{4}}\right]\\\
f_{2}(z)\;=\frac{1}{2i}\left[e^{L_{1}(z)+H_{1}(z)-L_{1}(c)+B_{1}}-e^{L_{2}(z)+H_{2}(z)-L_{2}(c)+B_{3}}\right].\end{cases}$

Case 3. Let $\alpha_{4}-\alpha_{3}=\eta\in\mathbb{C}$ and
$\alpha_{2}-\alpha_{1}$ be non-constant. Then, (3.3) yields that $p_{2}(z)$ is
a constant in $\mathbb{C}$.

Then first equation of (3.18) yields that

$\displaystyle
ie^{\alpha_{1}(z+c)-\alpha_{3}(z)}+ie^{\alpha_{2}(z+c)-\alpha_{3}(z)}=p_{1}(z)-e^{\eta}p_{2}(z).$
(3.32)

From (3.32), we observe that
$T\left(r,e^{\alpha_{1}(z+c)-\alpha_{3}(z)}\right)=T\left(r,e^{\alpha_{2}(z+c)-\alpha_{3}(z)}\right)+O(1)$.

We also note that

$\displaystyle
N\left(r,e^{\alpha_{1}(z+c)-\alpha_{3}(z)}\right)=N\left(r,\frac{1}{e^{\alpha_{1}(z+c)-\alpha_{3}(z)}}\right)=S\left(r,e^{\alpha_{1}(z+c)-\alpha_{3}(z)}\right)$

and

$\displaystyle
N\left(r,\frac{1}{e^{\alpha_{1}(z+c)-\alpha_{3}(z)}-w}\right)=N\left(r,\frac{1}{e^{\alpha_{2}(z+c)-\alpha_{3}(z)}}\right)=S\left(r,e^{\alpha_{2}(z+c)-\alpha_{3}(z)}\right),$

where $w=-i(p_{1}-e^{eta}p_{2})$. Now, by the second fundamental theorem of
Nevanlinna for several complex variables, we obtain

$\displaystyle T\left(r,e^{\alpha_{1}(z+c)-\alpha_{3}(z)}\right)$
$\displaystyle\leq\overline{N}\left(r,e^{\alpha_{1}(z+c)-\alpha_{3}(z)}\right)+\overline{N}\left(r,\frac{1}{e^{\alpha_{1}(z+c)-\alpha_{3}(z)}}\right)+\overline{N}\left(r,\frac{1}{e^{\alpha_{1}(z+c)-\alpha_{3}(z)}-w}\right)$
$\displaystyle+S\left(r,e^{\alpha_{1}(z+c)-\alpha_{3}(z)}\right)$
$\displaystyle\leq
S\left(r,e^{\alpha_{1}(z+c)-\alpha_{3}(z)}\right)+S\left(r,e^{\alpha_{2}(z+c)-\alpha_{3}(z)}\right).$

This implies that $\alpha_{1}(z+c)-\alpha_{3}(z)$ is a constant in
$\mathbb{C}$. Therefore, $\alpha_{2}(z+c)-\alpha_{3}(z)$ is also constant.
Thus, $\alpha_{2}(z)-\alpha_{1}(z)$ becomes constant, which is a
contradiction.

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

###### Proof of Theorem 2.3.

Let $(f_{1}(z),f_{2}(z))$ be a pair of finite order transcendental entire
solution of (2.9).

Then, by similar argument as in Theorem 2.1, we obtain

$\displaystyle\begin{cases}f_{1}^{(k)}(z)=\displaystyle\frac{1}{2}\left[e^{p_{1}(z)}+e^{-p_{1}(z)}\right],\\\
f_{2}(z+c)-f_{2}(z)=\displaystyle\frac{1}{2i}\left[e^{p_{1}(z)}-e^{-p_{1}(z)}\right],\\\
f_{2}^{(k)}(z)=\displaystyle\frac{1}{2}\left[e^{p_{2}(z)}+e^{-p_{2}(z)}\right],\\\
f_{1}(z+c)-f_{1}(z)=\displaystyle\frac{1}{2i}\left[e^{p_{2}(z)}-e^{-p_{2}(z)}\right],\end{cases}$
(3.33)

