Solutions to some generalized Fermat-type differential-difference equations

1.   Introduction and results

We first assume that the reader is familiar with the basic results and notations of the Nevanlinna theory and difference Nevanlinna theory with one complex variable, which can be found in ^[3,7,10,22]. In the past thirty years, there were lots of research focusing on the solutions of Fermat-type differential-difference equations; readers can refer to ^{[1,4,5,6,9,11,12,15,16,17,18,19,20,21]}.

A. Wiles ^[13] in 1995 pointed out: The Fermat equation $x^{m}+y^{m} = 1$ does not admit nontrivial solutions in rational numbers as $m \ge 3$, and this equation possesses nontrivial rational solutions as $m = 2$. About sixty years ago, Gross ^[2] 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 $f = \cos \alpha(z)$, $g = \sin\alpha(z)$, where $\alpha(z)$ is an entire function; for $m > 2$, there are no nonconstant entire solutions. In 2009, Liu ^[7] 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 ^[8] studied that the Fermat-type equation $f'(z)^{2}+[f(z+c)-f(z)]^{2} = 1$ has the transcendental entire solutions with finite order. In 2018, Zhang ^[23] generalized Liu's ^[7,8] theorem, and obtained

Theorem 1.1. 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 $R(z)$ is a nonzero rational function and $c$ is a nonzero constant.

Theorem 1.2. 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 $R(z)$ is a nonzero rational function and $c$ is a nonzero constant, then $R(z)$ is a nonzero constant, and $f$ is of the form

where $c_{1}$, $c_{2}$ are two nonzero constants such that $16c_{1}c_{2} = R(z)$, $b$ is a constant, and $k$ is an integer.

In 2020, Wang et al. ^[14] promoted Zhang's ^[23] form and obtained

Theorem 1.3. Let $\alpha(\ne0)$, $\beta \in \mathbb{C}$, $k$ be 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 $\deg_{z} R \le2$, or $R(z)$ is a nonzero constant. Furthermore

($I$) If $R(z)$ is a nonconstant polynomial and $\deg_{z}R\le 2$, then $f$ is of the form

where $R(z) = (as_{1}(z)+m_{1})(-as_{2}(z)+m_{2})$, $a\ne 0$, $b\in \mathbb{C}$, and $a, b, c, \alpha, \beta$ satisfy $\alpha \ne \pm \beta$, $a = -i(\alpha +\beta)$, $c = \frac{(2k+1)\pi}{a}i$, $i\alpha c = -1$ or $a = i(\alpha -\beta)$, $c = \frac{2k\pi}{a}i$, $i\alpha c = 1$, where $s_{j} = m_{j}z+n_{j}$, $m_{j}$, $n_{j}\in \mathbb{C}(j = 1, 2)$;

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

$R(z) = -a^{2}n_{1}n_{2}$, $a\ne 0$, $b\in \mathbb{C}$, 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 $\alpha \ne \pm\beta$, then $d = 0$ and $a = -i(\alpha+\beta)$, $c = \frac{(2k+1)\pi }{a}i$ or $a = i(\alpha -\beta), c = \frac{2k\pi }{a}i$.

Theorem 1.4. Let $\alpha(\ne0)$, $\beta \in \mathbb{C}$, $k$ be 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 (1.1) has no finite-order transcendental meromorphic solution with finitely many poles;

($II$) If $\alpha \ne \pm \beta$, and Eq (1.1) has a finite-order transcendental meromorphic solution $f$ with finitely many poles, then $R(z)$ must be a nonzero polynomial with $\deg_{z} R \le 1$. Furthermore,

($II_{1}$) if $R(z)$ is a nonzero polynomial of degree one, then $f(z)$ is of the form

where $a^{4} = \alpha^{2}-\beta^{2}$, $b\in \mathbb{C}$, $c = \dfrac{\log\frac{a^{2}+i\beta}{i\alpha}+2k\pi i}{a}$, $e^{ac} = \frac{2a}{i\alpha c}\ne \pm1$ and $R(z) = a^{3}n_{2}[as_{1}(z)+2m_{1}]$, $s_{1}(z) = m_{1}z+n_{1}$, or $f(z)$ is of the form

where $a^{4} = \alpha^{2}-\beta^{2}$, $b\in \mathbb{C}$, $c = \dfrac{\log\frac{-a^{2}+i\beta}{i\alpha}+(2k+1)\pi i}{a}$, $e^{ac} = \frac{i\alpha c}{2a}\ne \pm1$ and $R(z) = a^{3}n_{1}[as_{2}(z)-2m_{2}]$, $s_{2}(z) = m_{2}z+n_{2}$;

($II_{2}$) if $R(z)$ is a nonzero constant, then $f(z)$ is of the form

where $a, b, c, \alpha, \beta, c_{1}, c_{2}, R(z)$ satisfy $a^{4} = \alpha^{2}-\beta^{2}, b\in \mathbb{C}$, $c = \dfrac{\log\frac{a^{2}+i\beta}{i\alpha}+2k\pi i}{a}$ and $R(z) = a^{4}c_{1}c_{2}$.

