# Solutions of Fermat-type partial differential-difference equations in
$\mathbb{C}^{n}$

Goutam Haldar  Goutam Haldar, Department of Mathematics, Malda College, Malda
- 732101, West Bengal, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

For two meromorphic functions $f$ and $g$, the equation $f^{m}+g^{m}=1$ can be
regarded as Fermat-type equations. Using Nevanlinna theory for meromorphic
functions in several complex variables, the main purpose of this paper is to
investigate the properties of the transcendental entire solutions of Fermat-
type difference and partial differential-difference equations in
$\mathbb{C}^{n}$. In addition, we find the precise form of the transcendental
entire solutions in $\mathbb{C}^{2}$ with finite order of the Fermat-type
partial differential-difference equation

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

and

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

where $P(z_{1},z_{2})$ is a polynomial in $\mathbb{C}^{2}$. Moreover, one of
the main results of the paper significantly improved the result of Xu and Cao
[Mediterr. J. Math. (2018) 15:227 , 1-14 and Mediterr. J. Math. (2020) 17:8,
1-4].

###### Key words and phrases:

Functions of Several complex variables, meromorphic functions, transcendental
entire functions, Fermat-type equations, Nevanlinna theory.

###### 2010 Mathematics Subject Classification:

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

##  1\. Introduction

This paper mainly deals with the existence of transcendental entire solutions
with finite order of Fermat-type difference and partial differential-
difference equations in several complex variables. We adopt the standard
notations of the Nevanlinna theory of meromorphic functions in one variable
(see [10, 15, 40]). In the past several decades, considerable attention had
been paid to the existence of entire solutions for Fermat-type equation
$x^{n}+y^{m}=1$. Specially, in $1995$, Wiles and Taylor [30, 32] pointed out
that the Fermat-type equation $x^{n}+y^{m}=1$, where $m,n\in\mathbb{N}$ 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 Gross [5, 6], Montel [25].

In $1970$, Yang [37] considered the following functional equation

$\displaystyle f^{n}+g^{m}=1$ (1.1)

and proved the following interesting result.

######  Theorem A.

[37] There are no non-constant entire solutions of the functional equation
$(\ref{e1.1})$ if $m$, $n$ are positive integers satisfying $1/m+1/n<1$.

In recent years, looking for non-constant entire as well as meromorphic
solutions of Fermat-type difference as well as differential-difference
equations have been studied extensively after the development of the
difference analogous lemma of the logarithmic derivative by Halburd and
Korhonen [7, 8], and Chiang and Feng [4], independently. As a result,
successively a lot of investigations done by many researchers in this
direction (see [3, 9, 13, 18, 20, 21, 22, 23, 24, 29, 31, 38, 41]).

In view of Theorem A, Liu et. al. [22] have showed that there is no finite
order transcendental entire solution of Fermat-type difference equation
$f^{n}(z)+f^{m}(z+c)=1$ when $n>m>1$ or $n=m>2$, and for the case $n=m=2$,
$f(z)$ must be of the form $f(z)=\sin(Az+B)$, where $c(\neq 0)$,
$B\in\mathbf{C}$ and $A=(4k+1)\pi/2c$, $k$ is an integer. Later, in $2013$,
Liu and Yang [19] extended this result by considering the Fermat-type
difference equation $f^{2}(z)+P^{2}(z)f^{2}(z+c)=Q(z)$ where $P(z)$ and $Q(z)$
are two non-zero polynomials.

In 2018, Xu and Cao [33] extended the above result to several complex
variables as follows.

######  Theorem B.

[33] Let $c=(c_{1},c_{2},\ldots,c_{n})\in\mathbb{C}^{n}\setminus\\{0\\}$. Then
any non-constant entire solution
$f:\mathbb{C}^{n}\rightarrow\mathbb{P}^{1}(\mathbb{C})$ with finite order of
the Fermat-type difference equation $f(z)^{2}+f(z+c)^{2}=1$ has the form of
$f(z)=\cos(L(z)+B)$, 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 $\mathbb{C}$.

It would be interesting of finding the solutions as well as the precise form
of the solutions of the equation considered in Theorem B. Henceforth,
motivated by the results of Xu and Cao [33], and Liu and Yang [19], in this
paper, we are mainly interested to generalize Theorem B by considering the
following equation

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

where $P(z)$ and $Q(z)$ (with $Q(0)\neq 0$) are two non-zero polynomials in
$\mathbb{C}^{n}$. It is easy to see that (1.2) is a general setting of the
equation $f(z)^{2}+f(z+c)^{2}=1$. Consequently, we prove the following result.

######  Theorem 1.1.

Let $f(z)$ be a transcendental entire function defined on $\mathbb{C}^{n}$ and
$c=(c_{1},c_{2},\ldots,c_{n})\in\mathbb{C}^{n}$. If $f(z)$ is a solution of
(1.2) such that $T(r,P)=o(T(r,f))$ and $T(r,Q)=o(T(r,f))$, then $P(z)=\pm 1$,
$Q(z)$ reduces to a constant and $f(z)$ is of the form $f(z)=\cos(L(z)+B)$ ,
where $L(z)=a_{1}z_{1}+\cdots+a_{n}z_{n}$ such that $L(c)=-\pi/2-2k\pi$
$(k\in\mathbb{Z})$, and $B$ is a constant on $\mathbb{C}$.

Besides finding the solutions of Fermat-type difference equations, Fermat-type
differential-difference equations are also studied by many researchers. For
example, Xu and Cao [33] have investigated the entire solutions of Fermat-type
partial differential-difference equation

$\displaystyle\left(\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}\right)^{n}+f^{m}(z_{1}+c_{1},z_{2}+c_{2})=1$ (1.3)

and obtained the following interesting result for the functions in
$\mathbb{C}^{2}$.

######  Theorem C.

[33] Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$. Then the Fermat-
type partial differential-difference equation $(\ref{e1.3})$ does not have any
transcendental entire solution with finite order, where $m$ and $n$ are two
distinct positive integers.

In $2020$, Xu and Wang [35] generalized Theorem C by considering the following
Fermat-type partial differential-difference equation

$\displaystyle\left(\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}+\frac{\partial f(z_{1},z_{2})}{\partial
z_{2}}\right)^{n}+f^{m}(z_{1}+c_{1},z_{2}+c_{2})=1$ (1.4)

and proved the following result.

######  Theorem D.

[35] Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$ and $m$, $n$ be
two positive integers. If the Fermat-type partial differential-difference
equation (1.4) satisfies one of the following conditions:

  1. (i)

$m>n$;

  2. (ii)

$n>m\geq 2$,

then (1.4) does not have any finite order transcendental entire solutions.

As far as we know, there are few results about the complex difference and
complex difference equations for the functions in several complex variables.
Furthermore, it appears that the Fermat-type mixed partial differential-
difference equations in several complex variables has not been addressed in
the literature before. In order to generalize and also to establish a result
which combines Theorem C and Theorem D, in our investigation, we consider the
following partial differential-difference equation

$\displaystyle(\partial^{I}f(z_{1},z_{2})+\partial^{J}f(z_{1},z_{2}))^{n}+f^{m}(z_{1}+c_{1},z_{2}+c_{2})=1,$
(1.5)

where

$\displaystyle\partial^{I}f(z_{1},z_{2})=\displaystyle\frac{\partial^{|I|}f(z_{1},z_{2})}{\partial
z_{1}^{\alpha_{1}}\partial
z_{2}^{\alpha_{2}}}\;\;\mbox{and}\;\;\partial^{J}f(z_{1},z_{2})=\displaystyle\frac{\partial^{|J|}f(z_{1},z_{2})}{\partial
z_{1}^{\beta_{1}}\partial z_{2}^{\beta_{2}}}$

with $I=(\alpha_{1},\alpha_{2})$ and $J=(\beta_{1},\beta_{2})$ are two multi-
index, where $\alpha_{1},\alpha_{2}$, $\beta_{1}$ and $\beta_{2}$ are non-
negative integers and $c_{1},c_{2}\in\mathbb{C}$. We denote by $\mid I\mid$ to
denote the length of $I$, that is, $\mid I\mid=\alpha_{1}+\alpha_{2}$.
Similarly, for $J$ also. Throughout this paper, we denote
$z+w=(z_{1}+w_{1},z_{2}+w_{2})$ for any $z=(z_{1},z_{2})$, $w=(w_{1},w_{2})$
and $c=(c_{1},c_{2})$.