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

Differentiating fourth equation of (3.33) $k$ times, we get

$\displaystyle
f_{1}^{(k)}(z+c)-f_{1}^{(k)}(z)=\frac{1}{2i}\left[G_{1}e^{p_{2}(z)}-G_{2}e^{-p_{2}(z)}\right],$
(3.34)

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

Now in view of the first equation of (3.33) and (3.34), we obtain

$\displaystyle-
ie^{p_{1}(z+c)+p_{2}(z)}-ie^{-p_{1}(z+c)+p_{2}(z)}+ie^{p_{1}(z)+p_{2}(z)}+ie^{-p_{1}(z)+p_{2}(z)}$
$\displaystyle+G_{1}e^{2p_{2}(z)}=G_{2}.$ (3.35)

Similarly in view of the second and third equations of (3.33), we obtain

$\displaystyle-
ie^{p_{2}(z+c)+p_{1}(z)}-ie^{-p_{2}(z+c)+p_{1}(z)}+ie^{p_{2}(z)+p_{1}(z)}+ie^{-p_{2}(z)+p_{1}(z)}$
$\displaystyle+G_{3}e^{2p_{1}(z)}=G_{4},$ (3.36)

where
$G_{3}=\left(p_{1}^{\prime}\right)^{k}+O_{k}\left(p_{1}^{(k)},\ldots,p_{1}^{\prime}\right)$
and
$G_{4}=\left(-p_{1}^{\prime}\right)^{k}+R_{k}\left(p_{1}^{(k)},\ldots,p_{1}^{\prime}\right)$,
where $O_{k}$ is a differential polynomial in $p_{1}(z)$ of degree less than
$k$ in which $p_{1}^{\prime}$ appears in the product with at least one more
derivative of $p_{1}$ of order greater than or equal to $2$. 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 (3) and (3)
yield

$\displaystyle\begin{cases}-ie^{\eta}e^{p_{1}(z+c)+p_{1}(z)}-ie^{\eta}e^{p_{1}(z)-p_{1}(z+c)}+e^{\eta}(i+G_{1}e^{\eta})e^{2p_{1}(z)}=G_{2}-ie^{\eta},\\\
-ie^{\eta}e^{p_{1}(z+c)+p_{1}(z)}-ie^{-\eta}e^{p_{1}(z)-p_{1}(z+c)}+(ie^{\eta}+G_{3})e^{2p_{1}(z)}=G_{4}-ie^{-\eta}.\end{cases}$
(3.37)

Note that $i+G_{1}e^{\eta}$ and $G_{2}-ie^{\eta}$ can not be zero
simultaneously. Otherwise, from the first equation of (3.37), we get
$p_{1}(z+c)$, and hence $p_{1}(z)$ is constant, a contradiction.

Next, suppose that $i+G_{1}e^{\eta}\not\equiv 0$ and $G_{2}-ie^{\eta}\equiv
0$. Then the first equation of (3.37) reduces to

$\displaystyle
ie^{p_{1}(z+c)-p_{1}(z)}+ie^{-(p_{1}(z)+p_{1}(z+c))}=i+G_{1}e^{\eta}.$ (3.38)

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 (3.38), 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 $w=1-iG_{1}e^{\eta}$.

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 $i+G_{1}e^{\eta}\equiv 0$
and $G_{2}-ie^{\eta}\not\equiv 0$. Hence, $i+G_{1}e^{\eta}\not\equiv 0$ and
$G_{2}-ie^{\eta}\not\equiv 0$.

In a similar manner, we can show that $ie^{\eta}+G_{3}\not\equiv 0$ and
$G_{4}-ie^{-\eta}\not\equiv 0$.

