\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 159, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/159\hfil Fermat type differential and difference equations]
{Fermat type differential and difference equations}

\author[K. Liu, X. Dong \hfil EJDE-2015/159\hfilneg]
{Kai Liu, Xianjing Dong}

\address{Kai Liu \newline
Department of Mathematics,
Nanchang University,
Nanchang, Jiangxi 330031, China}
\email{liukai418@126.com, liukai@ncu.edu.cn}

\address{Xianjing Dong \newline
Department of Mathematics,
Nanchang University,
Nanchang, Jiangxi 330031, China}
\email{xjdong05@126.com}

\thanks{Submitted January 6, 2015. Published June 15, 2015.}
\subjclass[2010]{39B32, 34M05, 30D35}
\keywords{Fermat equations; entire solutions; finite order}

\begin{abstract}
 This article we explore the relationship between the number
 of differential and difference operators with the existence of meromorphic
 solutions of Fermat type differential and difference equations.
 Some Fermat differential and difference equations of certain types are also
 considered.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{question}[theorem]{Question}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

Some classical results on the meromorphic solutions of Fermat type
functional equation
\begin{equation}\label{e1}
f(z)^{n}+g(z)^{n}=1
\end{equation}
can be stated as follows.
Gross \cite{gross} proved that \eqref{e1} has no transcendental
meromorphic solutions when $n\geq4$, and Montel \cite{montel} showed
that \eqref{e1} has no transcendental entire solutions when $n\geq3$.
Iyer \cite{iyer} concluded that if
$n=2$, then \eqref{e1} has the entire solutions
$f(z)=\sin(h(z))$ and $g(z)=\cos(h(z))$, where $h(z)$ is any entire
function, no other solutions exist.
We remark that \eqref{e1} can be rewritten as
\begin{equation}\label{e1bianxing}
f(z)^{n}+[g_{1}(z)+\dots+g_{m}(z)]^{n}=1,
\end{equation}
it is easy to see that there is no relationship between the number of
$m$ with the existence of meromorphic solutions of \eqref{e1bianxing},
 since that $n$ and functional fields are the determinants of the existence
of solutions. In this paper, we will explore the corresponding problem when
$g_{i}(z), i=1,2,\dots,m$, are some differential or difference operators
of $f(z)$. Our aim is to explore the relationship between the number of
differential or difference operators with the existence of meromorphic
solutions of Fermat type differential or difference equations.

Let us begin from a result given in \cite{liping}.
Yang and Li considered the entire solutions of Fermat type differential
equations. Here, we rewrite the original theorem as follows for
the goal of this article


\begin{theorem} \label{thmA}
 Let $n$ be a positive integer, $b_{0}, b_{1},\ldots, b_{n-1}$ be constants, 
$b_{n}$ be a non-zero constant and let $L(f)=\sum_{k=0}^{n}b_{k}f^{(k)}$. 
Then the transcendental meromorphic solution of the following equation
\begin{equation}\label{eq1}
f^{2}+L(f)^{2}=1
\end{equation}
must have the form $f(z)=\frac{1}{2}\left(Pe^{\lambda z
}+\frac{1}{P}e^{-\lambda z}\right)=ch(\lambda z+A)$, where $e^{A}=P$, $P$ 
is a non-zero constant and $\lambda$ satisfies the following equations:
\begin{equation}\label{eq2}
\sum_{k=0}^{n}b_{k}\lambda^{k}=\frac{1}{i}, \quad
\sum_{k=0}^{n}b_{k}(-\lambda)^{k}=-\frac{1}{i}.
\end{equation}
\end{theorem}

In fact, all the transcendental meromorphic solutions of \eqref{eq1}
should be transcendental entire solutions. Let $L(f)=f^{(n)}$. 
From $\eqref{eq2}$ of Theorem \ref{thmA}, we see easily that if $n$ is an odd,
then \eqref{eq1} has transcendental entire solutions. If $n$ is an even,
then \eqref{eq1} has no transcendental entire solutions.
Some related results on $f^{2}+R(z)(f^{(n)})^{2}=1$ also can be found in
 \cite{tang}, where $R(z)$ is a rational function and $n$ is an odd. 
Let $L(f)=b_{n}f^{(n)}+b_{n+1}f^{(n+1)}$. Yang and Li also obtained 
the following result.


