Let $f(z)$ be a meromorphic function in the complex plane. We assume that the reader is familiar with standard symbols and fundamental results of Nevanlinna theory [5,16]. As we all know that Nevanlinna theory was extensively applied to considering the growth, value distribution, and solvability of meromorphic solutions of differential equations [6]. Recently, difference analogues of Nevanlinna theory were established, which also be used to consider the corresponding properties of meromorphic solutions on difference equations or $q$-difference equations, such as [2, 4, 7-12, 14, 17].

Let us recall the classical Fermat type equation

$ \begin{equation} f(z)^{2}+g(z)^{2}=1. \end{equation} $

(1.1)

Equation (1.1) 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. However, the above result fails to give more precise informations when $g(z)$ has a special relationship with $f(z)$. Yang and Li [15] first considered the entire solutions of the Fermat type differential equation

$ \begin{equation} f(z)^{2}+f'(z)^{2}=1, \end{equation} $

(1.2)

and they proved the following result.

Theorem A  [15, Theorem 1] The transcendental meromorphic solutions of (1.2) must satisfy $f(z)=\frac{1}{2}\left(Pe^{-i z}+\frac{1}{P}e^{iz}\right)=\sin(z+B)$, where $P$ is non-zero constant and $e^{iB}=\frac{i}{P}$.

Tang and Liao [13] further investigated the entire solutions of a generalization of (1.2) as follows

$ \begin{equation} f(z)^{2}+P(z)^{2}f^{(k)}(z)^{2}=Q(z), \end{equation} $

(1.3)

where $P(z), Q(z)$ are non-zero polynomials and obtained the next result.

Theorem B  [13, Theorem 1] If the differential equation (1.3) has a transcendental meromorphic solution $f$, then $P(z)\equiv A$, $Q(z)\equiv B$, $k$ is an odd and $f(z)=b\sin(az+d)$, where $a, b, d$ are constants such that $Aa^{k}=\pm1$, $b^{2}=B$.

Recently, the difference analogues of Nevanlinna theory were used to consider the solutions properties of Fermat type difference equations. Liu, Cao and Cao [8] investigated the finite order entire solutions of the difference equation

$ \begin{equation} f(z)^{2}+f(z+c)^{2}=1, \end{equation} $

(1.4)

here and in the following, $c$ is a non-zero constant and $P(z), Q(z)$ are non-zero polynomial, unless otherwise specified. The result can be stated as follows.

Theorem C  [8, Theorem 1.1] The transcendental entire solutions with finite order of $(1.4)$ must satisfy $f(z)=\sin(Az+B)$, where $B$ is a constant and $A=\frac{(4k+1)\pi}{2c}$, $k$ is an integer.

Furthermore, Liu and Yang [10] considered a generalization of (1.4) as follows

$ \begin{equation} f(z)^{2}+P(z)^{2}f(z+c)^{2}=Q(z), \end{equation} $

(1.5)

and obtained the following result.

Theorem D  Let $P(z)$, $Q(z)$ be non-zero polynomials. If the difference equation (1.5) admits a transcendental entire solution of finite order, then $P(z)\equiv\pm1$ and $Q(z)$ reduces to a constant $q$.

If an equation includes the $q$-difference $f(qz)$ and the derivatives of $f(z)$ or $f(z+c)$, then this equation can be called $q$-difference differential equation. Liu and Cao [11] considered the entire solutions on Fermat type $q$-difference differential equation

$ \begin{equation} f'(z)^{2}+f(qz)^{2}=1, \end{equation} $

(1.6)

and obtained the following result.

Theorem E  [11, Theorem 3.1]The transcendental entire solutions with finite order of (1.6) must 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 $q\not=\pm1$.

By comparing with the above five theorems, we state the following questions which will be considered in this paper.

Question 1  From Theorem A to Theorem E, we remark that the order of all transcendental entire solutions with finite order of different equations are equal to one. Hence, considering a generalization of equation (1.6), such as

$ \begin{equation} f'(z)^{2}+P(z)^{2}f(qz)^{2}=Q(z), \end{equation} $

(1.7)

it is natural to ask if the finite order of the entire solutions of (1.7) is equal to one or not?

