Theory and Modern Applications

# Meromorphic functions that share a polynomial with their difference operators

## Abstract

In this paper, we prove the following result: Let f be a nonconstant meromorphic function of finite order, p be a nonconstant polynomial, and c be a nonzero constant. If f, $$\Delta _{c}f$$, and $$\Delta_{c}^{n}f$$ ($$n\ge 2$$) share ∞ and p CM, then $$f\equiv \Delta_{c}f$$. Our result provides a difference analogue of the result of Chang and Fang in 2004 (Complex Var. Theory Appl. 49(12):871–895, 2004).

## 1 Introduction and main results

In this paper, we use the base notations of the Nevanlinna theory of meromorphic functions which are defined as follows [9, 18, 19].

Let f be a meromorphic function. Throughout this paper, a meromorphic function always means meromorphic in the whole complex plane.

### Definition 1

$$m(r,f)=\frac{1}{2\pi } \int_{0}^{2\pi }\log^{+} \bigl\vert f \bigl(re^{i\theta } \bigr) \bigr\vert \,d \theta .$$

$$m(r,f)$$ is the average of the positive logarithm of $$\vert f(z) \vert$$ on the circle $$\vert z \vert =r$$.

### Definition 2

$$\begin{gathered} N(r,f) = \int_{0}^{r}\frac{n(t,f)-n(0,f)}{t}\,dt+n(0,f)\log r, \\ \overline{N}(r,f)= \int_{0}^{r} \frac{\overline{n}(t,f)-\overline{n}(0,f)}{t}\,dt+ \overline{n}(0,f) \log r, \end{gathered}$$

where $$n(t,f)$$ ($$\overline{n}(t,f)$$) denotes the number of poles of f in the disc $$\vert z \vert \le t$$, multiples poles are counted according to their multiplicities (ignore multiplicity). $$n(0,f)$$ ($$\overline{n}(0,f)$$) denotes the multiplicity of poles of f at the origin (ignore multiplicity).

$$N(r,f)$$ is called the counting function of poles of f, and $$\overline{N}(r,f)$$ is called the reduced counting function of poles of f.

### Definition 3

$$T(r,f)=m(r,f)+N(r,f).$$

$$T(r,f)$$ is called the characteristic function of f. It plays a cardinal role in the whole theory of meromorphic functions.

### Definition 4

Let f be a meromorphic function. The order of growth of f is defined as follows:

$$\rho (f)=\overline{\lim_{r\to \infty }}\frac{\log^{+} T(r,f)}{\log r}.$$

If $$\rho (f)< \infty$$, then we say that f is a meromorphic function of finite order.

### Definition 5

Let a, f be two meromorphic functions. If $$T(r,a)=S(r,f)$$, where $$S(r,f)=o(T(r,f))$$, as $$r\to \infty$$ outside of a possible exceptional set of finite logarithmic measure. Then we say that a is a small function of f. And we use $$S(f)$$ to denote the family of all small functions with respect to f.

### Definition 6

Let f and g be two meromorphic functions, and p be a polynomial. We say that f and g share p CM, provided that $$f(z)-p(z)$$ and $$g(z)-p(z)$$ have the same zeros counting multiplicity. And if f and g have the same poles counting multiplicity, then we say that f and g shareCM.

In this paper, we also use some known properties of the characteristic function $$T(r,f)$$ as follows [9, 18, 19].

### Property 1

Let $$f_{j}$$ ($$j=1,2,\ldots,q$$) be q meromorphic functions in $$\vert z \vert < R$$ and $$0< r< R$$. Then

$$T \Biggl(r,\prod_{j=1}^{q}T(r,f_{j}) \Biggr)\le \sum_{j=1}^{q}T(r,f_{j}), \quad\quad T \Biggl(r, \sum_{j=1}^{q}f_{j} \Biggr)\le \sum_{j=1}^{q}T(r,f_{j})+ \log q$$

hold for $$1\le r< R$$.

### Property 2

Suppose that f is meromorphic in $$\vert z \vert < R$$ ($$R\le \infty$$) and a is any complex number. Then, for $$0< r< R$$, we have

$$T \biggl(r,\frac{1}{f-a} \biggr)=T(r,f)+O(1).$$

Property 2 is the first fundamental theorem.

### Property 3

Suppose that f is a nonconstant meromorphic function and $$a_{1}, a _{2},\ldots, a_{n}$$ are $$n\ge 3$$ distinct values in the extended complex plane. Then

$$(n-2)T(r,f)< \sum_{j=1}^{n}\overline{N} \biggl(r,\frac{1}{f-a_{j}} \biggr)+S(r,f).$$

Property 3 is the second fundamental theorem. For more properties about $$T(r,f)$$, please see [9, 18, 19].

For a meromorphic function $$f(z)$$, we define its shift by $$f_{c}(z) = f(z + c)$$ and its difference operators by

$$\Delta_{c}f(z)=f(z+c)-f(z),\quad\quad \Delta_{c}^{n}f(z)= \Delta _{c}^{n-1} \bigl(\Delta_{c}f(z) \bigr). \bigl(\Delta_{c}f(z) \bigr).$$

In [10] the following result was proved.

### Theorem 1

Let f be a nonconstant meromorphic function, and a be a nonzero finite complex number. If f, $$f'$$, and $$f''$$ share a CM, then $$f\equiv f'$$.

In 2001, Li and Yang [12] considered the case when f, $$f'$$, and $$f^{(n)}$$ share one value.

### Theorem 2

Let f be an entire function, a be a finite nonzero constant, and $${n\ge 2}$$ be a positive integer. If f, $$f'$$, and $$f^{(n)}$$ share a CM, then f assumes the form

$$f(z)=be^{wz}-\frac{a(1-w)}{w},$$
(1.1)

where b, w are two nonzero constants satisfying $$w^{n-1}=1$$.

### Remark 1

It is easy to see that the functions in ( 1.1 ) really share value a, since when $$b\neq 0$$ and $$w^{n-1}=1$$, from $$f^{(j)}(z)=a$$, $$j=0,1,n$$, it follows that $$bwe^{wz}=a$$ for each $$j=0,1,n$$. So, the functions $$f^{(j)}-a$$, $$j=0, 1, n$$, have the same zeros counting multiplicity.

In 2004, Chang and Fang [1] considered the case when f, $$f'$$, and $$f^{(n)}$$ share a small function.