As a matter of fact, we prove the next result for any order Fermat-type
partial differential-difference equation (1.5).

######  Theorem 1.2.

Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$ and $m$, $n$ be two
positive integers. If the Fermat-type partial differential-difference equation
(1.5) satisfies one of the following conditions:

  1. (i)

$m>n$;

  2. (ii)

$n>m\geq 2$,

then (1.5) does not have any finite order transcendental entire solutions.

In this regard, we recall here a result of Xu and Cao [33] in which the
authors have investigated the solutions of the following Fermat-type partial
differential-difference equation

$\displaystyle\left(\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+f^{2}(z_{1}+c_{1},z_{2}+c_{2})=1$ (1.6)

and proved that the finite order transcendental entire solutions of (1.6) must
assume the form $f(z_{1},z_{2})=\sin(Az_{1}+B)$, where $A\in\mathbb{C}$ be
such that $Ae^{iAc_{1}}=1$, and $B$ is a constant on $\mathbb{C}$ (see [33,
Theorem 1.2]). The authors concluded that, all entire solutions
$f(z_{1},z_{2})$ of (1.6) must be a function of one variable $z_{1}$.

######  Remark 1.1.

By the following example one can observe that the concluding part of the
result of Xu and Cao is not true in general.

######  Example 1.1.

Let $f(z_{1},z_{2})=\sin(z_{1}+z_{2})$ or $\cos(z_{1}+z_{2})$. Choose,
$c=(c_{1},c_{2})\in\mathbb{C}^{2}$ such that $c_{1}+c_{2}=k\pi$ with
$k\in\mathbb{Z}$. Then it is easy to see that $f(z_{1},z_{2})$ is a solution
of the equation $(\ref{e2.3})$ but $f$ is not a function of $z_{1}$ only.

In 2020, Xu and Cao [34] corrected Theorem 1.2 of [33] and obtained the
following result.

######  Theorem E.

[34] Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$. Then any
transcendental entire solution with finite order of the Fermat-type partial
differential-difference equation $(\ref{e2.3})$ has the form
$f(z_{1},z_{2})=\sin(Az_{1}+Bz_{2}+H(z_{2}))$, Where $A,B$ are constant on
$\mathbb{C}$ satisfying $A^{2}=1$ and $Ae^{i(Ac_{1}+Bc_{2})}=1$, and
$H(z_{2})$ is a polynomial in one variable $z_{2}$ such that $H(z_{2})\equiv
H(z_{2}+c_{2})$. In the special case whenever $c_{2}\neq 0$, we have
$f(z_{1},z_{2})=\sin(Az_{1}+Bz_{2}+Constant)$.

After a detailed study of the proof of result of Xu and Cao [33, Theorem 1.2]
and [34, Theorem 1.1], a lacuna is found when the authors were trying to come
up with a specific form of solution. For a transcendental entire function
$f(z_{1},z_{2})$ with finite order satisfying the relation

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

the authors assumed that

$\displaystyle\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}+if(z_{1}+c_{1},z_{2}+c_{2})=e^{ip(z_{1},z_{2})}$

and

$\displaystyle\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}-if(z_{1}+c_{1},z_{2}+c_{2})=e^{-ip(z_{1},z_{2})},$

whereas the correct assumption will be

$\displaystyle\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}+if(z_{1}+c_{1},z_{2}+c_{2})=\beta_{1}e^{\alpha p(z_{1},z_{2})}$

and

$\displaystyle\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}-if(z_{1}+c_{1},z_{2}+c_{2})=\beta_{2}e^{-\alpha p(z_{1},z_{2})},$

where $\alpha$, $\beta_{1}$ and $\beta_{2}$ are complex constants in one
variable such that $\beta_{1}\beta_{2}=1$, and $p(z_{1},z_{2})$ is a
polynomial. Therefore, in order to find out the complete characterization of
the solutions of (1.6) that Xu and Cao considered in their investigations, it
is required to answer the following question.

######  Question 1.1.

What could be the precise form of solutions of the Fermat-type partial
differential-difference equation $(\ref{e2.3})$?

We prove the following result which answers Question 1.1 completely.

######  Theorem 1.3.

Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$. Then any
transcendental entire solution with finite order of the partial differential-
difference equation $(\ref{e2.3})$ is of the form

$\displaystyle
f(z_{1},z_{2})=\frac{A}{2i}\left(\beta_{1}e^{\alpha\left(\displaystyle\frac{i}{\alpha}Az_{1}+Bz_{2}+h(z_{2})\right)}-\beta_{2}e^{-\alpha\left(\displaystyle\frac{i}{\alpha}Az_{1}+Bz_{2}+h(z_{2})\right)}\right),$

where $A$, $B$, $\alpha$, $\beta_{1}$, $\beta_{2}$ are complex constants in
one variable with $A^{2}=1$, $\beta_{1}\beta_{2}=1$, and $h$ is a polynomial
in $z_{2}$ satisfying the relation
$h(z_{2})-h(z_{2}+c_{2})=\frac{i}{\alpha}Ac_{1}+Bc_{2}+\frac{i}{\alpha}\log
A$, and $Ae^{iAc_{1}+\alpha Bc_{2}}=1$. In the special case whenever
$c_{2}\neq 0$, $h$ reduces to a constant and the form of the solution is

$\displaystyle
f(z_{1},z_{2})=\frac{A}{2i}\left(\beta_{1}e^{\alpha\left(\displaystyle\frac{i}{\alpha}Az_{1}+Bz_{2}+\text{constant}\right)}-\beta_{2}e^{-\alpha\left(\displaystyle\frac{i}{\alpha}Az_{1}+Bz_{2}+\text{constant}\right)}\right).$

######  Remark 1.2.

Let $A=1$, $\alpha=i$, $\beta_{1}=\beta_{2}=1$. Choose $c_{1}$, $c_{2}$ in
$\mathbb{C}$ such that $Ae^{i(Ac_{1}+Bc_{2})}=1$, and $h$ is a polynomial in
$z_{2}$ only such that $h(z_{2})=h(z_{2}+c_{2})$. Then from Theorem 1.3, it is
clear that the solution of the partial differential-difference equation (1.6)
is

$\displaystyle f(z_{1},z_{2})=\sin(Az_{1}+Bz_{2}+h(z_{2})).$

Hence, Theorem 1.3 is a significant improvement of the result of Xu and Cao
[33, Theorem 1.2].

######  Example 1.2.

Let $\beta_{1}=\beta_{2}=1$, $\alpha=1$, $A=B=1$ and $h(z_{2})=7$, where $h$
is defined in Theorem 1.3. Choose $c=(c_{1},c_{2})$ in $\mathbb{C}^{2}$ such
that $e^{ic_{1}+c_{2}}=1$, $c_{2}\neq 0$. Then it can be easily verified that

$\displaystyle f(z_{1},z_{2})=-i\sinh(iz_{1}+z_{2}+7)$

is a solution of the Fermat-type partial differential-difference equation
(1.6).

######  Example 1.3.

Let $\beta_{1}=\beta_{2}=1$, $\alpha=1$, $A=1$, $B=3$ and
$h(z_{2})=z_{2}^{3}+z_{2}^{2}+13$, where $h$ is defined in Theorem 1.3. Choose
$c=(c_{1},0)$ in $\mathbb{C}^{2}$ such that $e^{ic_{1}}=1$. Then it can be
easily verified that

$\displaystyle f(z_{1},z_{2})=-i\sinh(iz_{1}+3z_{2}+z_{2}^{3}+z_{2}^{2}+13)$

is a solution of the Fermat-type partial differential-difference equation
(1.6).

