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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 59, 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 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/59\hfil Fermat type $q$-difference differential 
equations]
{Entire solutions of Fermat type $q$-difference differential equations}

\author[K. Liu, T. B. Cao \hfil EJDE-2013/59\hfilneg]
{Kai Liu, Ting-Bin Cao}  % in alphabetical order

\address{
Department of Mathematics\\
Nanchang University\\
Nanchang, Jiangxi, 330031, China}
\email[Kai Liu]{liukai418@126.com}
\email[Ting-Bin Cao]{tbcao@ncu.edu.cn}

\thanks{Submitted November 25, 2012. Published February 26, 2013.}
\thanks{Partially supported by grant 11101201 from the NSFC}
\subjclass[2000]{39B32, 34M05, 30D35}
\keywords{$q$-difference equations; $q$-difference differential equations;
 \hfill\break\indent entire solutions; finite order}

\begin{abstract}
 In this article, we describe the finite-order transcendental
 entire solutions of Fermat type $q$-difference and $q$-difference 
 differential  equations. In addition, we investigate the similarities 
 and other properties among those solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks



\section{Introduction}

Fermat's last theorem \cite{w} states that do not exist nonzero
rational numbers $x, y$ and an integer
$n\geq3$ such that $x^{n}+y^{n} = 1$. The
equation $x^{2}+y^{2}=1$ does admit nontrivial rational solutions.
Replacing $x, y$ in above equation by entire or meromorphic functions $f,g$,
Fermat type functional equations were studied by Gross \cite{gross, gross22}
and many others thereafter, such as \cite{montel, yang}. Yang \cite{yang}
investigated the Fermat type functional equation
\begin{equation}\label{hahaha}
a(z)f(z)^{n}+b(z)g(z)^{m}=1,
 \end{equation}
where $a(z),b(z)$ are small functions with respect to $f(z)$.
Recall that $\alpha(z)\not\equiv0, \infty$ is a small
function with respect to $f(z)$, if $T(r,\alpha)=S(r,f)$, where
$S(r,f)$ is used to denote any quantity satisfying
$S(r,f)=o(T(r,f))$, and $r\rightarrow\infty$ outside of a possible
exceptional set of finite logarithmic measure. In fact, Yang
\cite[Theorem 1]{yang} obtained the following result.


\begin{theorem} \label{thmA}
 Let $m, n$ be positive integers satisfying
$\frac{1}{m} + \frac{1}{n}<1$. Then there are no nonconstant entire
solutions $f(z)$ and $g(z)$ that satisfy \eqref{hahaha}.
\end{theorem}

The above theorem implies that there is no nonconstant entire solutions with the
assumption of $n>2$, $m>2$ in \eqref{hahaha}. However, when $m=n=2$ and
$g(z)$ has a specific relationship with $f(z)$ in \eqref{hahaha},
the problem that can we obtain the accurate expressions of entire solutions
is deserve to be considered.
This article is devoted to considering Fermat type functional equations in the
cases where an entire function $f(z)$ together with one the following:
derivative $f'(z)$, shift $f(z+c)$ or $q$-difference $f(qz)$.
This article is organized as follows. In Section 2, we mainly consider
Fermat type $q$-difference equations. Some similarities or
other properties among the Fermat functional equations of different types
can be found in the section.
In Section 3, some results on entire solutions of
Fermat type $q$-difference differential equations are given.
In this article, we assume that the reader is familiar with standard
symbols and fundamental results of Nevanlinna theory
\cite{hayman, yang and yi}.

\section{Fermat type $q$-difference equations}

Let us recall some results on Fermat type differential equations.
Yang and Li \cite{liping} considered the entire
solutions of
\begin{equation}\label{che}
f(z)^{2}+f'(z)^{2}=1.
\end{equation}
In fact, they considered a generalization of above equation.
The solutions of \eqref{che} can be described as follows.


\begin{theorem}[{\cite[Theorem 1]{liping}}] \label{thmB}
 Transcendental meromorphic solutions of
\eqref{che} satisfy $f(z)=\frac{1}{2}\left(Pe^{-i
z}+\frac{1}{P}e^{i z}\right)$, where $P$ is a nonzero
constant.
\end{theorem}

Let $e^{A}=P$ in Theorem \ref{thmB}. The solutions of \eqref{che} also can be
written as $f(z)=\sin(z+Ai+\frac{\pi}{2})$. In addition,
they obtained the following result \cite[Theorem 3]{liping}.


\begin{theorem} \label{thmC}
 Let $a_{1}$, $a_{2}$, $a_{3}$ be
nonzero meromorphic functions in the complex plane $\mathbb{C}$.
Then a necessary condition for the differential equation
\begin{equation}\label{tou}
a_{1}f^{2}+a_{2}(f')^{2}=a_{3}
\end{equation}
 to have a transcendental
meromorphic solution
satisfying $T(r,a_{k})=S(r,f)$, $k=1,2,3$, is $\frac{a_{3}}{a_{1}}\equiv
a$, where $a$ is a constant.
\end{theorem}


