Theory and Modern Applications

# Sharing values of q-difference-differential polynomials

## Abstract

This paper is devoted to the uniqueness of q-difference-differential polynomials of different types. Using the idea of common zeros and common poles (Chin. Ann. Math., Ser. A 35:675–684, 2014), we improve the conditions of the former theorems and obtain some new results on the uniqueness of q-difference-differential polynomials of meromorphic functions.

## Introduction and main results

In this paper, a meromorphic function is assumed meromorphic in the whole complex plane. We assume that the reader is familiar with the basic symbols and fundamental results of Nevanlinna theory; see, for example, [2, 3, 10]. We say that two meromorphic functions f and g share a point a CM (IM) if $$f(z)-a$$ and $$g(z)-a$$ have the same zeros counting multiplicities (ignoring multiplicities). The logarithmic density of the set E is defined by

\begin{aligned} \limsup_{r\rightarrow\infty}\frac{1}{\log r} \int_{[1,r]\cap E}\frac{1}{t}\,dt. \end{aligned}

Denote by $$S(r,f)$$ a quantity of $$o(T(r,f))$$ as $$r\rightarrow\infty$$ outside a possible exceptional set E of logarithmic density 0.

Yang and Hua  obtained an important result on the uniqueness when the differential polynomials $$f^{n}f'$$ and $$g^{n}g'$$ share one value CM. Recently, many studies are devoted to the uniqueness of difference and q-difference polynomials; see [46, 1114]. Zhang  obtained the following result.

### Theorem A

()

Let$$f(z)$$and$$g(z)$$be transcendental entire functions of zero order, and letn, m, dbe positive integers. If$$n \geq m+5d$$and$$f(z)^{n}(f(z)^{m}-1)\prod_{i=1}^{d}f(q_{i}z)$$and$$g(z)^{n} (g(z)^{m}-1)\prod_{i=1}^{d}g(q_{i}z)$$share 1 CM, then$$f(z)\equiv tg(z)$$, $$t^{n+d}=t^{m}=1$$.

Liu, Liu, and Cao  and Zhang and Korhonen  obtained the following two theorems.

### Theorem B

([4, Theorem 1.5])

Let$$f(z)$$and$$g(z)$$be transcendental zero-order entire functions, and letmbe a positive integer. If$$n \geq m+5$$and$$f(z)^{n}(f(z)^{m}-a)f(qz+c)$$and$$g(z)^{n}(g(z)^{m}-a)g(qz+c)$$share a nonzero polynomial$$p(z)$$CM, then$$f(z)\equiv g(z)$$.

### Theorem C

([11, Theorem 5.1])

Let$$f(z)$$and$$g(z)$$be transcendental zero-order meromorphic functions. If$$n\geq8$$and$$f(z)^{n}f(qz)$$and$$g(z)^{n}g(qz)$$share 1 andCM, then$$f(z)\equiv tg(z)$$, $$t^{n+1}=1$$.

Zhao and Zhang  proved the following theorem.

### Theorem D

([13, Theorem 1.4])

Let$$f(z)$$and$$g(z)$$be transcendental zero-order entire functions, and letkbe a positive integer. If$$n \geq2k+6$$and$$(f(z)^{n}f(qz+c))^{(k)}$$and$$(g(z)^{n}g(qz+c))^{(k)}$$share 1 CM, then$$f(z)\equiv tg(z)$$, where$$t^{n+1}=1$$.

Wang and Ye  improved the conditions of Theorems B and C to $$n \geq m+4$$ and $$n \geq6$$, respectively, by using the idea of common zeros and common poles. Here we give the main idea of common zeros and common poles. Let f, g be two nonconstant meromorphic functions. Denote by $$\overline{n}_{0}(r)$$ or $$\overline {n}_{1}(r)$$ the numbers of common zeros or poles of fg and g, ignoring multiplicities. Let p, q be positive integers. We assume that the Laurent series of f and g at $$z_{0}$$ are as follows:

$$f(z)=\frac{1}{(z-z_{0})^{p}}f_{1}(z), \qquad g(z)=(z-z_{0})^{q}g_{1}(z),$$

where $$f_{1}(z)$$ and $$g_{1}(z)$$ are analytic functions at $$z_{0}$$, and $$f_{1}(z_{0})\neq0$$, $$g_{1}(z_{0})\neq0$$; the other cases can be discussed in a similar way. So $$z_{0}$$ is a zero of $$g(z)$$ with multiplicities q. If $$q>p$$, then $$z_{0}$$ is a zero of $$f(z)g(z)$$ with multiplicity $$q-p$$, and thus the contribution to $$\overline{n}_{0}(r)$$ is 1 at $$z_{0}$$. If $$q\leq p$$, then $$z_{0}$$ is a pole of $$f(z)g(z)$$ with multiplicity $$p-q$$ or an analytic point of $$f(z)g(z)$$, and thus the contribution to $$\overline{n}_{0}(r)$$ is 0 at $$z_{0}$$. A similar method can be discussed for $$\overline{n}_{1}(r)$$. As usual, denote by $$\overline{N}_{0}(r)$$ or $$\overline{N}_{1}(r)$$ the counting functions of the common zeros or poles of fg and g, ignoring multiplicities. Thus we have $$\overline{N}(r,\frac{1}{fg})\leq\overline{N}(r,\frac {1}{f})+\overline{N}_{0}(r)$$ and $$\overline{N}(r,fg)\leq\overline{N}(r,f)+\overline{N}_{1}(r)$$. In this paper, we continue to consider the uniqueness of q-difference-differential polynomials. Firstly, we improve the condition $$n\geq m+5d$$ in Theorem A to $$n\geq m+d+3$$ in Theorem 1.1. Set

\begin{aligned} L(z,f)=\prod_{i=1}^{d}f(q_{i}z+c_{i}), \end{aligned}

where $$c_{i}$$ and $$q_{i}\neq0$$ ($$i=1,\ldots, d$$) are constants, and d is a positive integer.

### Theorem 1.1

Let$$f(z)$$and$$g(z)$$be transcendental zero-order entire functions, and letmbe a positive integer. If$$n \geq m+d+3$$and$$f(z)^{n}(f(z)^{m}-1)L(z,f)$$and$$g(z)^{n}(g(z)^{m}-1)L(z,g)$$share 1 CM, then$$f(z)\equiv c_{1}g(z)$$, $$c_{1}^{n+d}=c_{1}^{m}=1$$.

In the following theorem, we improve the condition $$n\geq2k+6$$ in Theorem D to $$n\geq6$$.

### Theorem 1.2

Let$$f(z)$$and$$g(z)$$be transcendental zero-order meromorphic functions, and letkbe a positive integer. If$$n \geq6$$and$$(f(z)^{n}f(qz+c))^{(k)}$$and$$(g(z)^{n}g(qz+c))^{(k)}$$share 1 andCM, then$$f(z)\equiv c_{2}g(z)$$, $$c_{2}^{n+1}=1$$.

We also consider the following theorems for q-difference polynomials of different types. The following theorem is also an improvement of Theorem C.

### Theorem 1.3

Let$$f(z)$$and$$g(z)$$be transcendental zero-order meromorphic functions, and letsbe a positive integer. If$$n\geq(d+1)s+4$$and$$f(z)^{n}L(z,f)^{s}$$and$$g(z)^{n}L(z,g)^{s}$$share 1 andCM, then$$f(z)\equiv c_{3}g(z)$$, $$c_{3}^{n+sd}=1$$.

### Theorem 1.4

Let$$f(z)$$and$$g(z)$$be transcendental zero-order meromorphic functions, $$q, c\in\mathbb{C}$$, and$$q\neq0$$. If$$n\geq7$$, and$$f(z)^{n}(f(qz+c)-f(z))$$and$$g(z)^{n}(g(qz+c)-g(z))$$share 1 andCM, then

$$f(z)^{n}\bigl(f(qz+c)-f(z)\bigr)= g(z)^{n} \bigl(g(qz+c)-g(z)\bigr).$$

If$$\frac{g(z)}{g(qz+c)}$$is transcendental with only finitely many zeros, then$$f(z)\equiv c_{4}g(z)$$, where$$c_{4}^{n+1}=1$$.

## Lemmas

Combining [11, Theorem 1.1] and [1, Theorem 2.1], we easily get the following lemma.

### Lemma 2.1

Let$$f(z)$$be a transcendental zero-order meromorphic function, $$q, c\in \mathbb{C}$$, and$$q\neq0$$. Then

\begin{aligned} T\bigl(r,f(qz+c)\bigr)=T(r,f)+S(r,f) \end{aligned}

on a set of logarithmic density 1.

