1 Introduction

Throughout the paper, we assume that the reader is familiar with the standard notations and fundamental results of Nevanlinna theory of meromorphic functions (see [15, 22, 32]). In what follows, a meromorphic function always means meromorphic in the whole complex plane \( \mathbb {C} \), and c always means a nonzero complex constant. For a meromorphic function f, we define its shift by \( f(z+c) \) and its difference operators by

$$\begin{aligned} \Delta _cf(z)=f(z+c)-f(z),\; \Delta ^n_cf(z)=\Delta ^{n-1}_c\left( \Delta _cf(z)\right) ,\; n\in \mathbb {N},\; n\ge 2. \end{aligned}$$

Two non-constant meromorphic functions f and g share the value \( a\in \mathbb {C}\cup \{\infty \} \), if \( f^{-1}(a)=g^{-1}(a)\). We say that f and g share the value a CM (Counting Multiplicities) if in addition to the sharing of values if \( f(z_0)=a \) with multiplicity p implies \( g(z_0)=a \) with multiplicity p. If we do not consider the multiplicities, then f and g are said to share the value a IM (Ignoring Multiplicities). When \( a=\infty \), the zeros of \( f-a \) means the poles of f.

The uniqueness theory of entire and meromorphic functions in the purview of sharing values (using value distribution theory of Nevanlinna [29]) has grown up to an extensive subfield of the value distribution theory. Interested readers are referred to the articles [9,10,11, 13, 21, 23, 28, 30, 31] and references therein. After the development of the difference analogue lemma of logarithmic derivatives, by Halburd and Korhonen [14] in 2006, and Chiang and Feng [6], in 2008, independently, the research findings dealing with the sharing value problems between the shifts \( f(z+c) \) or with the difference operators \( \Delta _cf \) of meromorphic functions f, gets a new dimension in the literature of meromorphic functions. The reader is referred the articles [16, 17] in this regard. In 2013, Jiang and Chen [20] studied two shared value problems for meromorphic functions and proved that if \( \Delta _cf \) and f share a, b CM, then \( f(z+c)=2f(z) \).

Recently, Barki et al. [5] have established result of for higher order difference operator \( \Delta ^n_cf \) considering two shared values.

Theorem 1.1

[5] Let f be a non-constant meromorphic of finite order such that \( N(r,f)=S(r,f) \), let c be a constant such that \( \Delta ^n_cf\not \equiv 0 \) and let a, b be two non-zero distinct finite complex constants. If \( \Delta ^n_cf \) and f share a, b CM, then \( \Delta ^n_cf=f(z) \).

However, in [5], Barki et al. have obtained the following corollary generalizing the result of Jiang and Chen [20]..

Corollary 1.2

[5] Let f be a non-constant meromorphic of finite order such that \( N(r,f)=S(r,f) \), let c be a constant such that \( \Delta ^n_cf\not \equiv 0 \) and let a, b be two non-zero distinct finite complex constants. If \( \Delta ^n_cf \) and f share a, b CM, then \( f(z+nc)=2^nf(z) \).

We define the linear difference polynomial of a meromorphic function f as follows (see [3]):

$$\begin{aligned} L^n_c(f)=a_nf(z+nc)+\cdots +a_1f(z+c)+a_0f(z), \end{aligned}$$

where \( a_n(\ne 0), \ldots , a_1, a_0\in \mathbb {C} \). It is easy to see that if \( a_j=\left( {\begin{array}{c}n\\ j\end{array}}\right) (-1)^{n-j} \), then \( L^n_c(f)=\Delta ^n_cf \). Clearly, \( L^n_c(f) \) is a general setting of the difference operator \( \Delta ^n_cf \).

In a number of articles, the higher order difference operators \( \Delta ^n_cf \) have been studied extensively (for example, see [4, 7, 8, 18, 19, 24,25,26,27]) in view of sharing values with entire or meromorphic functions f but less investigations done till date with the linear difference polynomials \( L^n_c(f) \). In fact, what could be the precise form of a meromorphic function f, when sharing values with its linear difference polynomials \( L^n_c(f) \), is completely unknown. In 2019, succeeding in particular with \( n=1 \) in order to find the specific form of the function, Ahamed [1] investigated value sharing problem (CM sharing) between \( L^1_c(f) \) and f and obtained \( L^1_c(f)\equiv f \), and showed that f assumes the following form

$$\begin{aligned} f(z)= {\left\{ \begin{array}{ll} \left( \displaystyle \frac{1-a_0}{a_1}\right) ^{{z}/{c}}\pi _1(z),\;\;\text {when} &{} a_0+a_1\ne 1 \\ \;\;\;\pi _2(z), \;\;\text {when}\;\; a_0+a_1= 1, \end{array}\right. } \end{aligned}$$

for \( \pi _1, \;\pi _2\in {\mathcal {M}}_c \).

Remark 1.3

In order to obtain the relation \( L^1_c(f)\equiv f \) with the specific form of the function, by the following two examples, it was shown in [1] that the nature of 2 CM sharing of values cannot be replaced by \( 1\;CM+1\;IM \) or \( 2\;IM \).