Here, we will show the different properties among results on the existence
of Fermat type differential equations, difference equations and $q$-difference
equations. If we replace $f'$ with $f(z+c)$ in
\eqref{tou}, Theorem \ref{thmC} is not valid for difference equation
$$
a_{1}f^{2}+a_{2}f(z+c)^{2}=a_{3}.
$$
For example, the equation
$$
(z+c)^{2}f(z)^{2}+z^{2}f(z+c)^{2}=z^{2}(z+c)^{2}
$$
has an
entire solution $f(z)=z\sin z$ with the order $\rho(f)=1$,
where $c=\frac{\pi}{2}$. Here
$\frac{a_{3}}{a_{1}}=z^{2}$ is not a constant.

Theorem\ref{thmC} is  not valid for the $q$-difference equation
\begin{equation}\label{chehao}
a_{1}f^{2}+a_{2}f(qz)^{2}=a_{3}.
\end{equation}
The equation
$f(z)^{2}+f(-z)^{2}=z^{2}$ has a transcendental entire solution
\begin{equation}\label{1023}
f(z)=\frac{ze^{z+\frac{\pi}{4}i}+ze^{-z-\frac{\pi}{4}i}}{2},
\end{equation}
 here $\frac{a_{3}}{a_{1}}=z^{2}$.
A natural question is that how to describe the entire solutions
of Fermat type difference equations or
$q$-difference equations. The difference
analogue of the logarithmic derivative lemma \cite{chiang, R.G} for
meromorphic function with finite order has been developed to study
difference equations \cite{R.G, hal}, also can be used to consider Fermat
type difference equations \cite{liuk4,liucaocao,liukkk}. One of the results
can be stated as follows.

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


Li \cite{li} considered the meromorphic solutions of $f^{2}+a(f')^{2}=1$,
where $a$ is a nonzero function. Tang and Liao \cite{tang} investigated
the entire solutions of a generalization of \eqref{che} as follows
\begin{equation}\label{ctan}
f(z)^{2}+P(z)^{2}f^{(k)}(z)^{2}=Q(z),
\end{equation}
 where $P(z),Q(z)$ are nonzero polynomials and obtained the following result.


\begin{theorem}[{\cite[Theorem 1]{tang}}] \label{thmE}
 Let $P(z)$, $Q(z)$ be nonzero polynomials.
If \eqref{ctan} has a transcendental
meromorphic solution $f(z)$, then $P(z)\equiv A$, $Q\equiv B$, $k=2n+1$
for some nonnegative integer $n$ and $f(z)= b\sin(az+c)$, where $a,
b, c$ are constants such that $Aa^{k}=\pm1$, $b^{2}=B$.
\end{theorem}


In a recent article \cite{liu}, Liu considered an improvement of 
Theorem \ref{thmD}
and obtained the following result which can be seen as the difference
analogue of Theorem \ref{thmE}.

\begin{theorem}[{\cite[Theorem 2.1]{liu}}] \label{thmF}
Let $P(z)$, $Q(z)$ be nonzero polynomials.
If the difference equation
\begin{equation}\label{yaohao}
f(z)^{2}+P(z)^{2}f(z+c)^{2}=Q(z)
\end{equation}
admits a transcendental entire solution of finite order, then
$P(z)\equiv\pm1$ and $Q(z)$ reduces to a constant $q$.
 Thus, $f(z)=\sqrt{q}\sin(Az+B)$, where $B$ is a constant and
$A=\frac{(4k+1)\pi}{2c}$, $k$ is an integer, $c$ is a nonzero constant.
\end{theorem}

In \cite[Theorem1.1]{bhkm}, a $q$-difference analogue of
the logarithmic derivative lemma was given. Similarly, as the finite-order
solutions play a key role in complex difference equations,
solutions of order zero are in focus for $q$-difference equations.
A natural idea is how to describe the entire solutions with zero order
of Fermat type $q$-difference equations. However, we will consider
the entire solutions with finite order, not limited to zero order
in the following, we mainly study the entire solutions of Fermat type
$q$-difference equations
\begin{equation}\label{19}
f(z)^{2}+P(z)f(qz)^{2}=Q(z),
\end{equation}
where $P(z), Q(z)$ are nonzero polynomials, and $|q|=1$.
We obtain the following result.

\begin{theorem}\label{thm1}
Let $P(z)$, $Q(z)$ be nonzero polynomials and $|q|=1$.
If the q-difference equation
\begin{equation}\label{5656}
f(z)^{2}+P(z)^{2}f(qz)^{2}=Q(z)
\end{equation}
admits a transcendental entire solution of finite order, then $P(z)$
must be a constant $P$. This solution can be written as
$$
f(z)=\frac{Q_{1}(z)e^{p(z)}+Q_{2}(z)e^{-p(z)}}{2}
$$
satisfying one of the following conditions:
\begin{itemize}
\item[(i)] $q$ satisfies $p(qz)=p(z)$ and $Q_{1}(z)-iPQ_{1}(qz)\equiv0$,
$Q_{2}(z)+iPQ_{2}(qz)\equiv0$,
$P^{4}Q(q^{2}z)=Q(z)$;