### Lemma 2.2

()

Let$$f(z)$$be a zero-order meromorphic function$$q, c\in\mathbb{C}$$, and$$q\neq0$$. Then

\begin{aligned} m \biggl(r,\frac{f(qz+c)}{f(z)} \biggr)=S(r,f) \end{aligned}

on a set of logarithmic density 1.

### Lemma 2.3

Iffis a transcendental zero-order entire function, then

\begin{aligned} T\bigl(r, f(z)^{n}\bigl(f(z)^{m}-1\bigr)L(z,f) \bigr)=(m+n+d)T(r,f)+S(r,f) \end{aligned}

on a set of logarithmic density 1.

### Proof

Set $$F(z)=f(z)^{n}(f(z)^{m}-1)L(z,f)$$. By Lemma 2.2 and the standard Valiron–Mohon’ko theorem, if f is a transcendental zero-order entire function, then

\begin{aligned} (n+m+d)T(r,f) =& T\bigl(r,f^{n+d}\bigl(f^{m}-1\bigr) \bigr)+S(r,f) \\ =& m\bigl(r,f^{n+d}\bigl(f^{m}-1\bigr)\bigr)+S(r,f) \\ \leq& m \biggl(r,\frac{f^{n+d}(f^{m}-1)}{f^{n}(f^{m}-1)L(z,f)} \biggr)+m(r,F)+S(r,f) \\ \leq& m \biggl(r,\frac{f^{d}}{L(z,f)} \biggr)+m(r,F)+S(r,f) \\ \leq& T(r,F)+S(r,f) \end{aligned}

on a set of logarithmic density 1. On the other hand, combining Lemma 2.1 with the fact that f is a transcendental zero-order function, we have

\begin{aligned} T(r,F) \leq& T\bigl(r,f^{n}\bigl(f^{m}-1\bigr)\bigr)+T \bigl(r,L(z,f)\bigr) \\ \leq& (n+m+d)T(r,f)+S(r,f) \end{aligned}

on a set of logarithmic density 1. □

## Proofs of theorems

### Proof of Theorem 1.1

Let $$F(z)=f^{n}(f^{m}-1)L(z,f)$$ and $$G(z)=g^{n}(g^{m}-1)L(z,g)$$. Since $$F(z)$$ and $$G(z)$$ share 1 and ∞ CM, we have that $$\frac{F-1}{G-1}=B$$, that is,

\begin{aligned} F=BG+1-B, \end{aligned}
(1)

where B is a nonzero constant.

If $$B\neq1$$, then from the second main theorem of Nevanlinna theory, Lemma 2.1, and Lemma 2.3 we obtain

\begin{aligned} (n+m+d)T(r,f) =&T(r,F)+S(r,f) \\ \leq& \overline{N}(r,F)+\overline{N} \biggl(r,\frac{1}{F} \biggr) + \overline{N} \biggl(r,\frac{1}{F-1+B} \biggr)+S(r,f) \\ \leq& \overline{N} \biggl(r,\frac{1}{f^{n}} \biggr)+\overline{N} \biggl(r,\frac{1}{f^{m}-1} \biggr) +\overline{N} \biggl(r, \frac{1}{L(z,f)} \biggr) +\overline{N} \biggl(r,\frac{1}{G} \biggr)+S(r,f) \\ \leq& (m+d+1) \bigl(T(r,f)+T(r,g)\bigr)+S(r,f)+S(r,g). \end{aligned}
(2)

Using the same method, we have

$$(n+m+d)T(r,g)\leq(m+d+1) \bigl(T(r,f)+T(r,g) \bigr)+S(r,f)+S(r,g).$$
(3)

Combining (2) with (3), we have

\begin{aligned} (n-m-d-2) \bigl(T(r,f)+T(r,g)\bigr)\leq S(r,f)+S(r,g), \end{aligned}

which contradicts to $$n\geq m+d+3$$. Thus $$B=1$$, and from (1) we have

$$f^{n}\bigl(f^{m}-1\bigr)L(z,f)=g^{n} \bigl(g^{m}-1\bigr)L(z,g).$$
(4)