Example 1.4

[1] Let \( f(z)={\left( e^z+e^{-z}\right) }/{2} \), and for \( c\ne k\pi i \), \( k\in \mathbb {Z} \), let

$$\begin{aligned} {L}^1_c(f)=\frac{2e^c}{e^{2c}-1}f(z+c)-\frac{2}{e^{2c}-1}f(z). \end{aligned}$$

A simple computation shows that \( {L}^1_c(f)=e^z \). It is easy to see that f and \( {L}^1_c(f) \) share the values 1 and \( -1\) IM but f neither in the specific form nor satisfies \( {L}^1_c(f)\equiv f. \)

Example 1.5

[1] Let \( f(z)=1+\left( e^z-1\right) ^2 \) and \( c\in \mathbb {C} \) be such that \( e^c=-1 \), let

$$\begin{aligned} {L}^1_c(f)=\frac{1}{4}f(z+c)-\frac{1}{4}f(z). \end{aligned}$$

We see that \( {L}^1_c(f)=e^z \). We verify that f and \( {L}^1_c(f) \) share the values 2 CM and 1 IM, but note that f has neither the specific form nor satisfies \( {L}^1_c(f)\equiv f. \)

It was shown in [1] that similar results can be obtained for meromorphic functions also.

Theorem 1.6

[1] Let f be transcendental meromorphic function of finite order, and a, b be two distinct constants. If \( L^1_c(f)(\not \equiv (d_1e^{\alpha }+d_2)/(d_3e^{\beta }+d_4)) \), where \( d_j\in \mathbb {C} \) and \( \alpha , \beta \) are polynomials in z, and f share a, b and \( \infty \) CM, then \( L^1_c(f)\equiv f \). Furthermore, f must be of the following form

$$\begin{aligned} f(z)= {\left\{ \begin{array}{ll} \left( \displaystyle \frac{1-a_0}{a_1}\right) ^{z/c}\pi _1(z),\quad \text{ when }\; a_1+a_0\ne 1\\ \pi _2(z),\quad \text{ when }\; a_1+a_0=1 , \end{array}\right. } \end{aligned}$$

where \( \pi _1, \pi _2\in {\mathcal {M}}_c \).

To establish an improved version of Theorem 1.6, it is natural to raise the following question.

Question 1.7

Can we prove a result replacing “\( L^1_c(f)(\not \equiv (d_1e^{\alpha }+d_2)/(d_3e^{\beta }+d_4)) \) and f share a, b and \( \infty \) CM" in Theorem 1.6 by “\( L^1_c(f)(\not \equiv 0) \) and f share a and b CM"?

In case of investigating the general meromorphic solutions to the difference equation \( L^n_c(f)\equiv f \), recently, Ahamed [2] posed the following conjecture.

Conjecture 1

[2] Let f be a non-constant meromorphic function such that \( L^n_c(f)\equiv f \), then f assumes the form

$$\begin{aligned} f(z)=\alpha _n^{z/c}g_n(z)+\cdots +\alpha _1^{z/c}g_1(z), \end{aligned}$$

where \( g_j\; (j=1, 2, \ldots , n) \) are periodic meromorphic functions of period c.

In [3], Banerjee and Ahamed, also posed the following question with the similar query.

Question 1.8

Let \( L_n(f,\Delta )=L^n_c(f)-\left( \sum _{j=0}^{n}a_j\right) f(z) \) and t be a non-zero constant. What would be the general meromorphic solution of the difference equation \( L_n(f,\Delta )\equiv tf \)?

In this paper, our aim is two fold. In view of Remark 1.3, in one hand, instead of considering \( 1\;CM+1\;IM \) or \( 2\;IM \) sharing of values, however, we shall be interested to deal with the \( 2\;CM \) sharing problems between meromorphic functions f and their linear difference polynomials \( L^n_c(f) \). On the other hand, in order to give the positive answers of Conjecture 1 and Question 1.8, our aim is to find the general solutions to the difference equation which is obtained as a conclusion of our main result. In fact, we obtain a corollary of the main result to give a complete answer of Question 1.7.

2 Main results

We define the class \( {\mathcal {M}}_c \) by

$$\begin{aligned} {\mathcal {M}}_c:=\{f : f\; \text{ is } \text{ meromorphic } \text{ in }\; \mathbb {C}\;\text{ and }\; f(z+c)=f(z)\}. \end{aligned}$$

It can be shown that \( {\mathcal {M}}_c \) is a field of meromorphic functions of period c. We state the main result of this paper.