\item[(ii)] $q$ satisfies $p(qz)+p(z)=2a_{0}$, and
$iPQ_{1}(qz)e^{2a_{0}}\equiv-Q_{2}(z)$,
$iPQ_{2}(qz)\equiv Q_{1}(z)e^{2a_{0}}$,
$P^{4}Q(q^{2}z)=Q(z)$, $e^{8a_{0}}=1$,
\end{itemize}
where $Q(z)=Q_{1}(z)Q_{2}(z)$ and $p(z)$ is a nonconstant polynomial.
\end{theorem}

Before giving some examples, we want to explain why we assume the condition
 $|q|=1$. From the proof of Theorem \ref{thm1} blew, it is easy
to find that the entire solutions with zero order of $\eqref{5656}$
must be polynomials. In addition, Bergweiler, Ishizaki and Yanagihara \cite{be}
considered the linear $q$-difference equation
\begin{equation}\label{1234}
\sum_{j=0}^{n}a_{j}(z)f(c^{j}z)=Q(z),
\end{equation}
where $Q(z)$ and $a_{j}(z)$ $(j=0,1,\ldots,n)$ are polynomials without
common zeros, $a_{n}(z)a_{0}(z)\not=0$ and $0<|c|<1$.
They proved that any meromorphic solution $f(z)$ of \eqref{1234}
satisfies $T(r,f)=O((\log r)^{2})$, which implies that the order of
$f(z)$ is zero. Thus, assume that $f(z)$ is an entire solution of \eqref{5656}.
Let $F(z)=f(z)^{2}$. If $|q|<1$, then \eqref{5656} changes
 into $F(z)+P(z)^{2}F(qz)=1$, we have $\rho(F)=0$, implies that $\rho(f)=0$,
thus $f(z)$ should be a polynomial. If $|q|>1$, thus $|\frac{1}{q}|<1$,
we assume that $G(z)=f(qz)^{2}$. Thus, \eqref{5656} takes into
$G(\frac{1}{q}z)+P(z)^{2}G(z)=1$. Using the observation (see \cite[p. 2]{be}),
 we have
$T(r,f(z))\leq T(|q|r,f(z))=T(r,f(qz))+O(1)=\frac{1}{2}T(r,G(z))+O(1)
=O((\log r)^{2})+O(1)$. We also get $\rho(f)=0$,
thus $f(z)$ should be a polynomial. So, when considering the transcendental
entire solution of \eqref{5656}, we need the condition $|q|=1$.



\begin{example} \label{examp1}\rm
 If $P(z)\equiv i$, $q=-1$ and $Q(z)\equiv z^{3}$, then
$f(z)^{2}-f(-z)^{2}=z^{3}$ has a transcendental entire
solution $f(z)=\frac{ze^{p(z)}+z^{2}e^{-p(z)}}{2}$, where $p(z)$ is
a polynomial satisfies $p(-z)=p(z)$, which implies that we can
assume that
$p(z)=a_{2n}z^{2n}+a_{2n-2}z^{2(n-1)}+\ldots+a_{2}z^{2}+a_{0}$.
Thus, $\rho(f)=2n$.
\end{example}


\begin{example} \label{examp2}\rm
 If $P(z)\equiv1$, $q=-i$ and $Q(z)\equiv z^{4}$, then
$f(z)^{2}+f(-iz)^{2}=z^{4}$ has a transcendental entire
solution $f(z)=\frac{ze^{p(z)}+z^{3}e^{-p(z)}}{2}$, where $p(z)$ is
a polynomial satisfies $p(-iz)=p(z)$, which implies that we can
assume that
$p(z)=a_{4n}z^{4n}+a_{4n-4}z^{4(n-1)}+\ldots+a_{4}z^{4}+a_{0}$.
Thus, $\rho(f)=4n$.
\end{example}

\begin{example} \label{examp3} \rm
If $P(z)\equiv-1$, $q=-1$ and $Q(z)\equiv
z^{2}$, then $f(z)^{2}+f(-z)^{2}=z^{2}$ has a transcendental entire
solution $f(z)=\frac{ze^{p(z)}+ze^{-p(z)}}{2}$, where $p(z)$ is a
polynomial satisfies $p(-z)+p(z)=4ki\pi+\frac{i\pi}{2}$, which
implies that we can assume that
$p(z)=a_{2n+1}z^{2n+1}+a_{2n-1}z^{2n-1}+\ldots+a_{1}z+2ki\pi+\frac{i\pi}{4}$,
where $k\in\mathbb{Z}$. Thus, $\rho(f)=2n+1$. We also remark that
\eqref{1023} is a solution of above equation.
\end{example}