Let $$h(z)=\frac{f(z)}{g(z)}$$. So $$\frac{L(z,f)}{L(z,g)}=\prod_{i=1}^{d}\frac{f(q_{i}z+c_{i})}{g(q_{i}z+c_{i})} =\prod_{i=1}^{d}h(q_{i}z+c_{i})=L(z,h)$$, and then (4) can be written as

$$g^{m}\bigl(h^{n+m}L(z,h)-1 \bigr)=h^{n}L(z,h)-1.$$
(5)

Next, we prove that $$h(z)\equiv c_{1}$$ and $$c_{1}^{n+d}=c_{1}^{m}=1$$, where $$c_{1}$$ is a constant. Assume on the contrary that $$h(z)$$ is not a constant. From Lemma 2.1 we have

$$T\bigl(r,h^{n+m}L(z,h)\bigr)\leq(n+m+d)T(r,h)+S(r,h).$$

We also have

\begin{aligned} (n+m)T(r,h) =&T\bigl(r,h^{n+m}\bigr) \\ \leq& T \biggl(r,\frac{1}{h^{n+m}L(z,h)} \biggr) +T\bigl(r,L(z,h)\bigr))+O(1) \\ \leq& T \bigl(r,h^{n+m}L(z,h) \bigr) +dT(r,h)+S(r,h). \end{aligned}

Since $$n\geq m+d+3$$, from the last two inequalities it follows that $$S(r,h^{n+m}L(z,h))=S(r,h)$$. Denote by $$\overline{N}_{1}(r)$$ the counting function of the common poles of $$h^{n+m}L(z,h)$$ and $$L(z,h)$$ ignoring multiplicities. Then

$$\overline{N}\bigl(r,h^{n+m}L(z,h)\bigr)\leq\overline{N}(r,h)+ \overline{N}_{1}(r).$$

Here we should remark that the poles of $$L(z,h)$$ may be the zeros of h and the zeros of $$L(z,h)$$ may be the poles of h. Similarly, denote by $$\overline{N}_{0}(r)$$ the counting function of the common zeros of $$h^{n+m}L(z,h)$$ and $$L(z,h)$$ ignoring multiplicities, and then

$$\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)} \biggr)\leq\overline {N}\biggl(r, \frac{1}{h}\biggr)+\overline{N}_{0}(r).$$

From the second main theorem of Nevanlinna theory and the last two inequalities we have

\begin{aligned} T\bigl(r,h^{n+m}L(z,h)\bigr)&\leq\overline{N} \bigl(r,h^{n+m}L(z,h)\bigr) +\overline{N} \biggl(r, \frac{1}{h^{n+m}L(z,h)} \biggr) \\ &\quad+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr)+{S \bigl(r,h^{n+m}L(z,h)\bigr)} \\ &\leq\overline{N}(r,h)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{h} \biggr) +\overline{N}_{0}(r) \\ &\quad+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr)+S(r,h). \end{aligned}
(6)

On the other hand,

\begin{aligned}& (n+m)m(r,h)=m\bigl(r,h^{n+m}\bigr)\leq m \bigl(r,h^{n+m}L(z,h)\bigr)+m \biggl(r,\frac {1}{L(z,h)} \biggr)+O(1), \end{aligned}
(7)
\begin{aligned}& \begin{aligned}[b](n+m)N(r,h)&=N\bigl(r,h^{n+m}\bigr)= N \biggl( \frac{h^{n+m}L(z,h)}{L(z,h)} \biggr) \\ &\leq N\bigl(r,h^{n+m}L(z,h)\bigr)+N \biggl(r,\frac{1}{L(z,h)} \biggr)-\overline {N}_{1}(r)-\overline{N}_{0}(r).\end{aligned} \end{aligned}
(8)

From (7) and (8) we get

\begin{aligned} (n+m)T(r,h)&\leq T\bigl(r,h^{n+m}L(z,h)\bigr)+T \biggl(r,\frac{1}{L(z,h)} \biggr) \\ &\quad-\overline{N}_{1}(r)-\overline{N}_{0}(r)+O(1). \end{aligned}
(9)

From (6) and (9) we get

\begin{aligned} (n+m)T(r,h) \leq& \overline{N}(r,h)+\overline{N} \biggl(r, \frac {1}{h} \biggr)+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr) \\ &{}+T \biggl(r,\frac{1}{L(z,h)} \biggr)+S(r,h) \\ \leq& (d+2)T(r,h)+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr)+S(r,h). \end{aligned}
(10)

