The existence and forms of solutions for some Fermat-type differential-difference equations

1.   Introduction and main results

The main purpose of this article is to deal with the existence of solutions for several complex differential-difference equations of Fermat type. The basic results and the standard symbols of Nevanlinna theory will be used in this paper (see ^[8,28,30]). A. Wiles and R. Taylor ^[23,24] in 1995 pointed out: The Fermat equation $x^m+y^m = 1$ does not admit nontrivial solutions in rational numbers as $m\geq 3$, and this equation possesses nontrivial rational solutions as $m = 2$. About sixty years ago, Gross ^[4] investigated the existence of solutions for the Fermat-type functional equation $f^m+g^m = 1$, and obtained: For $m = 2$, the entire solutions are $f = \cos a(z), g = \sin a(z)$, where $a(z)$ is an entire function; for $m > 2$, there are no nonconstant entire solutions.

In the last twenty years, Nevanlinna theory (especially the difference analogues such as logarithmic derivative lemma, Tumura-Clunie theorem, etc.) has played an important role in studying the properties of solutions for complex difference equations, complex differential-difference equations, and there were a number of results about the existence and the form of solutions for some equations (including ^{[1,2,3,5,6,7,9,10,11,12,13,17,18,19,20,21,22,25,26,27]}). In 2009, Liu ^[14] proved that the Fermat type equation $f(z)^2 +[f(z+c)-f(z)]^2 = a^2$ has no nonconstant entire solutions of finite order, where $a$ is a nonzero constant. In 2012, Liu et al. ^[15] pointed out that equation $f(z)^2 + f(z + c)^2 = 1$ has a transcendental entire solution of finite order. Furthermore, they also obtained that equation $f'(z)^2 + [f(z + c)-f(z)]^2 = 1$ admits the finite order transcendental entire solutions with the form $f(z) = 1/2 \sin(2z + Bi)$, where $c = (2k+1)\pi$, and $B$ is a constant (see ^[15]).

In 2018, Zhang ^[31] further discussed the existence of solutions for some Fermat type differential-difference equations, which forms are more general than those given by Liu ^[14], Liu etal. ^[15], and obtained

Theorem A (see [31,Theorem 1.3]). Let $f$ be a transcendental meromorphic function with finitely many poles and $\sigma(f) < \infty$. Then $f$ can not be a solution of the difference equation

where $R$ is a nonzero rational function and $c$ is a nonzero constant.

Theorem B (see [31,Theorem 1.4]). Let $f$ be a transcendental meromorphic function with finitely many poles and $\sigma(f) < \infty$. If $f$ is a solution of the differential-difference equation

where $R$ is a nonzero rational function and $c$ is a nonzero constant, then $R$ is a nonzero constant and $f$ is of form

where $c_1, c_2$ are two nonzero constants such that $16c_1c_2 = R$, $b$ is a constant and $k$ is an integer.

In this paper, we proceed to study the existence and the form of the solutions for some differential-difference equations with more general form than the previous form given by Liu, Liu et al. and Zhang ^[14,15,31]. Our results are listed as follows.

Theorem 1.1. Let $c$ be a nonzero constant, $R$ be a nonzero rational function, and $\alpha, \beta \in \mathbb{C}$ satisfy $\alpha^2-\beta^2\neq1$. Then the following difference equation of Fermat type

has no finite order transcendental meromorphic solutions with finitely many poles.

Theorem 1.2. Let $c(\neq0), \alpha(\neq0), \beta \in \mathbb{C}$ and $P(z), Q(z)$ be nonzero polynomials satisfying one of two following cases

(i) $\deg_zR\geq 1$ or $\deg_zQ\geq 1$;

(ii) $P, Q$ are two constants and $P^2(\alpha^2-\beta^2)\neq1$. Then the following Fermat-type difference equation

has no transcendental entire solutions with finite order.