Next, we are interested to find solutions of Fermat-type partial differential-
difference equations. Henceforth, we consider the following equations

$\displaystyle\left(\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}\right)^{2}+(f(z_{1}+c_{1},z_{2}+c_{2})-f(z_{1},z_{2}))^{2}=1$ (1.7)

and

$\displaystyle f^{2}(z_{1},z_{2})+P^{2}(z_{1},z_{2})\left(\frac{\partial
f(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}-\frac{\partial
f(z_{1},z_{2})}{\partial z_{1}}\right)^{2}=1,$ (1.8)

where $P(z_{1},z_{2})$ is a non-zero polynomial in $\mathbb{C}^{2}$. For
solutions of the system of Fermat-type partial differential-difference
equations analogue to (1.7), we refer to the article [36] and references
therein.

We prove the following result finding the precise form of the solutions of
(1.7).

######  Theorem 1.4.

Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$. Then any
transcendental entire solution with finite order of the Fermat-type partial
differential-difference equation $(\ref{e2.4})$ is of the form

$\displaystyle
f(z_{1},z_{2})=-\frac{1}{4i}\left(\beta_{1}e^{\alpha\left(\displaystyle\frac{-2i}{\alpha}z_{1}+Bz_{2}+h(z_{2})\right)}-\beta_{2}e^{-\alpha\left(\displaystyle\frac{-2i}{\alpha}z_{1}+Bz_{2}+h(z_{2})\right)}\right),$

where $B$, $\alpha$, $\beta_{1}$, $\beta_{2}$ are complex constants in one
variable with $\beta_{1}\beta_{2}=1$, and $h$ is a polynomial in $z_{2}$ only
satisfying the relation

$\displaystyle
h(z_{2})-h(z_{2}+c_{2})=-\frac{2ic_{1}}{\alpha}+Bc_{2}+\frac{1}{\alpha}\log(-1).$

In the special case when $c_{2}\neq 0$, $h$ must be constant and the solution
is of the form

$\displaystyle
f(z_{1},z_{2})=-\frac{1}{4i}\left(\beta_{1}e^{\alpha\left(\displaystyle\frac{-2i}{\alpha}z_{1}+Bz_{2}+C)\right)}-\beta_{2}e^{-\alpha\left(\displaystyle\frac{-2i}{\alpha}z_{1}+Bz_{2}+C\right)}\right),$

where $B$ and $C$ are complex constant such that $e^{2ic_{1}-\alpha
Bc_{2}}=-1$.

Next, we list some examples to exhibit the existence of solution of the
Fermat-type partial differential-difference equation (1.7).

######  Example 1.4.

Let $\beta_{1}=\beta_{2}=1$, $\alpha=i$, $B=3$. Choose $c=(c_{1},0)$ in
$\mathbb{C}^{2}$ such that $e^{2ic_{1}}=-1$. Then from Theorem 1.4, it is
clear that

$\displaystyle
f(z_{1},z_{2})=-\frac{1}{2}\sin(-2z_{1}+3z_{2}+z_{2}^{5}+z_{2}^{3}+1)$

is a solution of the Fermat-type partial differential-difference equation
(1.7).

######  Example 1.5.

Let $\beta_{1}=\beta_{2}=1$, $\alpha=i$, $B=5$. Choose $c=(c_{1},c_{2})$ in
$\mathbb{C}^{2}$ such that $e^{2ic_{1}-5c_{2}}=-1$, $c_{2}\neq 0$. Then from
Theorem 1.4, it is clear that

$\displaystyle f(z_{1},z_{2})=-\frac{1}{2}\sin(-2z_{1}+5z_{2}+11)$

is a solution of the Fermat-type partial differential-difference equation
(1.7).

######  Example 1.6.

Let $\beta_{1}=\beta_{2}=1$, $\alpha=1$, $B=1$ and $h(z_{2})=10$, where $h$ is
defined in Theorem 1.4. Choose $c=(c_{1},c_{2})$ in $\mathbb{C}^{2}$ such that
$e^{-(-2ic_{1}+c_{2})}=-1$, $c_{2}\neq 0$. Then one can easily verify that

$\displaystyle f(z_{1},z_{2})=-\frac{1}{2i}\sinh(-2iz_{1}+z_{2}+10)$

is a solution of the Fermat-type partial differential-difference equation
(1.7).

######  Example 1.7.

Let $\beta_{1}=\beta_{2}=1$, $\alpha=1$, $B=15$ and
$h(z_{2})=z_{2}^{6}+z_{2}^{4}+z_{2}^{2}+1$, where $h$ is defined in Theorem
1.4. Choose $c=(c_{1},0)$ in $\mathbb{C}^{2}$ such that $e^{2ic_{1}}=-1$. Then
one can easily verify that

$\displaystyle
f(z_{1},z_{2})=-\frac{1}{2i}\sinh(-2iz_{1}+15z_{2}+z_{2}^{6}+z_{2}^{4}+z_{2}^{2}+1)$

is a solution of the Fermat-type partial differential-difference equation
(1.7).

We prove the next result finding the precise form of the solutions of (1.8).

######  Theorem 1.5.

Let $c=(c_{1},c_{2})$ be a constant in $\mathbb{C}^{2}$. Then any
transcendental entire solution with finite order of the Fermat-type partial
differential-difference equation $(\ref{e2.5})$ is of the form

$\displaystyle
f(z_{1},z_{2})=\frac{1}{2}\left(\beta_{1}e^{\alpha(Az_{1}+Bz_{2}+C)}+\beta_{2}e^{-\alpha(Az_{1}+Bz_{2}+C)}\right),$

where $A$, $B$, $C$ and $\alpha$ are complex constants in one variable such
that $e^{\alpha(Ac_{1}+Bc_{2})}=-1$, and $P(z_{1},z_{2})$ reduces to a
constant $-1/2i\alpha A$.

Next, we list some examples to exhibit the existence of solution of the
Fermat-type partial differential-difference equation (1.8).

######  Example 1.8.

Let $\beta_{1}=\beta_{2}=1$, $\alpha=i$, $A=2$, $B=5$, $C=12$ and
$P(z_{1},z_{2})=1/4$. Choose $c=(c_{1},c_{2})$ in $\mathbb{C}^{2}$ such that
$e^{i(2c_{1}+5c_{2})}=-1$. Then from Theorem 1.5, it is clear that

$\displaystyle f(z_{1},z_{2})=\cos(2z_{1}+5z_{2}+12)$

is a solution of the Fermat-type partial differential-difference equation
(1.8).

######  Example 1.9.

Let $\beta_{1}=\beta_{2}=1$, $\alpha=1$, $A=B=C=1$ and $P(z_{1},z_{2})=-1/2i$.
Choose $c=(c_{1},c_{2})$ in $\mathbb{C}^{2}$ such that $e^{c_{1}+c_{2}}=-1$.
Then from Theorem 1.5, it is clear that

$\displaystyle f(z_{1},z_{2})=\cosh(z_{1}+z_{2}+1)$

is a solution of the Fermat-type partial differential-difference equation
(1.8).

##  2\. Key Lemmas

In this section, we present some necessary lemmas which will play key role to
prove the main results of this paper.

######  Lemma 2.1.

[12] 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 2.2.

[16, 27, 28] 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 2.3.

[26] 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 2.4.