Inspired by ^[14], can $f'$ and $f''$ in Theorems 1.3 and 1.4 be replaced by $f^{(k)}$? In this paper, we consider this question. Our results are listed as follows.

Theorem 1.5. Let $\alpha(\ne0)$, $\beta \in \mathbb{C}$, $k\in \mathbb{Z}^{+}$, and $k$ be an odd number. 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 polynomial with $\deg_{z} R = 1$, or $R(z)$ is a nonzero constant. Let $s_{j}(z) = m_{j}z+n_{j}$, $m_{j}$, $n_{j}\in \mathbb{C}(j = 1, 2)$.

Case I: If $R(z)$ is a polynomial with $\deg_{z}R = 1$, then $f$ is of the form

where $R(z) = -n_{2}a^{2k-1}[as_{1}(z)+km_{1}]$, $m_{1}\ne 0$, $m_{2} = 0$, $a\ne 0$, $b\in \mathbb{C}$, and $a, b, c, \alpha, \beta$ satisfy $\alpha \ne \pm \beta$, $a^{k} = -i(\alpha +\beta)$, $c = \frac{(2l+1)\pi}{a}i$, $l\in \mathbb{Z}$, $i\alpha c = -ka^{k-1} = -1$ or $a^{k} = i(\alpha -\beta)$, $c = \frac{2l\pi}{a}i$, $l\in \mathbb{Z}$, $i\alpha c = ka^{k-1} = 1$; or $R(z) = n_{1}a^{2k-1}[-as_{2}(z)+km_{2}]$, $m_{1} = 0$, $m_{2}\ne 0$, $a\ne 0$, $b\in \mathbb{C}$, and $a, b, c, \alpha, \beta$ satisfy $\alpha \ne \pm \beta$, $a^{k} = -i(\alpha +\beta)$, $c = \frac{(2l+1)\pi}{a}i$, $l\in \mathbb{Z}$, $i\alpha c = ka^{k-1} = -1$ or $a^{k} = i(\alpha -\beta)$, $c = \frac{2l\pi}{a}i$, $l\in \mathbb{Z}$, $i\alpha c = -ka^{k-1} = 1$.

Case II: If $R(z)$ is a nonzero constant, then $m_{j} = 0(j = 1, 2)$, $f$ is of the form

where $R(z) = -a^{2k}n_{1}n_{2}, a\ne 0, b\in \mathbb{C}$, and $a, b, c, \alpha, \beta$ satisfy the following cases:

($II_{1}$) if $\alpha = \beta$, then $a^{k} = -2\alpha i$, $c = \frac{(2l+1)\pi }{a}i$ and $c_{0}\in \mathbb{C}$, $l\in \mathbb{Z}$;

($II_{2}$) if $\alpha = -\beta$, then $a^{k} = 2\alpha i$, $c = \frac{2l\pi }{a}i$ and $c_{0} = 0$, $l\in \mathbb{Z}$;

($II_{3}$) if $\alpha \ne \pm\beta$, then $c_{0} = 0$ and $a^{k} = -i(\alpha+\beta)$, $c = \frac{(2l+1)\pi }{a}i$, $l\in \mathbb{Z}$ or $a^{k} = i(\alpha -\beta)$, $c = \frac{2l\pi }{a}i$, $l\in \mathbb{Z}$.

Remark 1. When $k = 1$, Theorem 1.5 becomes Theorem 1.3.

Remark 2. In ^[14], we find that Case (I) of Theorem 1.3 is not accurate, including the corresponding examples. On page 695, line 11 of ^[14], substituting (5.10) into the first equation in (5.3), we have $R_{2}(z) = -as_{2}(z)+m_{2}$. Meanwhile, substituting (5.10) into the second equation in (5.3), we have $R_{2}(z) = -as_{2}(z)-m_{2}$. From this, we have $m_{2} = 0$, which is a contradiction. Therefore, if $f'(z)^{2}+[\alpha f(z+c)-\beta f(z)]^{2} = R(z)$ has a solution, then $\deg_{z}R\le 1$.