Theorem 2.1

Let f be a non-constant meromorphic of finite order such that \( N(r,f)=S(r,f) \), let c be a constant such that \( L^n_c(f)\not \equiv 0 \) with \( \sum _{j=0}^{n}a_j=0 \) and let a, b be two non-zero distinct finite complex constants. If \( L^n_c(f) \) and f share a, b CM, then \( L^n_c(f)=f \). Furthermore,

  1. (i)

    \( f(z)=\displaystyle \rho _1^{z/c}\pi _1(z)+\rho _2^{z/c}\pi _2(z)+\cdots +\rho _n^{z/c}\pi _n(z), \) where \( \pi _j(z)\in {\mathcal {M}}_c\; (j=1, 2, \ldots , n) \) and \( \rho _j \) \( (j=1, 2, \ldots , n) \) are distinct roots of the equation

    $$\begin{aligned} a_nw^n+\cdots +a_1w+a_0-1=0. \end{aligned}$$

    In particular, if \( \rho _j\in \{0, 1\} \) be such that atleast one of \( \rho _j \)’s are non-zero, then \( f\in {\mathcal {M}}_c \).

  2. (ii)

    \(\qquad \quad \qquad \quad \qquad \quad \qquad \quad \displaystyle f(z)=\left( \sum _{m_1=1}^{N_1-1}z^{m_1}\pi _{m_1}(z)\right) \sigma _1^{z/c}+\cdots +\left( \sum _{m_q=1}^{N_q-1}z^{m_q}\pi _{m_q}(z)\right) \sigma _q^{z/c}\),

    where \( \pi _j\in {\mathcal {M}}_c \) and \( \sigma _1, \sigma _2 \), \( \ldots \), \( \sigma _q \) are multiple roots, of respective multiplicities \( N_1, N_2, \ldots , N_q \), of the equation

    $$\begin{aligned} a_nw^n+\cdots +a_1w+a_0-1=0. \end{aligned}$$

Remark 2.2

The following observations are clear.

  1. (i)

    Clearly, part-(i) of Theorem 2.1 answered the conjecture of Ahamed [2] completely.

  2. (ii)

    In particular, if \( \sum _{j=0}^{n}a_j=0 \), then it is easy to see that part-(i) of Theorem 2.1 is the complete answer of Question 1.8.

  3. (iii)

    In view of Example 1.5 and the following example, in Theorem 2.1, we see that \( 2\;CM \) sharing cannot be replaced by \( 1\;CM+1\;IM \).

Example 2.3

Let \( f(z)=1+(e^z-1)^2 \) and c be so chosen that \( e^c\ne \pm 1 \) and

$$\begin{aligned} {\left\{ \begin{array}{ll} a_0=\displaystyle \frac{e^{2c}}{\left( e^{2c}-1\right) -\left( e^{2c}+1\right) \left( e^{c}-1\right) }\\ a_1=\displaystyle \frac{1-e^{4c}}{\left( e^{2c}-1\right) \left( \left( e^{2c}-1\right) -\left( e^{2c}+1\right) \left( e^{c}-1\right) \right) }\\ a_2=\displaystyle \frac{1}{\left( e^{2c}-1\right) \left( \left( e^{2c}-1\right) -\left( e^{2c}+1\right) \left( e^{c}-1\right) \right) }. \end{array}\right. } \end{aligned}$$

A simple computation shows that \( L^2_c(f)=e^z \) with \( \sum _{j=0}^{2}a_j=0 \) and \( N(r, f)=S(r,f) \). It is easy to verify that \( L^2_c(f) \) and f share the values \( 1\; IM \) and \( 2\; CM \) but conclusion of Theorem 2.1 fails to hold.

We obtain the following corollary of Theorem 2.1 which gives a more compact form of Theorem 1.1.

Corollary 2.4

Let f be a non-constant meromorphic of finite order such that \( N(r,f)=S(r,f) \), let c be a constant such that \( \Delta ^n_cf\not \equiv 0 \) and let a, b be two non-zero distinct finite complex constants. If \( \Delta ^n_cf \) and f share a, b CM, then \( \Delta ^n_cf\equiv f \). Furthermore,

$$\begin{aligned} f(z)=\sum _{k=0}^{n}\lambda ^{z/c}_k\pi _k(z),\; \text{ where }\; \pi _k\in {\mathcal {M}}_c \end{aligned}$$
(2.1)

and \( \lambda _k=1+e^{2ki\pi /n} \), \( k=0, 1, \ldots , n-1 \).

Remark 2.5

It is easy to verify that function in (2.1) satisfies \( f(z+nc)=2^nf(z) \) of Corollary 1.2.

We obtain the following corollary of Theorem 2.1 in case of when \( n=1 \).

Corollary 2.6

Let f be transcendental meromorphic function of finite order, and a, b be two distinct constants. If \( L^1_c(f)(\not \equiv 0) \) and f share a and b CM, then \( L^1_c(f)\equiv f \). Furthermore,

$$\begin{aligned} f(z)= {\left\{ \begin{array}{ll} \left( \displaystyle \frac{1-a_0}{a_1}\right) ^{z/c}\pi _1(z),\quad \text{ when }\; a_1+a_0\ne 1\\ \pi _2(z),\quad \text{ when }\; a_1+a_0=1 , \end{array}\right. } \end{aligned}$$

where \( \pi _1, \pi _2\in {\mathcal {M}}_c \).

Remark 2.7

Clearly, Corollary 2.6 is the complete answer of Question 1.7.

3 Key lemmas

In this section, we present some lemmas which will play key roles to prove the main results of this paper.