Theorem 1.3. Let $\alpha(\neq0), \beta \in \mathbb{C}$, and $k$ is an integer. Let $f$ be a transcendental meromorphic solution of difference-differential equation of Fermat type

where $R(z)$ is a nonzero rational function and $c$ is a nonzero constant. If $f$ is of finite order and has finitely many poles, then $i\alpha c = \pm1$ and $R(z)$ is a nonconstant polynomial with $\deg_zR\leq 2$ or $R(z)$ is a nonzero constant. Furthermore,

$(i)$ If $R(z)$ is a nonconstant polynomial and $\deg_zR\leq 2$, then $f$ is of the form

where $R(z) = -(as_1(z)+m_1)(as_2(z)+m_2)$, $a\neq 0, b\in \mathbb{C}$, and $a, b, c, \alpha, \beta$ satisfy $\alpha\neq\pm\beta$, $a = -i(\alpha+\beta)$, $c = \frac{(2k+1)\pi}{a}i$, $i\alpha c = -1$ or $a = i(\alpha-\beta)$, $c = \frac{2k\pi}{a}i$, $i\alpha c = 1$, where $s_j(z) = m_jz+n_j$, $m_j, n_j\in \mathbb{C}$$(j = 1, 2)$;

$(ii)$ If $R(z)$ is a nonzero constant, then $f$ is of the form

$R = -a^2n_1n_2$, $a\neq 0, b\in \mathbb{C}$, and $a, b, c, \alpha, \beta$ satisfy the following cases:

$(ii_1)$ if $\alpha = \beta$, then $a = -2\alpha i$, $c = \frac{(2k+1)\pi}{a}i$ and $d\in \mathbb{C}$;

$(ii_2)$ if $\alpha = -\beta$, then $a = 2\alpha i$, $c = \frac{2k\pi}{a}i$ and $d = 0$;

$(ii_3)$ if $\alpha\neq\pm\beta$, then $d = 0$ and $a = -i(\alpha+\beta)$, $c = \frac{(2k+1)\pi}{a}i$ or $a = i(\alpha-\beta)$, $c = \frac{2k\pi}{a}i$.

Next, we give some examples to explain the existence of solutions for Eq. (1.3) in the above cases.

● For Case (ⅰ), let $s_1(z) = 1$, $s_2 = z+1$, $c = \pi i$, $a = 1$ and $b\in \mathbb{C}$. That is

Thus, $f(z)$ satisfies Eq. (1.3) with $c = \pi i$, $\alpha = \frac{1}{\pi}$, $\beta = i-\frac{1}{\pi}$ and $R(z) = -z$;

Let $s_1(z) = z+1$, $s_2 = z-1$, $a = 1$ and $b\in \mathbb{C}$. That is

Thus, $f(z)$ satisfies Eq. (1.3) with $c = \pi i$, $\alpha = \frac{1}{\pi}$, $\beta = i-\frac{1}{\pi}$ and $R(z) = -z(z+2)$;

● For Case $(ii_1)$, let $n_1(z) = 1$, $n_2 = 2$, $a = 1$ and $b, d\in \mathbb{C}$. That is

Thus, $f(z)$ satisfies Eq. (1.3) with $c = \pi i$, $\alpha = \frac{i}{2}$, $\beta = \frac{i}{2}$ and $R(z) = -2$;

For Case $(ii_2)$, let $n_1(z) = 1$, $n_2 = 1$, $a = 2$ and $b\in \mathbb{C}$. That is

Thus, $f(z)$ satisfies Eq. (1.3) with $c = \pi i$, $\alpha = -i$, $\beta = i$ and $R(z) = -4$;

For Case $(ii_3)$, let $n_1(z) = 2$, $n_2 = 1$, $a = 1$ and $b\in \mathbb{C}$. That is

Thus, $f(z)$ satisfies Eq. (1.3) with $c = 2\pi i$, $\alpha = 1$, $\beta = 1+i$ and $R(z) = -2$.

Theorem 1.4. Let $\alpha(\neq0), \beta \in \mathbb{C}$, and $k$ is an integer. Let $f$ be a transcendental meromorphic solution of difference-differential equation of Fermat type

where $R(z)$ is a nonzero rational function and $c$ is a nonzero constant.

(i) If $\alpha = \pm\beta$, then Eq. (1.4) has no finite order transcendental meromorphic solutions with finitely many poles;

(ii) If $\alpha\neq \pm\beta$, and Eq. (1.4) has a finite order transcendental meromorphic solution $f$ with finitely many poles, then $R(z)$ must be a nonzero polynomial with $\deg_zR\leq 1$. Furthermore,