### Theorem 3

Let f be an entire function, a be a nonzero small function of f, and $$n\ge 2$$ be a positive integer. If f, $$f'$$, and $$f^{(n)}$$ share a CM, then $$f\equiv f'$$.

Recently, value distribution in difference analogue of meromorphic functions has become a subject of some interest, see, e.g., [28, 11].

In 2012 and 2014, Chen et al. [2, 3] considered difference analogue of Theorem 1 and Theorem 2, and established the following result.

### Theorem 4

Let f be a nonconstant entire function of finite order, and a (0) $$\in S(f)$$ be a periodic entire function with period c. If f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ ($$n\ge 2$$) share a CM, then $$\Delta_{c}f\equiv \Delta_{c}^{n}f$$.

For other related results, the reader is referred to the references due to Latreuch, El Farissi, Belaïdi [11], El Farissi, Latreuch, Asiri [5], El Farissi, Latreuch, Belaïdi and Asiri [6].

### Remark 2

There are examples in [3] which show that the conclusion $$\Delta_{c}f\equiv \Delta_{c}^{n}f$$ in Theorem 4 cannot be replaced by $$f\equiv \Delta_{c}f$$, and the condition $$a(z)\not \equiv 0$$ is necessary.

By Theorems 3 and 4, it is natural to ask: Can we provide a difference analogue of Theorem 3? Or, can we delete the condition that ‘$$a(z)$$ is a periodic entire function with period c’ in Theorem 4?

In this paper, we study the problem and prove the following result.

### Theorem 5

Let f be a nonconstant meromorphic function of finite order, and p be a nonconstant polynomial. If f, $$\Delta_{c}f$$, and $$\Delta_{c} ^{n}f$$ ($$n\ge 2$$) share p andCM, then $$f\equiv \Delta_{c}f$$.

If f is an entire function, then f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ have no poles, obviously f, $$\Delta_{c}f$$ and $$\Delta_{c}^{n}f$$ share ∞ CM. By Theorem 5, we consequently get the following result.

### Corollary 1

Let f be a nonconstant entire function of finite order, and $$n \ge 2$$ be a positive integer. If f, $$\Delta_{c}f$$, and $$\Delta_{c} ^{n}f$$ share z CM, then $$f\equiv \Delta_{c}f$$.

### Example 1

Let A, a, b, c be four finite nonzero complex numbers satisfying $$a\neq b$$, n (≥2) $$\in \mathbb{N}$$ satisfying $$[e^{Ac}-1]^{n-1}=1$$, $$e^{Ac}-1=\frac{a}{a-b}$$, and $$g(z)$$ be a periodic entire function with period c, and let $$f(z)=g(z)e^{Az}+b$$. By simple calculation, we obtain

$$\Delta_{c}^{n}f(z)=\Delta_{c}f(z)= \bigl[e^{Ac}-1 \bigr]f(z)+a \bigl[1-e^{Ac}+1 \bigr].$$

It is easy to see that f, $$\Delta_{c}f$$, and $$\Delta_{c} ^{n}f$$ ($$n\ge 2$$) share a CM, and $$f\neq \Delta_{c}f$$ when $$e^{Ac} \neq 2$$. This example shows that ‘$$p(z)$$ cannot be a constant’ in Theorem 5.

### Example 2

Let A, b, c be three nonzero finite complex numbers satisfying $$e^{Ac}=1$$, and $$f(z)=e^{Az}+b$$, $$p(z)=b$$. It is easy to see that f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ share $$p(z)$$ CM. But $$\Delta_{c}f\equiv 0 \not \equiv f$$. This example also shows that ‘$$p(z)$$ cannot be a constant’ in Theorem 5.

### Example 3

Let A, c be two nonzero finite complex numbers satisfying $$e^{Ac}=2$$ and $$f(z)=e^{Az}\cot (\frac{\pi z}{c})$$. By simple calculation, we obtain

$$f(z)=\Delta_{c}f=\Delta_{c}^{n}f=e^{Az} \cot \biggl(\frac{\pi z}{c} \biggr).$$

Obviously, for any polynomial p, f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ share p and ∞ CM. This example satisfies Theorem 5.

In Examples 1 and  2 , we have $$\Delta_{c}f\equiv tf+a(1-t)$$ and $$f(z)=e^{Az+B}+a$$, respectively, when f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ ($$n\ge 2$$) share a nonzero constant a CM. Hence we posed the following problem.

### Problem 1

Assume that f is a nonconstant entire function of finite order, a is a nonzero constant, and that f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ ($$n\ge 2$$) share a CM. Whether or not, one of the following two cases occurs:

1. (1)

$$\Delta_{c}f\equiv tf+a(1-t)$$, where t is a constant satisfying $$t^{n-1}=1$$,

2. (2)

$$f(z)=e^{Az+B}+a$$, where A (≠0), B are two constants satisfying $$e^{Ac}=1$$.

## 2 Some lemmas

### Lemma 1

([4, 7])

Let f be a meromorphic function of finite order, and c be a nonzero complex constant. Then

$$T \bigl(r,f(z+c) \bigr)=T(r,f)+S(r,f).$$

### Lemma 2

([7, 8])

Let $$c\in \mathbb{C}$$, k be a positive integer, and f be a meromorphic function of finite order. Then

$$m \biggl( r,\frac{\Delta_{c}^{k}f(z)}{f(z)} \biggr) =S(r,f).$$

### Lemma 3

([18, 19])

Let $$n\ge 2$$ be a positive integer. Suppose that $$f_{i}(z)$$ ($$i = 1,2,\ldots,n$$) are meromorphic functions and $$g_{i}(z)$$ ($$i = 1,2,\ldots,n$$) are entire functions satisfying

1. (i)

$$\sum_{i=1}^{n}f_{i}(z)e^{g_{i}(z)}\equiv 0$$,

2. (ii)

the orders of $$f_{i}$$ are less than those of $$e^{g_{k}-g_{l}}$$ for $$1\le i \le n$$, $$1\le k < l \le n$$.

Then $$f_{i}(z)\equiv 0$$ ($$i=1,2,\ldots,n$$).

The following lemma is well known.

### Lemma 4

Let the function f satisfy the following difference equation:

$$f(w+1)=\alpha (w)f(w)+\beta (w)$$

in the complex plane.

Then the following formula holds:

$$f(w+k)=f(w)\prod_{j=0}^{k-1} \alpha (w+j)+\sum_{l=0}^{k-1}\beta (w+l) \prod_{j=l+1}^{k-1}\alpha (w+j)$$
(2.1)

for every $$w\in \mathbb{C}$$ and $$k\in \mathbb{N}^{+}$$.