[1, 39] Let $f$ be a non-constant meromorphic function on $\mathbb{C}^{n}$ and
let $I=(\alpha_{1},\ldots,\alpha_{n})$ be a multi-index with length
$|I|=\sum_{j=1}^{n}\alpha_{j}$. Assume that $T(r_{0},f)\geq e$ for some
$r_{0}$. Then

$\displaystyle m\left(r,\frac{\partial^{I}f}{f}\right)=S(r,f)$

holds for all $r\geq r_{0}$ outside a set $E\subset(0,+\infty)$ of finite
logarithmic measure, $\displaystyle\int_{E}\frac{dt}{t}<\infty$, where
$\partial^{I}f=\displaystyle\frac{\partial^{|I|}f}{\partial
z_{1}^{\alpha_{1}}\ldots\partial z_{2}^{\alpha_{2}}}$.

######  Lemma 2.5.

[2, 14] Let $f$ be a non-constant meromorphic function with finite order on
$\mathbb{C}^{n}$ such that $f(0)\neq 0,\infty$, and let $\epsilon>0$. Then for
$c\in\mathbb{C}^{n}$,

$\displaystyle
m\left(r,\frac{f(z+c)}{f(z)}\right)+m\left(r,\frac{f(z)}{f(z+c)}\right)=S(r,f)$

holds for all $r\geq r_{0}$ outside a set $E\subset(0,+\infty)$ of finite
logarithmic measure, $\displaystyle\int_{E}\frac{dt}{t}<\infty$.

##  3\. Proof of the main results

###### Proof of Theorem 1.1.

Let
$P(z)=\sum_{|I|=0}^{p}a_{\alpha_{1},\ldots,\alpha_{n}}z_{1}^{\alpha_{1}}\ldots
z_{n}^{\alpha_{n}}$ and
$Q(z)=\sum_{|J|=0}^{q}a_{\beta_{1},\ldots,\beta_{n}}z_{1}^{\beta_{1}}\ldots
z_{n}^{\beta_{n}}$ be two polynomials in $\mathbb{C}^{n}$, where
$I=(\alpha_{1},\ldots,\alpha_{n})$, $J=(\beta_{1},\ldots,\beta_{n})$ be two
multi-index with $|I|=\sum_{j=0}^{n}\alpha_{j}$ and
$|I|=\sum_{j=0}^{n}\beta_{j}$, and $\alpha_{j}$, $\beta_{j}$ are non-negative
integers.

Suppose $f$ is a finite order transcendental entire solution of
$(\ref{e2.1})$. We write $(\ref{e2.1})$ as

$\displaystyle h_{1}(z)h_{2}(z)=Q(z),$ (3.1)

where $h_{1}(z)=f(z)+iP(z)f(z+c)$ and $h_{2}(z)=f(z)-iP(z)f(z+c)$.

Since $Q(0)\neq 0$, hence we must have $h_{1}(0)\neq 0$ and $h_{2}(0)\neq 0$.
Now in view of $(\ref{e4.1})$ and Lemma 2.2, it is easy to see that

$\displaystyle f(z)+iP(z)f(z+c)=\chi_{{}_{1}}(z)e^{\chi_{{}_{2}}(z)};$ (3.2)
$\displaystyle f(z)-iP(z)f(z+c)=\chi_{{}_{3}}(z)e^{-\chi_{{}_{2}}(z)},$ (3.3)

where $\chi_{{}_{1}}(z)$, $\chi_{{}_{3}}(z)$ are canonical functions of
$h_{1}$ and $h_{2}$, respectively and $\chi_{{}_{2}}(z)$ is entire on
$\mathbb{C}^{n}$ such that $\chi_{{}_{1}}(z)\chi_{{}_{3}}(z)=Q(z)$.

Since $f$ is a transcendental entire function of finite order, it follows from
$(\ref{e4.2})$ and $(\ref{e4.3})$ that $e^{\chi_{{}_{2}}(z)}$ is of finite
order, and hence by Lemma 2.3, it is easy to see that $\chi_{{}_{2}}(z)$ is a
polynomial in $\mathbb{C}^{n}$.

Solving $(\ref{e4.2})$ and $(\ref{e4.3})$, we obtain

$\displaystyle
f(z)=\frac{1}{2}\left(\chi_{{}_{1}}(z)e^{\chi_{{}_{2}}(z)}+\chi_{{}_{3}}(z)e^{-\chi_{{}_{2}}(z)}\right);$
(3.4) $\displaystyle
f(z+c)=\frac{1}{2iP(z)}\left(\chi_{{}_{1}}(z)e^{\chi_{{}_{2}}(z)}-\chi_{{}_{3}}(z)e^{-\chi_{{}_{2}}(z)}\right).$
(3.5)

Combining $(\ref{e4.4})$ and $(\ref{e4.5})$ yields that

$\displaystyle f(z+c)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\chi_{{}_{1}}(z+c)e^{\chi_{{}_{2}}(z+c)}+\chi_{{}_{3}}(z+c)e^{-\chi_{{}_{2}}(z+c)}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2iP(z)}\left(\chi_{{}_{1}}(z)e^{\chi_{{}_{2}}(z)}-\chi_{{}_{3}}(z)e^{-\chi_{{}_{2}}(z)}\right).$

Therefore, a simple computation shows that

$\displaystyle\frac{iP(z)\chi_{{}_{1}}(z+c)e^{\chi_{{}_{2}}(z+c)+\chi_{{}_{2}}(z)}}{-\chi_{{}_{3}}(z)}$
$\displaystyle+$
$\displaystyle\frac{iP(z)\chi_{{}_{1}}(z+c)e^{\chi_{{}_{2}}(z)-\chi_{{}_{2}}(z+c)}}{-\chi_{{}_{3}}(z)}$
(3.6)
$\displaystyle+\frac{\chi_{{}_{1}}(z)e^{2\chi_{{}_{2}}(z)}}{\chi_{{}_{3}}(z)}=1.$

Now $\chi_{{}_{2}}(z)$ being a non-constant polynomial, it is easy to see that
both

$\displaystyle\frac{iP(z)\chi_{{}_{1}}(z+c)e^{\chi_{{}_{2}}(z+c)+\chi_{{}_{2}}(z)}}{-\chi_{{}_{3}}(z)}\;\;\mbox{and}\;\;\frac{\chi_{{}_{1}}(z)e^{2\chi_{{}_{2}}(z)}}{\chi_{{}_{3}}(z)}$

are non-constants.

Since $T(r,P)=o(T(r,f))$ and $T(r,Q)=o(T(r,f))$, by Lemma 2.1, we obtain

$\displaystyle
iP(z)\chi_{{}_{3}}(z+c)e^{\chi_{{}_{2}}(z)-\chi_{{}_{2}}(z+c)}=-\chi_{{}_{3}}(z).$
(3.7)

Thus, we may write $\chi_{{}_{2}}(z)=L(z)+b$, where
$L(z)=a_{1}z_{1}+a_{2}z_{2}+\cdots+a_{n}z_{n}$, $a_{i},b\in\mathbb{C}$, $1\leq
i\leq n$. Therefore, $(\ref{e4.7})$ reduces to the following form

$\displaystyle
iP(z)\chi_{{}_{3}}(z+c)e^{-\sum_{i=1}^{n}a_{i}c_{i}}=-\chi_{{}_{3}}(z).$ (3.8)

Similarly, from $(\ref{e4.6})$, we easily obtain

$\displaystyle
iP(z)\chi_{{}_{1}}(z+c)e^{\sum_{i=1}^{n}a_{i}c_{i}}=\chi_{{}_{1}}(z).$ (3.9)

Combining 3.8 and $(\ref{e4.9})$ yields

$\displaystyle
P^{2}(z)\chi_{{}_{1}}(z+c)\chi_{{}_{3}}(z+c)=\chi_{{}_{1}}(z)\chi_{{}_{3}}(z)$

which can be written as

$\displaystyle P^{2}(z)Q(z+c)=Q(z).$

Since $Q(z)$ is a non-zero polynomial, we must have $P^{2}(z)=1$ and hence
$P(z)=\pm 1$. Consequently, $Q(z)$ reduces to a constant $q$, say. Hence,
$(\ref{e2.1})$ takes the form $g^{2}(z)+g^{2}(z+c)=1$, where
$g(z)=f(z)/\sqrt{q}$. Therefore, with the help of [33, Theorem 1.4], we easily
obtain conclusion of the theorem. ∎