Next, we will provide some examples to explain the existence of solutions to Eq (1.2) in the above situation.

Example 1.1. For Case $I$, the transcendental entire function

satisfies

where $k = 3$, $s_{1}(z) = z+1$, $s_{2}(z) = 1$, $c = \sqrt{3}\pi i$, $a = \frac{\sqrt{3}}{3}$, $b\in \mathbb{C}$, $\alpha = \frac{1}{\sqrt{3}\pi}$, $\beta = \frac{\sqrt{3}}{9}i-\frac{1}{\sqrt{3}\pi}$ and $R(z) = -\frac{\sqrt{3}}{9}[\frac{\sqrt{3}}{9}(z+1)+1]$.

The transcendental entire function

satisfies

where $k = 3$, $s_{1}(z) = 1$, $s_{2}(z) = z+1$, $c = \sqrt{3}\pi$, $a = \frac{\sqrt{3}}{3}i$, $b\in \mathbb{C}$, $\alpha = \frac{i}{\sqrt{3}\pi}$, $\beta = \frac{\sqrt{3}}{9}-\frac{i}{\sqrt{3}\pi}$ and $R(z) = -\frac{\sqrt{3}}{9}i[\frac{\sqrt{3}}{9}i(z+1)-1]$.

Example 1.2. For Case $II_{1}$, the transcendental entire function

satisfies

where $k = 3$, $n_{1} = 1$, $n_{2} = 2$, $a = 1$, $b, c_{0} \in \mathbb{C}$, $c = \pi i$, $\alpha = \frac{i}{2}$, $\beta = \frac{i}{2}$ and $R(z) = -2$.

Example 1.3. For Case $II_{2}$, the transcendental entire function

satisfies

where $k = 3$, $n_{1} = 1$, $n_{2} = 1$, $a = 2$, $b\in \mathbb{C}$, $c = \pi i$, $\alpha = -4i$, $\beta = 4i$ and $R(z) = -64$.

Example 1.4. For Case $II_{3}$, the transcendental entire function

satisfies

where $k = 3$, $n_{1} = 2$, $n_{2} = 1$, $a = 1$, $b \in \mathbb{C}$, $c = 2\pi i$, $\alpha = 1$, $\beta = 1+i$ and $R(z) = -2$.

Theorem 1.6. Let $\alpha(\ne0)$, $\beta \in \mathbb{C}$, $k\in \mathbb{Z}^{+}$ and $k$ be an even number. Let $f$ be a transcendental meromorphic solution of Eq (1.2).

If $f$ is of finite order and has finitely many poles, then $R(z)$ is a polynomial with $\deg_{z} R = 1$, or $R(z)$ is a nonzero constant.

Case I: If $\alpha = \pm \beta$, then Eq (1.2) has no-finite order transcendental meromorphic solution with finitely many poles;

Case II: If $\alpha \ne \pm \beta$, and Eq (1.2) has a finite-order transcendental meromorphic solution $f$ with finitely many poles, then $R(z)$ must be a nonzero polynomial with $\deg_{z} R \le 1$. Let $s_{j} = m_{j}z+n_{j}$, $m_{j}$, $n_{j}\in \mathbb{C}(j = 1, 2)$,

($II_{1}$) if $R(z)$ is a polynomial with $\deg_{z}R = 1$, then $f(z)$ is of the form

where $a^{2k} = \alpha^{2}-\beta^{2}, b\in \mathbb{C}$, $c = \dfrac{\log\frac{a^{k}+i\beta}{i\alpha}+2l\pi i}{a}$, $l\in \mathbb{Z}$, $e^{ac} = \frac{ka^{k-1}}{i\alpha c}\ne \pm1$ and $R(z) = n_{2}a^{2k-1}[as_{1}(z)+km_{1}]$, or $f(z)$ is of the form

where $a^{2k} = \alpha^{2}-\beta^{2}$, $b\in \mathbb{C}$, $c = \dfrac{\log\frac{a^{k}+i\beta}{i\alpha}+2l\pi i}{a}$, $l\in \mathbb{Z}$, $e^{ac} = \frac{i\alpha c}{ka^{k-1}}\ne \pm1$ and $R(z) = n_{1}a^{2k-1}[as_{2}(z)-km_{2}]$.