Lemma 3.1

[32] Let f be a non-constant meromorphic function, \( a_j\; (j=1, 2, \ldots , q) \) be q distinct complex numbers. Then

$$\begin{aligned} m\left( r,\sum _{j=1}^{q}\frac{1}{f-a_j}\right) =\sum _{j=1}^{n}m\left( r,\frac{1}{f-a_j}\right) +O(1). \end{aligned}$$
(3.1)

Lemma 3.2

[6, 14] Let f be a meromorphic function of finite order and c be a non-zero constant. Then

$$\begin{aligned} m\left( r,\frac{f(z+c)}{f(z)}\right) +m\left( r,\frac{f(z)}{f(z+c)}\right) =S(r,f). \end{aligned}$$
(3.2)

Lemma 3.3

[5] Let f be a non-constant meromorphic function in \( \mathbb {C} \). Let \( a_j\; (j=1, 2, \ldots , q) \) be q-distinct complex numbers. Then

$$\begin{aligned} \sum _{j=1}^{q}m\left( r,\frac{1}{f-a_j}\right) \le m\left( r,\frac{1}{f^{(k)}}\right) +S(r,f). \end{aligned}$$
(3.3)

To prove our main result, we prove the following lemma for linear difference polynomial \( L^n_c(f) \).

Lemma 3.4

Let f be a meromorphic function and \( L^n_c(f) \) be its linear difference polynomial such that \( \sum _{j=0}^{n}a_j=0 \). Then

$$\begin{aligned} \sum _{j=1}^{q}m\left( r,\frac{1}{f-a_j}\right) \le m\left( r,\frac{1}{L^n_c(f)}\right) +S(r,f). \end{aligned}$$
(3.4)

Proof

In view of Lemmas 3.1 and 3.2, a simple computation shows that

$$\begin{aligned} \sum _{j=1}^{q}m\left( r,\frac{1}{f-a_j}\right)&=m\left( r,\frac{L^n_c(f)}{L^n_c(f)(f-a_j)}\right) +O(1)\\&\le m\left( r,\frac{1}{\L ^n_c(f)}\right) +m\left( r,\sum _{j=1}^{q}\frac{\L ^n_c(f)}{f-a_j}\right) +O(1)\\&\le m\left( r,\frac{1}{\L ^n_c(f)}\right) +\sum _{j=1}^{q}m\left( r,\frac{\L ^n_c(f)}{f-a_j}\right) +O(1)\\&=m\left( r,\sum _{k=0}^{n}a_k\left( \frac{f(z+kc)-a_j}{f(z)-a_j}\right) +\frac{a_j}{f-a_j}\sum _{k=0}^{n}a_k\right) \\&\quad +m\left( r,\frac{1}{\L ^n_c(f)}\right) \\&\le m\left( r,\frac{1}{\L ^n_c(f)}\right) +\sum _{k=0}^{n}m\left( r,\frac{f(z+kc)-a_j}{f(z)-a_j}\right) +S(r,f)\\&\le m\left( r,\frac{1}{\L ^n_c(f)}\right) +S(r,f). \end{aligned}$$

This completes the proof. \(\square \)

Lemma 3.5

[16] Let f be a meromorphic function with order \( \sigma (f)=\sigma <\infty \), and c be a fixed non-zero complex number, then for each \( \epsilon >0 \),

$$\begin{aligned} T(r, f(z+c))=T(r,f)+O\left( r^{\sigma -1+\epsilon }\right) +S(r.f). \end{aligned}$$

4 Proof of the main results

This section is devoted to the detailed discussion on our proof of the main results.

Proof of Theorem 2.1

Since \( L^n_c(f) \) and f share the values a, b CM and \( N(r,f)=S(r,f) \), then by the Second Fundamental Theorem of Nevanlinna, we obtain

$$\begin{aligned} T(r,f)&\le N(r,f)+N\left( r,\frac{1}{f-a}\right) +N\left( r,\frac{1}{f-b}\right) +S(r,f)\nonumber \\&=N\left( r,\frac{1}{L^n_c(f)-a}\right) +N\left( r,\frac{1}{L^n_c(f)-b}\right) +S(r,f)\nonumber \\&\le N\left( r,\frac{1}{L^n_c(f)-f}\right) +S(r,f) \end{aligned}$$
(4.1)

In view of the First Fundamental Theorem of Nevanlinna, a simple computation shows that

$$\begin{aligned} N\left( r,\frac{1}{L^n_c(f)-f}\right) +S(r,f)&=N\left( r,\frac{f}{L^n_c(f)-f}\right) +S(r,f)\nonumber \\&= N\left( r,\frac{1}{\frac{L^n_c(f)}{f}-1}\right) +S(r,f)\le T\left( r,\frac{1}{\frac{L^n_c(f)}{f}-1}\right) +S(r,f)\nonumber \\&\le T\left( r,\frac{L^n_c(f)}{f}\right) +S(r,f)\nonumber \\&\le m\left( r,\frac{L^n_c(f)}{f}\right) +N\left( r,\frac{L^n_c(f)}{f}\right) +S(r,f)\nonumber \\&\le N\left( r,\frac{L^n_c(f)}{f}\right) +S(r,f)\le N\left( r,\frac{1}{f}\right) +N\left( r,L^n_c(f)\right) +S(r,f)\nonumber \\&\le N\left( r,\frac{1}{f}\right) +2^n N(r,f)+S(r,f)= N\left( r,\frac{1}{f}\right) +S(r,f)\nonumber \\&\le T\left( r,\frac{1}{f}\right) +S(r,f)=T(r,f)+S(r,f). \end{aligned}$$
(4.2)