$(ii_1)$ if $R(z)$ is a nonzero polynomial of degree one, then $f(z)$ is of the form

where $a^4 = \alpha^2-\beta^2$, $b\in \mathbb{C}$, $c = \frac{\log\frac{a^2+i\beta}{i\alpha}+2k\pi i}{a}$, $e^{ac} = \frac{2a}{i\alpha c}\neq \pm1$ and $R = a^3n_2[as_1(z)+2m_1]$, $s_1(z) = m_1z+n_1$, or $f(z)$ is of the form

where $a^4 = \alpha^2-\beta^2$, $b\in \mathbb{C}$, $c = \frac{\log\frac{-a^2+i\beta}{i\alpha}+(2k+1)\pi i}{a}$, $e^{ac} = \frac{i\alpha c}{2a}\neq \pm1$ and $R = a^3n_1[as_2(z)-2m_2]$, $s_2(z) = m_2z+n_2$;

$(ii_2)$ if $R(z)$ is a nonzero constant, then $f(z)$ is of the form

where $a, b, c, \alpha, \beta, c_1, c_2, R$ satisfy $a^4 = \alpha^2-\beta^2$, $b\in \mathbb{C}$, $c = \frac{\log\frac{a^2+i\beta}{i\alpha}+2k\pi i}{a}$ and $R = a^4c_1c_2$;

The following examples show that the existence of solutions for complex differential-difference equation of Theorem 1.4 $(ii_1)$ and $(ii_2)$.

Example 1.1. Let $s_1(z) = z$, $n_2 = 1$, $a = 1$ and $b\in \mathbb{C}$. And let $c_0$ be a solution of equation $e^{2c}(1-c) = 1$, $\alpha = \frac{2}{ic_0e^{c_0}}$, and $\beta = \frac{2-c_0}{ic_0}$. Then it follows that $i\alpha e^{c_0}-i\beta = 1$, $i\alpha e^{-c_0}-i\beta = 1$, $\alpha^2-\beta^2 = 1$, and $i\alpha c_0e^{c_0} = 2$. Thus, we can deduce that

satisfies the following equation

Example 1.2. Let $c_1 = 1$, $c_2 = 1$, $e^c = \sqrt{5}-2$, $a = 1, b\in \mathbb{C}$, $\alpha = \frac{i}{2}$, and $\beta = \frac{\sqrt{5}}{2}i$. Thus

satisfies the following equation

2.   Some Lemmas

To prove our theorems, we require the following lemmas.

Lemma 2.1. ([31,Lemma 2.1]). If $f$ is a nonconstant rational function and it satisfies the following differential-difference equation

where $\eta$ and $c$ are two nonzero constants, then $\eta c = 1$ and $f$ is a polynomial of degree one.

Lemma 2.2. Let $c, a, \alpha$ be three nonzero constants satisfying $\eta c\neq n-1$ and $n\geq 1$ be an integer. If $f$ is a nonconstant rational solution of the following differential-difference equation

Then $i\alpha c = n\eta^{n-1}$ and $f$ is a polynomial of degree one.

Proof. We firstly prove that $f(z)$ has no poles. On the contrary, suppose that $z_0$ is a pole of $f$. Since (2.1) can be rewritten (2.1) in the following form

It is easy to see that $z_0+c$ is also a pole of $f(z)$ by comparing the order of pole $z_0$ on both sides of Eq. (2.2). By the cyclic utilization of this operation, we can get that a sequence poles of $f(z)$ are $z_0+2c$, $z_0+3c$, $\ldots$, $z_0+tc$, $\cdots$, this is impossible since $f(z)$ is a nonconstant rational function. Hence, $f(z)$ is a polynomial.

Let $f(z)$ be a polynomial of the form $f(z) = a_kz^k+a_{k-1}z^{k-1}+\cdots+a_0$, where $k\geq 1$ and $a_k(\neq0), a_{k-1}, \ldots, a_0$ are constants. Then

and

Substituting (2.3), (2.4) into Eq. (2.2), we obtain

which means $\left\{ \begin{aligned} &i\alpha c = n\eta^{n-1}, \\ & \frac{1}{2}a_kk(k-1)\eta^{n-2}(\eta c-n+1) = 0. \end{aligned} \right.$ In view of $\eta c\neq n-1$, thus it follows that $i\alpha c = n\eta ^{n-1}$ and $k = 1$.

Noting that $f(z)$ is a polynomial of degree one if $k = 1$, then the conclusions of this lemma are proved.

Therefore, this completes the proof of this lemma.

Lemma 2.3. Let $\alpha, a, c$ be three nonzero constants. If $R_1, R_2$ are two nonconstant rational functions satisfying the following differential-difference equations

Then $e^{ac} = \pm1$ and $R_1, R_2$ are two polynomials of degree one.