\begin{remark} \label{rmk1}\rm
 From Examples \ref{examp1}--\ref{examp3}, we find that the order
of entire solutions of $q$-difference equation \eqref{5656} can be
large enough which is different from the growth of entire solutions
on \eqref{ctan} and \eqref{yaohao}.
\end{remark}

\begin{corollary}\label{coro1}
If $P(z)$, $Q(z)$ are nonconstant polynomials, then there does not
exist transcendental entire solutions of finite order of q-difference equation
\begin{equation}\label{565655}
f(z)^{2}+P(z)^{2}f(qz)^{2}=Q(z).
\end{equation}
\end{corollary}

\begin{theorem}\label{66}
Let $P(z)$, $Q(z)$ be nonzero polynomials.
Then the q-difference equation
\begin{equation}\label{5656pp}
f(z)^{2}+zP(z)^{2}f(qz)^{2}=Q(z)
\end{equation}
has no transcendental entire
solutions of finite order.
\end{theorem}

\begin{proof}  Assume that $f(z)$ is a transcendental
entire solutions of finite order. Let $z=z_{1}^{2}$ and
$F(z_{1})=f(z_{1}^{2})$, $M(z_{1})=P(z_{1}^{2})$,
$N(z_{1})=Q(z_{1}^{2})$. Then $F(z_{1})$ is entire function in
$z_{1}$, and $M(z_{1})$, $N(z_{1})$ are polynomials in $z_{1}$.
Thus, from \eqref{5656pp}, we have the following equation
\begin{equation}\label{69}
F(z_{1})^{2}+z_{1}^{2}M(z_{1})^{2}F(\sqrt{q}z_{1})^{2}=N(z_{1})
\end{equation}
From Theorem \ref{thm1}, we obtain $z_{1}^{2}M(z_{1})^{2}$ reduce to a
constant, which is impossible.
\end{proof}

If we replace $f(qz)$ with $f(qz)-f(z)$,  then we obtain the
following result.

\begin{theorem}\label{thm2}
If $P(z)$ is a nonzero polynomial with nonzero constant term, then
there is no finite order entire solution $f(z)$ satisfying
\begin{equation}\label{ds}
f(z)^{2}+P(z)^{2}[f(qz)-f(z)]^{2}=Q(z).
\end{equation}
\end{theorem}

The following result plays an important part in the proofs of our
theorems.

\begin{lemma}[{\cite[Theorem 1.62]{yang and yi}}]\label{lemd}
Let $f_{j}(z)$ be meromorphic functions, $f_{k}(z)$
 be not constant functions $(k=1,2,\dots, n-1)$, satisfying $\sum_{j=1}^{n}f_{j}=1$ and
$n\geq3$. If $f_{n}(z)\not\equiv0$ and
\begin{eqnarray*}
\sum_{j=1}^{n}N(r,\frac{1}{f_{j}})+(n-1)
\sum_{j=1}^{n}\overline{N}(r,f_{j})<(\lambda+o(1))T(r,f_{k}),
\end{eqnarray*}
where $\lambda<1$ and $k=1,2,\dots, n-1$, then $f_{n}(z)\equiv1$.
\end{lemma}

\begin{lemma}\label{leml}
Let $p(z)$ be a nonzero polynomial with degree $n$. If $p(qz)-p(z)$
is a constant, then $q^{n}=1$ and $p(qz)\equiv p(z)$. If
$p(qz)+p(z)$ is a constant, then $q^{n}=-1$ and
$p(qz)+p(z)\equiv2a_{0}$, where $a_{0}$ is the constant term of
$p(z)$.
\end{lemma}

\begin{proof}  Assume that
$p(z)=a_{n}z^{n}+\dots+a_{1}z+a_{0}$. Then
$$p(qz)-p(z)=a_{n}(qz)^{n}+\dots+a_{1}qz-a_{n}z^{n}+\dots-a_{1}z,$$
we obtain $a_{n}(qz)^{n}=a_{n}z^{n}$, which implies that $q^{n}=1$.

From $
p(qz)+p(z)=a_{n}(qz)^{n}+\dots+a_{1}qz+a_{n}z^{n}+\dots+a_{1}z+2a_{0}
$, we obtain $a_{n}(qz)^{n}+a_{n}z^{n}=0$, which implies that
$q^{n}=-1$. We have completed the proof.
\end{proof}