($II_{2}$) if $R(z)$ is a nonzero constant, then $f(z)$ is of the form

where $a, b, c, \alpha, \beta, c_{1}, c_{2}$, $R(z)$ satisfy $a^{2k} = \alpha^{2}-\beta^{2}$, $b\in \mathbb{C}$, $l\in \mathbb{Z}$, $c = \dfrac{\log\frac{a^{k}+i\beta}{i\alpha}+2l\pi i}{a}$ and $R(z) = a^{2k}n_{1}n_{2}$.

Remark 3. When $k = 2$, Theorem 1.6 becomes Theorem 1.4.

Next, we will provide some examples to explain the existence of solutions to Eq (1.2) in the above situation.

Example 1.5. For Case $II_{1}$, let $c_{0}$ be a solution of equation $e^{2c}(2-c) = 2$, the transcendental entire function

satisfies

where $k = 4$, $s_{1}(z) = z$, $n_{2} = 1$, $a = 1$, $b\in \mathbb{C}$, $\alpha = \dfrac{4}{ic_{0}e^{c_{0}}}$, $\beta = \dfrac{4-c_{0}}{ic_{0}}$, $\alpha^{2}-\beta^{2} = 1$, $c_{0} = \log\dfrac{1+i\beta}{i \alpha}$, $e^{ac_{0}} = \dfrac{4}{i\alpha c_{0}}$ and $R(z) = z+4$.

Example 1.6. For Case $II_{2}$, the transcendental entire function

satisfies

where $k = 4$, $n_{1} = 1$, $n_{2} = 1$, $e^{c} = \sqrt{5}-2$, $a = 1$, $b\in \mathbb{C}$, $\alpha = \frac{i}{2}$, $\beta = \frac{\sqrt{5}}{2}i$, $\alpha^{2}-\beta^{2} = 1$, $c = \log(\sqrt{5}-2)$ and $R(z) = 1$.

In 2024, Long and Qin ^[9] studied this equation

and obtained

Theorem 1.7. There is no transcendental entire solution with finite order of the equation

where $P(z)$ is a non-constant polynomial, and $Q(z)$ is a non-zero polynomial.

Motivating from Theorems 1.5 and 1.6, we replace $f(z+c)$ by the $\alpha f(z+c)-\beta f(z)$ in Theorem 1.7 and obtain

Theorem 1.8. There is no transcendental entire solution with finite order of the equation

where $k\in \mathbb{Z}^{+}$, $\alpha(\ne0)$, $\beta \in \mathbb{C}$, $P(z)$ is a non-constant polynomial and $Q(z)$ is a non-zero polynomial.

2.   Some lemmas

We can use these lemmas to prove our theorems.

Lemma 2.1. Let $c$, $a$, $\alpha$ be three nonzero constants, $k \in \mathbb{Z}^{+}$ be an odd number, and $ac\ne \pm (k-1)$. If $R_{1}, R_{2}$ are two nonzero rational functions satisfying the following differential-difference equations

Then $i \alpha c = ka^{k-1}$ and $R_{i}$ are nonzero polynomials with $\deg_{z} R_{i} \le 1$ $(i = 1, 2)$.

Wang et al. ^[14] proved the case of $k = 1$, below we prove the case of $k\ge3$.

Proof. First, we prove that $R_{1}(z)$ has no poles. On the contrary, suppose that $z_{0}$ is a pole of $R_{1}(z)$. We can write (2.1) in a new form

It is easy to see that $z_{0}+c$ is also a pole of $R_{1}(z)$ by comparing the order of pole $z_{0}$ on both sides of (2.2). By recycling this operation, we obtain a sequence of poles of $R_{1}(z)$ that are $z_{0}+2c$, $z_{0}+3c$, $\cdots$, $z_{0}+tc$, this is impossible since $R_{1}(z)$ is a nonzero rational function. Then, $R_{1}(z)$ is a polynomial. Similarly, it can be inferred that $R_{2}(z)$ also is a polynomial.

Therefore, $R_{1}(z)$ and $R_{2}(z)$ are two nonzero polynomials. Let

where $a_{j}\in \mathbb{C}$, $j\in \{0, 1, \cdots, p\}$, $b_{j}\in \mathbb{C}$, $j\in \{0, 1, \cdots, t\}$, $a_{p}\ne 0$, $b_{t}\ne 0$. Substituting $R_{1}(z)$ and $R_{2}(z)$ into (2.1) and comparing the coefficients of $z^{p-1}$, $z^{p-2}$, $z^{t-1}$, and $z^{t-2}$ on both sides of these two equations, we have

which means

It follows that $i\alpha c = ka^{k-1}$, $p = 0$ or $1$ and $t = 0$ or $1$. Therefore, this completes the proof of Lemma 2.1. □

Lemma 2.2. Let $c$, $a$, $\alpha$ be three nonzero constants and $k \in \mathbb{Z}^{+}$ is an even number. If $R_{1}, R_{2}$ are two nonzero rational functions satisfying the following differential-difference equations:

Then $e^{ac} = \pm 1$ and $R_{i}$ are nonzero polynomials with $\deg_{z} R_{i} \le 1$ $(i = 1, 2)$.