Therefore, in view of Lemma 3.1 and (3.37), we obtain

$\displaystyle\begin{cases}-ie^{\eta}e^{p_{1}(z)-p_{1}(z+c)}=G_{2}-ie^{\eta},\\\
-ie^{-\eta}e^{p_{1}(z)-p_{1}(z+c)}=G_{4}-ie^{-\eta}.\end{cases}$ (3.39)

By (3.37) and (3.39), we get

$\displaystyle\begin{cases}ie^{p_{1}(z+c)-p_{1}(z)}=i+G_{1}e^{\eta},\\\
ie^{\eta}e^{p_{1}(z+c)-p_{1}(z)}=G_{3}+ie^{\eta}.\end{cases}$ (3.40)

Since $p_{1}(z)$ is a non-constant polynomial in $\mathbb{C}^{2}$, it follows
from (3.40) that $p_{1}(z+c)-p_{1}(z)$, $G_{1},G_{2},G_{3},G_{4}$ are all
constants in $\mathbb{C}$. Therefore, we must have
$p_{1}(z)=L(z)+H(z_{2})+\beta$, where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$H(z_{2})$ is a polynomial in $z_{2}$, only,
$\alpha_{1},\alpha_{2},\beta\in\mathbb{C}$. Thus,
$p_{2}(z)=L(z)+H(z_{2})+\beta+\eta$.

Therefore, in view of (3.39), (3.40), we obtain

$\displaystyle\begin{cases}-ie^{\eta}e^{-L(c)}=(-\alpha_{1})^{k}-ie^{\eta},\;\;-ie^{\eta}e^{-L(c)}=(-\alpha_{1})^{k}-ie^{-\eta},\\\
ie^{L(c)}=i+\alpha_{1}^{k}e^{\eta},\;\;ie^{\eta}e^{L(c)}=\alpha_{1}^{k}+ie^{\eta}.\end{cases}$
(3.41)

Now from the first and second equations of (3.41), we easily obtain
$e^{2\eta}=1$. From the first and third equations, we get

$\displaystyle(-1)^{k}e^{\eta}\alpha_{1}^{2k}+i\left[(-1)^{k}-1\right]\alpha_{1}^{k}=0.$
(3.42)

If $k$ is even, the in view of (3.42) and the fact that $\alpha_{1}\neq 0$, we
easily obtain a contradiction. Hence, $k$ must be odd and therefore, from
(3.42), we get

$\displaystyle\alpha_{1}^{k}=-2ie^{-\eta}.$

Thus from the first equation of (3.41), we obtain

$\displaystyle e^{L(c)}=\frac{ie^{\eta}}{\alpha_{1}^{k}+ie^{\eta}}.$

If $e^{\eta}=1$, then $\alpha_{1}^{k}=-2i$, and if $e^{\eta}=-1$, then
$\alpha_{1}^{k}=2i$.

Thus, from (3.33), we obtain

$\displaystyle
f_{1}(z)=\frac{e^{L(z)+H(z_{2})+\beta}-e^{-(L(z)+H(z_{2})+\beta)}}{2\alpha_{1}^{k}},$
$\displaystyle
f_{2}(z)=\frac{e^{L(z)+H(z_{2})+\beta+\eta}-e^{-(L(z)+H(z_{2})+\beta+\eta)}}{2\alpha_{1}^{k}}.$

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, (3) and (3) yield

$\displaystyle\begin{cases}e^{p_{1}(z+c)-p_{1}(z)}+e^{-(p_{1}(z+c)+p_{1}(z))}+(ie^{\eta}G_{1}-1)e^{-2p_{1}(z)}=1+ie^{-\eta}G_{2},\\\
-ie^{\eta}e^{p_{1}(z)-p_{1}(z+c)}-ie^{-\eta}e^{p_{1}(z+c)+p_{1}(z)}+(G_{3}+ie^{-\eta})e^{2p_{1}(z)}=G_{4}-ie^{\eta}.\end{cases}$
(3.43)

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

Therefore, by Lemma 3.1 and (3.43), we have

$\displaystyle
e^{p_{1}(z+c)-p_{1}(z)}=1+ie^{-\eta}G_{2}\;\;\text{and}\;\;-ie^{\eta}e^{p_{1}(z)-p_{1}(z+c)}=G_{4}-ie^{\eta}.$
(3.44)

From (3.43) and (3.44), we obtain

$\displaystyle
e^{-p_{1}(z+c)+p_{1}(z)}=1-ie^{\eta}G_{1}\;\;\text{and}\;\;e^{p_{1}(z+c)-p_{1}(z)}=1-ie^{\eta}G_{3}.$
(3.45)