\begin{theorem}[{\cite[Theorem 2]{liping}}] \label{thmB} 
Let $b_{n}$ and $b_{n+1}$ be non-zero constants. Then
\begin{equation}\label{eq3}
f^{2}+[b_{n}f^{(n)}+b_{n+1}f^{(n+1)}]^{2}=1
\end{equation}
has no transcendental meromorphic solutions.
\end{theorem}

It is not difficult to see that $b_{n+1}f^{(n+1)}$ can be replaced by 
$b_{n+2k+1}f^{(n+2k+1)}$ in Theorem \ref{thmB}, where $k$ is a non-negative integer.
However, if $f^{(n+1)}$ is replaced by $f^{(n+2)}$ or $f^{(n+2k)}$, 
then $\eqref{eq3}$ can admit some transcendental entire solutions 
from Example \ref{examp1} below.

\begin{example} \label{examp1}\rm
The function $f(z)=\cos z$ satisfies 
$$
f^{2}+[\frac{3}{2}f'+\frac{1}{2}f''']^{2}=1,
$$ 
and $f(z)=\sin z$ solve
$$
f^{2}+[\frac{1}{2}f'-\frac{1}{2}f''']^{2}=1.
$$ 
In fact, a necessary condition of existence of transcendental solutions of
 $$
f^{2}+[b_{n}f^{(n)}+b_{n+1}f^{(n+2k)}]^{2}=1
$$ 
is that $n$ must be an odd. Furthermore, the necessary condition 
of existence of solutions of $f^{2}+[b_{n}f^{(n)}+b_{m}f^{(m)}]^{2}=1$ 
is that $m,n$ are odds, which can be obtained by \eqref{eq2} of 
Theorem \ref{thmA}.
\end{example}

Let us consider the case of 
$L(f)=b_{n}f^{(n)}+b_{n+1}f^{(n+1)}+b_{n+2}f^{(n+2)}$. 
From Theorem \ref{thmA}, we see that if $n$ is an even, then
\begin{equation}\label{eq4}
f^{2}+[b_{n}f^{(n)}+b_{n+1}f^{(n+1)}+b_{n+2}f^{(n+2)}]^{2}=1
\end{equation}
may admit some transcendental solutions. If $n$ is odd, 
then \eqref{eq4} has no transcendental solutions. From the above statements, 
we conclude that the necessary conditions for \eqref{eq1} to admit
transcendental solutions are as follows:
\begin{itemize}
\item[(i)] $k$ is an odd (it is also a sufficient condition), provided 
that $L(f)=f^{(k)}$,

\item[(ii)] $m+n$ is an even and $m,n$ are odds, provided that 
$L(f)=b_{n}f^{(n)}+b_{m}f^{(m)}$,

\item[(iii)] $n+p+q$ is an odd, provided that 
$L(f)=b_{n}f^{(n)}+b_{p}f^{(p)}+b_{q}f^{(q)}$.
\end{itemize}


We give two notations for more convenient statements. 
Denoting $D(L(f))$ by the number of differential operators in 
$L(f)$, for example $D(L(f))=n+1$, if $L(f)=\sum_{k=0}^{n}b_{k}f^{(k)}$ 
and $b_{k}\neq0, k=0,1,\dots,n$. Denoting $W(L(f))$ by the sum of order
 of all differential operators in $L(f)$, for example $W(L(f))=\frac{n(n+1)}{2}$, 
if $L(f)=\sum_{k=0}^{n}b_{k}f^{(k)}$ and $b_{k}\neq0, k=0,1,\dots,n$. 
So far, we have obtained that if  $D(L(f))$ is odd (even), the necessary 
condition of existence of transcendental solutions of \eqref{eq1} is that 
$W(L(f))$ is odd (even), provided that $L(f)$ has no more than three terms. 
A natural question happens as follows.

\begin{question}\label{qio} \rm
Can we get the parity of $D(L(f))$ and $W(L(f))$ are same for any $L(f)$, 
provided that there exist transcendental solutions of \eqref{eq1}?
\end{question}

Unfortunately, Question \ref{qio} is not true for $L(f)$ with four terms by 
Example \ref{examp2} below.

\begin{example} \label{examp2} \rm
The function $f(z)=\cos(-\sqrt{\frac{b_{2}}{b_{4}}}z+Ai)$  is an entire 
solution of 
$$
f^{2}+[b_{1}f'+b_{2}f''+b_{3}f'''+b_{4}f^{(4)}]^{2}=1,
$$ 
where $b_{1}b_{4}+b_{2}b_{3}=b_{4}\sqrt{\frac{b_{2}}{b_{4}}}$. 
Here $D(L(f))=4$ and $W(L(f))=10$. The function $f(z)=\sin z$ solves 
the equation 
$$
f^{2}+[f'+\frac{3}{2}f''+2f^{(4)}+\frac{1}{2}f^{(6)}]^{2}=1.
$$ 
Here, $D(L(f))=4$ and $W(L(f))=13$.
\end{example}