Formula (2.1) has many applications. For example, many solvable difference equations are essentially solved by using it (see [1517]), and by using such obtained formulas the behavior of their solution can be studied (see, for example, recent papers [13, 14]; see also many related references therein). As another simple application, by using a linear change of variables, the following corollary is obtained:

### Corollary 2

Let the function f satisfy the following difference equation:

$$f(w+c)=\alpha (w)f(w)+\beta (w)$$

in the complex plane.

Then the following formula holds:

$$f(w+kc)=f(w)\prod_{j=0}^{k-1} \alpha (w+jc)+\sum_{l=0}^{k-1}\beta (w+lc) \prod_{j=l+1}^{k-1}\alpha (w+jc)$$

for every $$w\in \mathbb{C}$$ and $$k\in \mathbb{N}^{+}$$.

From the ideas of Chang and Fang [1] and Chen and Li [3], we prove the following lemma.

### Lemma 5

Let f be a nonconstant meromorphic function of finite order, p (0) be a polynomial, and $$n\ge 2$$ be an integer. Suppose that

$$\frac{\Delta_{c}^{n}f(z)-p(z)}{f(z)-p(z)}=e^{\alpha (z)},\quad\quad \frac{ \Delta_{c}f(z)-p(z)}{f(z)-p(z)}=e^{\beta (z)},$$
(2.2)

where α and β are two polynomials, and that

$$T \bigl(r,e^{\alpha } \bigr)+T \bigl(r,e^{\beta } \bigr)=S(r,f).$$
(2.3)

Then $$\Delta_{c}f\equiv tf+b(1-t)$$, where t, b are constants satisfying $$t^{n-1}=1$$ and $$b\neq 0$$. Moreover, if $$t\neq 1$$, then $$p(z)\equiv b$$.

### Proof

Firstly, we prove that f cannot be a rational function. Otherwise, suppose that $$f(z)=P(z)/Q(z)$$, where $$P(z)$$ and $$Q(z)$$ are two co-prime polynomials. It follows from (2.2) that $$f(z)$$ and $$\Delta_{c}f(z)$$ share ∞ CM. We claim that $$Q(z)$$ is a constant. Otherwise, suppose that there exists $$z_{0}$$ such that $$Q(z_{0}+c)=0$$. Since $$f(z)$$ and $$\Delta_{c}f(z)$$ share ∞ CM, and

$$\Delta_{c}f(z)=\frac{P(z+c)}{Q(z+c)}-\frac{P(z)}{Q(z)}= \frac{P(z+c)Q(z)-P(z)Q(z+c)}{Q(z)Q(z+c)}.$$
(2.4)

We deduce that all zeros of $$Q(z+c)$$ must be the zeros of $$Q(z)$$. Otherwise, suppose that there exists $$z_{1}$$ such that $$Q(z_{1}+c)=0$$ but $$Q(z_{1})\neq 0$$, then it follows from (2.4) that $$z_{1}+c$$ is a pole of $$\Delta_{c}f$$ but not the pole of f, which contradicts with $$f(z)$$ and $$\Delta_{c}f(z)$$ share ∞ CM. Then we get

$$Q(z_{0}+c)=0\quad \Rightarrow \quad Q(z_{0})=0 \quad \Rightarrow\quad Q(z_{0}-c)=0\quad \Rightarrow \quad \cdots \quad \Rightarrow \quad Q(z_{0}-lc)=0.$$

This implies that $$Q(z)$$ has infinitely many zeros, which is a contradiction. Thus, the claim is proved.

Then f is a nonconstant polynomial, suppose that

$$f(z)=a_{k}z^{k}+a_{k-1}z^{k-1}+\cdots +a_{1}z+a_{0}, \quad\quad p(z)=b_{m}z ^{m}+\cdots +b_{1}z+b_{0}.$$

Then we get $$\Delta_{c}f(z)=f(z+c)-f(z)$$, and obviously, $$\deg \Delta_{c}^{n}f(z)\le \deg \Delta_{c}f(z) < \deg f(z)$$. Then it follows from (2.2) that $$\alpha (z)$$, $$\beta (z)$$ are constants, and we let $$e^{\alpha (z)}=a$$, $$e^{\beta (z)}=b$$. So we have

$$\Delta_{c}^{n}f(z)-p(z)=a \bigl(f(z)-p(z) \bigr), \quad \quad \Delta_{c}f(z)-p(z)=b \bigl(f(z)-p(z) \bigr).$$

Then we get $$\deg f=k\le \deg p=m$$ since $$\deg \Delta_{c}^{n}f(z) \le \deg \Delta_{c}f(z) < \deg f(z)$$. If $$k< m$$, then we get $$a=b=1$$. This implies $$f\equiv \Delta_{c}f(z) \equiv \Delta_{c}^{n}f$$, which contradicts with $$\deg \Delta_{c}f(z) < \deg f(z)$$. If $$k= m$$ then we get $$a=b=b_{k}/(a_{k}-b_{k})$$, $$\Delta_{c}f(z) \equiv \Delta_{c}^{n}f$$, and hence $$f(z)$$ is a constant, which is a contradiction.

Hence, f is a transcendental meromorphic function. Thus $$T(r,p)=S(r,f)$$.

Next, we consider two cases.

Case 1. $$\beta (z)$$ is a nonconstant polynomial. It follows from the second equation in (2.2) that $$\Delta_{c}f(z)=e^{\beta (z)}(f(z)-p(z))+p(z)$$, and that

$$f(z+c)=a_{1}(z)f(z)+b_{1}(z),$$

where $$a_{1}(z)=e^{\beta (z)}+1$$, $$b_{1}(z)=p(z)[1-e^{\beta (z)}]$$.