Proof. First, we prove that $R_{1}(z)$ has no poles. On the contrary, suppose that $z_{0}$ is a pole of $R_{1}(z)$. We can write (2.5) in a new form

Similar to the proof of Lemma 2.1, we can get $R_{1}(z)$ and $R_{2}(z)$ are two nonzero polynomials. Substituting $R_{1}(z)$ and $R_{2}(z)$ into (2.5) and comparing the coefficients of $z^{p-1}$, $z^{p-2}$, $z^{t-1}$ and $z^{t-2}$ on both sides of such two equations, it yields

which means

It follows that $e^{ac} = \pm 1$, $p = 0$ or $1$, and $t = 0$ or $1$. Therefore, this completes the proof of Lemma 2.2. □

Lemma 2.3. ^[14] Let $R$ be a nonconstant rational function and $p(z) = az+b(a\ne 0)$. 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

Lemma 2.4. ^[22] Suppose that $f_{1}, f_{2}, \cdots, f_{n}(n\ge 2)$ are meromorphic functions and $g_{1}, g_{2}, \cdots, g_{n}$ are entire functions satisfying the following conditions:

(i) $\sum\limits_{j = 1}^{n}f_{j}e^{g_{i}}\equiv 0$;

(ii) $g_{j}-g_{k}$ are not constants for $1\le j < k\le n$;

(iii) For $1\le j\le n, 1\le h < k\le n, T(r, f_{j}) = o\{ T(r, e^{g_{h}-g_{k}}) \} (r\to \infty, r\notin E)$, where $E$ is a set of (0, $\infty$) with finite linear measure.

Then $f_{j}\equiv 0$ $(j = 1, 2, \cdots, n)$.

Lemma 2.5. ^[22] Let $f$ be a meromorphic function of finite order $\rho (f)$. Write

near $z = 0$ and let $\{ a_{1}, a_{2, \cdots }\}$ and $\{ b_{1}, b_{2, \cdots }\}$ be the zeros and poles of $f$ in $\mathbb{C} \verb|\| \{0\}$, 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 the degree $\le \rho (f)$.

Lemma 2.6. ^[22] Suppose that $f_{1}(z), f_{2}(z), \cdots, f_{n}(z)$, $(n\ge 3)$ are meromorphic functions that are not constants except for $f_{n}(z)$. Furthermore, let

If $f_{n}(z)\not\equiv 0$ and

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

3.   Proof of Theorem 1.5

Proof. Suppose that Eq (1.2) admits a finite order transcendental meromorphic solution $f(z)$ with finitely many poles. We can rewrite Eq (1.2) in the following form:

Since $f(z)$ has finitely many poles and $R(z)$ is a nonzero rational function, then $f^{(k)}(z)+i(\alpha f(z+c)-\beta f(z))$ and $f^{(k)}(z)-i(\alpha f(z+c)-\beta f(z))$ both have finitely many poles and zeros. Thus, in view of Lemma 2.5, (3.1) can be written as

where $R_{1}, R_{2}$ are two nonzero rational functions such that $R_{1}R_{2} = R$ and $p(z)$ is a nonzero polynomial. By solving the above equations system, we have

In view of the second equation of (3.3), it follows that

where $A_{1} = R'_{1}+R_{1}p'$, $B_{1} = R'_{2}-R_{2}p'$, $A_{k} = A'_{k-1}+A_{k-1}p'$ and $B_{k} = B'_{k-1}-B_{k-1}p'$. Substituting the first equation of system (3.3) into (3.4), it yields that

By Lemma 2.4, it follows from (3.5) that

Since $R_{1}, R_{2}$ are two nonzero rational functions, which implies that $p(z)$ is a polynomial of degree one. Let $p(z) = az+b$, $a\ne 0$, $b\in \mathbb{C}$. Substituting $p(z)$, $A_{k}$, and $B_{k}$ into (3.6), and letting $|z| \to \infty$, thus, we can conclude from Lemma 2.3 that

Two equations of (3.7), which mean that

Hence, it yields $e^{ac} = \pm 1$.