If we add the condition that all $|b_{i}|=1$, 
Question \ref{qio} is also not true, which can be seen by the 
Example \ref{examp3}.

\begin{example} \label{examp3} \rm
Assume that $\lambda$ satisfies 
\begin{gather*} %\label{tian}
 b_{5}\lambda^{5}+b_{4}\lambda^{4}+b_{1}\lambda+b_{0}=\frac{1}{i},\\
b_{5}[-\lambda]^{5}+b_{4}[-\lambda]^{4}+b_{1}[-\lambda]+b_{0}=-\frac{1}{i}\,.
\end{gather*}
Then $f(z)=ch(\lambda z+A)$ solves
 $f^{2}+[b_{5}f^{(5)}+b_{4}f^{(4)}+b_{1}f'+b_{0}f]^{2}=1$, where $A$ is a constant.
 If $t$ satisfies
\begin{gather*}
b_{5}t^{5}+b_{4}t^{4}+b_{1}t^{2}+b_{0}=\frac{1}{i},\\
b_{5}[-t]^{5}+b_{4}[-t]^{4}+b_{2}[-t]^{2}+b_{0}=-\frac{1}{i},
   \end{gather*}
then $f(z)=ch(t z+B)$ solves 
$f^{2}+[b_{5}f^{(5)}+b_{4}f^{(4)}+b_{2}f''+b_{0}f]^{2}=1$, where $B$ is a constant.
\end{example}

We always considered that there is just one term $f(z)^{2}$ in the beginning 
place of \eqref{eq1}, we proceed to consider the following equation
 \begin{equation}\label{ethur}
 [f+f']^{2}+[f+f'']^{2}=1,
 \end{equation}
and obtain the following result.

 \begin{theorem}\label{thth1}
 The equation \eqref{ethur} has no transcendental meromorphic solutions.
 \end{theorem}

Furthermore, we want to explore when the following equation 
\begin{equation}\label{ethur11}
 [af+bf']^{2}+[cf+df'']^{2}=1,
 \end{equation} can admit a transcendental meromorphic solution, 
where $a, b, c, d$ are constants. We obtain the following result.

 \begin{theorem}\label{th2}
If \eqref{ethur11} admits transcendental meromorphic solutions $f(z)$, 
then one of the following holds:
\begin{itemize}
\item[(i)] $a=0, c=0,$ $b\neq0$, $d\neq0$ and 
$f(z)=\frac{id}{b^{2}}sh(\frac{bi}{d}z-B)+A$, where $A$ is a constant.

\item[(ii)] $a,b,c,d$ are non-zero constants and 
$\frac{c}{a}=\frac{1-e^{2p}}{1+e^{2p}}i$, then 
$f(z)=De^{-\frac{a}{b}z}+\frac{e^{p}+e^{-p}}{2a}$, where $p$ is a constant, 
$D$ is a non-zero constant.

\item[(iii)] $a=0,$ $b\neq0$, $c\neq0$ and  
$f(z)=\frac{bi+d}{bc}sh(e^{\frac{c}{bi+d}z-A})+B$, where 
$d+\frac{dc}{b^{2}i+bd}=0$ and $A, B$ are constants.
\end{itemize}
 \end{theorem}

Recently, using the difference analogues of Nevanlinna theory, 
the meromorphic solutions of complex difference equations or differential-difference 
equations with certain types also be considered, such as 
\cite{liuk4, liukkk, liucaocao, liu, lc, qiyang}. The first author and 
his colleagues considered Fermat type difference equations, such 
as Liu, Cao and Cao \cite{liucaocao} investigated the finite order entire
solutions of
\begin{equation}\label{cheche}
f(z)^{2}+f(z+c)^{2}=1.
\end{equation} 
Here and in the following, $c$ is a non-zero constant, unless otherwise 
specified. The result can be stated as follows.

\begin{theorem}[{\cite[Theorem 1.1]{liucaocao}}] \label{thmC}
The transcendental entire solutions with
finite order of \eqref{cheche} must satisfy $f(z)=\sin(Az+B)$,
where $B$ is a constant and $A=\frac{(4k+1)\pi}{2c}$, with $k$ an
integer.
\end{theorem}