###### Proof of Theorem 1.2.

We discuss the whole proof into the following two cases. Case 1: Let $m>n$.
Since $f$ is entire, using Lemma 2.4, we obtain

$\displaystyle T(r,f(z_{1},z_{2}))$ $\displaystyle=$ $\displaystyle
m(r,z_{1},z_{2})$ $\displaystyle\leq$ $\displaystyle
m\left(r,\frac{f(z_{1},z_{2})}{f(z_{1}+c_{1},z_{2}+c_{2})}\right)+m(r,f(z_{1}+c_{1},z_{2}+c_{2}))+\log
2$ $\displaystyle\leq$ $\displaystyle m(r,f(z_{1}+c_{1},z_{2}+c_{2}))+S(r,f)$
$\displaystyle\leq$ $\displaystyle T(r,f(z_{1}+c_{1},z_{2}+c_{2}))+S(r,f).$

By the Mohon’ko theorem for functions in $\mathbb{C}^{2}$ (see [11]), an easy
and straight forward computation shows that

$\displaystyle mT(r,f(z_{1},z_{2}))$ $\displaystyle\leq$ $\displaystyle
mT(r,f(z_{1}+c_{1},z_{2}+c_{2}))+S(r,f)$ $\displaystyle=$ $\displaystyle
T(r,f^{m}(z_{1}+c_{1},z_{2}+c_{2}))+S(r,f)$ $\displaystyle\leq$ $\displaystyle
T(r,(\partial^{I}f+\partial^{J}f)^{n}-1)+S(r,f)$ $\displaystyle\leq$
$\displaystyle nT(r,\partial^{I}f+\partial^{J}f)+S(r,f)$ $\displaystyle=$
$\displaystyle nm(r,\partial^{I}f+\partial^{J}f)+S(r,f)$ $\displaystyle\leq$
$\displaystyle
nm\left(r,\frac{\partial^{I}f+\partial^{J}f}{f}\right)+nm(r,f(z_{1},z_{2}))+S(r,f)$
$\displaystyle\leq$ $\displaystyle
nm\left(r,\frac{\partial^{I}f}{f}\right)+m\left(r,\frac{\partial^{J}f}{f}\right)+nm(r,f(z_{1},z_{2}))+S(r,f)$
$\displaystyle\leq$ $\displaystyle nT(r,f(z_{1},z_{2}))+S(r,f).$

Therefore, we must have

$\displaystyle(n-m)T(r,f(z_{1},z_{2}))\leq S(r,f),$

which contradicts to the fact that $f$ is transcendental and also $m>n$.

Case 2: Suppose $n>m\geq 2$. In this case, it is easy to see that $1/m+1/n<1$
and $m>n/(n-1)$. Let $d_{1},d_{2},\ldots,d_{n}$ are the roots of the equation
$w^{n}=1$. Then by the Second Fundamental Theorem of Nevalinna, and the
equation $(\ref{e22.5})$, a simple computation shows that

$\displaystyle(n-1)T(r,\partial^{I}f+\partial^{J}f)$ $\displaystyle\leq$
$\displaystyle\overline{N}(r,\partial^{I}f+\partial^{J}f)+\sum_{j=1}^{n}\overline{N}\left(r,\frac{1}{\partial^{I}f+\partial^{J}f-d_{j}}\right)$
$\displaystyle+S(r,\partial^{I}f+\partial^{J}f)$ $\displaystyle\leq$
$\displaystyle\overline{N}\left(r,\frac{1}{(\partial^{I}f+\partial^{J}f)^{n}-1}\right)+S(r,\partial^{I}f+\partial^{J}f)$
$\displaystyle\leq$
$\displaystyle\overline{N}\left(r,\frac{1}{f(z_{1}+c_{1},z_{2}+c_{2})}\right)+S(r,\partial^{I}f+\partial^{J}f)$
$\displaystyle\leq$ $\displaystyle T(r,f(z_{1}+c_{1},z_{2}+c_{2}))+S(r,f).$

On the other hand, by Mohon’ko theorem for the functions in $\mathbf{C}^{2}$
[11] and (1.5), we obtain

$\displaystyle mT(r,f(z_{1}+c_{1},z_{2}+c_{2}))$ $\displaystyle=$
$\displaystyle T(r,f^{m}(z_{1}+c_{1},z_{2}+c_{2}))+S(r,f)$ $\displaystyle=$
$\displaystyle T(r,(\partial^{I}f+\partial^{J}f)^{n}-1)+S(r,f)$
$\displaystyle=$ $\displaystyle nT(r,\partial^{I}f+\partial^{J}f)+S(r,f).$

Thus we have

$\displaystyle
mT(r,f(z_{1}+c_{1},z_{2}+c_{2}))\leq\frac{n}{n-1}T(r,f(z_{1}+c_{1},z_{2}+c_{2}))+S(r,f)$

and this can be written as

$\displaystyle\left(m-\frac{n}{n-1}\right)T(r,f(z_{1}+c_{1},z_{2}+c_{2}))\leq
S(r,f),$

which is not possible since $f$ is transcendental and $m>{n}/{(n-1)}$. This
completes the proof. ∎

###### Proof of Theorem 1.3.

First we assume that $f$ is a finite order transcendental entire solution of
$(\ref{e2.3})$. Rewriting equation $(\ref{e2.3})$, we obtain

$\displaystyle\left(\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}+if(z_{1}+c_{1},z_{2}+c_{2})\right)\left(\frac{\partial
f(z_{1},z_{2})}{\partial z_{1}}-if(z_{1}+c_{1},z_{2}+c_{2})\right)=1.$

From this equation, it is easy to see that both $\frac{\partial
f(z_{1},z_{2})}{\partial z_{1}}+if(z_{1}+c_{1},z_{2}+c_{2})$ and
$\frac{\partial f(z_{1},z_{2})}{\partial z_{1}}-if(z_{1}+c_{1},z_{2}+c_{2})$
do not have any zeros in $\mathbb{C}^{2}$.

Hence, in view of Lemma 2.2, we may assume

$\displaystyle\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}+if(z_{1}+c_{1},z_{2}+c_{2})=\beta_{1}e^{\alpha p(z_{1},z_{2})}$

and

$\displaystyle\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}-if(z_{1}+c_{1},z_{2}+c_{2})=\beta_{2}e^{-\alpha p(z_{1},z_{2})},$

where $p(z_{1},z_{2})$ is a non-constant entire function in $\mathbb{C}^{2}$,
$\alpha$, $\beta_{1}$, $\beta_{2}$ are complex constants such that
$\beta_{1}\beta_{2}=1$.

From these last two equations, a simple computation shows that

$\displaystyle\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}=\frac{\beta_{1}e^{\alpha p(z_{1},z_{2})}+\beta_{2}e^{-\alpha
p(z_{1},z_{2})}}{2}$ (3.10)

and

$\displaystyle f(z_{1}+c_{1},z_{2}+c_{2})=\frac{\beta_{1}e^{\alpha
p(z_{1},z_{2})}-\beta_{2}e^{-\alpha p(z_{1},z_{2})}}{2i}.$ (3.11)

Using Lemma 2.3, it is easy to see that $p(z_{1},z_{2})$ must be a non-
constant polynomial in $\mathbb{C}^{2}$. Hence, it follows from
$(\ref{e4.11})$ that $p(z_{1},z_{2})$ is a non-constant polynomial in
$\mathbb{C}^{2}$.