If $e^{ac} = 1$, then $a^{k} = i \alpha -i\beta$. Thus, we can rewrite (3.6) in the following form:

If $R_{j}(j = 1, 2)$ are two nonzero rational functions, then in view of Lemma 2.1, it follows that $i\alpha c = ka^{k-1}$ and $R_{i}$ are nonzero polynomials with $\deg_{z} R_{i} \le 1$ $(i = 1, 2)$. In view of $R = R_{1}R_{2}$, thus $R$ is a nonzero polynomial with $\deg_{z} R \le 2$.

If $e^{ac} = -1$, then $a^{k} = -i\alpha -i\beta$. Thus, we can rewrite (3.6) in the following form

Similar to the discussion above, we can obtain that $i\alpha c = -ka^{k-1}$ and $R_{i}$ are nonzero polynomials with $\deg_{z} R_{i} \le 1$ $(i = 1, 2)$. In view of $R = R_{1}R_{2}$, thus $R$ is a nonzero polynomial with $\deg_{z}R \le 2$.

Hence, we can obtain that $i\alpha c = \pm ka^{k-1}$, $R$ is a nonzero polynomial with $\deg_{z} R \le 2$.

Suppose that $R(z)$ is a nonzero polynomial with $\deg_{z}R\le 2$, then in view of the first equation of (3.3), it follows that $f(z)$ is of the form

where $s_{j}(z) = m_{j}z+n_{j}$, $m_{j}$, $n_{j}\in \mathbb{C}(j = 1, 2)$ and $c_{0}, \cdots, c_{k-1}$ are constants.

If $\deg_{z}R = 2$, then it follows that $m_{j}\ne 0(j = 1, 2)$.

If $i\alpha c = ka^{k-1}$ and $a^{k} = i(\alpha -\beta)$, then $e^{ac} = 1$, i.e., $c = \frac{2l\pi i}{a}$, $l\in \mathbb{Z}$. According to (3.8), if $\alpha = \beta$, we have $a = 0$, it is a contradiction. If $\alpha = -\beta$, then $a^{k} = 2i\alpha$. Combining $i\alpha c = ka^{k-1}$, $a^{k} = 2i\alpha$ and $e^{ac} = 1$, we have $1 = e^{ac} = e^{2k}$, it is a contradiction. Hence, $\alpha \ne \pm \beta$. Substituting (3.11) into the second equation of (3.3), it follows that $c_{0} = \cdots = c_{k-1} = 0$, we have

Substituting (3.12) into the first equation of (3.3), it yields

Substituting (3.12) into the second equation of (3.3), it yields

Comparing (3.13) and (3.14), we have $ka^{k-1} = 1$ and $ka^{k-1} = -1$, it is a contradiction.

If $i\alpha c = -ka^{k-1}$ and $a^{k} = -i(\alpha +\beta)$, then $e^{ac} = -1$, i.e., $c = \frac{(2l+1)\pi i}{a}$, $l\in \mathbb{Z}$. Similar to the discussion above, we can obtain a contradiction. Therefore, there are two categories below:

Case I: If $\deg_{z}R = 1$, one of $m_{1}$ and $m_{2}$ is zero, without loss of generality, assume $m_{2} = 0$. Substituting (3.12) into (3.3), it follows that $R_{1}$ is a polynomial of degree one and $R_{2}$ is a constant, where $i\alpha c = ka^{k-1} = 1$ and $a^{k} = i(\alpha -\beta)$ or $i\alpha c = -ka^{k-1} = -1$ and $a^{k} = -i(\alpha +\beta)$. Similar to the discussion above, it is easy to prove that $\alpha \ne \pm \beta$ and $c_{0} = \cdots = c_{k-1} = 0$.

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

where $R(z) = -n_{2}a^{2k-1}[as_{1}(z)+km_{1}]$, $m_{1}\ne 0$, $a\ne 0, b\in \mathbb{C}$, and $a, b, c, \alpha, \beta$ satisfy $\alpha \ne \pm \beta$, $a^{k} = -i(\alpha +\beta)$, $c = \frac{(2l+1)\pi i}{a}$, $l\in \mathbb{Z}$, $i\alpha c = -ka^{k-1} = -1$, or $a^{k} = i(\alpha -\beta)$, $c = \frac{2l\pi i}{a}$, $l\in \mathbb{Z}$, $i\alpha c = ka^{k-1} = 1$.