Proof. Firstly, (2.5) can be written in the following form

Similar to the argument in Lemma 2.2, we can prove that $R_1, R_2$ are two nonconstant polynomials. Let

where $a_j, b_j\in \mathbb{C}$, $a_k\neq 0, b_t\neq 0$, $k\geq1$ and $t\geq1$. Substituting $R_1(z), R_2(z)$ into (2.5), and comparing the coefficients of $z^{k-1}, z^{k-2}, z^{t-1}$ and $z^{t-2}$ both sides of such two equations, it yields

which means

Since $k\geq 1$ and $t\geq 1$, it follows $e^{ac} = \pm1$ and $k = 1, s = 1$. Therefore, this completes the proof of Lemma 2.3

Lemma 2.4. Let $R$ be a nonconstant rational function and $p(z) = az+b$ $(a\neq0)$. Denote $A_1 = R'+Rp'$, $A_n = A_{n-1}'+A_{n-1}p'$, $B_1 = R'-Rp'$, $B_n = B_{n-1}'+B_{n-1}(-p)'$. Then

Proof. We use the mathematical induction to prove it. When $k = 1$, since $R$ is a nonconstant rational function and $p' = a$, then $\lim\limits_{|z|\rightarrow\infty}\frac{A_1'}{R} = \lim\limits_{|z|\rightarrow\infty}\frac{R''+R'a}{R} = 0$ and $\lim\limits_{|z|\rightarrow\infty}\frac{A_1}{R} = \lim\limits_{|z|\rightarrow\infty}\frac{R'+Rp'}{R} = a$.

Suppose that $\lim\limits_{|z|\rightarrow\infty}\frac{A_k'}{R} = 0, \; \; \; \lim\limits_{|z|\rightarrow\infty}\frac{A_k}{R} = a^k$. Thus, $\lim\limits_{|z|\rightarrow\infty}\frac{A_{k+1}'}{R} = \lim\limits_{|z|\rightarrow\infty}\frac{A_k'+A_k'a}{R} = 0$ and $\lim\limits_{|z|\rightarrow\infty}\frac{A_{k+1}}{R} = \lim\limits_{|z|\rightarrow\infty}\frac{A_{k}'+A_{k}p'}{R} = a^{k+1}$. Hence, we have $\lim\limits_{|z|\rightarrow\infty}\frac{A_n'}{R} = 0$ and $\lim\limits_{|z|\rightarrow\infty}\frac{A_n}{R} = a^n$.

Similar to the above argument, we can prove that $\lim\limits_{|z|\rightarrow\infty}\frac{B_n'}{R} = 0$ and $\lim\limits_{|z|\rightarrow\infty}\frac{B_n}{R} = (-a)^n$.

Therefore, this completes the proof of Lemma 2.4.

Lemma 2.5. ([30,Theorem 1.51]). Suppose that $f_1, f_2, \ldots, f_n(n\geq 2)$ are meromorphic functions and $g_1, g_2, \ldots$, $g_n$ are entire functions satisfying the following conditions

$(i)$ $\sum_{j = 1}^n f_je^{g_j}\equiv 0$;

$(ii)$ $g_j-g_k$ are not constants for $1\leq j < k\leq n$;

$(iii)$ For $1\leq j\leq n, 1\leq h < k\leq n$, $T(r, f_j) = o\{T(r, e^{g_h-g_k})\}$ $(r\rightarrow\infty, r\not\in E)$, where $E$ is a set of $r\in (0, \infty)$ with finite linear measure.

Then $f_j\equiv0$ $(j = 1, 2, \ldots, n)$.

Lemma 2.6. (see [30,Theorem 2.7]). Let $f$ be a meromorphic function of finite order $\rho(f)$. Write

near $z = 0$ and let $\{a_1, a_2, \ldots\}$ and $\{b_1, b_2, \ldots\}$ be the zeros and poles of $f$ in $\mathbb{C}\backslash\{0\}$, respectively. Then

where $P_1(z)$ and $P_2(z)$ are the canonical products of $f$ formed with the non-null zeros and poles of $f$, respectively, and $Q(z)$ is a polynomial of degree $\leq \rho(f)$.