Since $$n\geq m+d+3$$, the value 1 is not the Picard exceptional value of $$h^{n+m}L(z,h)$$ from (10). Furthermore, we prove $$h^{n+m}L(z,h)\equiv1$$ and $$h(z)\equiv c_{1}$$ is a nonzero constant. If $$h^{n+m}L(z,h)\not\equiv1$$,then since 1 is not the Picard exceptional value of $$h^{n+m}L(z,h)$$, there exists a point $$z_{0}$$ satisfying $$h(z_{0})^{n+m}L(z,h(z_{0}))=1$$. From the condition that $$g(z)$$ is an entire function and (5) we have $$h(z_{0})^{m}=1$$ and

$$\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr) \leq \overline{N} \biggl(r,\frac{1}{h^{m}-1} \biggr) \leq mT(r,h)+S(r,h).$$
(11)

Substituting (11) into (10), we get a contradiction to $$n\geq m+d+3$$, so $$h^{n+m}L(z,h)\equiv1$$, that is, $$h^{n+m}=\frac {1}{L(z,h)}$$. From Lemma 2.1 we have

\begin{aligned} (n+m)T(r,h)=T\bigl(r,L(z,h)\bigr)\leq dT(r,h)+S(r,h), \end{aligned}

which also contradicts to $$n\geq m+d+3$$, so $$h(z)\equiv c_{1}$$, where $$c_{1}$$ is a nonzero constant, that is, $$f(z)\equiv c_{1}g(z)$$, and $$L(z,h)=c_{1}^{d}$$. From (5) we can get $$c_{1}^{m}=c_{1}^{n+d}=1$$. Thus the theorem is proved. □

### Proof of Theorem 1.2

Let $$F(z)=f(z)^{n}f(qz+c)$$ and $$G(z)=g(z)^{n}g(qz+c)$$. From the condition in Theorem 1.2 we know that $$F^{(k)}$$ and $$G^{(k)}$$ share 1 and ∞ CM, so

\begin{aligned} \frac{F^{(k)}-1}{G^{(k)}-1}=C, \end{aligned}

where C is a nonzero constant, that is,

$$F^{(k)}=CG^{(k)}-C+1.$$
(12)

Integrating both sides of (12), we have

$$F=CG+\frac{1-C}{k!}z^{k}+p_{1}(z),$$
(13)

where $$p_{1}(z)$$ is a polynomial of degree at most $$k-1$$. Denote $$\frac {1-C}{k!}z^{k}+p_{1}(z)$$ by $$p(z)$$. If $$p(z)\not\equiv0$$, then by the second main theorem of Nevanlinna theory, Lemma 2.1, and (13) we obtain

\begin{aligned} T(r,F) \leq& \overline{N}(r,F)+\overline{N} \biggl(r, \frac{1}{F} \biggr) +\overline{N} \biggl(r,\frac{1}{F-p} \biggr)+S(r,f) \\ \leq& \overline{N}(r,f)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{f} \biggr) +\overline{N}_{0}(r)+\overline{N} \biggl(r,\frac{1}{G} \biggr)+S(r,f) \\ \leq& 2T(r,f)+2T(r,g)+\overline{N}_{1}(r)+\overline{N}_{0}(r)+S(r,f)+S(r,g), \end{aligned}
(14)

where $$\overline{N}_{0}(r)$$ denotes the counting function ignoring multiplicities of the common zeros of $$F(z)$$ and $$f(qz+c)$$, and $$\overline{N}_{1}(r)$$ denotes the counting function ignoring multiplicities of the common poles of $$F(z)$$ and $$f(qz+c)$$. On the other hand,

\begin{aligned}& nm(r,f)=m\bigl(r,f^{n}\bigr)\leq m(r,F)+m \biggl(r, \frac{1}{f(qz+c)} \biggr)+O(1). \end{aligned}
(15)
\begin{aligned}& \begin{aligned}[b]nN(r,f)&=N\bigl(r,f^{n}\bigr)=N \biggl( \frac{F(z)}{f(qz+c)} \biggr) \\ &\leq N(r,F)+N \biggl(r,\frac{1}{f(qz+c)} \biggr)-\overline{N}_{1}(r)- \overline {N}_{0}(r).\end{aligned} \end{aligned}
(16)

From (15), (16), and Lemma 2.1 we have

$$(n-1)T(r,f)\leq T(r,F)-\overline{N}_{1}(r)- \overline{N}_{0}(r)+O(1).$$
(17)