If $m_{1} = 0$, similar to the discussion above, it is easy to prove that $f(z)$ is of the form

where $R(z) = n_{1}a^{2k-1}[-as_{2}(z)+km_{2}]$, $m_{2}\ne 0$, $a\ne 0, b\in \mathbb{C}$, and $a, b, c, \alpha, \beta$ satisfy $\alpha \ne \pm \beta$, $a^{k} = -i(\alpha +\beta)$, $c = \frac{(2l+1)\pi i}{a}$, $l\in \mathbb{Z}$, $i\alpha c = ka^{k-1} = -1$, or $a^{k} = i(\alpha -\beta)$, $c = \frac{2l\pi i}{a}$, $l\in \mathbb{Z}$, $i\alpha c = -ka^{k-1} = 1$.

Case II: If $R(z)$ is a nonzero constant, by using the first equation of (3.3), it follows that $f(z)$ is of the form

where $n_{1}, n_{2}\in \mathbb{C}$ and $c_{0}, \cdots, c_{k-1}\in \mathbb{C}$. Substituting (3.15) into the second equation of (3.3), it yields $R = -a^{2k}n_{1}n_{2}$.

($II_{1}$) If $\alpha = \beta$, in view of (3.8), it follows that $e^{ac} = \pm 1$. If $e^{ac} = 1$, then $a = 0$, as $i\alpha (e^{ac}-1) = a^{k}$, a contradiction. Thus, $e^{ac} = -1$. Hence, it follows that $c = \frac{(2l+1) \pi i}{a}$, $l\in \mathbb{Z}$, $a^{k} = -2i\alpha$, and $c_{0}\in \mathbb{C}$, $c_{1} = \cdots = c_{k-1} = 0$.

($II_{2}$) If $\alpha = -\beta$, in view of (3.8), it follows that $e^{ac} = \pm 1$. If $e^{ac} = -1$, then $a = 0$, as $i\alpha (e^{ac}+1) = a^{k}$, a contradiction. Thus, $e^{ac} = 1$. Hence, it follows that $c = \frac{2l \pi i}{a}$, $l\in \mathbb{Z}$, $a^{k} = 2i\alpha$, and $c_{0} = \cdots = c_{k-1} = 0$.

($II_{3}$) If $\alpha \ne \pm \beta$, substituting (3.15) into the second equation of (3.3), it yields $c_{0} = \cdots = c_{k-1} = 0$. In view of (3.8), it follows that $e^{ac} = \pm 1$. If $e^{ac} = 1$, it follows that $c = \frac{2l\pi i}{a}$ and $a^{k} = i(\alpha -\beta)$, $l\in \mathbb{Z}$. If $e^{ac} = -1$, it follows that $c = \frac{(2l+1) \pi}{a}$ and $a^{k} = -i(\alpha +\beta)$, $l\in \mathbb{Z}$. Therefore, this completes the proof of Theorem 1.5. □

4.   Proof of Theorem 1.6

Proof. Similar to the method of proving Theorem 1.5, we can obtain the expression (3.8).

When $k$ is an even number, two equations of (3.7), which mean that

Hence, it follows $a^{2k} = \alpha^{2}-\beta^{2}$.

Case I: If $\alpha = \pm \beta$, this is a contradiction with $a\ne 0$.

Case II: If $\alpha \ne \pm \beta$. Substituting $p(z) = az+b$ and (4.1) into (3.6), it yields

Suppose that $R_{1}, R_{2}$ are nonzero rational functions; in view of Lemma 2.2, we can conclude that $e^{ac} = \pm 1$ and $R_{i}$ are nonzero polynomials with $\deg_{z} R_{i} \le 1$ $(i = 1, 2)$. In view of $R = R_{1}R_{2}$, thus $R$ is a nonzero polynomial with $\deg_{z} R \le 2$. Set $\deg_{z}R_{1} = p$ and $\deg_{z}R_{2} = t$.

When $p = 1$ and $t = 1$, if $e^{ac} = 1$, then from (4.1), it follows that $i\alpha -i\beta = a^{k}$ and $i\alpha -i\beta = -a^{k}$, a contradiction. If $e^{ac} = -1$, then from (4.1), it follows that $-i\alpha -i\beta = a^{k}$ and $-i\alpha -i\beta = -a^{k}$, a contradiction. Hence, there is at most a polynomial of degree one in $R_{1}$ and $R_{2}$.