When $f(z+c)$ is replaced by $f(z+c)-f(z)$ in  \eqref{cheche}, that is
\begin{equation}\label{cheche21}
f(z)^{2}+[f(z+c)-f(z)]^{2}=1,
\end{equation} 
Liu \cite{liuk4} proved the following result.

\begin{theorem}[{\cite[Proposition 5.3]{liuk4}}] \label{thmD}
There is no transcendental entire solutions with
finite order of \eqref{cheche21}.
\end{theorem}

Let $a_{0}, a_{1},\ldots,a_{n}$ be non-zero constants. If
\begin{equation}\label{cheche3}
f(z)^{2}+[a_{0}f(z)+a_{1}f(z+c_{1})+\ldots+a_{n}f(z+c_{n})]^{2}=1
\end{equation}
admits transcendental entire solutions with
finite order, we hope that $n$ is an odd. However, it is not true 
again by Example \ref{examp4} below.

\begin{example} \label{examp4}\rm
 The function $f(z)=\sin z$ is a solution of
\[
f(z)^{2}+[2f(z)+f(z+\frac{\pi}{2})+f(z+\pi)]^{2}=1
\]
and also a solution of
\[
f(z)^{2}+[2f(z)+f(z+\frac{\pi}{2})+\frac{1}{2}f(z+\pi)+\frac{3}{2}f(z+3\pi)]^{2}=1.
\]
\end{example}

If we put the additional condition that $|a_{i}|=1$ for $i=0,1,2,\dots,n$, 
then Question \ref{qio} is also not true;
 we will construct an example as follows.


\begin{example} \label{examp5}\rm
 The function $f(z)=\sin z$ is a solution of
\[
f(z)^{2}+[if(z)+f(z+c_{1})+f(z+c_{2})+f(z+c_{3})]^{2}=1,
\]
where $e^{c_{1}}=A$, $e^{c_{2}}=B$, $e^{c_{3}}=C$ and
\begin{equation}\label{jintian}
\begin{gathered}
 A+B+C=0,\\
\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=-2i,
\end{gathered}
\end{equation}
it is also a solution of
\[
f(z)^{2}+[if(z)+f(z+t_{1})+f(z+t_{2})+f(z+t_{3})+f(z+t_{4})]^{2}=1,
\]
where $e^{t_{1}}=A_{1}$, $e^{t_{2}}=B_{1}$, $e^{t_{3}}=C_{1}$, 
$e^{t_{4}}=D_{1}$ and
\begin{equation}\label{jintianjou}
\begin{gathered}
 A_{1}+B_{1}+C_{1}+D_{1}=0,\\
\frac{1}{A_{1}}+\frac{1}{B_{1}}+\frac{1}{C_{1}}+\frac{1}{D_{1}}=-2i.
   \end{gathered}
 \end{equation}
\end{example}

As a generalization of \eqref{cheche21}, we want to consider when the following 
equation \begin{equation}\label{cheche2}
D^{2}f(z)^{2}+[Af(z+c)+Bf(z)]^{2}=1
\end{equation}
admits transcendental entire solutions, we get the next result.

\begin{theorem}\label{th1}
Let $A,B$ be constants. If there are transcendental entire solutions with
finite order of \eqref{cheche2}, then $A^{2}=B^{2}+D^{2}$.
\end{theorem}

\begin{example} \label{examp6}\rm
 We see that $f(z)=\sin(az+b)$ is an entire solution of 
$$
f(z)^{2}+[-\sqrt{2}f(z+c)+f(z)]^{2}=1,
$$ 
here $D=1$, $B=1$, $A=-\sqrt{2}$ and $c=\frac{\pi}{4a}$.
\end{example}

Finally, we also  consider a  difference equation similar to \eqref{ethur},
and get the following result.