Therefore, in view of (4.1) and (4.2), we obtain

$$\begin{aligned} T(r,f)&=N\left( r,\frac{1}{f}\right) +S(r,f)\nonumber \\&=N\left( r,\frac{1}{f-a}\right) +N\left( r,\frac{1}{f-b}\right) +S(r,f). \end{aligned}$$
(4.3)

We define

$$\begin{aligned} \Psi (f):=\frac{\left( L^n_c(f)\right) ^{\prime }}{L^n_c(f)-a}-\frac{f^{\prime }}{f-a}. \end{aligned}$$
(4.4)

By the logarithmic derivative lemma, it is easy to see that

$$\begin{aligned} m(r,\Psi (f))&\le m\left( r,\frac{\left( L^n_c(f)\right) ^{\prime }}{L^n_c(f)-a}\right) +m\left( r,\frac{f^{\prime }}{f-a}\right) +O(1)\\ {}&=S(r,L^n_c(f))+S(r,f). \end{aligned}$$

On the other hand, a simple computation shows that

$$\begin{aligned} T\left( r,L^n_c(f)\right)&=m\left( r,L^n_c(f)\right) +N\left( r,L^n_c(f)\right) +S(r,f)\nonumber \\&\le m\left( r,\frac{L^n_c(f)}{f}\right) +m(r,f)+\sum _{j=0}^{n}N(r,f)+S(r,f)\nonumber \\&\le m(r,f)+2^n N(r,f)+S(r,f)\nonumber \\&=m(r,f)+S(r,f)\le T(r,f)+S(r,f). \end{aligned}$$
(4.5)

In view of Lemma 3.5, it can be shown that \( S\left( r,L^n_c(f)\right) =S(r,f) \), thus it follows that

$$\begin{aligned} m(r,\Psi )=S(r,f). \end{aligned}$$
(4.6)

Since \( \Psi \) is the logarithmic derivative of \( (L^n_c(f)-a)/(f-a) \), the poles of \( \Psi \) derive from the poles and zeros of \( (L^n_c(f)-a)/(f-a) \). Again, \( L^n_c(f) \) and f share a CM, then \( (L^n_c(f)-a)/(f-a) \) has no zeros and has atmost N(r, f) poles. Therefore, it follows that

$$\begin{aligned} N(r,\Psi )=N(r,f)=S(r,f). \end{aligned}$$
(4.7)

Hence, in view of (4.6) and (4.7), it is easy to see that

$$\begin{aligned} T(r,\Psi )=m(r,\Psi )+N(r,\Psi )+O(1)=S(r,f). \end{aligned}$$

On the other hand, we see that

$$\begin{aligned} \frac{\Psi }{f-a}=\frac{\left( L^n_c(f)\right) ^{\prime }}{L^n_c(f)-a}\frac{L^n_c(f)}{f-b}-\frac{f^{\prime }}{(f-a)(f-b)}. \end{aligned}$$
(4.8)

We suppose that \( \Psi \not \equiv 0 \). Then, from (4.8), we obtain

$$\begin{aligned} m\left( r,\frac{1}{f-b}\right)&\le m\left( r,\frac{1}{\Psi }\right) +m\left( r,\frac{\Psi }{f-b}\right) \\&\le m\left( r,\frac{1}{\Psi }\right) +m\left( r,\frac{\left( L^n_c(f)\right) ^{\prime }}{L^n_c(f)-a}\right) +m\left( r,\frac{L^n_c(f)}{f-b}\right) \\&\quad +m\left( r,\frac{f^{\prime }}{(f-a)(f-b)}\right) +S(r,f)\\&\le m\left( r,\frac{1}{\Psi }\right) +m\left( r,\frac{\left( L^n_c(f)\right) ^{\prime }}{a}\left( \frac{1}{L^n_c(f)-a}-\frac{1}{L^n_c(f)}\right) \right) \\&\quad +m\left( r,\frac{L^n_c(f)}{f-b}\right) +m\left( r,\frac{f^{\prime }}{(a-b)}\left( \frac{1}{f-a}-\frac{1}{f-b}\right) \right) +S(r,f)\\&\le T(r,\Psi )+S(r,L^n_c(f))+S(r,f)=S(r,f) \end{aligned}$$

which shows that

$$\begin{aligned} m\left( r,\frac{1}{f-b}\right) =S(r,f). \end{aligned}$$

By the First Fundamental Theorem of Nevanlinna, we obtain

$$\begin{aligned} T(r,f)=N\left( r,\frac{1}{f-b}\right) +S(r,f). \end{aligned}$$
(4.9)

