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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.

1 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 [9] 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 [12] obtained the following result.

Theorem A

([12])

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 [4] and Zhang and Korhonen [11] 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 [13] 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 [8] 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\(c_{4}^{n+1}=1\).

2 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

([7])

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. □

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

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

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Li, J., Liu, K. Sharing values of q-difference-differential polynomials. Adv Differ Equ 2020, 212 (2020). https://doi.org/10.1186/s13662-020-02668-z

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