Question 2  From Theorem B to Theorem E, the existence of finite order entire solutions of (1.3) and (1.5) forces the polynomial $P(z)$ reduce to a constant. Is it also remain valid for equation (1.7)?

However, Examples 1 and 2 below show that Questions 1 and 2 are false in generally.

Example 1  Entire function $f(z)=\sin z^{n}$ solves

$ f'(z)^{2}+n^{2}z^{2(n-1)}f(qz)^{2}=n^{2}z^{2(n-1)}, $

where $q$ satisfies $q^{n}=1$. It implies that the solutions order of (1.7) may take arbitrary numbers and $P(z)^{2}=n^{2}z^{2(n-1)}$ is not a constant.

Example 2  We can construct a general solution from Example 1. Entire function $f(z)=\sin (h(z))$ solves

$ f'(z)^{2}+[h'(z)]^{2}f(qz)^{2}=[h'(z)]^{2}, $

where $q$ satisfies $q^{n}=1$ and $h(z)$ is a non-constant polynomial.

Example 3  Function $f(z)=\sinh z$ is also an entire solution of $f'(z)^{2}-f(qz)^{2}=1$ and $f(z)=\cosh z$ is an entire solution of $f'(z)^{2}-f(qz)^{2}=-1$, where $q=-1$.

From Example 1 to Example 3, we also remark that if $P(z)^{2}=\pm1$, the transcendental entire solutions $f(z)$ are of order one, if $P(z)=nz^{(n-1)}$, the transcendental entire solutions $f(z)$ are of order $n$. Hence, it is reasonable to conjecture that the order of entire solutions of $(1.7)$ is equal to $\rho(f)=1+\deg{P(z)}$. In this paper, we will answer the above conjecture and obtain the following result.

Theorem 1.1  If $|q|>1$, then the entire solution of $(1.7)$ should be a polynomial. If there exists a finite order transcendental entire solution $f$ of $(1.7)$, then $\rho(f)=1+\deg{P(z)}$ and $|q|=1$.

In the following, we will consider another $q$-difference differential equation

$ \begin{equation} f'(z+c)^{2}+P(z)^{2}f(qz)^{2}=Q(z), \end{equation} $

(1.8)

and obtain the following result.

Theorem 1.2  If $|q|>1$, then the entire solution of $(1.8)$ should be a polynomial. If there exist a finite order transcendental entire solution $f$ of $(1.8)$, then $\rho(f)=1+\deg{P(z)}$ and $|q|=1$.

Example 4  Function $f(z)=\sin z$ is an entire solution of $f'(z+c)^{2}+f(qz)^{2}=1$, where $c=\pi$ and $q=-1$.

Finally, we consider other $q$-difference equation

$ \begin{equation} f(z+c)^{2}+P(z)^{2}f(qz)^{2}=Q(z). \end{equation} $

(1.9)

Theorem 1.3  If $|q|>1$, then the entire solution $f(z)$ of $(1.9)$ should be a polynomial.

If $P(z)^{2}=1$ in $(1.9)$, the following example shows that we can not give the precise expression of finite order entire solution and the order of $f(z)$ does not satisfy $\rho(f)=1+\deg{P(z)}$ and $|q|=1$.

Example 5   [11] 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. $

Remark 1  The proofs of Theorem 1.2 and Theorem 1.3 are similar as the proof of Theorem 1.1. Hence we will not give the details here.

For the proofs of Theorems 1.1, 1.2 and 1.3, we need the following results.

Lemma 2.1  [3, Lemma 3.1] Let $\Phi: (1, \infty)\rightarrow (0, \infty)$ be a monotone increasing function, and let $f$ be a nonconstant meromorphic function. If for some real constant $\alpha\in(0, 1)$, there exist real constants $K_{1}>0$ and $K_{2}\geq1$ such that

$ T(r, f)\leq K_{1}\Phi(\alpha r)+K_{2}T(\alpha r, f)+S(\alpha r, f), $

then

$ \rho(f)\leq\frac{\log K_{2}}{-\log \alpha}+\lim\sup\limits_{r\rightarrow\infty}\frac{\log\Phi(r)}{\log r}. $

Lemma 2.2  [11, Lemma 2.15] Let $p(z)$ be a non-zero 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 $p(z)$.