Therefore, it follows from (4.3) and (4.9) that

$$\begin{aligned} N\left( r,\frac{1}{f-a}\right) =S(r,f). \end{aligned}$$

By the assumption, since \( L^n_c(f) \) and f share the value a CM, then it is easy to see that

$$\begin{aligned} N\left( r,\frac{1}{L^n_c(f)-a}\right) =N\left( r,\frac{1}{f-a}\right) =S(r,f). \end{aligned}$$

Again, since \( L^n_c(f) \) and f share the value b CM, then by using (4.5) and (4.9), by First Fundamental Theorem of Nevanlinna, we obtain

$$\begin{aligned} m\left( r,\frac{1}{L^n_c(f)-b}\right)&+N\left( r,\frac{1}{L^n_c(f)-b}\right) =T(r,L^n_c(f))+S(r,f)\\&\le T(r,f)+S(r,f)\le N\left( r,\frac{1}{f-b}\right) +S(r,f)\\&=N\left( r,\frac{1}{L^n_c(f)-b}\right) +S(r,f) \end{aligned}$$

which in turn shows that

$$\begin{aligned} m\left( r,\frac{1}{L^n_c(f)-b}\right) =S(r,f). \end{aligned}$$

In view of Lemma 3.3, we obtain

$$\begin{aligned}&m\left( r,\frac{1}{L^n_c(f)}\right) +m\left( r,\frac{1}{L^n_c(f)-a}\right) +m\left( rm\frac{1}{L^n_c(f)-b}\right) \nonumber \\&\qquad \le m\left( r,\frac{1}{\left( L^n_c(f)\right) ^{\prime }}\right) +S(r,f). \end{aligned}$$
(4.10)
$$\begin{aligned}&m\left( r,\frac{1}{f-a}\right) +m\left( r,\frac{1}{f-b}\right) \le m\left( r,\frac{1}{L^n_c(f)}\right) +S(r,f). \end{aligned}$$
(4.11)

By using (4.3) and the previous inequalities for counting functions, it is easy to obtain

$$\begin{aligned}&N\left( r,\frac{1}{L^n_c(f)-a}\right) +N\left( r,\frac{1}{L^n_c(f)-b}\right) \le N\left( r,\frac{1}{L^n_c(f)-b}\right) +S(r,f). \end{aligned}$$
(4.12)
$$\begin{aligned}&N\left( r,\frac{1}{f-a}\right) +N\left( r,\frac{1}{f-b}\right) \le T(r,f)+S(r,f). \end{aligned}$$
(4.13)

Therefore, in view of (4.10) to (4.13), a simple computation shows that

$$\begin{aligned}&T\left( r,\frac{1}{L^n_c(f)-a}\right) +T\left( r,\frac{1}{L^n_c(f)-b}\right) +T\left( r,\frac{1}{f-a}\right) +T\left( r,\frac{1}{f-b}\right) \nonumber \\&\quad =m\left( r,\frac{1}{L^n_c(f)-a}\right) +m\left( r,\frac{1}{L^n_c(f)-b}\right) +N\left( r,\frac{1}{L^n_c(f)-a}\right) \nonumber \\&\quad \quad +N\left( r,\frac{1}{L^n_c(f)-b}\right) +m\left( r,\frac{1}{f-a}\right) +m\left( r,\frac{1}{f-b}\right) +N\left( r,\frac{1}{f-a}\right) \nonumber \\&\quad \quad +N\left( r,\frac{1}{f-b}\right) \nonumber \\&\quad \le m\left( r,\frac{1}{L^n_c(f)}\right) +m\left( r,\frac{1}{L^n_c(f)-a}\right) \nonumber \\&\quad \quad +m\left( r,\frac{1}{L^n_c(f)-b}\right) +N\left( r,\frac{1}{L^n_c(f)-b}\right) +T(r,f)+S(r,f)\nonumber \\&\quad \le m\left( r,\frac{1}{\left( L^n_c(f)\right) ^{\prime }}\right) +T\left( r,\frac{1}{L^n_c(f)-b}\right) +T(r,f)+S(r,f). \end{aligned}$$
(4.14)

Since \( N(r,f)=S(r,f) \), it is easy to see that

$$\begin{aligned} T\left( r,\frac{1}{L^n_c(f)-b}\right)&=m(r,\left( L^n_c(f)\right) ^{\prime }) +N\left( r,\left( L^n_c(f)\right) ^{\prime }\right) +O(1)\nonumber \\&=m\left( r,L^n_c(f)\right) +m\left( r,\frac{\left( L^n_c(f)\right) ^{\prime }}{L^n_c(f)}\right) +N\left( r,L^n_c(f)\right) \nonumber \\ {}&\quad +\overline{N}\left( r,L^n_c(f)\right) +S(r,f)\nonumber \\&\le m\left( r,L^n_c(f)\right) +S(r,f)\le T(r,L^n_c(f))+S(r,f). \end{aligned}$$
(4.15)