\begin{theorem}\label{kp}
Let $a_{1}, a_{2},a_{3},a_{4}$ be non-zero constants. If
\begin{equation}\label{cheche3b}
[a_{1}f(z+c)+a_{2}f(z)]^{2}+[a_{3}f(z+c)+a_{4}f(z)]^{2}=1
\end{equation}
admits transcendental entire solutions with
finite order, then $a_{1}^{2}+a_{3}^{2}=a_{2}^{2}+a_{4}^{2}$ and 
$f(z)=\frac{a_{3}\cos(aiz+bi)+a_{1}\sin(aiz+bi)}{a_{2}a_{3}-a_{1}a_{4}}$, 
where $a$ is non-zero constant and $b$ is a constant.
\end{theorem}

\begin{example} \label{examp7}\rm
 It is easy to see that
\begin{equation}\label{checheche}
[f(z+c)+f(z)]^{2}+[f(z+c)-f(z)]^{2}=1
\end{equation}
can admit a transcendental entire solution of 
$f(z)=\frac{\sqrt{2}}{2}\sin(z+\frac{\pi}{4})=\frac{\cos z+\sin z}{2}$,
 which implies that $a=-i$ and $b=0$ in Theorem \ref{kp}.
\end{example}

\section{Proof of Theorem \ref{thth1}}

For the proof of Theorem \ref{thth1}, we need the following lemma.

\begin{lemma}[{\cite[Theorem 1.46]{yang and yi}}] \label{y}
Suppose that $f(z)$ is a transcendental meromorphic function and $h(z)$ 
is a non-constant entire function. Then 
$$
\lim_{r\to\infty}\frac{T(r,f(h))}{T(r,h)}=\infty.
$$
\end{lemma}

Now, we give the proof of Theorem \ref{thth1}. Using the classical results of 
Fermat equations in the beginning of the introduction, we have
 \begin{equation}\label{ki}
   \begin{gathered}
   f+f'=\sin(h(z)),\\
  f+f''=\cos(h(z)).
\end{gathered}
  \end{equation}
It is a system of differential equations, however we do not know what 
is $h(z)$. Hence, it is not convenient to slove using the basic theory of 
differential equations. Here, we use the following method. 
Firstly, from \eqref{ki}, we get
 \begin{equation}\label{kikl}
 f'-f''=\sin(h(z))-\cos(h(z)).
 \end{equation}
Then taking the first order derivative of first equation of \eqref{ki}, we have
\begin{equation}\label{ki0}
f'+f''=h'(z)\cos(h(z)).
\end{equation}
Thus, combining the second equation of \eqref{ki} and \eqref{ki0}, we have
\begin{equation}\label{kipp}
 f-f'=\cos(h(z))-h'(z)\cos(h(z)),
\end{equation}
then taking derivative of the above equation, we have
  \begin{equation}\label{kipp2ss2}
 f'-f''=-h'(z)\sin(h(z))-h''(z)\cos(h(z))+[h'(z)]^{2}\sin(h(z)).
 \end{equation}
Combining \eqref{kikl} and  \eqref{kipp2ss2}, we have
\begin{equation}\label{kipp22}
\tan(h(z))=\frac{h''(z)-1}{(h'(z))^{2}-h'(z)-1},
 \end{equation}
which implies that $h(z)$ should be a constant from Lemma \ref{y}. 
By a simple computation, we have $f(z)$ should be a constant. 
So, there is no transcendental entire solutions.

\section{Proof of Theorem \ref{th2}}

The method is a  factorization. Here, we give the details.
 We easily get
\begin{equation}\label{thur11pp}
af+bf'=\frac{e^{p(z)}+e^{-p(z)}}{2}
 \end{equation}
 and
 \begin{equation}\label{thur11ppoo}
 cf+df''=\frac{e^{p(z)}-e^{-p(z)}}{2i},
 \end{equation}
where $p(z)$ is a transcendental entire function. Taking the first derivative 
of \eqref{thur11pp}, we have
 \begin{equation}\label{e62}
af'+bf''=\frac{p'(z)e^{p(z)}-p'(z)e^{-p(z)}}{2}.
 \end{equation}
Combining \eqref{thur11ppoo} with \eqref{e62}, we have
\begin{equation}\label{e63}
bcf-adf'=\frac{e^{p(z)}-e^{-p(z)}}{2}[-bi-p'(z)d].
 \end{equation}
Combining \eqref{e63} with \eqref{thur11pp}, we have
\begin{equation}\label{e64}
 [a^{2}d+b^{2}c]f=\frac{ade^{p(z)}+ade^{-p(z)}}{2}
+\frac{e^{p(z)}-e^{-p(z)}}{2}[-b^{2}i-p'(z)bd]
\end{equation}
and
\begin{equation}\label{e65}
 [a^{2}d+b^{2}c]f'=\frac{bce^{p(z)}+bce^{-p(z)}}{2}
-\frac{e^{p(z)}-e^{-p(z)}}{2}[-bai-p'(z)ad].
\end{equation}
Next, we talk about two cases.
\smallskip