By Corollary  2 , it is easy to get, for any $$k\in \mathbb{N^{+}}$$,

$$f(z+kc)=a_{k}(z)f(z)+b_{k}(z),$$
(2.5)

where

$$a_{k}(z)=\prod_{j=0}^{k-1} \bigl(e^{\beta (z+jc)}+1 \bigr), \quad\quad b_{k}(z)=\sum _{l=0}^{k-1}b_{1}(z+lc)\prod _{j=l+1}^{k-1}a_{1}(z+jc).$$
(2.6)

It follows from (2.5) and (2.6) that

\begin{aligned}[b] \Delta_{c}^{n}f(z) &=\sum _{i=0}^{n}(-1)^{n-i}C_{n}^{i}f(z+ic) \\ &= \sum_{i=0}^{n}(-1)^{n-i}C_{n}^{i} \bigl[a_{i}(z)f(z)+b_{i}(z) \bigr] \\ &= \Biggl[ (-1)^{n}+ \sum_{i=1}^{n}(-1)^{n-i}C_{n}^{i} \prod_{j=0}^{i-1} \bigl(e^{\beta (z+jc)}+1 \bigr) \Biggr] f(z)+ \sum_{i=0}^{n}(-1)^{n-i}C_{n}^{i}b_{i}(z), \\ &=\mu_{n}(z)f(z)+\nu_{n}(z), \end{aligned}
(2.7)

where

$$\begin{gathered} \mu_{n}(z) =\prod_{j=0}^{n-1}e^{\beta (z+jc)}+ \sum_{t=0}^{n-1} \lambda_{n-1,t}\prod _{j=0,j\neq t}^{n-1}e^{\beta (z+jc)}+\cdots + \sum _{t=0}^{n-1}\lambda_{1,t}e^{\beta (z+tc)}, \\ \nu_{n}(z)=\sum_{i=0} ^{n}(-1)^{n-i}C_{n}^{i}b_{i}(z). \end{gathered}$$
(2.8)

In particular, $$\lambda_{1,t}=(-1)^{n-1-t}C_{n-1}^{t}$$, which implies

$$\sum_{t=0}^{n-1}\lambda_{1,t}e^{\beta (z+tc)}= \Delta_{c}^{n-1}e^{ \beta (z)}.$$
(2.9)

By ( 2.3 ), ( 2.8 ), and Lemma  1 , it is easy to get

$$T(r,\mu_{n})+T(r,\nu_{n})=S(r,f).$$
(2.10)

Since $$\beta (z)$$ is a nonconstant polynomial, we have that

$$\beta (z)=l_{m}z^{m}+l_{m-1}z^{m-1}+ \cdots +l_{0},$$

where $$l_{i}$$ ($$0\le i \le m$$) are constants satisfying $$l_{m}\neq 0$$ and $$m\ge 1$$. Obviously, for any $$j\in \{0, 1,\ldots, n-1\}$$, we have

$$\beta (z+jc)=l_{m}z^{m}+(l_{m-1}+ml_{m}jc)z^{m-1}+ \cdots +\sum_{t=0} ^{m}l_{t}(jc)^{t}.$$
(2.11)

From (2.8) and (2.11), we get

\begin{aligned}[b] \mu_{n}(z)&= e^{nl_{m}z^{m}+P_{n,0}(z)}+\lambda_{n-1,0}e^{(n-1)l_{m}z ^{m}+P_{n-1,0}(z)}+ \cdots +\lambda_{n-1,n-1}e^{(n-1)l_{m}z^{m}+P_{n-1,n-1}(z)} \\ &\quad {} +\cdots +\lambda_{1,0}e^{l_{m}z^{m}+P_{1,0}(z)}+\cdots + \lambda_{1,n-1}e^{l_{m}z^{m}+P_{1,n-1}(z)}, \end{aligned}
(2.12)

where $$P_{i,j}(z)$$ are polynomials with degree less than m for $$i\in \{1, 2,\ldots,n\}$$, $$j\in \{0,1,\ldots, {C_{n}^{i}-1}\}$$.

It follows from the first equation in (2.2) that

$$\Delta_{c}^{n}f(z)-e^{\alpha (z)}f(z)=p(z) \bigl(1-e^{\alpha (z)} \bigr).$$
(2.13)

From (2.7) and (2.13), we have

$$\bigl(\mu_{n}(z)-e^{\alpha (z)} \bigr)f(z)=p(z) \bigl(1-e^{\alpha (z)} \bigr)-\nu_{n}(z).$$
(2.14)

From ( 2.3 ), ( 2.10 ) and since $$T(r,p)=S(r,f)$$, we have that

$$T(r,p)+T \bigl(r,e^{\alpha } \bigr)+T \bigl(r,e^{\beta } \bigr)+T(r,\mu_{n})+T(r,\nu_{n})=S(r,f).$$
(2.15)

If $$\mu_{n}(z)-e^{\alpha (z)}\not \equiv 0$$, by ( 2.14 ), ( 2.15 ), Property  1 , and Property  2 , we obtain

$$T(r,f)=T \biggl(r,\frac{p(z)(1-e^{\alpha (z)})-\nu_{n}(z)}{\mu_{n}(z)-e^{ \alpha (z)}} \biggr)=S(r,f),$$

Hence $$\mu_{n}(z)-e^{\alpha (z)}\equiv 0$$. Combining this with (2.12), we get

\begin{aligned}[b] &e^{P_{n,0}(z)}e^{nl_{m}z^{m}}+ \bigl(\lambda_{n-1,0}e^{P_{n-1,0}(z)}+ \cdots +\lambda_{n-1,n-1}e^{P_{n-1,n-1}(z)} \bigr)e^{(n-1)l_{m}z^{m}} \\ &\quad{} +\cdots + \bigl( \lambda_{1,0}e^{P_{1,0}(z)}+\cdots + \lambda_{1,n-1}e^{P_{1,n-1}(z)} \bigr)e ^{l_{m}z^{m}}-e^{\alpha (z)} \equiv 0. \end{aligned}
(2.16)

Next, we consider three subcases.

Case 1.1. $$\deg \alpha (z)> m$$. Then, for any $$1\le i \le n$$, $$1\le k< j \le n$$, we have

$$\rho \bigl(e^{\alpha (z)-il_{m}z^{m}} \bigr)=\rho \bigl(e^{\alpha (z)} \bigr)=\deg \alpha (z)> m, \quad\quad \rho \bigl(e^{jl_{m}z^{m}-kl_{m}z^{m}} \bigr)=m.$$

Since $$P_{i,j}(z)$$ are polynomials with degree less than m for $$i\in \{1, 2,\ldots,n\}$$, $$j\in \{0,1,\ldots, C_{n}^{i}-1\}$$, then for $$i=1,2,\ldots,n-1$$,

$$\rho \Biggl( \sum_{j=0}^{C_{n}^{i}-1} \lambda_{i,j}e^{P_{i,j}(z)} \Biggr) \le m-1, \quad\quad \rho \bigl( e^{P_{n,0}(z)} \bigr) \le m-1.$$

By ( 2.16 ) and using Lemma 3, we obtain $$e^{P_{n,0}}\equiv 0$$, which is a contradiction.