Thus, from (4.5), (4.14) and (4.15), by First Fundamental Theorem of Nevanlinna, we obtain

$$\begin{aligned} 2T(r,L^n_c(f))+2T(r,f)&=T\left( r,\frac{1}{L^n_c(f)-a}\right) +T\left( r,\frac{1}{L^n_c(f)-b}\right) \\&\quad +T\left( r,\frac{1}{f-a}\right) +T\left( r,\frac{1}{f-b}\right) \\&\le T\left( r,L^n_c(f)\right) +T\left( r,\frac{1}{L^n_c(f)-b}\right) +T(r,f)+S(r,f)\\&\le 2T(r, L^n_c(f))+T(r,f)+S(r,f) \end{aligned}$$

which shows that \( T(r,f)\le S(r,f) \), a contradiction. Therefore, we must have \( \Psi \equiv 0 \), that is

$$\begin{aligned} \frac{\left( L^n_c(f)\right) ^{\prime }}{L^n_c(f)-a}=\frac{f^{\prime }}{f-a}. \end{aligned}$$
(4.16)

By integrating (4.16), we obtain

$$\begin{aligned} \frac{L^n_c(f)-a}{f-a}=B_1, \end{aligned}$$
(4.17)

where \( B_1 \) is a non-zero constant. On the other hand, since \( L^n_c(f) \) and f share the value b CM, proceeding in a similar way, it can be shown that

$$\begin{aligned} \frac{L^n_c(f)-b}{f-b}=B_2, \end{aligned}$$
(4.18)

where \( B_2 \) is a non-zero constant.

We now discuss the following two cases:

Case 1. If \( B_1=1 \) or \( B_2=1 \), then from (4.17) and (4.18), we easily obtain \( L^n_c(f)\equiv f \). This can be written as

$$\begin{aligned} a_nf(z+nc)+\cdots +a_1f(z+c)+(a_0-1)f(z)=0, \end{aligned}$$
(4.19)

where \( a_n\ne 0 \) and \( a_0\ne 1 \). This equation, being a linear difference equation with constant coefficients, is always explicitly solvable. To show this, we prove here a formula similar to the Leibnitz rule. Henceforth, we put \( f(z)=w(z)x(z) \). Then by the principle of Mathematical Induction, it can be shown that

$$\begin{aligned} f(z+mc)=w(z+mc)\sum _{k=0}^{m}\left( {\begin{array}{c}m\\ k\end{array}}\right) \Delta ^kx(z),\; (m=1. 2, \ldots ) \end{aligned}$$
(4.20)

In order to find the explicit solution of (4.19), we define

$$\begin{aligned}b_k= {\left\{ \begin{array}{ll} a_k,\; \text{ if }\; k\ne 0\\ a_0-1,\; \text{ if }\; k= 0. \end{array}\right. } \end{aligned}$$

Then (4.19) can be written as

$$\begin{aligned} b_nf(z+nc)+\cdots +b_1f(z+c)+b_0f(z)=0, \end{aligned}$$
(4.21)

where \( b_n\ne 0 \) and \( b_0\ne 0 \). We put \( f(z)=\rho ^{z/c}x(z) \). Substituting this into (4.21), and taking account of (4.20), we obtain

$$\begin{aligned} \sum _{m=0}^{n}b_mf(z+mc)&=\sum _{m=0}^{n}b_m\left( \rho ^{(z+m)/c}\sum _{k=0}^{m}\left( {\begin{array}{c}m\\ k\end{array}}\right) \Delta ^kx(z)\right) \\&=\sum _{k=0}^{n}\Delta ^kx(z)\left( \sum _{m=k}^{n}\left( {\begin{array}{c}m\\ k\end{array}}\right) b_m\rho ^{(z+m)/c}\right) \\&=\sum _{k=0}^{n}\frac{\rho ^{(z+k)/c}}{k!}\Delta ^kx(z)\left( \sum _{m=k}^{n}\prod _{j=0}^{k-1}(m-j) b_m\rho ^{m-k}/c\right) \\&=\sum _{k=0}^{n}\frac{\rho ^{(z+k)/c}}{k!}\Delta ^kx(z)f^{(k)}(\rho )=0, \end{aligned}$$

where \( f^{(k)}(\rho ) \) denotes the k-th derivative

$$\begin{aligned} f(\rho )=b_n\rho ^n+b_{n-1}\rho ^{n-1}+\cdots +b_1\rho +b_0. \end{aligned}$$
(4.22)

We call the equation \( f(\rho )=0 \), the characteristic equation. Let \( \rho \) be a root of the equation (4.22) and take \( x(z)=1 \). Then it is easy to see that \( f(z)=\rho ^{z/c} \) becomes a solution of (4.21).