Substituting (14) into (17), we obtain

$$(n-3)T(r,f)\leq2T(r,g)+S(r,f)+S(r,g).$$
(18)

Using the same method, we also get

$$(n-3)T(r,g)\leq2T(r,f)+S(r,f)+S(r,g).$$
(19)

Combining (18) with (19), we have

\begin{aligned} (n-5) \bigl(T(r,g)+T(r,f)\bigr)\leq S(r,f)+S(r,g), \end{aligned}

which contradicts to $$n\geq6$$, and thus $$p(z)\equiv0$$. Since the degree of $$p_{1}(z)$$ is at most $$k-1$$, we have $$C=1$$ and $$p_{1}(z)\equiv 0$$. From (13) we get

\begin{aligned} f^{n}f(qz+c)=g^{n}g(qz+c). \end{aligned}

Assume that $$h(z)=\frac{f(z)}{g(z)}$$. Then $$h(qz+c)h(z)^{n}=1$$, that is, $$h(z)^{n}=\frac{1}{h(qz+c)}$$, and from Lemma 2.1 we have

\begin{aligned} nT(r,h)=T\bigl(r,h(qz+c)\bigr)\leq T(r,h)+S(r,h), \end{aligned}

which also contradicts to $$n\geq6$$, so $$h(z)$$ is a nonzero constant, say $$c_{2}$$. So $$f(z)\equiv c_{2}g(z)$$, and $$c_{2}^{n+1}=1$$. Thus the theorem is proved. □

### Proof of Theorem 1.3

Since $$f(z)$$ and $$g(z)$$ are transcendental zero-order meromorphic functions and $$f(z)^{n}L(z,f)^{s}$$ and $$g(z)^{n}L(z,g)^{s}$$ share 1 and ∞ CM, we have

$$\frac{f(z)^{n}L(z,f)^{s}-1}{g(z)^{n}L(z,g)^{s}-1}=E,$$
(20)

where E is a nonzero constant. Then (20) can be rewritten as

$$Eg(z)^{n}L(z,g)^{s}=f(z)^{n}L(z,f)^{s}-1+E.$$
(21)

Let $$F(z)=f(z)^{n}L(z,f)^{s}$$ and $$G(z)=g(z)^{n}L(z,g)^{s}$$. We affirm that $$E=1$$. On the contrary, assume that $$E\neq1$$. Using the second main theorem of Nevanlinna theory and Lemma 2.1 for (21), we get

\begin{aligned} T(r,F) \leq& \overline{N}(r,F)+\overline{N} \biggl(r, \frac{1}{F} \biggr)+\overline{N} \biggl(r,\frac{1}{F-1+E} \biggr) +S(r,f) \\ \leq& \overline{N}(r,f)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{f} \biggr) +\overline{N}_{0}(r)+\overline{N} \biggl(r,\frac{1}{G} \biggr)+S(r,f) \\ \leq& 2T(r,f)+\overline{N} \biggl(r,\frac{1}{g} \biggr)+\overline{N} \biggl(r,\frac{1}{L(z,g)} \biggr) +\overline{N}_{1}(r) + \overline{N}_{0}(r)+S(r,f) \\ \leq& 2T(r,f)+(d+1)T(r,g)+\overline{N}_{1}(r)+\overline {N}_{0}(r)+S(r,f)+S(r,g), \end{aligned}
(22)

where $$\overline{N}_{0}(r)$$ denotes the counting function ignoring multiplicities of the common zeros of $$F(z)$$ and $$L(z,f)$$, and $$\overline{N}_{1}(r)$$ denotes the counting function ignoring multiplicities of the common poles of $$F(z)$$ and $$L(z,f)$$.

Since $$F(z)=f(z)^{n}L(z,f)^{s}$$, we have

\begin{aligned}& nm(r,f)=m\bigl(r,f^{n}\bigr)\leq m(r,F)+sm \biggl(r, \frac{1}{L(z,f)} \biggr)+O(1), \\& \begin{aligned}nN(r,f)&=N\bigl(r,f^{n}\bigr)=N \biggl(r,F(z) \biggl[ \frac{1}{L(z,f)} \biggr]^{s} \biggr) \\ &\leq N(r,F)+sN \biggl(r,\frac{1}{L(z,f)} \biggr)-\overline{N}_{1}(r)- \overline{N}_{0}(r).\end{aligned} \end{aligned}