Lemma 2.3  [2, Theorem 2.1] Let $f(z)$ be transcendental meromorphic function of finite order $\rho$. Then for any $\varepsilon>0$, we have

$ \begin{equation} T(r, f(z+c))=T(r, f)+O(r^{\rho-1+\varepsilon})+O(\log r)=T(r, f)+S(r, f). \end{equation} $

(2.1)

Lemma 2.4  [16, Theorem 1.62] Let $f_{j}(z)$ be meromorphic functions, $f_{k}(z)$ $(k=1, 2, \cdots, n-1)$ be not constants, satisfying $\sum\limits_{j=1}^{n}f_{j}=1$ and $n\geq3$. If $f_{n}(z)\not\equiv0$ and

$ \begin{eqnarray*}\sum\limits_{j=1}^{n}N(r, \frac{1}{f_{j}})+(n-1)\sum\limits_{j=1}^{n}\overline{N}(r, f_{j}) < (\lambda+o(1))T(r, f_{k}), \end{eqnarray*} $

where $\lambda < 1$ and $k=1, 2, \cdots, n-1$, then $f_{n}(z)\equiv1$.

If $|q|>1$ and $f(z)$ is an entire solution of $(1.7)$, we use the observation (see [1]) that

$ T(r, f(qz))=T(|q|r, f(z))+O(1) $

holds for any meromorphic function $f$ and any constant $q$. If $f(z)$ is a transcendental entire function, then from (1.7) and Valiron-Mohon'ko theorem, we have

$ T(|q|r, f(z))=T(r, f(qz))+O(1)\leq T(r, f'(z))+S(r, f)\leq T(r, f(z))+S(r, f). $

Let $\alpha=\frac{1}{|q|}$ and $|q|>1$. Then we have

$ T(|q|\alpha r, f(z))\leq T(\alpha r, f(z))+S(\alpha r, f(z)). $

Hence, we have $T(r, f(z))\leq T(\alpha r, f(z))+S(\alpha r, f(z))$. From Lemma 2.1, we have $\rho(f)=0$. Combining Hadamard factorization theorem, we have $f'(z)+iP(z)f(qz)=Q_{1}(z)$ and $f'(z)-iP(z)f(qz)=Q_{2}(z)$, thus $f'(z)=\frac{Q_{1}(z)+Q_{2}(z)}{2}$ is a polynomial, which is a contradiction with $f(z)$ is a transcendental entire function. Thus $f(z)$ should be a polynomial.

Assume that $f(z)$ is a transcendental entire solution of (1.7) with finite order, then

$ \begin{equation} [f'(z)+iP(z)f(qz)][f'(z)-iP(z)f(qz)]=Q(z). \end{equation} $

(3.1)

Thus both $f'(z)+iP(z)f(qz)$ and $f'(z)-iP(z)f(qz)$ have finitely many zeros. Combining (3.1) with the Hadamard factorization theorem, we assume that

$ f'(z)+iP(z)f(qz)=Q_{1}(z)e^{h(z)} $

and

$ f'(z)-iP(z)f(qz)=Q_{2}(z)e^{-h(z)}, $

where $h(z)$ is a non-constant polynomial provided that $f(z)$ is of finite order transcendental and $Q_{1}(z)Q_{2}(z)=Q(z)$, where $Q_{1}(z)$, $Q_{2}(z)$ are non-zero polynomials. Thus we have

$ \begin{equation} f'(z)=\frac{Q_{1}(z)e^{h(z)}+Q_{2}(z)e^{-h(z)}}{2} \end{equation} $

(3.2)

and

$ \begin{equation} f(qz)=\frac{Q_{1}(z)e^{h(z)}-Q_{2}(z)e^{-h(z)}}{2iP(z)}. \end{equation} $

(3.3)

From (3.2), we have

$ \begin{equation} f'(qz)=\frac{Q_{1}(qz)e^{h(qz)}+Q_{2}(qz)e^{-h(qz)}}{2}. \end{equation} $

(3.4)