From $(\ref{e4.10})$ and $(\ref{e4.11})$, we obtain the following

$\displaystyle\frac{\partial f(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}$
$\displaystyle=$ $\displaystyle\frac{\beta_{1}e^{\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}+\beta_{2}e^{-\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}}{2}$ $\displaystyle=$
$\displaystyle\alpha\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}\frac{\beta_{1}e^{\alpha p(z_{1},z_{2})}+\beta_{2}e^{-\alpha
p(z_{1},z_{2})}}{2i}.$

Therefore, an elementary computation shows that

$\displaystyle\frac{\beta_{1}\alpha}{i\beta_{2}}\frac{\partial
p(z_{1},z_{2})}{\partial z_{1}}e^{\alpha
p(z_{1},z_{2})+ip(z_{1}+c_{1},z_{2}+c_{2})}$ $\displaystyle+$
$\displaystyle\frac{\alpha}{i}\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}e^{\alpha p(z_{1}+c_{1},z_{2}+c_{2})-ip(z_{1},z_{2})}$ (3.12)
$\displaystyle-\frac{\beta_{1}}{\beta_{2}}e^{2\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}=1.$

From the above equation, we see that $\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}$ is non-zero polynomial. For otherwise, it follows from $(\ref{e4.12})$
that $e^{2\alpha p(z_{1}+c_{1},z_{2}+c_{2})}=-\beta_{2}^{2}=\text{constant}$,
which implies that $p(z_{1},z_{2})$ is constant, a contradiction. Hence, both
$e^{2\alpha p(z_{1}+c_{1},z_{2}+c_{2})}$ and
$\frac{\beta_{1}\alpha}{i\beta_{2}}\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}e^{\alpha p(z_{1},z_{2})+\alpha p(z_{1}+c_{1},z_{2}+c_{2})}$ are non-
constants.

Also notice that

$\displaystyle N(r,e^{\alpha
2p(z_{1}+c_{1},z_{2}+c_{2})})=N\left(r,\frac{1}{e^{\alpha
2p(z_{1}+c_{1},z_{2}+c_{2})}}\right)=S(r,f),$

$\displaystyle N\left(r,\frac{\beta_{1}\alpha}{i\beta_{2}}\frac{\partial
p(z_{1},z_{2})}{\partial z_{1}}e^{\alpha p(z_{1},z_{2})+\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}\right)$ $\displaystyle=$ $\displaystyle
N\left(r,\frac{i\beta_{2}}{\frac{\beta_{1}\alpha\partial p(z)}{\partial
z_{1}}e^{\alpha p(z_{1},z_{2})+\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}}\right)=S(r,f).$

Therefore, by Lemma 2.1, we obtain

$\displaystyle-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial z_{1}}e^{\alpha
p(z_{1}+c_{1},z_{2}+c_{2})-\alpha p(z_{1},z_{2})}=1,$

which implies that

$\displaystyle-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial z_{1}}=e^{\alpha
p(z_{1},z_{2})-\alpha p(z_{1}+c_{1},z_{2}+c_{2})}.$ (3.13)

From $(\ref{e4.12})$ and $(\ref{e4.13})$, it is easy to obtain

$\displaystyle-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial z_{1}}=e^{\alpha
p(z_{1}+c_{1},z_{2}+c_{2})-\alpha p(z_{1},z_{2})}.$ (3.14)

We observe that L.H.S. and R.H.S. of (3.13) are respectively, polynomial (non-
transcendental) and transcendental entire function. Therefore, the only
possibility is that $\alpha p(z_{1},z_{2})-\alpha p(z_{1}+c_{1},z_{2}+c_{2})$
and hence $e^{\alpha p(z_{1},z_{2})-\alpha p(z_{1}+c_{1},z_{2}+c_{2})}$ is a
constant. Assume that

$\displaystyle-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial z_{1}}=e^{\alpha
p(z_{1},z_{2})-\alpha p(z_{1}+c_{1},z_{2}+c_{2})}=A,$ (3.15)

where $A$ is a non-zero complex constant. Then from (3.14) , we obtain

$\displaystyle-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial z_{1}}=e^{\alpha
p(z_{1}+c_{1},z_{2}+c_{2})-\alpha p(z_{1},z_{2})}=\frac{1}{A}.$ (3.16)

From (3.15) and (3.16), it can be easily seen that $A^{2}=1$. Again in view of
(3.15), we may assume that $p(z_{1},z_{2})=iAz_{1}/\alpha+g(z_{2})$, where
$g(z_{2})$ is a polynomial in $z_{2}$ only.

From (3.15), we get

$\displaystyle p(z_{1},z_{2})-p(z_{1}+c_{1},z_{2}+c_{2})=\frac{1}{\alpha}\log
A.$

This implies that

$\displaystyle
g(z_{2})-g(z_{2}+c_{2})=\frac{i}{\alpha}Ac_{1}+\frac{1}{\alpha}\log A.$

We may write $g(z_{2})=Bz_{2}+h(z_{2})$, where $B$ is a complex constant in
one variable and $h$ is a polynomial in one variable $z_{2}$ of degree greater
than one. Then we easily obtain

$\displaystyle
h(z_{2})-h(z_{2}+c_{2})=\frac{i}{\alpha}Ac_{1}+Bc_{2}+\frac{1}{\alpha}\log A.$
(3.17)

Now if $c_{2}=0$, then from (3.17), we obtain $Ae^{iAc_{1}}=1$. If $c_{2}\neq
0$, then from (3.17), it is clear that $h(z_{2})$ must be constant, and in
that case we obtain $Ae^{iAc_{1}+\alpha Bc_{2}}=1$. Since $A^{2}=1$, we have
$iAc_{1}+\alpha Bc_{2}=k\pi$, where $k$ is an integer.

Hence, keeping in view of (3.11), (3.17), we get after simple calculation

$\displaystyle
f(z_{1},z_{2})=\frac{A}{2i}\left(\beta_{1}e^{\alpha\left(\displaystyle\frac{i}{\alpha}Az_{1}+Bz_{2}+h(z_{2})\right)}-\beta_{2}e^{-\alpha\left(\displaystyle\frac{i}{\alpha}Az_{1}+Bz_{2}+h(z_{2})\right)}\right).$

In the special case whenever $c_{2}\neq 0$, $h$ reduces to a constant and the
form of the solution is

$\displaystyle
f(z_{1},z_{2})=\frac{A}{2i}\left(\beta_{1}e^{\alpha\left(\displaystyle\frac{i}{\alpha}Az_{1}+Bz_{2}+\text{constant}\right)}-\beta_{2}e^{-\alpha\left(\displaystyle\frac{i}{\alpha}Az_{1}+Bz_{2}+\text{constant}\right)}\right).$

This completes the proof. ∎

###### Proof of Theorem 1.4.

Assume that $f$ is a finite order transcendental entire solution of
$(\ref{e2.4})$. Let us denote $f(z_{1}+c_{1},z_{2}+c_{2})-f(z_{1},z_{2})$ by
$\Delta_{c}f(z_{1},z_{2})$. Rewriting equation $(\ref{e2.4})$ we have

$\displaystyle\left(\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}+i\Delta_{c}f(z_{1},z_{2})\right)\left(\frac{\partial
f(z_{1},z_{2})}{\partial z_{1}}-i\Delta_{c}f(z_{1},z_{2})\right)=1.$

Then by the same argument in the proof of Theorem 1.3, we see that

$\displaystyle\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}=\frac{\beta_{1}e^{\alpha p(z_{1},z_{2})}+\beta_{2}e^{-\alpha
p(z_{1},z_{2})}}{2}$ (3.18)

and

$\displaystyle
f(z_{1}+c_{1},z_{2}+c_{2})-f(z_{1},z_{2})=\frac{\beta_{1}e^{\alpha
p(z_{1},z_{2})}-\beta_{2}e^{-\alpha p(z_{1},z_{2})}}{2i}.$ (3.19)

Differentiating $(\ref{e4.17})$ partially with respect to $z_{1}$, we obtain

$\displaystyle\frac{\partial f(z_{1}+c_{1},z_{2}+c_{2})}{\partial
z_{1}}=\left(1-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}\right)\frac{\beta_{1}e^{\alpha p(z_{1},z_{2})}+\beta_{2}e^{-\alpha
p(z_{1},z_{2})}}{2}.$ (3.20)