Since $p_{1}(z)$ is a non constant polynomial, it follows from (3.44) and
(3.45) that $G_{1}$, $G_{2}$, $G_{3}$, $G_{4}$ and $p_{1}(z+c)-p_{1}(z)$ are
all constants. As $p_{1}(z+c)-p_{1}(z)$ is constant, we must have
$p_{1}(z)=L(z)+H(z_{2})+\beta$, where $L(z)=\alpha_{1}z_{1}+\alpha_{2}z_{2}$,
$H(z_{2})$ is a polynomial in $z_{2}$, only,
$\alpha_{1},\alpha_{2},\beta\in\mathbb{C}$. Thus,
$p_{2}(z)=-(L(z)+H(z_{2})+\beta)+\eta$. Since $G_{1}$, $G_{2}$, $G_{3}$,
$G_{4}$ are all constants, in view of the definitions of them, we see that
$G_{2}=G_{3}=\alpha_{1}^{k}$ and $G_{1}=G_{4}=(-\alpha_{1})^{k}$

Therefore, (3.44) and (3.45) yield that

$\displaystyle\begin{cases}e^{L(c)}=1+ie^{-\eta}\alpha_{1}^{k},\;\;-ie^{\eta}e^{-L(c)}=(-\alpha_{1})^{k}-ie^{\eta},\\\
e^{-L(c)}=1-ie^{\eta}(-\alpha_{1})^{k},\;\;e^{L(c)}=1-ie^{\eta}\alpha_{1}^{k}.\end{cases}$
(3.46)

From the above two equations of (3.46), we easily obtain

$\displaystyle(-1)^{k}ie^{-\eta}\alpha_{1}^{k}+1+(-1)^{k}=0.$ (3.47)

If $k$ is odd, then in view of (3.47), we can get a contradiction. If $k$ is
even, then from (3.47), we get

$\displaystyle\alpha_{1}^{k}=2ie^{\eta}.$

From first and fourth equations of (3.46), we obtain

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

Therefore, from (3.33), we have

$\displaystyle
f_{1}(z_{1},z_{2})=\frac{e^{L(z)+H(z_{2})+\beta}+e^{-(L(z)+H(z_{2})+\beta)}}{2\alpha_{1}^{k}},$
$\displaystyle
f_{2}(z_{1},z_{2})=\frac{e^{-(L(z)+H(z_{2})+\beta)+\eta}+e^{L(z)+H(z_{2})+\beta-\eta}}{2\alpha_{1}^{k}}.$

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 3.1 and (3), we get $-ie^{-p_{2}(z+c)+p_{1}(z)}=G_{4}$. 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 3.1, we obtain from (3) and (3) that

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

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.∎

##  4\. An open problem

For further investigation, we pose an open problem as follows. What could be
the entire solutions of the system of Fermat type differential-difference
equation

$\displaystyle\begin{cases}\left(\frac{\partial^{k}f_{1}}{\partial
z_{1}^{k}}\right)^{2}+[f_{2}(z+c)-f_{2}(z)]^{2}=e^{g_{1}(z_{1},z_{2})},\\\
\left(\frac{\partial^{k}f_{2}}{\partial
z_{1}^{k}}\right)^{2}+[f_{1}(z+c)-f_{1}(z)]^{2}=e^{g_{2}(z_{1},z_{2})},\end{cases}$

where $g_{1}(z_{1},z_{2})$ and $g_{2}(z_{1},z_{2})$ are any two polynomials in
$\mathbb{C}^{2}$?