Case 1.2. $$\deg \alpha (z)< m$$. Then, for any $$1\le i \le n$$, $$1\le k< j \le n$$, we have

$$\rho \bigl(e^{\alpha (z)-il_{m}z^{m}} \bigr)=\rho \bigl(e^{-il_{m}z^{m}} \bigr)= m, \quad \quad \rho \bigl(e^{jl_{m}z^{m}-kl_{m}z^{m}} \bigr)=m.$$

Since $$P_{i,j}(z)$$ are polynomials with degree less than m for $$i\in \{1, 2,\ldots,n\}$$, $$j\in \{0,1,\ldots, C_{n}^{i}-1\}$$, then for $$i=1,2,\ldots,n-1$$,

$$\rho \Biggl( \sum_{j=0}^{C_{n}^{i}-1} \lambda_{i,j}e^{P_{i,j}(z)} \Biggr) \le m-1, \quad\quad \rho \bigl( e^{P_{n,0}(z)} \bigr) \le m-1.$$

By ( 2.16 ) and using Lemma 3, we obtain $$e^{P_{n,0}}\equiv 0$$, which is a contradiction.

Case 1.3. $$\deg \alpha (z)= m$$. Set $$\alpha (z)=dz^{m}+\alpha^{*}(z)$$, where $$d\neq 0$$ and $$\deg \alpha^{*}(z)< m$$. Rewrite (2.16) as

\begin{aligned}[b] &e^{P_{n,0}(z)}e^{nl_{m}z^{m}}+ \bigl(\lambda_{n-1,0}e^{P_{n-1,0}(z)}+ \cdots +\lambda_{n-1,n-1}e^{P_{n-1,n-1}(z)} \bigr)e^{(n-1)l_{m}z^{m}} \\ &\quad{} +\cdots + \bigl( \lambda_{1,0}e^{P_{1,0}(z)}+\cdots + \lambda_{1,n-1}e^{P_{1,n-1}(z)} \bigr)e ^{l_{m}z^{m}}-e^{\alpha^{*}(z)}e^{dz^{m}} \equiv 0. \end{aligned}
(2.17)

If $$d\neq jl_{m}$$, for any $$j=1,2,\ldots,n$$, we have

$$\rho \bigl(e^{dz^{m}-jl_{m}z^{m}} \bigr)=\rho \bigl(e^{(d-jl_{m})z^{m}} \bigr)=m,\quad \quad \rho \bigl(e^{\alpha^{*}(z)} \bigr)< m.$$

Combining this with (2.17), by using Lemma 3, we get a contradiction.

If $$d= jl_{m}$$, for some $$j=1,2,\ldots,n-1$$, without loss of generality, we assume that $$j=1$$, then (2.17) can be rewritten as

\begin{aligned} &e^{P_{n,0}(z)}e^{nl_{m}z^{m}}+ \bigl(\lambda_{n-1,0}e^{P_{n-1,0}(z)}+ \cdots +\lambda_{n-1,n-1}e^{P_{n-1,n-1}(z)} \bigr)e^{(n-1)l_{m}z^{m}} \\ &\quad{} +\cdots + \bigl( \lambda_{1,0}e^{P_{1,0}(z)}+\cdots + \lambda_{1,n-1}e^{P_{1,n-1}(z)}-e ^{\alpha^{*}(z)} \bigr)e^{l_{m}z^{m}} \equiv 0. \end{aligned}

And then, by using the same argument as above, we get a contradiction.

Hence, $$d= nl_{m}$$. Rewrite (2.17) as

\begin{aligned} &\bigl(e^{P_{n,0}(z)}-e^{\alpha^{*}(z)} \bigr)e^{nl_{m}z^{m}}+ \bigl(\lambda_{n-1,0}e ^{P_{n-1,0}(z)}+\cdots +\lambda_{n-1,n-1}e^{P_{n-1,n-1}(z)} \bigr)e^{(n-1)l _{m}z^{m}} \\ &\quad{} +\cdots + \bigl(\lambda_{1,0}e^{P_{1,0}(z)}+\cdots +\lambda _{1,n-1}e^{P_{1,n-1}(z)} \bigr)e^{l_{m}z^{m}}\equiv 0. \end{aligned}

Using the same argument as in Case 1.1 and using Lemma 3, we obtain

$$\lambda_{1,0}e^{P_{1,0}(z)}+\cdots +\lambda_{1,n-1}e^{P_{1,n-1}(z)} \equiv 0.$$
(2.18)

Then, by (2.9) and (2.18), we get

$$\sum_{t=0}^{n-1}\lambda_{1,t}e^{\beta (z+tc)}= \Delta_{c}^{n-1}e^{ \beta (z)}=\sum _{t=0}^{n-1}(-1)^{n-1-t}C_{n-1}^{t}e^{\beta (z+tc)} \equiv 0.$$
(2.19)

If $$m\ge 2$$, then for any $$t=0,1,\ldots,n-1$$, we have

$$\beta (z+tc)=l_{m}z^{m}+(l_{m-1}+ml_{m}tc)z^{m-1}+q_{t}(z),$$
(2.20)

where $$q_{t}(z)$$ are polynomials with $$\deg q_{t}(z)< m-1$$.

From (2.19) with (2.20), we get

\begin{aligned} &e^{q_{n-1}(z)}e^{l_{m}z^{m}+(l_{m-1}+ml_{m}(n-1)c)z^{m-1}}-(n-1)e^{q _{n-2}(z)}e^{l_{m}z^{m}+(l_{m-1}+ml_{m}(n-2)c)z^{m-1}} \\ &\quad{}+\cdots +(-1)^{n-1}e ^{q_{0}(z)}e^{l_{m}z^{m}+(l_{m-1}+ml_{m}c)z^{m-1}}\equiv 0. \end{aligned}

Using the same argument as Case 1.1, we obtain a contradiction.