Now keeping in view of (3.18) and (3.20), we obtain after simple calculation
that

$\displaystyle\frac{\beta_{1}}{\beta_{2}}\left(1-i\alpha\frac{\partial
p(z_{1},z_{2})}{\partial z_{1}}\right)e^{\alpha p(z_{1},z_{2})+\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}$ $\displaystyle+\left(1-i\alpha\frac{\partial
p(z_{1},z_{2})}{\partial z_{1}}\right)e^{-\alpha p(z_{1},z_{2})+\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}$
$\displaystyle-\frac{\beta_{1}}{\beta_{2}}e^{2\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}=1.$ (3.21)

From the above equation, we see that $\left(1-i\alpha\frac{\partial
p(z_{1},z_{2})}{\partial z_{1}}\right)$ is non-zero polynomial. For otherwise,
it follows from (3) that $e^{2\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}=-\beta_{1}^{2}=\text{constant}$, which implies
that $p(z_{1},z_{2})$ is constant, a contradiction. Hence, both
$-\frac{\beta_{1}}{\beta_{2}}e^{2\alpha p(z_{1}+c_{1},z_{2}+c_{2})}$ and
$\frac{\beta_{1}}{\beta_{2}}\left(1-i\alpha\frac{\partial
p(z_{1},z_{2})}{\partial z_{1}}\right)\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}e^{\alpha p(z_{1},z_{2})+\alpha p(z_{1}+c_{1},z_{2}+c_{2})}$ are non-
constants.

We also observe that

$\displaystyle N\left(r,e^{\alpha
2p(z_{1}+c_{1},z_{2}+c_{2})}\right)=N\left(r,\frac{1}{e^{\alpha
2p(z_{1}+c_{1},z_{2}+c_{2})}}\right)=S(r,f),$

$\displaystyle N\left(r,\left(1-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}\right)e^{\alpha ip(z_{1},z_{2})+\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}\right)$
$\displaystyle=N\left(r,\frac{1}{\left(1-i\alpha\frac{\partial
p(z_{1},z_{2})}{\partial z_{1}}\right)e^{\alpha ip(z_{1},z_{2})+\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}}\right)=S(r,f)$

and

$\displaystyle N\left(r,\left(1-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}\right)e^{-\alpha ip(z_{1},z_{2})+\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}\right)$
$\displaystyle=N\left(r,\frac{1}{\left(1-i\alpha\frac{\partial
p(z_{1},z_{2})}{\partial z_{1}}\right)e^{-\alpha ip(z_{1},z_{2})+\alpha
p(z_{1}+c_{1},z_{2}+c_{2})}}\right)=S(r,f).$

By Lemma 2.1, we easily obtain

$\displaystyle\left(1-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}\right)e^{ip(z_{1}+c_{1},z_{2}+c_{2})-ip(z_{1},z_{2})}=1,$ (3.22)

which implies that

$\displaystyle 1-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}=e^{ip(z_{1},z_{2})-ip(z_{1}+c_{1},z_{2}+c_{2})}.$ (3.23)

Again, using (3) and (3.22), we obtain

$\displaystyle 1-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}=e^{\alpha p(z_{1}+c_{1},z_{2}+c_{2})-ip(z_{1},z_{2})}.$ (3.24)

Now we observe that L.H.S. of $(\ref{e4.21})$ is a polynomial, whereas R.H.S.
of it is transcendental. Therefore, we conclude that $\alpha
p(z_{1},z_{2})-\alpha p(z_{1}+c_{1},z_{2}+c_{2})$, and thus $e^{\alpha
p(z_{1},z_{2})-\alpha p(z_{1}+c_{1},z_{2}+c_{2})}$ must be constant.

Let

$\displaystyle 1-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}=e^{ip(z_{1},z_{2})-ip(z_{1}+c_{1},z_{2}+c_{2})}=A,$ (3.25)

where $A$ is a non-zero complex constant in one variable.

Then from (3.24), we get

$\displaystyle 1-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial
z_{1}}=e^{\alpha p(z_{1}+c_{1},z_{2}+c_{2})-ip(z_{1},z_{2})}=\frac{1}{A}.$
(3.26)

Therefore, it follows from (3.25) and (3.26) that $A^{2}=1$.

Again, since $1-i\alpha\frac{\partial p(z_{1},z_{2})}{\partial z_{1}}=A$, we
may assume that

$\displaystyle p(z_{1},z_{2})=\frac{A-1}{\alpha}iz_{1}+g(z_{2}),$

where $g$ is a polynomial in one variable $z_{2}$ only.

From (3.25), it follows that
$p(z_{1},z_{2})-p(z_{1}+c_{1},z_{2}+c_{2})=\frac{1}{\alpha}\log A$. This
implies that

$\displaystyle
g(z_{2})-g(z_{2}+c_{2})=\frac{(A-1)i}{\alpha}c_{1}+\frac{1}{\alpha}\log A.$

Let $g(z_{2})=Bz_{2}+h(z_{2})$, where $h$ is a polynomial in $z_{2}$ only with
degree greater than 1 or constant and $B$ is a complex number in one variable.

Then

$\displaystyle
p(z_{1},z_{2})=\frac{(A-1)i}{\alpha}z_{1}+Bz_{2}+h(z_{2})\;\;\text{and}$
$\displaystyle
h(z_{2})-h(z_{2}+c_{2})=\frac{(A-1)i}{\alpha}z_{1}+Bz_{2}+\frac{1}{\alpha}\log
A.$ (3.27)

If $c_{2}=0$, then from (3.27), it follows that $Ae^{i(A-1)}=1$. If $c\neq 0$,
then by (3.27), we conclude that $h$ must be constant and
$Ae^{i(A-1)c_{1}+\alpha Bc_{2}}=1$.

From (3.18), we obtain

$\displaystyle
f(z_{1},z_{2})=\frac{1}{2i(A-1)}\left(\beta_{1}e^{i(A-1)z_{1}+\alpha
Bz_{2}+\alpha h(z_{2})}-\beta_{2}e^{-(i(A-1)z_{1}+\alpha Bz_{2}+\alpha
h(z_{2}))}\right),$

where $A\neq 1$.

Since $A^{2}=1$, it must be $A=-1$. Therefore, the form of the function is

$\displaystyle
f(z_{1},z_{2})=-\frac{1}{4i}\left(\beta_{1}e^{\alpha\left(\displaystyle\frac{-2i}{\alpha}z_{1}+Bz_{2}+h(z_{2})\right)}-\beta_{2}e^{-\alpha\left(\displaystyle\frac{-2i}{\alpha}z_{1}+Bz_{2}+h(z_{2})\right)}\right),$

where $h$ satisfies the relation

$\displaystyle
h(z_{2})-h(z_{2}+c_{2})=-\frac{2i}{\alpha}z_{1}+Bz_{2}+\frac{1}{\alpha}\log(-1).$

In the special case whenever $c_{2}\neq 0$, $h$ is constant by (3.27) and the
form of the function is of the form

$\displaystyle
f(z_{1},z_{2})=-\frac{1}{4i}\left(\beta_{1}e^{\alpha\left(\displaystyle\frac{-2i}{\alpha}z_{1}+Bz_{2}+C\right)}-\beta_{2}e^{-\alpha\left(\displaystyle\frac{-2i}{\alpha}z_{1}+Bz_{2}+C\right)}\right),$

where $B$, $C$ are complex constants such that $e^{2iz_{1}-Bz_{2}}=-1$.