\begin{proof}[Proof of Theorem \ref{thm1}]
Assume that $f(z)$ is a transcendental
entire solution of finite order of \eqref{5656}, then
\begin{equation}\label{pp}
[f(z)+iP(z)f(qz)][f(z)-iP(z)f(qz)]=Q(z).
\end{equation}
Thus, both $f'(z)+iP(z)f(qz)$ and $f'(z)-iP(z)f(qz)$ have finitely
many zeros. Combining \eqref{pp} with the Hadamard factorization
theorem, we assume that
$$
f(z)+iP(z)f(qz)=Q_{1}(z)e^{p(z)}
$$
and
$$
f(z)-iP(z)f(qz)=Q_{2}(z)e^{-p(z)},
$$
where $p(z)$ is a nonconstant polynomial, otherwise $f(z)$ is
a polynomial, and $Q(z)=Q_{1}(z)Q_{2}(z)$, where $Q_{1}(z)$,
$Q_{2}(z)$ are nonzero polynomials. Thus, we have
\begin{equation}\label{men1}
f(z)=\frac{Q_{1}(z)e^{p(z)}+Q_{2}(z)e^{-p(z)}}{2}
\end{equation}
 and
\begin{equation}\label{menmen222}
f(qz)=\frac{Q_{1}(z)e^{p(z)}-Q_{2}(z)e^{-p(z)}}{2iP(z)}.
\end{equation}
Combining \eqref{men1} with \eqref{menmen222}, we obtain
$$
f(qz)=\frac{Q_{1}(qz)e^{p(qz)}+Q_{2}(qz)e^{-p(qz)}}{2}
=\frac{Q_{1}(z)e^{p(z)}-Q_{2}(z)e^{-p(z)}}{2iP(z)}.
$$
Thus,
\begin{equation}\label{363891}
\frac{iP(z)Q_{1}(qz)e^{p(qz)+p(z)}}{-Q_{2}(z)}
+\frac{iP(z)Q_{2}(qz)e^{p(z)-p(qz)}}{-Q_{2}(z)}
+\frac{Q_{1}(z)e^{2p(z)}}{Q_{2}(z)}=1.
\end{equation}
Since $p(z)$ is not a constant, then  $p(qz)-q(z)$ and $p(qz)+q(z)$
are not constants simultaneously. From Lemma \ref{lemd}, we have one of
$p(qz)-q(z)$ and $p(qz)+q(z)$ must be a constant. The following, we
will discuss two cases.

\noindent\textbf{Case 1.} Suppose that $p(qz)-q(z)$ is a constant.
 From Lemma \ref{leml}, then $p(qz)-p(z)\equiv0$. It implies that $q^{n}=1$. Thus,
$$
(Q_{1}(z)-iP(z)Q_{1}(qz))e^{2p(z)}=Q_{2}(z)+iP(z)Q_{2}(qz)
$$
follows
from \eqref{363891}. Since $p(z)$ is not a constant, then we have
$$
Q_{1}(z)-iP(z)Q_{1}(qz)\equiv0,
$$
and
$$
Q_{2}(z)+iP(z)Q_{2}(qz)\equiv0.
$$
Thus, $P(z)$ must be a constant $P$. From above two equations, we have
$P^{4}Q(q^{2}z)=Q(z)$, where $Q(z)=Q_{1}(z)Q_{2}(z)$.


\noindent\textbf{Case 2.}
Suppose that $p(qz)+p(z)$ is a constant. Thus, from
Lemma \ref{leml}, we obtain
\begin{equation}\label{3640}
iP(z)Q_{1}(qz)e^{2a_{0}}\equiv-Q_{2}(z).
\end{equation}
From \eqref{363891}, we obtain
\begin{equation}\label{364}
iP(z)Q_{2}(qz)\equiv Q_{1}(z)e^{2a_{0}}.
\end{equation}
Combining \eqref{3640} with \eqref{364}, we have
\begin{equation}\label{369}
P(z)P(qz)Q_{2}(q^{2}z)\equiv Q_{2}(z),
\end{equation}
which implies that $P(z)$ must be a constant $P$. We also have
\begin{equation}\label{36911}
P(z)P(qz)Q_{1}(q^{2}z)\equiv Q_{1}(z).
\end{equation}
Thus, from \eqref{369} and \eqref{36911}, we have
$P^{4}Q(q^{2}z)=Q(z)$ and
$$\frac{Q_{1}(q^{2}z)}{Q_{2}(q^{2}z)}=\frac{Q_{1}(z)}{Q_{2}(z)}.$$
Combining \eqref{3640} with \eqref{364}, we also have
$$\frac{Q_{1}(qz)}{Q_{2}(qz)}e^{4a_{0}}=-\frac{Q_{2}(z)}{Q_{1}(z)}.$$
Hence,
$\frac{Q_{1}(q^{2}z)}{Q_{2}(q^{2}z)}e^{4a_{0}}=-\frac{Q_{2}(qz)}{Q_{1}(qz)}=\frac{1}{e^{4a_{0}}}\frac{Q_{1}(z)}{Q_{2}(z)}$,
thus $e^{8a_{0}}=1$.
\end{proof}

For the proof of Theorem \ref{thm2}, we need the following two results.