3.   The proof of Theorem 1.1

Proof. Suppose that Eq. (1.1) admits a finite order transcendental meromorphic solution $f(z)$ with finitely many poles. We can rewrite Eq. (1.1) as

Since $f(z)$ has finitely many poles and $R$ is a nonzero rational function, then $f(z)+i(\alpha f(z+c)-\beta f(z))$ and $f(z)-i(\alpha f(z+c)-\beta f(z))$ both have finitely many poles and zeros. Thus, in view of Lemma 2.6, (3.1) can be written as

where $R_1, R_2$ are two nonzero rational functions such that $R_1R_2 = R$ and $p(z)$ is a nonconstant polynomial. By solving the above equations system, we have

Substituting the first equation of system (3.3) into the second equation of system (3.3), we get

By Lemma 2.5, it yields from (3.4) that

Since $R_1, R_2$ are two nonzero rational functions and $f$ is of finite order, we obtain that $p(z)$ is a polynomial of degree one. Otherwise, assume that $\deg_z p(z)\geq 2$. Thus, in view of (3.5), we have

which imply $1 = \rho\left(e^{p(z+c)-p(z)}\right) = \rho\left(\frac{i\beta R_1(z)+R_1(z)}{i\alpha R_1(z+c)}\right) = 0$, a contradiction. Hence, $\deg_z p(z) = 1$. Let $p(z) = az+b$, $a\neq 0, b\in \mathbb{C}$. Substituting this into (3.5), we can deduce

Thus, it yields that $\alpha^2-\beta^2 = 1$, a contradiction.

Therefore, this completes the proof of Theorem 1.1.

4.   The proof of Theorem 1.2

Proof. Suppose that $f(z)$ is a transcendental entire solution with finite order of Eq. (1.2). Similar to the above argument as in the proof of Theorem 1.1, it follows that

where $p(z)$ is a nonconstant polynomial and $Q_1(z)Q_2(z) = Q(z)$, $Q_1(z), Q_2(z)$ are nonzero polynomials. In view of (4.1), it yields that

Since $p(z)$ is a nonconstant polynomial, then by Lemma 2.5, we conclude that

which implies that $p(z)$ is a polynomial of degree one. Set $p(z) = az+b$, $a\neq 0, b\in \mathbb{C}$. Thus, it follows from (4.3) that

which means that

Hence, it yields

Set $\deg_zP = p$ and $\deg_z Q = q$, then $p\geq 0$, $q\geq 0$ and $p, q\in \mathbb{N}_+$.

Case 1. $p\geq1$ and $\alpha = \pm\beta$. If $q\geq 1$. Since $\alpha\neq 0$, thus, by comparing the order both sides of Eq. (4.4), it follows $2p+q-1 = q$, that is, $p = \frac{1}{2}$, a contradiction. If $q = 0$, that is, $Q(z)$ is a constant. Thus, we can conclude from (4.4) that $Q(z)\equiv0$, a contradiction.

Case 2. $p\geq1$ and $\alpha\neq\pm\beta$. If $q\geq 1$. Since $\alpha\neq 0$, thus, by comparing the order both sides of Eq. (4.4), it follows $2p+q = q$, that is, $p = 0$, a contradiction. If $q = 0$, that is, $Q(z)$ is a constant. Thus, we can conclude from (4.4) that $P(z)$ is a constant, a contradiction with $p\geq 1$.

Case 3. $p = 0$ and $\alpha = \pm\beta$. Thus, $P(z) = K(\neq0)$. If $q\geq 1$. Since $\alpha\neq 0$, thus, by comparing the order both sides of Eq. (4.4), it follows $q-1 = q$, a contradiction. If $q = 0$, it follows $Q(z)\equiv0$, a contradiction.

Case 4. $p = 0$ and $\alpha\neq\pm\beta$. If $q\geq 1$, set $P(z) = K(\neq0)$, $Q(z) = b_qz^q+b_{q-1}z^{q-1}+\cdots+b_0$, $b_q\neq 0, b_{q-1}, \ldots, b_0$ are constants. By comparing the coefficients of $z^q, z^{q-1}$ both sides of (4.4), it yields that

which implies that $\alpha^2qca_q = 0$, a contradiction. If $q = 0$, then $K^2(\alpha^2-\beta^2) = 1$, a contradiction.

Therefore, this completes the proof of Theorem 1.2.