\noindent\textbf{Case 1:} If $a^{2}d+b^{2}c=0$, then  from \eqref{e64} we have
\begin{equation}\label{e66}
e^{2p(z)}[ad-b^{2}i-p'(z)bd]=-ad-b^{2}i-p'(z)bd.
\end{equation}
Subcase 1.1. If $a=0$, $d\not=0$ and $b^{2}i+p'(z)bd\not=0$, then $c=0$ follows. 
Thus, it implies $e^{2p(z)}=1$, that is $p(z)$ reduces to a constant, which 
is a contradiction with $f$ is transcendental. If $a=0$, $d\not=0$ and 
$b^{2}i+p'(z)bd=0$, then $c=0$ and $b\not=0$ ($a,b$ are not zeros simultaneously) 
and $p(z)=-\frac{bi}{d}z+B$, where $B$ is a constant, which implies that 
$f(z)=\frac{di}{b^{2}}sh(\frac{bi}{d}z-B)+A$, where $A$ is a constant.
\\
Subcase 1.2. If $a\not=0$, $d=0$ and $b^{2}i+p'(z)bd\not=0$, then 
$b\not=0$ and $c=0$ follows, a contradiction.
If $a\not=0$, $d=0$ and $b^{2}i+p'(z)bd=0$, then $b=0$, thus $f$ should 
be a constant from $\eqref{ethur11}$.
\\
Subcase 1.3. If $ad\neq0$, then $bc\neq0$, thus $p(z)$ should be a constant 
for avoiding a contradiction. From a simple computation, we have 
$f(z)=De^{-\frac{a}{b}z}+\frac{e^{p}+e^{-p}}{2a}$ and 
$\frac{c}{a}=\frac{1-e^{2p}}{1+e^{2p}}i$.
\smallskip

\noindent\textbf{Case 2:} If $a^{2}d+b^{2}c\neq0$, then taking derivative for 
\eqref{e64} and combining with  \eqref{e65}, we have
\begin{equation}\label{e67}
e^{2p(z)}[iba+p''(z)bd+\phi]=iba+p''(z)bd-\phi,
\end{equation}
where $p'(z)b^{2}i+[p'(z)]^{2}bd+bc=\phi$,
hence we have either $iba+p''(z)bd=0$ and $\phi=0$ or $p(z)$ should be a constant.
\\
Subcase 2.1. If $iba+p''(z)bd=0$ and $\phi=0$, we assume that $b=0$, 
form $\eqref{thur11pp}$  and \eqref{thur11ppoo}, we have
\begin{equation}\label{e68}
e^{2p(z)}[\frac{dp''(z)}{a}+\frac{d}{a}p'(z)^{2}+\frac{c}{a}
-\frac{1}{i}]=\frac{dp''(z)}{a}-\frac{d}{a}p'(z)^{2}-\frac{c}{a}-\frac{1}{i},
\end{equation}
hence either $p(z)$ is a constant, then $f(z)$ is a constant or 
$T(r,e^{2p})=S(r,e^{2p})$ which is impossible. Assume that $b\neq0$, 
we can get $p'(z)=-\frac{c}{bi+d}$ and $a=0$, by a simple computation, 
we have $f(z)=\frac{bi+d}{bc}sh(e^{\frac{c}{bi+d}z-A})+B$, where 
$d+\frac{dc}{b^{2}i+bd}=0$ and $A, B$ are constants.
\\
Subcase 2.2. If $p(z)$ is a constant $p$, we have $e^{2p}(iba+bc)=iba-bc$ 
from $\eqref{e67}$, if $b=0$, then $f$ should be a constant. If $b\neq0$, we have
$e^{2p}=\frac{ia-c}{ia+c}$, then $f(z)=De^{-\frac{a}{b}z}+\frac{e^{p}+e^{-p}}{2a}$, 
then substitute $f(z)$ into \eqref{thur11pp} and \eqref{thur11ppoo}, we have 
$a^{2}d+b^{2}c=0$, which is a contradiction of the Case 2.