\begin{lemma}\label{lem2}
Let $Q(z)$ be a nonzero polynomial, $|t|\not=1$.
If $Q(qz)\equiv tQ(z)$, then $Q(z)$ should be reduce to a monomial and
$Q(z)=a_{n}z^{n}$.
\end{lemma}

\begin{proof} Obviously, we have $Q(0)=0$. Assume that
$$
Q(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+\ldots+a_{1}z,
$$
where $a_{n}\not=0$. Then,
$$
Q(qz)=a_{n}(qz)^{n}+a_{n-1}(qz)^{n-1}+a_{n-2}(qz)^{n-2}+\ldots+a_{1}qz.
$$
Since $Q(qz)\equiv tQ(z)$, then $a_{n}(qz)^{n}=ta_{n}z^{n}$,
thus $q^{n}=t$. From $a_{n-1}(qz)^{n-1}=ta_{n-1}z^{n-1}$, we have
either $a_{n-1}=0$ or $q^{n-1}=t$. If $q^{n-1}=t$, combining with
$q^{n}=t$, we have $q=1$, which is impossible. Thus, $a_{n-1}=0$.
From $a_{n-2}(qz)^{n-2}=ta_{n-2}z^{n-2}$, then $a_{n-2}=0$ or
$q^{n-2}=t$. If $q^{n-2}=t$, combining with $q^{n}=t$, we have
$q^{2}=1$, hence $t^{2}=q^{2n}=1$, thus $|t|=1$, a contradiction
with the condition.  Using this method, we can get
$a_{j}=0(j=1,2,\ldots,n-1)$.
\end{proof}

\begin{lemma}\label{lem3}
Let $P(z)$ be a polynomial with nonzero constant term. If
$P(z)$ and $Q(z)$ satisfy $P(z)^{2}Q(qz)=(1+P(z)^{2})Q(z)$,
then $P(z)$ must be a constant and $Q(z)$ must be a monomial.
\end{lemma}

\begin{proof}
 Obviously, $Q(0)=0$. Assume that
$P(z)^{2}=b_{m}z^{m}+b_{m-1}z^{m-1}+\ldots+b_{1}z+b_{0}$, where $b_{0}\not=0$ and
$Q(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+\ldots+a_{1}z$. From
$P(z)^{2}Q(qz)=(1+P(z)^{2})Q(z)$, if $P(z)^{2}\equiv b_{0}$, we obtain
 $Q(qz)=\frac{1+b_{0}}{b_{0}}Q(z)$, from Lemma \ref{lem2},
we obtain $Q(z)$ should be reduce to a monomial $Q(z)=a_{n}z^{n}$.
If $P(z)^{2}$ is not a constant, that is $b_{m}\not=0$, then
$P(z)^{2}Q(qz)=(1+P(z)^{2})Q(z)$, thus
$b_{m}q^{n}z^{n+m}=b_{m}z^{n+m}$, $q^{n}=1$ follows,
and $b_{0}q^{n}z^{n}=(1+b_{0})z^{n}$, we obtain $b_{0}=(1+b_{0})$
which is a contradiction.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
Assume that $f(z)$ is a transcendental
entire solution of finite order of \eqref{ds}.
Similar idea as the beginning of the proof of Theorem \ref{thm1}, we can obtain
\begin{equation}\label{f1}
f(z)=\frac{Q_{1}(z)e^{h(z)}+Q_{2}(z)e^{-h(z)}}{2}
\end{equation}
 and
\begin{equation}\label{f2}
f(qz)-f(z)=\frac{Q_{1}(z)e^{h(z)}-Q_{2}(z)e^{-h(z)}}{2iP(z)},
\end{equation}
where $h(z)$ is a nonconstant polynomial with $\deg(h(z))=s$.
Thus, from the expression of $f(qz)$, we have
\begin{equation}\label{f3}
\frac{Q_{1}(qz)e^{h(qz)}+Q_{2}(qz)e^{-h(qz)}}{2}
=\frac{Q_{1}(z)(1+P(z)i)e^{h(z)}-Q_{2}(z)(1-P(z)i)e^{-h(z)}}{2iP(z)}.
\end{equation}
Hence, we obtain
\begin{equation}\label{f4}
\frac{Q_{1}(z)(1+P(z)i)}{iQ_{2}(qz)P(z)}e^{h(z)+h(qz)}
-\frac{Q_{2}(z)(1-P(z)i)}{iQ_{2}(qz)P(z)}e^{h(qz)-h(z)}
-\frac{Q_{1}(qz)}{Q_{2}(qz)}e^{2h(qz)}=1.
\end{equation}
If $P(z)\equiv i$, from \eqref{f4},
 \begin{equation}\label{f5}
\frac{2Q_{2}(z)}{Q_{2}(qz)}e^{h(qz)-h(z)}-\frac{Q_{1}(qz)}{Q_{2}(qz)}e^{2h(qz)}=1
\end{equation}
follows. It is easy to find that $2h(qz)$ and $h(qz)-q(z)$ are not constants
simultaneously. Thus, the equation \eqref{f5} is impossible.
If $P(z)\equiv -i$, from \eqref{f4}, we have
$\frac{2Q_{1}(z)}{Q_{2}(qz)}e^{h(qz)+h(z)}-\frac{Q_{1}(qz)}{Q_{2}(qz)}e^{2h(qz)}=1$,
and $2h(qz)$ and $h(qz)+q(z)$ are not constants simultaneously,
above equation also is impossible.