##  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] Courant, R., Hilbert, D.: Methods of Mathematical Physics, vol. II: Partial Differential Equations, Interscience (New York, 1962)
  * [6] Gao, L.Y.: Entire solutions of two types of systems of complex differential-difference equations, Acta Math. Sinica (Chin. Ser.) 59 (2016), 677–685
  * [7] Garabedian, P.R.: Partial Differential Equations, Wiley (New York, 1964).
  * [8] Gross, F.: On the equation $f^{n}(z)+g^{n}(z)=1$, Bull. Amer. Math. Soc. 72, 86–88 (1966)
  * [9] Gross, F.: On the equation $f^{n}(z)+g^{n}(z)=h^{n}(z)$, Amer. Math. Monthly. 73, 1093–1096 (1966)
  * [10] 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)
  * [11] Halburd, R.G., Korhonen, R.J.: Nevanlinna theory for the difference operator, Ann. Acad. Sci. Fenn. Math. 31, 463–478 (2006)
  * [12] 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
  * [13] 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
  * [14] Hayman, W.K.: Meromorphic Functions, The Clarendon Press, Oxford, 1964.
  * [15] 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)
  * [16] Iyer, G.: On certain functional equations, J. Indian. Math. Soc. 3, 312–315 (1939)
  * [17] Korhonen, R.J.: A difference Picard theorem for meromorphic functions of several variables, Comput. Methods Funct. Theory 12(1), 343–361 (2012)
  * [18] Laine, I.: Nevanlinna Theory and Complex Differential Equations, Walter de Gruyter, Berlin, (1993)
  * [19] Lelong, P.: Fonctionnelles Analytiques et Fonctions Enti‘eres (n variables), Presses de L’Universit´e de Montr´eal (1968)
  * [20] Li, B.Q.: On reduction of functional-differential equations, Complex Var. 31, 311–324 (1996)
  * [21] Li, B.Q.: On entire solutions of Fermat type partial differential equations, Int. J. Math. 15, 473–485 (2004)
  * [22] Li, B.Q.: On certain non-linear differential equations in complex domains, Arch. Math. 91, 344–353 (2008)
  * [23] Liu, M.L., Gao, L.Y.: Transcendental solutions of systems of complex differential-difference equations, Sci. Sin. Math. 49, 1–22 (2019)
  * [24] Liu, K., Yang, L.Z.: On Entire Solutions of Some Differential-Difference Equations, Comput. Methods Funct. Theory 13, 433–447 (2012)
  * [25] Liu, K., Cao, T.B., Cao, H.Z.: Entire solutions of Fermat-type differential-difference equations, Arch. Math. 99, 147–155 (2012)
  * [26] 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)
  * [27] Montel, P.: Lecons sur les familles de nomales fonctions analytiques et leurs applications, Gauthier-Viuars Paris, 135–136 (1927)
  * [28] P´olya, G.: On an integral function of an integral function, J. Lond. Math. Soc. 1, 12–15 (1926)
  * [29] Rauch, J.: Partial Differential Equations, Springer-Verlag (New York, 1991)
  * [30] Ronkin, L.I.: Introduction to the Theory of Entire Functions of Several Variables, Moscow: Nauka 1971 (Russian), American Mathematical Society, Providence (1974)
  * [31] Saleeby, E.G.: Entire and meromorphic solutions of Fermat-type partial differential equations, Analysis (Munich) 19, 369–376 (1999)
  * [32] 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)
  * [33] Stoll, W.: Holomorphic Functions of Finite Order in Several Complex Variables, American Mathematical Society, Providence, 1974
  * [34] 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)
  * [35] Taylor, R., Wiles, A.: Ring-theoretic properties of certain Hecke algebra, Ann. Math. 141, 553–572 (1995)
  * [36] Wiles, A.: Modular elliptic curves and Fermats last theorem, Ann. Math. 141, 443–551 (1995)
  * [37] Wu, L.L, Hu, P.C.: Entire Solutions on a System of Fermat Type Q-Difference-Differential Equations, Analysis Math. 48, 1257–1280 (2022)
  * [38] Xu, L., Cao, T.B.: Solutions of complex Fermat-type partial difference and differential-difference equations, Mediterr. J.Math. 15, 1–14 (2018)
  * [39] 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)
  * [40] 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
  * [41] Xu, X.Y. Wang, H.: Notes on the Existence of Entire Solutions for Several Partial Differential-Difference Equations, Bull. Iran. Math. Soc. 47, 1477–1489 (2021)
  * [42] 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)
  * [43] Zheng, X.M., Xu, X.Y.: Entire solutions of some Fermat type functional equations concerning difference and partial differential in $\mathbb{C}^{2}$, Anal. Math. https://doi.org/10.1007/s10476-021-0113-7