5.   The proof of Theorem 1.3

Proof. Suppose that Eq. (1.3) admits a finite order transcendental meromorphic solution $f(z)$ with finitely many poles. We can rewrite Eq. (1.3) in the following form

Since $f(z)$ has finitely many poles and $R$ is a nonzero rational function, then $f'(z)+i(\alpha f(z+c)-\beta f(z))$ and $f'(z)-i(\alpha f(z+c)-\beta f(z))$ both have finitely many poles and zeros. Thus, in view of Lemma 2.6, (5.1) can be written as

where $R_1, R_2$ are two nonzero rational functions such that $R_1R_2 = R$, and $p(z)$ is a nonconstant polynomial. By solving the above equations system, we have

In view of the second equation of (5.3), it follows that

where $A_1 = R_1'+R_1p'$ and $B_1 = R_2'-R_2p'$. Substituting the first equation of system (5.3) into (5.4), it yields that

By Lemma 2.5, it yields from (5.5) that

Since $R_1, R_2$ are two nonzero rational functions and $f$ is of finite order, similar to argument as in the proof of Theorem 1.1, it follows that $p(z)$ is a polynomial of degree one. Let $p(z) = az+b$, $a\neq 0, b\in \mathbb{C}$. Substituting $p(z), A_1, B_1$ into (5.6), and let $z\rightarrow\infty$, thus we can conclude from Lemma 2.4 that

which means that

Hence, it yields $e^{ac} = \pm1$.

If $e^{ac} = 1$, then $a = i\alpha-i\beta$. Thus, we can rewrite (5.6) in the following form

If $R_1, R_2$ are nonzero constants, then (5.8) holds and $R = R_1R_2$ is a constant.

If $R_j(j = 1, 2)$ is a nonzero rational function, then in view of Lemma 2.1, it follows that $i\alpha c = 1$ and $R_j(j = 1, 2)$ is a polynomial of $\deg_zR_j = 1$. In view of $R = R_1R_2$, thus $R$ is a nonconstant polynomial with $\deg_zR\leq 2$.

If $e^{ac} = -1$, then $a = -i\alpha-i\beta$. Thus, we can rewrite (5.6) in the following form

Like in the previous case, we can obtain that $i\alpha c = \pm1$ and $R(z)$ is a nonconstant polynomial with $\deg_zR\leq 2$ or $R(z)$ is a nonzero constant.

Hence, we can conclude that $R$ is a nonconstant polynomial with $\deg_zR\leq 2$ or $R$ is a nonzero constant.

(ⅰ) Suppose that $R(z)$ is a nonconstant polynomial with $\deg_zR\leq 2$, then in view of the first equation of (5.3), it follows that $f(z)$ is of the form

where $s_j(z) = m_jz+n_j, m_j, n_j\in \mathbb{C}$ $(j = 1, 2)$ and $\gamma\in \mathbb{C}$.

Case 1. If $\deg_zR = 2$, then it follows that $m_j\neq 0 (j = 1, 2)$. Substituting (5.10) into (5.3), it follows that $R(z) = -(as_1(z)+m_1)(as_2(z)+m_2)$, $i\alpha c = 1$ and $a = i(\alpha-\beta)$ or $i\alpha c = -1$ and $a = -i(\alpha+\beta)$.

If $i\alpha c = 1$ and $a = i(\alpha-\beta)$, then $e^{ac} = 1$, $i.e.$, $c = \frac{2k\pi i}{a}$. Obviously, $\alpha\neq \beta$ as $a\neq0$. If $\alpha = -\beta$, then $a = 2i\alpha$. Thus, since $\alpha\neq0$ and from (5.7), it follows $1 = e^{ac} = e^{2i\alpha c} = e^2$, a contradiction. Hence, $\alpha\neq \pm\beta$. Thus, substituting (5.10) into the second equation of (5.3), it follows $\gamma\equiv0$.

If $i\alpha c = -1$ and $a = -i(\alpha+\beta)$, then $e^{ac} = -1$, $i.e.$, $c = \frac{(2k+1)\pi i}{a}$. Obviously, $\alpha\neq -\beta$ as $a\neq0$. If $\alpha = \beta$, then $a = -2i\alpha$. Thus, since $\alpha\neq0$ and from (5.7), it follows $-1 = e^{ac} = e^{-2i\alpha c} = e^2$, a contradiction. Hence, $\alpha\neq \pm\beta$.

Case 2. If $\deg_zR = 1$, then one of $m_1, m_2$ is zero, without loss of generality, assuming that $m_1 = 0$. Substituting (5.10) into (5.3), it follows that $R_1$ is a constant and $R_2$ is a polynomial of degree one, and $i\alpha c = 1$ and $a = i(\alpha-\beta)$ or $i\alpha c = -1$ and $a = -i(\alpha+\beta)$. Similar to the argument as in Case 1, it is easy to prove that $\alpha\neq\pm\beta$ and $\gamma\equiv0$.