Taking first derivative of (3.3), we have

$ \begin{equation} f'(qz)=\frac{A(z)e^{h(z)}-B(z)e^{-h(z)}}{2iqP(z)^{2}}, \end{equation} $

(3.5)

where

$ \begin{equation} A(z)=P(z)Q_{1}'(z)+Q_{1}(z)[P(z)h'(z)-P'(z)] \end{equation} $

(3.6)

and

$ \begin{equation} B(z)=P(z)Q_{2}'(z)-Q_{2}(z)[P(z)h'(z)+P'(z)]. \end{equation} $

(3.7)

From (3.4) and (3.5), we have

$ \begin{equation} \frac{A(z)e^{h(qz)+h(z)}}{iqP(z)^{2}Q_{2}(qz)}-\frac{B(z)e^{h(qz)-h(z)}}{iqP(z)^{2}Q_{2}(qz)}-\frac{Q_{1}(qz)}{Q_{2}(qz)}e^{2h(qz)}\equiv1. \end{equation} $

(3.8)

Obviously, if $h(qz)$ is a constant, then $h(z)$ is a constant, thus $f(z)$ should be a polynomial. If $h(qz)$ is a non-constant entire function, then $h(qz)-h(z)$ and $h(qz)+h(z)$ are not constants simultaneously. The following, we will discuss two cases.

Case 1  If $h(qz)-h(z)$ is not a constant, from Lemma 2.4, we know that

$ \begin{equation} \frac{A(z)e^{h(qz)+h(z)}}{iqP(z)^{2}Q_{2}(qz)}\equiv1. \end{equation} $

(3.9)

Since $f(z)$ is a finite order entire solution, then $h(z)$ should satisfies $h(z)=a_{n}z^{n}+\cdots+a_{0}$ is a non-constant polynomial, thus $|q|=1$ follows for avoiding a contradiction. From Lemma 2.2, we have $h(qz)+h(z)=2a_{0}$. Hence, we have

$ \begin{equation} A(z)=iqP(z)^{2}Q_{2}(qz)e^{-2a_{0}}. \end{equation} $

(3.10)

In addition, from (3.8), we also get

$ \begin{equation} \frac{B(z)e^{h(qz)-h(z)}}{iqP(z)^{2}Q_{2}(qz)}+\frac{Q_{1}(qz)}{Q_{2}(qz)}e^{2h(qz)}\equiv 0, \end{equation} $

(3.11)

which implies that

$ \begin{equation} B(z)=-iqQ_{1}(qz)P(z)^{2}e^{2a_{0}}. \end{equation} $

(3.12)

Thus

$ \begin{equation} A(z)B(z)=q^{2}P(z)^{4}Q(qz). \end{equation} $

(3.13)

Substitute (3.6) and (3.7) into (3.13), we have

$ \begin{eqnarray} &&\{P(z)Q_{1}'(z)+Q_{1}(z)[P(z)h'(z)-P'(z)]\}\{P(z)Q_{2}'(z)-Q_{2}(z)[P(z)h'(z)+P'(z)]\}\nonumber\\ &=&q^{2}P(z)^{4}Q(qz). \end{eqnarray} $

(3.14)

Since $f(z)$ is a finite order entire solution, by comparing with the degree of both hand side of (3.14), we have

$ \deg(h(z))=1+\deg{P(z)}. $

It implies that $\rho(f)=1+\deg{P(z)}$.

Case 2  If $h(qz)+h(z)$ is not a constant, from Lemma 2.4, we know that

$ -\frac{B(z)e^{h(qz)-h(z)}}{iqP(z)^{2}Q_{2}(qz)}\equiv1. $

Hence $|q|=1$ follows for avoiding a contradiction. Assume that $h(z)=a_{n}z^{n}+\cdots+a_{0}$, thus $h(qz)=h(z)$. Hence we have

$ \begin{equation} -B(z)=iqP(z)^{2}Q_{2}(qz). \end{equation} $

(3.15)

In addition, from (3.8), we also get

$ \begin{equation} A(z)=iqQ_{1}(qz)P(z)^{2}. \end{equation} $

(3.16)

Thus, similar as the above, we also get $\rho(f)=1+\deg{P(z)}$.