Thus, we have $P(z)\not\equiv \pm i$. Since that $2h(z)$ is not a constant,
then we discuss two cases from Lemma \ref{lemd} in the following:

\noindent\textbf{Case 1.}
Suppose that $h(qz)-h(z)$ is a constant, which implies that $h(qz)=h(z)$
and $q^{s}=1$.
 Thus,
$$
Q_{1}(z)(1+P(z)i)\equiv iQ_{1}(qz)P(z)
$$
and
$$
Q_{2}(z)(1-P(z)i)\equiv-iQ_{2}(qz)P(z).
$$
 From above two equations, we have
\begin{equation}\label{y}
Q(z)(1+P(z)^{2})\equiv Q(qz)P(z)^{2}.
\end{equation}
Thus, we have $Q(0)=0$. If $P(z)$ is a constant, then equation
\eqref{y} can be written as $Q(qz)\equiv\frac{1+P^{2}}{P^{2}}Q(z)$, where
$|\frac{1+P^{2}}{P^{2}}|\not=1$. Using Lemma \ref{lem2}, we obtain
$Q(z)=a_{n}z^{n}$. Thus, $a_{n}(1+P^{2})=a_{n}q^{n}P^{2}$, which
implies that $|q|\not=1$, a contradiction with $q^{s}=1$. If $P(z)$
is a polynomial with nonzero constant term, from Lemma \ref{lem3}, which is impossible.


\noindent\textbf{Case 2.}
Suppose that $h(qz)+h(z)$ is a constant $2a_{0}$. We have
$$
Q_{1}(z)(1+iP(z))e^{2a_{0}}\equiv iQ_{2}(qz)P(z),
$$
and
$$
Q_{2}(z)(1-iP(z))\equiv-iQ_{1}(qz)P(z)e^{2a_{0}}
$$
From above two equations, we have
$$
Q(z)(1+P(z)^{2})\equiv Q(qz)P(z)^{2}.
$$
Similar discussions as the Case 1, we can get a contradiction.
Thus, we have completed the proof of Theorem \ref{thm2}.
\end{proof}

\section{Fermat type $q$-difference differential equations}

If an equation includes some of $f(z)$, $f(qz)$, $f^{(k)}(z)$,
$f(z+c)$, then this equation can be called $q$-difference
differential equation. Yang and Laine \cite{yang and L} considered
entire solutions of difference-differential equation. Liu, Cao and
Cao have considered Fermat type differential-difference equation
in \cite[Theorem 1.3]{liucaocao}, and obtained the following result.

\begin{theorem} \label{thmG}
Transcendental entire solutions with finite order
of the differential-difference equation
\begin{equation}\label{yu}
f'(z)^{2}+f(z+c)^{2}=1
\end{equation}
 satisfy $f(z)=\sin(z\pm Bi)$, where $B$ is a constant 
and $c=2k\pi$ or $c=2k\pi+\pi$, $k$ is an integer.
\end{theorem}

Next we will consider entire solutions for Fermat type $q$-difference 
differential equation, such as
\begin{gather}\label{2}
f'(z)^{2}+f(qz)^{2}=1, \\
\label{1025}
f(z+c)^{2}+f(qz)^{2}=1, \\
\label{1026}
f'(z+c)^{2}+f(qz)^{2}=1.
\end{gather}
Using a method similar to the one in this paper,  we can
consider some generalizations of above equations, such as 
$f'(z)^{2}+P(z)^{2}f(qz)^{2}=Q(z)$, but they will not be
considered here. We also find that the methods for the
proofs of the following two theorems are similar, so we just give
the details of proof of Theorem \ref{hhh}.
 
\begin{theorem}\label{jjj}
The transcendental entire solutions with finite order of \eqref{2}
satisfy $f(z)=\sin(z+B)$ when $q=1$, and $f(z)=\sin(z+k\pi)$ or
$f(z)=-\sin(z+k\pi+\frac{\pi}{2})$ when $q=-1$. There are no
transcendental entire solutions with finite order when $q\not=\pm1$.
\end{theorem}

\begin{theorem}\label{hhh}
The transcendental entire solutions with finite order of
\eqref{1026} satisfy $f(z)=\sin(z-Ai-c)$, where $2A-ic=2ki\pi$ or 
$f(z)=\sin(z-Ai+c)$, where $2A+ic=2ki\pi+i\pi$ when $q=-1$, and
$f(z)=\sin(-z-Ai)$, $c=2i\pi$ or $f(z)=\sin(-z-Ai+\pi)$, $c=2k\pi+\pi$ when
$q=1$.
\end{theorem}