Therefore, $f(z)$ is of the form

where $R(z) = -(as_1(z)+m_1)(as_2(z)+m_2)$, $a\neq 0, b\in \mathbb{C}$, and $a, b, c, \alpha, \beta$ satisfy $\alpha\neq\pm\beta$, $a = -i(\alpha+\beta)$, $c = \frac{(2k+1)\pi}{a}i$, $i\alpha c = -1$ or $a = i(\alpha-\beta)$, $c = \frac{2k\pi}{a}i$, $i\alpha c = 1$.

(ⅱ) If $R(z)$ is a nonzero constant, then in view of the first equation of (5.3), it follows that $f(z)$ is of the form

where $n_1, n_2\in \mathbb{C}$ and $\gamma\in \mathbb{C}$. Substituting (5.11) into the first equation of (5.3), it yields $R = -a^2n_1n_2$. For the sake of brevity and readability, we give the proof of Theorem 1.3 $(ii_1)$-$(ii_3)$ in Appendix A.

Therefore, this completes the proof of Theorem 1.3.

6.   The proof of Theorem 1.4

Proof. Suppose that Eq. (1.4) admits a finite order transcendental meromorphic solution $f(z)$ with finitely many poles. We can rewrite Eq. (1.4) as

Since $f(z)$ has finitely many poles and $R$ is a nonzero rational function, then $f''(z)+i(\alpha f(z+c)-\beta f(z))$ and $f''(z)-i(\alpha f(z+c)-\beta f(z))$ both have finitely many poles and zeros. Thus, (6.1) can be written as

where $R_1, R_2$ are two nonzero rational functions such that $R_1R_2 = R$ and $p(z)$ is a nonconstant polynomial. By solving the above equations system, we have

In view of the second equation of (6.3), it follows that

where $A_2 = A_1'+A_1p'$ and $B_2 = B_1'-B_1p'$. Substituting the first equation of system (6.3) into (6.4), it yields

By Lemma 2.5, it yields from (6.5) that

Since $R_1, R_2$ are two nonzero rational functions and $f$ is of finite order, we obtain that $p(z)$ is a polynomial of degree one. Let $p(z) = az+b$, $a\neq 0, b\in \mathbb{C}$. Substituting $p(z), A_2, B_2$ into (6.6), and let $z\rightarrow\infty$, thus we can conclude from Lemma 2.4 that

which means that

Hence, it follows $a^4 = \alpha^2-\beta^2$.

(ⅰ) If $\alpha = \pm\beta$, this is a contradiction with $a^4 = \alpha^2-\beta^2$. Therefore, this proves the conclusion of Theorem 1.3 (ⅰ).

(ⅱ) If $\alpha\neq \pm\beta$. Substituting $p(z) = az+b$ and (6.7) into (6.6), it yields

Suppose that $R_1, R_2$ are nonconstant rational functions, in view of Lemma 2.3 and (6.8), we can conclude that $e^{ac} = \pm1$ and $R_1, R_2$ are nonconstant polynomials of degree one. Set $\deg_zR_1 = k$ and $\deg_zR_2 = s$.

If $k = 1$ and $s = 1$, in view of Lemma 2.3, we obtain (2.7). If $e^{ac} = 1$, then from (6.7), it follows that $i\alpha-i\beta = a^2$ and $i\alpha-i\beta = -a^2$, a contradiction. If $e^{-ac} = 1$, then from (6.7), it follows that $-i\alpha-i\beta = a^2$ and $-i\alpha-i\beta = -a^2$, a contradiction. Hence, there is at most a polynomial of degree one in $R_1$ and $R_2$. For the sake of brevity and readability, we give the proof of Theorem 1.4 $(ii_1)$ and $(ii_2)$ in Appendix B.

Therefore, this completes the proof of Theorem 1.4.

Acknowledgments

We thank the referee(s) for reading the manuscript very carefully and making a number of valuable and kind comments which improved the presentation.

This work was supported by the National Natural Science Foundation of China (11561033, 11561031), the Natural Science Foundation of Jiangxi Province in China (20181BAB201001), and the Foundation of Education Department of Jiangxi (GJJ180734) of China.

Conflict of interest

The authors declare that none of the authors have any competing interests in the manuscript.