($II_{1}$) Suppose that $p = 1$, $t = 0$. In view of (3.3), it follows that $f$ is of the form

where $a\ne 0$, $b\in \mathbb{C}$, $s_{1}(z) = m_{1}z+n_{1}$, $m_{1}(\ne 0)$, $n_{1}$, $n_{2}\in \mathbb{C}$, and $P(z)$ is a polynomial of degree $k-1$. Since $\alpha \ne \beta$, then it yields from the second equation of (3.3) that $P(z) \equiv 0$. And by using the first equation in (4.2), it follows that $i\alpha e^{ac}c = ka^{k-1}$. Hence, $f(z)$ is of the form

where $a^{2k} = \alpha^{2}-\beta^{2}$, $b\in \mathbb{C}$, $c = \dfrac{\log \frac{a^{k}+i\beta}{i\alpha}+2l\pi i}{a}$, $l\in \mathbb{Z}$, $e^{ac} = \frac{ka^{k-1}}{i\alpha c}\ne \pm 1$ and $R = n_{2}a^{2k-1}[as_{1}(z)+km_{1}]$.

Suppose that $p = 0$, $t = 1$. Similar to the above argument as in ($II_{1}$), we obtain

where $a^{2k} = \alpha^{2}-\beta^{2}, b\in \mathbb{C}$, $c = \dfrac{\log \frac{a^{k}+i\beta}{i\alpha}+2l\pi i}{a}$, $l\in \mathbb{Z}$, $e^{ac} = \frac{i\alpha c }{ka^{k-1}}\ne \pm 1$ and $R = n_{1}a^{2k-1}[as_{2}(z)-km_{2}]$.

($II_{2}$) Suppose that $p = 0$, $t = 0$. By using (3.3), it follows that $f$ is of the form

where $a\ne 0$, $b\in \mathbb{C}$, $n_{1}$, $n_{2}\in \mathbb{C} \verb|\| \{0\}$, and $P(z)$ is a polynomial of degree $k-1$. Since $\alpha \ne \beta$, then it yields from the second equation of (3.3) that $P(z) \equiv 0$. Hence, $f(z)$ is of the form

where $a^{2k} = \alpha^{2}-\beta^{2}$, $b\in \mathbb{C}$, $c = \dfrac{\log \frac{a^{k}+i\beta}{i\alpha}+2l\pi i}{a}$, $l\in \mathbb{Z}$ and $R = a^{2k}n_{1}n_{2}$. Therefore, this completes the proof of Theorem 1.6. □

5.   Proof of Theorem 1.8

Proof. Suppose, on the contrary, to the assertion that there exists a transcendental entire solution $f$ of (1.4) with finite order. We aim for a contradiction. By using a similar reason as in the proof of Theorem 1.5, we obtain

and

where $h(z)$ is a non-constant polynomial, $Q_{1}(z)$ and $Q_{2}(z)$ are non-zero polynomials such that $Q_{1}(z)Q_{2}(z) = Q(z)$. Combining (5.1) and (5.2), we obtain

where

$M_{t}$ and $N_{t}$ are differential polynomials of $(h, h', \cdots, h^{(t)})$. Thus from (5.3), we get

It is easy to see that both $\dfrac{h_{1}(z)+\beta iP(z)^{k+1}Q_{1}(z)}{\alpha iP(z)^{k+1}Q_{2}(z+c)}e^{h(z)+h(z+c)}$ and $\dfrac{Q_{1}(z+c)}{Q_{2}(z+c)}e^{2h(z+c)}$ are not constants. Using Lemma 2.6, we obtain $-[h_{2}(z)-\beta iP(z)^{k+1}Q_{2}(z)]e^{h(z+c)-h(z)} = \alpha iP(z)^{k+1}Q_{2}(z+c)$, so $h(z) = Az+B$, $A$ is a non-zero constant, and $B$ is a constant. Thus, we obtain

Set $\deg(P(z)) = p$, $\deg(Q(z)) = q$, $\deg(Q_{1}(z)) = q_{1}$, $\deg(Q_{2}(z)) = q_{2}$ and $\deg(h(z)) = h$. By comparing the degree of both sides of (5.5), it is not difficult to find that the degree of the left hand-side is $(k + 1)p+q_{2}$ or $(k + 1)p+q_{2}-1$, and the degree of right-hand side is $kp+q_{2}-1$; this is a contradiction. Therefore, this completes the proof of Theorem 1.8. □

Author contributions

Zhiyong Xu: Conceptualization, Methodology, Writing-original draft; Junfeng Xu: Supervision, Writing-review and editing. Both of the authors have read and approved the final version of the manuscript for publication.

Acknowledgments

This research was supported by Fund of Education Department of Guangdong (Nos. 2022ZDZX1034, 2023GXJK517).

Conflict of interest

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