\begin{proof}  As in the beginning of the proof of
Theorem \ref{thm1}, we have
\begin{equation}\label{wan1}
f'(z+c)=\frac{e^{p(z)}+e^{-p(z)}}{2}
\end{equation}
 and
\begin{equation}\label{wan2}
f(qz)=\frac{e^{p(z)}-e^{-p(z)}}{2i}.
\end{equation}
Combining \eqref{wan1} with \eqref{wan2}, we obtain
\begin{equation}\label{36389}
\frac{p'(z+\frac{c}{q})e^{p(z+\frac{c}{q})+p(qz)}}{iq}
+\frac{p'(z+\frac{c}{q})e^{p(qz)-p(z+\frac{c}{q})}}{iq}-e^{2p(qz)}=1.
\end{equation}
From Lemma \ref{lemd}, if $p(z+\frac{c}{q})+p(qz)=B$, then we have
$\frac{p'(z+\frac{c}{q})e^{B}}{iq}=1$ and
$\frac{p'(z+\frac{c}{q})e^{-B}}{iq}=1$. Thus $e^{2B}=1$. If
$e^{B}=1$, then $p(z)=iqz+A-ic$, where $A$ is a
constant. Thus, from $p(z+\frac{c}{q})+p(qz)=2ki\pi$, we have 
$q=-1$ and $2A-ic=2ki\pi$, $k$ is
an integer. Hence $p(z)=-iz+A-ic$, and
$$
f(z)=\frac{e^{iz+A-ic}-e^{-iz-A+ic}}{2i}=\sin(z-Ai-c),
$$
where $2A-ic=2ki\pi$.
If $e^{B}=-1$, then $p(z)=-iqz+A+ic$, where $A$
is a constant. Thus, from $p(z+\frac{c}{q})+p(qz)=2ki\pi+i\pi$,
 we have $q=-1$ and $2A+ic=2ki\pi+i\pi$,
$k$ is an integer. Hence $p(z)=-iz+A+ic$, and
$$
f(z)=\frac{e^{iz+A+ic}-e^{-iz-A-ic}}{2i}=\sin(z-Ai+c), 
$$
where $2A+ic=2ki\pi+i\pi$.

If $p(qz)-p(z+\frac{c}{q})=D$, where $D$ is a constant. Then we have
$\frac{p'(z+\frac{c}{q})e^{D}}{iq}=1$ and
$\frac{p'(z+\frac{c}{q})e^{-D}}{iq}=1$. If $e^{D}=1$, thus
$p(z)=iqz+A-ic$, where $A$ is a constant. Thus, we have $q=1$ and
$c=2k\pi$, $k$ is an integer. Hence $p(z)=iz+A-2ki\pi$, and
$$
f(z)=\frac{e^{iz+A-2ki\pi}-e^{-iz-A+2ki\pi}}{2i}=\sin(-z-Ai).
$$
 If $e^{D}=-1$, thus $p(z)=-iqz+A+ic$, where $A$ is a constant. Thus, we
have $q=1$ and $c=2k\pi+\pi$, $k$ is an integer. Hence
$p(z)=-iz+A-2ki\pi-i\pi$, and
$$
f(z)=\frac{e^{-iz+A-2ki\pi-i\pi}-e^{iz-A+2ki\pi+i\pi}}{2i}=\sin(-z-Ai+\pi).
$$
\end{proof}

\begin{remark} \label{rmk2} \rm
Obviously, Theorem \ref{jjj} is an improvement of Theorem \ref{thmB}. 
However, it may be difficult to give all entire
solutions of \eqref{1025}. Because, we can not get the precise
expression of $p(z)$ satisfying $p(z+\frac{c}{q})-p(qz)=B$ or
$p(z+\frac{c}{q})+p(qz)=B$. If $c=0$, it is the special case of
Theorem \ref{thm1}. If $q=1$, it is the case of Theorem \ref{thmD}. Here, we can
construct entire solutions of \eqref{1025}. For example, if $q=-1$,
$c=\frac{\pi}{2}$, thus $f(z)=\sin z$ satisfies
$f(z+\frac{\pi}{2})^{2}+f(-z)^{2}=1$. If $q=\frac{1+i\sqrt{3}}{2}$,
$c=\frac{1-i\sqrt{3}}{2}$, and
$p(z)=\frac{1}{3}z^{3}+z^{2}+z+\frac{3i}{4}\pi+\frac{1}{3}+ki\pi$,
thus $p(z+\frac{c}{q})+p(qz)=\frac{3i\pi}{2}+2ki\pi$ and $k$ is an
integer. Thus,
$f(z)=\frac{e^{p(z-\frac{1-i\sqrt{3}}{2})}-e^{-p(z-\frac{1-i\sqrt{3}}{2})}}{2}$
satisfies
$f(z+\frac{1-i\sqrt{3}}{2})^{2}+f(\frac{1+i\sqrt{3}}{2}z)^{2}=1$.
\end{remark}

\subsection*{Acknowledgments}
The authors would like to thank the anonymous referees for their helpful
suggestions and comments.

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\end{document}