Hence $$m=1$$. Thus $$\beta (z)=l_{1}z+l_{0}$$, where $$l_{1}\neq 0$$. Then, for any $$n\ge 1$$, we deduce that

$$\Delta_{c}^{n-1}e^{\beta (z)}= \bigl(e^{l_{1}c}-1 \bigr)^{n-1}e^{\beta (z)}.$$

Hence, it follows from (2.19) that $$(e^{l_{1}c}-1)^{n-1}\equiv 0$$, which yields $$e^{l_{1}c}=1$$. Then, for any $$t\in \mathbb{N^{+}}$$, we have

$$e^{\beta (z+tc)}=e^{l_{1}z+tl_{1}c+l_{0}}=e^{l_{1}z+l_{0}} \bigl(e^{l_{1}c} \bigr)^{t}=e ^{\beta (z)}.$$
(2.21)

By the second equation in (2.2) and (2.21), we get

\begin{aligned}& \Delta_{c}f(z) =e^{\beta (z)}f(z)+p(z) \bigl(1-e^{\beta (z)} \bigr)=e^{\beta (z)}f(z)+ \bigl(1-e ^{\beta (z)} \bigr)b_{1}(z), \\& \begin{aligned} \Delta_{c}^{2}f(z)&=e^{\beta (z)}\Delta_{c}f(z)+ \Delta_{c}p(z) \bigl(1-e^{\beta (z)} \bigr) \\ &=e^{\beta (z)} \bigl[e^{\beta (z)}f(z)+p(z) \bigl(1-e ^{\beta (z)} \bigr) \bigr]+\Delta_{c}p(z) \bigl(1-e^{\beta (z)} \bigr) \\ &=e^{2\beta (z)}f(z)+ \bigl(1-e ^{\beta (z)} \bigr) \bigl[p(z)e^{\beta (z)}+ \Delta_{c}p(z) \bigr] \\ &=e^{2\beta (z)}f(z)+ \bigl(1-e ^{\beta (z)} \bigr)b_{2}(z). \end{aligned} \end{aligned}

By mathematical induction, it is easy to get, for any integer $$t\ge 2$$,

$$\Delta_{c}^{t}f(z)=e^{t\beta (z)}f(z)+ \bigl(1-e^{\beta (z)} \bigr)b_{t}(z),$$

where $$b_{1}(z)=p(z)$$, $$b_{t}(z)=p(z)e^{(t-1)\beta (z)}+\Delta_{c}b _{t-1}=\sum_{i=0}^{t-1}e^{(t-1-i)\beta (z)}\Delta_{c}^{i}p(z)$$.

Hence,

$$\Delta_{c}^{n}f(z)=e^{n\beta (z)}f(z)+ \bigl(1-e^{\beta (z)} \bigr)b_{n}(z),$$
(2.22)

where $$b_{n}(z)=\sum_{i=0}^{n-1}e^{(n-1-i)\beta (z)}\Delta_{c}^{i}p(z)$$.

From the first equation in (2.2) and (2.22), we have

$$\bigl(e^{\alpha (z)}-e^{n\beta (z)} \bigr)f(z)= \bigl(1-e^{\beta (z)} \bigr)b_{n}(z)-p(z) \bigl(1-e ^{\alpha (z)} \bigr).$$
(2.23)

If $$e^{\alpha (z)}-e^{n\beta (z)}\not \equiv 0$$, then by ( 2.15 ), ( 2.23 ), Property  1 , and Property  2 , we have

$$T(r,f)=T \biggl(r,\frac{(1-e^{\beta (z)})b_{n}(z)-p(z)(1-e^{\alpha (z)})}{e ^{\alpha (z)}-e^{n\beta (z)}} \biggr)=S(r,f),$$

Hence $$e^{\alpha (z)}-e^{n\beta (z)}\equiv 0$$. It follows from (2.22) and (2.23) that

\begin{aligned} \bigl(1-e^{\beta (z)} \bigr)b_{n}(z) &= \bigl(1-e^{\beta (z)} \bigr) \Biggl( \sum_{i=0}^{n-1}e ^{(n-1-i)\beta (z)}\Delta_{c}^{i}p(z) \Biggr) \\ &=\sum_{i=0}^{n-1}e ^{(n-1-i)\beta (z)} \Delta_{c}^{i}p(z)-\sum_{i=0}^{n-1}e^{(n-i)\beta (z)} \Delta_{c}^{i}p(z) \\ &=-p(z)e^{n\beta (z)}+\sum_{i=0}^{n-2}e^{(n-1-i) \beta (z)} \bigl( \Delta_{c}^{i}p(z)-\Delta_{c}^{i+1}p(z) \bigr) +\Delta _{c}^{n-1}p(z) \\ &\equiv p(z) \bigl(1-e^{\alpha (z)} \bigr). \end{aligned}

That is,

$$\sum_{i=0}^{n-2}e^{(n-1-i)\beta (z)} \bigl( \Delta_{c}^{i}p(z)-\Delta_{c} ^{i+1}p(z) \bigr)+\Delta_{c}^{n-1}p(z)-p(z)\equiv 0.$$
(2.24)

If $$p(z)$$ is a constant, as $$\Delta_{c}^{i}p(z)=0$$ for any $$i\in \mathbb{N^{+}}$$. It follows from (2.24) that

$$p(z)e^{(n-1)\beta (z)}-p(z)\equiv 0.$$

Hence, $$e^{(n-1)\beta (z)}\equiv 1$$, which is a contradiction.

If $$p(z)$$ is a nonconstant polynomial, then $$p(z)-\Delta_{c}^{i}p(z) \not \equiv 0$$ for any $$i\in \mathbb{N^{+}}$$. It follows from (2.24) that

$$e^{(n-1)\beta (z)} \bigl(p(z)-\Delta_{c}p(z) \bigr)\equiv -\sum _{i=1}^{n-2}e^{(n-1-i) \beta (z)} \bigl( \Delta_{c}^{i}p(z)- \Delta_{c}^{i+1}p(z) \bigr)-\Delta_{c}^{n-1}p(z)+p(z).$$

Thus we have

\begin{aligned} (n-1)T \bigl(r,e^{\beta } \bigr) &=T \bigl(r,e^{(n-1)\beta (z)} \bigl(p(z)-\Delta_{c}p(z) \bigr) \bigr)+S \bigl(r,e ^{\beta } \bigr) \\ &=T \Biggl(r,-\sum_{i=1}^{n-2}e^{(n-1-i)\beta (z)} \bigl(\Delta_{c} ^{i}p(z)-\Delta_{c}^{i+1}p(z) \bigr)-\Delta_{c}^{n-1}p(z)+p(z) \Biggr) \\ &\quad{} +S \bigl(r,e^{ \beta } \bigr) \\ &\le (n-2)T \bigl(r,e^{\beta } \bigr)+S \bigl(r,e^{\beta } \bigr), \end{aligned}