This completes the proof of the theorem. ∎

###### Proof of Theorem 1.5.

First we assume that $f$ is a finite order transcendental entire solution of
$(\ref{e2.5})$. Let us rewrite $(\ref{e2.5})$ as

$\displaystyle h_{1}(z_{1},z_{2})h_{2}(z_{1},z_{2})=1,$ (3.28)

where

$\begin{cases}\displaystyle
h_{1}(z_{1},z_{2})=f(z_{1},z_{2})+iP(z_{1},z_{2})\left(\frac{\partial
f(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}-\frac{\partial
f(z_{1},z_{2})}{\partial z_{1}}\right)\vspace{1.5mm}\\\ \displaystyle
h_{2}(z_{1},z_{2})=f(z_{1},z_{2})-iP(z_{1},z_{2})\left(\frac{\partial
f(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}-\frac{\partial
f(z_{1},z_{2})}{\partial z_{1}}\right).\end{cases}$

In view of $(\ref{e3.27})$ and Lemma 2.2, we assume that

$\displaystyle f(z_{1},z_{2})+iP(z_{1},z_{2})\left(\frac{\partial
f(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}-\frac{\partial
f(z_{1},z_{2})}{\partial z_{1}}\right)=\beta_{1}e^{\alpha H(z)}$ (3.29)

and

$\displaystyle f(z_{1},z_{2})-iP(z_{1},z_{2})\left(\frac{\partial
f(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}-\frac{\partial
f(z_{1},z_{2})}{\partial z_{1}}\right)=\beta_{2}e^{-\alpha H(z)},$ (3.30)

where $H(z)$ is an entire functions in $\mathbb{C}^{2}$ and $\alpha(\neq 0)$,
$\beta_{1}$, $\beta_{2}$ are complex constants such that
$\beta_{1}\beta_{2}=1$.

Solving $(\ref{e3.28})$ and $(\ref{e3.29})$, we obtain

$\displaystyle f(z_{1},z_{2})=\frac{\beta_{1}e^{\alpha
H(z_{1},z_{2})}+\beta_{2}e^{-\alpha H(z_{1},z_{2})}}{2}$ (3.31)

and

$\displaystyle P(z_{1},z_{2})\left(\frac{\partial
f(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}-\frac{\partial
f(z_{1},z_{2})}{\partial z_{1}}\right)=\frac{\beta_{1}e^{\alpha
H(z_{1},z_{2})}-\beta_{2}e^{-\alpha H(z_{1},z_{2})}}{2i}.$ (3.32)

Since $f(z_{1},z_{2})$ is a transcendental entire function of finite order, in
view of $(\ref{e3.30})$, it is easy to see that $H(z_{1},z_{2})$ is a
polynomial in $\mathbb{C}^{2}$.

Differentiating $(\ref{e3.30})$ partially with respect to $z_{1}$, we obtain

$\displaystyle\frac{\partial f(z_{1},z_{2})}{\partial
z_{1}}=\frac{\alpha}{2}\frac{\partial H(z_{1},z_{2})}{\partial
z_{1}}\left(\beta_{1}e^{\alpha H(z_{1},z_{2})}-\beta_{2}e^{-\alpha
H(z_{1},z_{2})}\right).$ (3.33)

From $(\ref{e3.31})$ and $(\ref{e3.32})$, we obtain

$\displaystyle\frac{\left(\alpha\frac{\partial H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)\beta_{1}}{-\alpha\beta_{2}\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}e^{\alpha H(z_{1},z_{2})+\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}$ $\displaystyle+\frac{\left(\alpha\frac{\partial
H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)}{\alpha\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}e^{-\alpha H(z_{1},z_{2})+\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}$
$\displaystyle+\frac{\beta_{1}}{\beta_{2}}e^{2\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}=1.$ (3.34)

Since $H(z_{1},z_{2})$ is a non-constant polynomial, it follows from (3) that
$\alpha\frac{\partial H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}$ can not be zero. For otherwise, from (3), we
obtain $e^{2\alpha H(z_{1}+c_{1},z_{2}+c_{2})}=-\beta_{1}^{2}=$ constant,
which implies that $H$ is constant, which is a contradiction.

Therefore, both

$\displaystyle\frac{\left(\alpha\frac{\partial H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)\beta_{1}}{-\alpha\beta_{2}\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}e^{\alpha H(z_{1},z_{2})+\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}\;\text{and}\;\frac{\beta_{1}}{\beta_{2}}e^{\alpha
2H(z_{1}+_{1},z_{2}+c_{2})}$

are not constants.

Also in view of Section 3.11 (page no. $301$) of [12], a simple computation
shows that

$\displaystyle N\left(r,\frac{\left(\alpha\frac{\partial
H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)\beta_{1}}{-\alpha\beta_{2}\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}e^{\alpha H(z_{1},z_{2})+\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}\right)$
$\displaystyle=N\left(r,\frac{-\alpha\beta_{2}\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}{\left(\alpha\frac{\partial
H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)\beta_{1}}e^{-\alpha
H(z_{1},z_{2})-\alpha H(z_{1}+c_{1},z_{2}+c_{2})}\right)=S(r,f),$

$\displaystyle N\left(r,\frac{\left(\alpha\frac{\partial
H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)}{\alpha\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}e^{-\alpha H(z_{1},z_{2})+\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}\right)$
$\displaystyle=N\left(r,\frac{\alpha\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}{\left(\alpha\frac{\partial
H(z_{1},z_{2})}{\partial z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)}e^{\alpha
H(z_{1},z_{2})-\alpha H(z_{1}+c_{1},z_{2}+c_{2})}\right)=S(r,f)$

and

$\displaystyle N\left(r,e^{2\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}\right)=N\left(r,e^{-2\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}\right)=S(r,f).$

Hence, applying Lemma 2.1, we obtain

$\displaystyle\frac{\left(\alpha\frac{\partial H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)}{\alpha\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}e^{-\alpha H(z_{1},z_{2})+\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}=1,$ (3.35)

which implies that

$\displaystyle\frac{\left(\alpha\frac{\partial H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)}{\alpha\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}=e^{\alpha H(z_{1},z_{2})-\alpha
H(z_{1}+c_{1},z_{2}+c_{2})}.$ (3.36)

Next, using $(\ref{e3.33})$ and $(\ref{e3.34})$, we obtain

$\displaystyle\frac{\left(\alpha\frac{\partial H(z_{1},z_{2})}{\partial
z_{1}}+\frac{1}{iP(z_{1},z_{2})}\right)}{\alpha\frac{\partial
H(z_{1}+c_{1},z_{2}+c_{2})}{\partial z_{1}}}=e^{\alpha
H(z_{1}+c_{1},z_{2}+c_{2})-\alpha H(z_{1},z_{2})}.$ (3.37)

Observe that L.H.S. of (3.36) is rational, whereas R.H.S. of the same is
transcendental entire. Therefore, $\alpha H(z_{1},z_{2})-\alpha
H(z_{1}+c_{1},z_{2}+c_{2})$ and hence
$e^{\alpha H(z_{1},z_{2})-\alpha H(z_{1}+c_{1},z_{2}+c_{2})}$ must be
constant.

Hence, we may assume that

$\displaystyle H(z)=Az_{1}+Bz_{2}+C,$ (3.38)

where $A,B,C$ are complex constants.

Using (3.35), (3.37) and (3.38), we obtain

$\displaystyle(\alpha AP(z_{1},z_{2}))^{2}=(\alpha AP(z_{1},z_{2})-i)^{2},$

which implies that $P(z_{1},z_{2})={-1}/{2i\alpha A}$=constant.

Also, from (3.35) and (3.38), we obtain that $e^{\alpha(Ac_{1}+Bc_{2})}=-1.$
This implies that $\alpha(Ac_{1}+Bc_{2})=(2k+1)\pi i$, where $k$ is an
integer.

Hence from $(\ref{e3.30})$, it is easy to see that

$\displaystyle
f(z_{1},z_{2})=\frac{1}{2}\left(\beta_{1}e^{\alpha(Az_{1}+Bz_{2}+C)}+\beta_{2}e^{-\alpha(Az_{1}+Bz_{2}+C)}\right).$

This completes the proof of the theorem. ∎

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

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