So

\begin{aligned} nT(r,f)&\leq T(r,F)+sT \biggl(r, \frac{1}{L(z,f)} \biggr)-\overline {N}_{1}(r)- \overline{N}_{0}(r)+O(1) \\ &\leq T(r,F)+dsT(r,f)-\overline{N}_{1}(r)-\overline{N}_{0}(r)+S(r,f). \end{aligned}
(23)

Substituting (22) into (23), we obtain

$$(n-ds-2)T(r,f)\leq(d+1)T(r,g)+S(r,f)+S(r,g).$$
(24)

Using the same method, we can get

$$(n-ds-2)T(r,g)\leq(d+1)T(r,f)+S(r,f)+S(r,g).$$
(25)

Combining (24) with (25), it follows

\begin{aligned} (n-d-ds-3) \bigl(T(r,g)+T(r,f)\bigr)\leq S(r,f)+S(r,g), \end{aligned}

which contradicts to $$n\geq(d+1)s+4$$, and thus $$E=1$$. From (21) we get

\begin{aligned} f(z)^{n}L(z,f)^{s}=g(z)^{n}L(z,g)^{s}. \end{aligned}

Let $$h(z)=\frac{f(z)}{g(z)}$$. So $$\frac{L(z,f)}{L(z,g)}=L(z,h)$$. Then

$$h(z)^{n}\bigl[L(z,h)\bigr]^{s}=1.$$
(26)

So $$nT(r,h)=sT(r,L(z,h))\leq sdT(r,h)+S(r,h)$$. Since $$n\geq(d+1)s+4$$, $$h(z)$$ must be a constant, say $$c_{3}$$, that is, $$f(z)\equiv c_{3}g(z)$$. Then from (26) it follows that $$c_{3}^{n+sd}=1$$. □

### Proof of Theorem 1.4

Letting $$L(z,f)=f(qz+c)-f(z)$$ and $$L(z,g)=g(qz+c)-g(z)$$, $$s=1$$ in Theorem 1.3, we obtain that if $$n\geq7$$, then

\begin{aligned} f(z)^{n}\bigl(f(qz+c)-f(z)\bigr)=g(z)^{n} \bigl(g(qz+c)-g(z)\bigr). \end{aligned}

Let $$h(z)=\frac{f(z)}{g(z)}$$ and $$H(z)=h(qz+c)h(z)^{n}$$. The last equation implies that

$$g(qz+c) \bigl(H(z)-1 \bigr)=g(z) \bigl(h(z)^{n+1}-1 \bigr).$$
(27)

We know that $$T(r,H)\leq(n+1)T(r,h)+S(r,h)$$ from the expression of $$H(z)$$ and Lemma 2.1. Thus $$S(r,H)=S(r,h)$$. Next, we prove that $$h(z)\equiv c_{4}$$, where $$c_{4}^{n+1}=1$$, when $$\frac {g(z)}{g(qz+c)}$$ is transcendental with only finitely many zeros. Obviously, $$h(z)$$ is neither a constant except $$c_{4}$$ nor a rational function from (27) since $$\frac{g(z)}{g(qz+c)}$$ is transcendental. Thus we assume that $$h(z)$$ is a transcendental meromorphic function.

First, we affirm that $$H(z)-1$$ has infinitely many zeros. Otherwise, by the second main theorem of Nevanlinna theory and Lemma 2.1

\begin{aligned} T\bigl(r,H(z)\bigr) \leq& \overline{N}\bigl(r,H(z)\bigr)+ \overline{N} \biggl(r,\frac {1}{H(z)} \biggr) +\overline{N} \biggl(r, \frac{1}{H(z)-1} \biggr)+S(r,H) \\ \leq& 2T\bigl(r,h(z)\bigr)+2T\bigl(r,h(qz+c)\bigr)+S(r,h) \\ \leq& 4T\bigl(r,h(z)\bigr)+S(r,h). \end{aligned}
(28)

From the Valiron–Mohon’ko theorem, Lemma 2.1, and (28) we obtain

\begin{aligned} nT(r,h)&=T\bigl(r,h(z)^{n}\bigr)\leq T\bigl(r,H(z) \bigr)+T \biggl(r,\frac{1}{h(qz+c)} \biggr)+S(r,h) \\ &\leq5T\bigl(r,h(z)\bigr)+S(r,h), \end{aligned}

which contradicts to $$n\geq7$$. Thus $$H(z)-1$$ has infinitely many zeros.