Case A. Let the characteristic equation (4.22) has n distinct roots, say \( \rho _j\; (j=1, 2, \ldots , n) \). Then we have the following n particular solutions

$$\begin{aligned} \rho _1^{z/c}, \rho _2^{z/c}, \ldots , \rho _n^{z/c}, \end{aligned}$$

of (4.21) which form a fundamental set of solutions of (4.21). One can compute the Casoratian with respect to them as

$$\begin{aligned} C\left( \rho _1, \rho _2, \ldots , \rho _n\right) =\left( \rho _1\rho _2\cdots \rho _n\right) ^{z/c}V\left( \rho _1, \rho _2, \ldots , \rho _n\right) , \end{aligned}$$

where \( V\left( \rho _1, \rho _2, \ldots , \rho _n\right) \) is the Vandermonde determinant

$$\begin{aligned} V\left( \rho _1, \rho _2, \ldots , \rho _n\right) =\prod _{1\le k<j\le n}\left( \rho _j-\rho _k\right) \ne 0. \end{aligned}$$

Therefore, the general solution of (4.21) can be expressed as

$$\begin{aligned} f(z)=\rho _1^{z/c}\pi _1(z)+\rho _2^{z/c}\pi _2(z)+\cdots +\rho _n^{z/c}\pi _n(z), \end{aligned}$$

where \( \pi _j(z)\; (j=1, 2, \ldots , n) \) are periodic functions with period c.

Case B. Let the characteristic equation (4.22) has multiple roots, say \( \sigma _1\), \(\sigma _2\), \(\ldots \), \(\sigma _q \) with respective multiplicities \( N_1, N_2, \ldots , N_q \). Then, as solutions corresponding to \( \sigma _j\; (j=1, 2, \ldots , q) \), we obtain \( N_j \) linearly independent solution

$$\begin{aligned} \sigma _j^{z/c}, z\sigma _j^{z/c}, \ldots , z^{N_j-1}\sigma _j^{z/c}. \end{aligned}$$

As a matter of course, we must have \( N_1+N_2+\ldots +N_q=n \). Then we have q-sets of solutions of the form

$$\begin{aligned} S_j=\bigg \{\sigma _j^{z/c}, z\sigma _j^{z/c}, \ldots , z^{N_j-1}\sigma _j^{z/c}\bigg \},\; (j=1, 2, \ldots , q) \end{aligned}$$
(4.23)

and a routine computation shows that (4.23) forms a fundamental set of solutions of (4.21). Thus the general solution is expressed by a linear combination of them with c-periodic functions as its coefficients. Hence, the general solution is

$$\begin{aligned} f(z)=\left( \sum _{m_1=1}^{N_1-1}z^{m_1}\pi _{m_1}(z)\right) \sigma _1^{z/c} +\cdots +\left( \sum _{m_q=1}^{N_q-1}z^{m_q}\pi _{m_q}(z)\right) \sigma _q^{z/c}, \end{aligned}$$

where \( \pi _j \)’s are periodic functions with period c.

Case 2. If \( B_1\ne 1 \) and \( B_2\ne 1 \), then it is easy to see from (4.17) and (4.18) that

$$\begin{aligned} \left( L^n_c(f)-a\right) -\left( L^n_c(f)-b\right) =(B_1f-B_1a)-(B_2f-B_2b) \end{aligned}$$

which can be written as

$$\begin{aligned} (B_1-B_2)f=(b-a)+(B_1a-B_2b). \end{aligned}$$
(4.24)

If \( B_1\ne B_2 \), then it is easy to see from (4.24) that f is a constant, which is a contradiction. Therefore, we must have \( B_1=B_2 \), hence, it follows from (4.24) that \( B_1=1=B_2 \), which leads to a contradiction. This completes the proof. \(\square \)

Proof of corollary 2.4

Let \( a_j=\left( {\begin{array}{c}n\\ j\end{array}}\right) (-1)^{n-j} \), where \( j=0, 1, 2, \ldots , n \). Then it is easy to see that \( L^n_c(f)=\Delta ^n_cf \) with \( \sum _{j=0}^{n}a_j=0 \). By the assumption, \( \Delta ^n_cf \) and f share the values a and b CM, hence following exactly the proof of Theorem 2.1, it can be easily shown that \( \Delta ^n_cf\equiv f \). This linear difference equation can be written as

$$\begin{aligned} \Delta ^n_cf(z)=\sum _{k=0}^{n}(-1)^{n-k}\left( {\begin{array}{c}n\\ k\end{array}}\right) f(z+kc)=f(z). \end{aligned}$$
(4.25)

We can solve (4.25) in terms of exponential functions with coefficients in \( {\mathcal {M}}_c \). Since, the distinct roots of \( \sum _{k=0}^{n}(-1)^{n-k}\left( {\begin{array}{c}n\\ k\end{array}}\right) \lambda ^k=1 \) are \( \lambda _k=1+e^{2k\pi i/n} \), where \( k=0, 1, \ldots , n-1 \) (see [8]), a routine computation then shows that

$$\begin{aligned} {\mathcal {S}}^{*}_{\lambda }:=\{\lambda _1, \lambda _2, \ldots , \lambda _n\} \end{aligned}$$

forms a fundamental set of solutions to the difference equation \( \Delta ^n_cf\equiv f \). Therefore, the general solution of (4.25) is

$$\begin{aligned} f(z)=\sum _{k=0}^{n}\lambda ^{z/c}_k\pi _k(z),\; \text{ where }\; \pi _k\in {\mathcal {M}}_c. \end{aligned}$$

This completes the proof. \(\square \)