\section{Proof of Theorem \ref{th1}}

We need the following result by Yang and Yi,  \cite[Theorem 1.56]{yang and yi}.

\begin{lemma}\label{yy} 
Let $f_{1},f_{2},f_{3}$ be meromorphic
functions such that $f_{1}$ is not a constant. If
$f_{1}+f_{2}+f_{3}=1$ and if
$$
\sum_{j=1}^{3}N(r,1/f_{j})+2\sum_{j=1}^{3}\overline{N}(r,f_{j})<(\lambda +o(1))T(r),
$$
where $\lambda <1$ and $T(r):=\max_{1\leq j\leq 3}T(r,f_{j})$, then
either $f_{2}=1$ or $f_{3}=1$.
\end{lemma}

Using the classical results of Fermat equations of in the beginning of the 
introductions, we get $Df(z)=\sin(h(z))$ and $Af(z+c)+Bf(z)=\cos(h(z))$. 
Thus, we have the equation
\[
A\sin(h(z+c))+B\sin(h(z))=D\cos(h(z)).
\]
So we have
\[
A\sin(h(z+c))=\sqrt{D^{2}+B^{2}}\sin(h(z)+\varphi),
\]
 where $\tan\varphi=-\frac{D}{B}$.
Thus, we have
\begin{equation*}\label{equ7}
\frac{e^{ih(z+c)}-e^{-ih(z+c)}}{2i}
=\frac{\sqrt{D^{2}+B^{2}}}{A}\frac{e^{ih(z)+i\varphi}-e^{-ih(z)-i\varphi}}{2i},
\end{equation*}
which implies that
\begin{equation}\label{equ8}
e^{2ih(z+c)}-\frac{\sqrt{D^{2}+B^{2}}}{A}e^{ih(z+c)+ih(z)
+i\varphi}+\frac{\sqrt{D^{2}+B^{2}}}{A}e^{ih(z+c)-ih(z)-i\varphi}=1.
\end{equation}
Using Lemma \ref{yy} and remark that $e^{2ih(z+c)}$ and 
$e^{ih(z+c)+ih(z)+i\varphi}$ are not constants, then we have
\begin{gather*}
\frac{\sqrt{D^{2}+B^{2}}}{A}e^{ih(z+c)-ih(z)-i\varphi}=1, \\
e^{2ih(z+c)}-\frac{\sqrt{D^{2}+B^{2}}}{A}e^{ih(z+c)+ih(z)+i\varphi}=0.
\end{gather*}
For avoiding a contradiction from the above two equations, we should have 
$h(z)=az+b$, thus, we get $A^{2}=B^{2}+D^{2}$. 
Then $ac=\varphi+2k\pi$ or $ac=\varphi+\pi+2k\pi$. 
If $Df(z)=\cos(h(z))=\sin(h(z)+\frac{\pi}{2}+2k\pi)$, using the similar 
method as above.

\section{Proof of Theorem \ref{kp}}

If one of $a_{1},a_{2},a_{3},a_{4}$ is zero, then using Theorem \ref{th1}, 
we get the conclusion. Next, we talk about all $a_{1},a_{2},a_{3},a_{4}$ 
being non-zero constants. Using the factorization, we obtain
\begin{gather}\label{15}
a_{1}f(z+c)+a_{2}f(z)=\frac{e^{p(z)}+e^{-p(z)}}{2}, \\
\label{16}
a_{3}f(z+c)+a_{4}f(z)=\frac{e^{p(z)}-e^{-p(z)}}{2i}.
 \end{gather}
 Then we  have the following two equations
 \begin{equation}\label{17}
\begin{gathered}
(a_{2}a_{3}-a_{1}a_{4})f(z)=a_{3}\frac{e^{p(z)}+e^{-p(z)}}{2}-a_{1}
\frac{e^{p(z)}-e^{-p(z)}}{2i},\\
(a_{2}a_{3}-a_{1}a_{4})f(z+c)=a_{2}\frac{e^{p(z)}-e^{-p(z)}}{2i}-a_{4}
\frac{e^{p(z)}+e^{-p(z)}}{2}.
   \end{gathered}
 \end{equation}

 If $a_{2}a_{3}-a_{1}a_{4}=0$, then we have $e^{2p(z)}=-1$, 
which implies that $a_{1}f(z+c)+a_{2}f(z)=0$ and $a_{3}f(z+c)+a_{4}f(z)=\pm1$, 
so $f(z)$ should be a constant.