Then we prove that $$H(z)\equiv1$$ and $$h(z)\equiv c_{4}$$ is a nonzero constant, that is, $$c_{4}^{n+1}=1$$. If $$H(z)\not\equiv1$$, then since $$H(z)-1$$ has infinitely many zeros, we can choose a point $$z_{0}$$ satisfying $$H(z_{0})=1$$, and $$z_{0}$$ is not the zero of $$\frac{g(z)}{g(qz+c)}$$. From (27) we have $$h(qz_{0}+c)=h(z_{0})$$. By Lemma 2.1

\begin{aligned} \overline{N} \biggl(r,\frac{1}{H(z)-1} \biggr) \leq \overline{N} \biggl(r,\frac{1}{h(qz+c)-h(z)} \biggr)\leq2T\bigl(r,h(z) \bigr)+S(r,h). \end{aligned}
(29)

Using the second main theorem of Nevanlinna theory, Lemma 2.1, and (29), we obtain

\begin{aligned} T\bigl(r,H(z)\bigr) \leq& \overline{N}\bigl(r,H(z)\bigr)+ \overline{N} \biggl(r,\frac{1}{H(z)} \biggr) +\overline{N} \biggl(r, \frac{1}{H(z)-1} \biggr)+S(r,h) \\ \leq& \overline{N}(r,h)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{h} \biggr)+\overline{N}_{0}(r) +\overline{N} \biggl(r,\frac{1}{H(z)-1} \biggr)+S(r,h) \\ \leq& 4T(r,h)+\overline{N}_{1}(r)+\overline{N}_{0}(r)+S(r,h), \end{aligned}
(30)

where $$\overline{N}_{0}(r)$$ denotes the counting function ignoring multiplicities of the common zeros of $$H(z)$$ and $$h(qz+c)$$, and $$\overline{N}_{1}(r)$$ denotes the counting function ignoring multiplicities of the common poles of $$H(z)$$ and $$h(qz+c)$$.

On the other hand, from $$H(z)=h(qz+c)h(z)^{n}$$ we have

\begin{aligned}& nm(r,h)=m\bigl(r,h^{n}\bigr)\leq m\bigl(r,H(z) \bigr)+m \biggl(r,\frac{1}{h(qz+c)} \biggr)+O(1), \end{aligned}
(31)
\begin{aligned}& nN(r,h)=N\bigl(r,h^{n}\bigr)\leq N\bigl(r,H(z) \bigr)+N \biggl(r,\frac{1}{h(qz+c)} \biggr)-\overline{N}_{1}(r)- \overline{N}_{0}(r). \end{aligned}
(32)

From (31), (32), and Lemma 2.1 we have

\begin{aligned}[b] nT(r,h)&\leq T\bigl(r,H(z)\bigr)+T \biggl(r,\frac{1}{h(qz+c)} \biggr)-\overline {N}_{1}(r)- \overline{N}_{0}(r)+O(1) \\ &\leq T\bigl(r,H(z)\bigr)+T(r,h)-\overline{N}_{1}(r)- \overline{N}_{0}(r)+S(r,h). \end{aligned}
(33)

Substituting (30) into (33), we obtain

\begin{aligned} nT(r,h)\leq5T(r,h)+S(r,h), \end{aligned}

which contradicts to $$n\geq7$$, so $$H(z)=h(qz+c)h(z)^{n}\equiv1$$, that is, $$h(z)^{n}=\frac{1}{h(qz+c)}$$. From Lemma 2.1 we have

\begin{aligned} nT(r,h)=T\bigl(r,h(qz+c)\bigr)\leq T(r,h)+S(r,h), \end{aligned}

which also contradicts to $$n\geq7$$, so $$h(z)$$ is a nonzero constant, say $$c_{4}$$, and from (27) we get $$c_{4}^{n+1}=1$$. Thus the theorem is proved. □

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### Acknowledgements

The authors thanks the anonymous referees for their valuable comments and suggestions, which improved the presentation of this manuscript.

### Availability of data and materials

Data sharing not applicable to this paper as no data sets were generated or analyzed during the current study.

## Funding

This work was supported by the NSFC (No. 11661052) and the outstanding youth scientist foundation plan of Jiangxi (No. 20171BCB23003).

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Correspondence to Kai Liu.

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