Case 2. $$\beta (z)=\beta \in \mathbb{C}$$ is a constant. By the second equation in (2.2), we get

\begin{aligned}& \Delta_{c}f(z) =e^{\beta }f(z)+p(z) \bigl(1-e^{\beta } \bigr), \\& \Delta_{c}^{2}f(z)=e ^{\beta }\Delta_{c}f(z)+ \Delta_{c}p(z) \bigl(1-e^{\beta } \bigr)=e^{\beta }\Delta _{c}f(z)+ \bigl(1-e^{\beta } \bigr)b_{2}(z). \end{aligned}

By mathematical induction, it is easy to get, for any integer $$t\ge 2$$,

$$\Delta_{c}^{t}f(z)=e^{(t-1)\beta }\Delta_{c}f(z)+ \bigl(1-e^{\beta } \bigr)b_{t}(z),$$

where $$b_{2}(z)=\Delta_{c}p(z)$$, $$b_{t}(z)=\Delta_{c}p(z)e^{(t-1) \beta }+\Delta_{c}b_{t-1}=\sum_{i=1}^{t-1}\Delta_{c}^{i}p(z)e^{(t-1-i) \beta }$$.

Hence,

$$\Delta_{c}^{n}f(z)=e^{(n-1)\beta }\Delta_{c}f(z)+ \bigl(1-e^{\beta } \bigr)b_{n}(z),$$
(2.25)

where $$b_{n}(z)=\sum_{i=1}^{n-1}\Delta_{c}^{i}p(z)e^{(n-1-i)\beta }$$.

Using the same argument as the above, it is easy to get $$e^{\alpha }=e ^{n\beta }$$. Then it follows from (2.2) and $$e^{\alpha }=e^{n\beta }$$ that

$$\Delta_{c}^{n}f(z)=e^{(n-1)\beta }\Delta_{c}f(z)+ \bigl(1-e^{(n-1)\beta } \bigr)p(z).$$
(2.26)

If $$\Delta_{c}f(z) \not \equiv \Delta_{c}^{n}f(z)$$, it follows from (2.26) that $$e^{(n-1)\beta }\neq 1$$. Combining (2.25) and (2.26), we have

$$\bigl(1-e^{\beta } \bigr)\sum_{i=1}^{n-1} \Delta_{c}^{i}p(z)e^{(n-1-i)\beta }= \bigl(1-e ^{\beta } \bigr)b_{n}(z)= \bigl(1-e^{(n-1)\beta } \bigr)p(z).$$
(2.27)

If $$p(z)$$ is a constant, then the left-hand side of equation (2.27) is equal to 0, and hence $$p(z)\equiv 0$$, which is a contradiction.

If $$p(z)$$ is a nonconstant polynomial, let $$d=\deg p(z)\ge 1$$, then the left-hand side of equation (2.27) is a polynomial with degree less than d, but the right-hand side of the equation is a polynomial with degree d, which is a contradiction.

Hence $$\Delta_{c}f(z)\equiv \Delta_{c}^{n}f(z)$$, and $$e^{(n-1)\beta }= 1$$.

If $$e^{\beta }\neq 1$$ and $$p(z)$$ is a nonconstant polynomial, then it follows from (2.25)–(2.26) that $$b_{n}(z)\equiv 0$$. Thus

$$\sum_{i=1}^{n-1}\Delta_{c}^{i}p(z)e^{(n-1-i)\beta } \equiv 0.$$
(2.28)

Let $$p(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots +a_{0}$$. It follows that $$\deg \Delta_{c}^{i}p(z)=m-i$$ if $$m\ge i$$. If $$m\ge 2$$, then the left-hand side of (2.28) is a polynomial with degree $$m-1\ge 1$$, which is a contradiction.

Hence $$m=1$$, that is, $$p(z)=a_{1}z+a_{0}$$. Thus $$\Delta_{c}p(z)=a_{1}c \neq 0$$. It follows from (2.28) that $$a_{1}ce^{(n-2)\beta }=0$$, which is a contradiction.

From the above discussion, we obtain that if $$e^{\beta }\neq 1$$, then $$p(z)$$ (≡b) is a nonzero constant, hence

$$\Delta_{c}f(z)=e^{\beta }f(z)+p(z) \bigl(1-e^{\beta } \bigr)=e^{\beta }f(z)+b \bigl(1-e^{\beta } \bigr)=tf(z)+b(1-t),$$

where $$t=e^{\beta }$$ satisfying $$t^{n-1}=1$$.

Thus, Lemma 5 is proved. □

### Lemma 6

Let f be an entire function of finite order $$\rho (f)$$ with zeros $$\{z_{1}, z_{2},\ldots \}\subset \mathbb{C}\backslash \{0\}$$ and a k-fold zero at the origin. Then

$$f(z)=z^{k}\alpha (z)e^{\beta (z)},$$

where α is the canonical product of f formed with the non-null zeros of f, and β is a polynomial of degree $$\le \rho (f)$$.

## 3 Proof of Theorem 5

### Proof

Since the order of f is finite, and f, $$\Delta_{c}f$$, $$\Delta_{c}^{n}f$$ share ∞ and $$p(z)$$ CM, obviously $$(\Delta _{c}^{n}f(z)-p(z))/(f(z)-p(z))$$ and $$(\Delta_{c}f(z)-p(z))/(f(z)-p(z))$$ have no zeros and poles. By Lemmas 1 and  6 , we have

$$\frac{\Delta_{c}^{n}f(z)-p(z)}{f(z)-p(z)}=e^{\alpha (z)}, \quad\quad \frac{ \Delta_{c}f(z)-p(z)}{f(z)-p(z)}=e^{\beta (z)},$$
(3.1)

where $$\alpha (z)$$ and $$\beta (z)$$ are two polynomials with degree $$\le \rho (f)$$.

Using the same discussion as in Lemma 5, we deduce that f cannot be a rational function. Hence, f is a transcendental meromorphic function, and $$T(r,p)=S(r,f)$$.

Set $$F(z):=f(z)-p(z)$$, then $$T(r,f)=T(r,F)+S(r,f)$$ and $$T(r,p)=S(r,F)$$.