 If $a_{2}a_{3}-a_{1}a_{4}\neq 0$, we have
 \begin{equation}\label{18}
\begin{aligned}
&a_{3}\frac{e^{p(z+c)}+e^{-p(z+c)}}{2}-a_{1}\frac{e^{p(z+c)}-e^{-p(z+c)}}{2i}\\
&=a_{2}\frac{e^{p(z)}-e^{-p(z)}}{2i}-a_{4}\frac{e^{p(z)}+e^{-p(z)}}{2}.
\end{aligned}
 \end{equation}
Furthermore, we get
 \begin{equation}\label{19}
 [a_{3}+a_{1}i]e^{p(z+c)+p(z)}+[a_{3}-a_{1}i]e^{p(z)-p(z+c)}
+[a_{4}+a_{2}i]e^{2p(z)}=a_{2}i-a_{4}.
 \end{equation}
From Lemma \ref{yy}, we have $p(z)-p(z+c)$ should be a constant, it implies 
that $p(z)=az+b$, where $a$ is a non-zero constant and we have
 \begin{gather}\label{20}
 [a_{3}+a_{1}i]e^{ac}+[a_{4}+a_{2}i]=0, \\
\label{21}
[a_{3}-a_{1}i]e^{-ac}=a_{2}i-a_{4},
 \end{gather}
which implies that $a_{1}^{2}+a_{3}^{2}=a_{2}^{2}+a_{4}^{2}$. 
Then $f(z)=\frac{a_{3}\cos(aiz+bi)+a_{1}\sin(aiz+bi)}{a_{2}a_{3}-a_{1}a_{4}}$ 
follows. Thus, we have the proof of Theorem \ref{kp}.

\section{Further discussions}

We remark that the finite order transcendental entire solutions should be 
of order one in Fermat differential or difference equations that we have considered. 
Of course, we always using the theory of factorization to get the expressions. 
We raise the following conjecture for the further studying on Fermat type 
differential or difference equations.

\begin{conjecture}\label{con1} \rm
If there exist transcendental entire solutions $f(z)$ of Fermat 
differential or difference equations $P(f)^{2}+Q(f)^{2}=1$, then the order 
$\sigma(f)=1$, where $P(f), Q(f)$ are two linear differential or difference 
polynomials with constant coefficients.
\end{conjecture}

Wiman-Valiron theory \cite{valiron} and the difference Wiman-Valiron theory \cite{chiang} can be used to get $\sigma(f)\geq1$ in some cases. However, we have no ideas to remove the case $\sigma(f)>1$.

Finally, let us remind some results on the equation
\begin{equation}\label{therehere}
f^{n}+g^{m}+h^{p}=1\,.
\end{equation}
Gundersen \cite{gundersen} summed up some results, and recent results can be 
found in \cite{yang}, \cite{zhang}. Considering Fermat type difference 
equation with three terms, such as
\begin{equation}\label{there}
f(z)^{n}+f(z+c_{1})^{m}+f(z+c_{2})^{p}=1.
\end{equation}
It is easy to see that the necessary conditions of existence of transcendental 
entire solutions of finite order is $m=n=p$. So the classical result in 
Fermat functional equations show that there is no any entire solutions when 
$m=n=p\geq7$. Here, we  consider when \eqref{there} can admit entire solutions. 
It is easy to see that  if $m=n=p=1$, then $f(z)=e^{z}+\frac{1}{3}$ and 
$e^{c_{1}}=\frac{1}{2}$ and $e^{c_{1}}=-\frac{3}{2}$ is a solution
 of $f(z)+f(z+c_{1})+f(z+c_{2})=1.$ We also can find an entire solution of 
 $f(z)^{2}+f(z+c_{1})^{2}+f(z+c_{2})^{2}=1$. 
For example, $f(z)=e^{z}+\frac{\sqrt{3}}{3}$ and $e^{c_{1}}=\frac{-1+\sqrt{3}i}{2}$ 
and $e^{c_{2}}=\frac{-1-\sqrt{3}i}{2}$.

For our further studying, we raise the following questions about the equation
\begin{equation}\label{there1}
f(z)^{n}+f(z+c_{1})^{n}+f(z+c_{2})^{n}=1,
\end{equation}
where $c_{1}$ and $c_{2}$ are non-zero constants.

\begin{question}\label{5l} \rm
Does there exist a transcendental entire solution with finite order 
of \eqref{there1}, provided that $3\leq n\leq6$.
\end{question}

\begin{question}\label{5lg} \rm
Does there exist a transcendental meromorphic solution with finite order 
of \eqref{there1}, provided that $3\leq n\leq8$.
\end{question}

\subsection*{Acknowledgments}
This work was  supported by the
NSFC (No. 11301260), by the NSF of Jiangxi (No.
20132BAB211003), and by the YFED of Jiangxi (No. GJJ13078) of China.
The first author also supported by China Scholarship Council(No. 201406825034).


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\end{document}