Obviously, we have

$$f(z)=F(z)+p(z),\quad\quad \Delta_{c}f(z)=\Delta_{c}F(z)+ \Delta_{c}p(z), \quad\quad \Delta_{c}^{n}f(z)= \Delta_{c}^{n}F(z)+\Delta_{c}^{n}p(z).$$

Rewrite (3.1) as

$$\frac{\Delta_{c}^{n}F(z)+\Delta_{c}^{n}p(z)-p(z)}{F(z)}=e^{\alpha (z)}, \quad\quad \frac{\Delta_{c}F(z)+\Delta_{c}p(z)-p(z)}{F(z)}=e^{\beta (z)}.$$
(3.2)

Since $$p(z)$$ is a nonconstant polynomial, it follows that $$\Delta_{c}^{n}p(z)-p(z)\not \equiv 0$$ and $$\Delta_{c}p(z)-p(z) \not \equiv 0$$. Set

$$\phi (z):=\frac{(p(z)-\Delta_{c}^{n}p(z))\Delta_{c}F(z)-(p(z)-\Delta _{c}p(z))\Delta_{c}^{n}F(z)}{F(z)}.$$
(3.3)

Next, we consider two cases.

Case 1. $$\phi (z)\not \equiv 0$$. Then, by $$T(r,p)=S(r,F)$$, Lemma 1, and Lemma 2, we get

$$m(r,\phi )=S(r,F).$$
(3.4)

By (3.2)–(3.3), we can rewrite $$\phi (z)$$ as

$$\phi (z)= \bigl(p(z)-\Delta_{c}^{n}p(z) \bigr)e^{\beta (z)}- \bigl(p(z)-\Delta_{c}p(z) \bigr)e ^{\alpha (z)}.$$
(3.5)

Since $$p(z)$$ is a polynomial, we deduce that $$N(r,\phi )=S(r,F)$$. Hence, we get

$$T(r,\phi )=m(r,\phi )+N(r,\phi )=S(r,F).$$
(3.6)

Since $$\phi (z)\not \equiv 0$$, by (3.5) we have

$$\bigl(p(z)-\Delta_{c}^{n}p(z) \bigr)\frac{e^{\beta (z)}}{\phi (z)}=1+ \bigl(p(z)-\Delta _{c}p(z) \bigr)\frac{e^{\alpha (z)}}{\phi (z)}.$$
(3.7)

Then by (3.6), (3.7), $$T(r,p)=S(r,F)$$, Property 2, and Property 3, we have

\begin{aligned} T \biggl( r, \bigl(p-\Delta_{c}^{n}p \bigr) \frac{e^{\beta }}{\phi } \biggr) &\le \overline{N} \biggl( r, \bigl(p- \Delta_{c}^{n}p \bigr)\frac{e^{\beta }}{\phi } \biggr) + \overline{N} \biggl( r,\frac{\phi }{(p-\Delta_{c}^{n}p)e^{\beta }} \biggr) \\ & \quad {} + \overline{N} \biggl( r, \frac{1}{(p-\Delta_{c}^{n}p)(e^{\beta }/\phi )-1} \biggr) +S \biggl( r, \bigl(p- \Delta_{c}^{n}p \bigr)\frac{e^{\beta }}{\phi } \biggr) \\ &= \overline{N} \biggl( r, \bigl(p-\Delta_{c}^{n}p \bigr) \frac{e^{\beta }}{\phi } \biggr) + \overline{N} \biggl( r,\frac{\phi }{(p-\Delta_{c}^{n}p)e^{\beta }} \biggr) \\ & \quad {} + \overline{N} \biggl( r,\frac{\phi }{(p-\Delta_{c}p)e^{\alpha }} \biggr) +S \biggl( r, \bigl(p- \Delta_{c}^{n}p \bigr)\frac{e^{\beta }}{\phi } \biggr) \\ &\le S(r,F)+S \biggl( r, \bigl(p-\Delta_{c}^{n}p \bigr) \frac{e^{\beta }}{\phi } \biggr) . \end{aligned}

Hence by ( 3.6 ), Property  1 , and the previous inequality, we get

$$T \bigl(r,e^{\beta } \bigr)=S(r,F).$$
(3.8)

Thus by (3.5)–(3.8), we have

$$T \bigl(r,e^{\alpha } \bigr)=T \biggl( r,\frac{(p-\Delta_{c}^{n}p)e^{\beta }-\phi }{p- \Delta_{c}p} \biggr) =S(r,F).$$
(3.9)

Hence, by Lemma 5 and since $$p(z)$$ is a nonconstant polynomial, we obtain $$f\equiv \Delta_{c}{f}$$.

Case 2. $$\phi (z)\equiv 0$$. That is,

$$\bigl(p(z)-\Delta_{c}^{n}p(z) \bigr)\Delta_{c}F(z)= \bigl(p(z)-\Delta_{c}p(z) \bigr)\Delta _{c}^{n}F(z).$$
(3.10)

By simple calculation, we can rewrite (3.10) as follows:

$$\bigl(p(z)-\Delta_{c}^{n}p(z) \bigr) \bigl( \Delta_{c}f(z)-p(z) \bigr)= \bigl(p(z)-\Delta_{c}p(z) \bigr) \bigl( \Delta_{c}^{n}f(z)-p(z) \bigr).$$
(3.11)

From (3.1) and (3.11), we get

$$\frac{\Delta_{c}^{n}f(z)-p(z)}{\Delta_{c}f(z)-p(z)}=e^{\alpha (z)- \beta (z)}=\frac{p(z)-\Delta_{c}^{n}p(z)}{p(z)-\Delta_{c}p(z)}.$$
(3.12)

Since $$p(z)$$ is a polynomial, it follows from (3.12) that $$e^{\alpha (z)-\beta (z)}$$ is a constant. Suppose that $$e^{\alpha (z)- \beta (z)}=A$$, then we get $$p(z)-\Delta_{c}^{n}p(z)=A(p(z)-\Delta_{c}p(z))$$. It follows that $$A=1$$ and $$p(z)$$ is a constant, which is a contradiction.

This completes the proof of Theorem  5 . □

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## Acknowledgements

Research supported by the NNSF of China (Grant No. 11371149; 11701188) and the Graduate Student Overseas Study Program from South China Agricultural University (Grant No. 2017LHPY003).

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Correspondence to Mingliang Fang.

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Deng, B., Liu, D., Gu, Y. et al. Meromorphic functions that share a polynomial with their difference operators. Adv Differ Equ 2018, 194 (2018). https://doi.org/10.1186/s13662-018-1645-4