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# Uniqueness of meromorphic functions concerning differential polynomials share one value

Journal of Inequalities and Applications20112011:133

https://doi.org/10.1186/1029-242X-2011-133

• Received: 24 July 2011
• Accepted: 6 December 2011
• Published:

## Abstract

In this paper, we study the uniqueness of meromorphic functions whose differential polynomial share a non-zero finite value. The results in this paper improve some results given by Fang (Math. Appl. 44, 828-831, 2002), Banerjee (Int. J. Pure Appl. Math. 48, 41-56, 2008) and Lahiri-Sahoo (Arch. Math. (Brno) 44, 201-210, 2008).

2010 Mathematics Subject Classification: 30D35

## Keywords

• Uniqueness
• Meromorphic functions
• Differential polynomials

## 1 Introduction and main results

In this paper, by meromorphic functions, we will always mean meromorphic functions in the complex plane. We adopt the standard notations in the Nevanlinna theory of meromorphic functions as explained in . It will be convenient to let E denote any set of positive real numbers of finite linear measure, not necessarily the same at each occurrence. For a non-constant meromorphic function h, we denote by T(r, h) the Nevanlinna characteristic of h and by S(r, h) any quantity satisfying S(r, h) = o{T(r, h)}, as r → ∞, r E.

Let f and g be two non-constant meromorphic functions and let a be a finite complex value. We say that f and g share a CM, provided that f - a and g - a have the same zeros with the same multiplicities. Similarly, we say that f and g share a IM, provided that f - a and g - a have the same zeros ignoring multiplicities. In addition, we say that f and g share ∞ CM, if 1/f and 1/g share 0 CM, and we say that f and g share ∞ IM, if 1/f and 1/g share 0 IM (see ). Suppose that f and g share a IM. Throughout this paper, we denote by ${\stackrel{̄}{N}}_{L}\left(r,\frac{1}{f-a}\right)$ the reduced counting function of those common a-points of f and g in |z| < r, where the multiplicity of each such a-point of f is greater than that of the corresponding a-point of g, and denote by ${N}_{11}\left(r,\frac{1}{f-a}\right)$ the counting function for common simple 1-point of both f and g. In addition, we need the following three definitions:

Definition 1.1 Let f be a non-constant meromorphic function, and let p be a positive integer and a C {∞}. Then by Np)(r, 1/(f - a)), we denote the counting function of those a-points of f (counted with proper multiplicities) whose multiplicities are not greater than p, by ${\overline{N}}_{p\right)}\left(r,1/\left(f-a\right)\right)$ we denote the corresponding reduced counting function (ignoring multiplicities). By N(p(r,1/(f - a)), we denote the counting function of those a-points of f (counted with proper multiplicities) whose multiplicities are not less than p, by ${\overline{N}}_{\left(p}\left(r,1/\left(f-a\right)\right)$ we denote the corresponding reduced counting function (ignoring multiplicities), where and what follows, ${N}_{p\text{)}}\text{(}r\text{,1/(}f\text{-}a\text{))},{\overline{N}}_{p\right)}\left(r,1/\left(f-a\right)\right),{N}_{\left(p}\text{(}r\text{,1/(}f\text{-}a\text{))},{\overline{N}}_{\left(p}\left(r,1/\left(f-a\right)\right)$ mean ${N}_{p\text{)}}\text{(}r\text{,}f\text{)},{\overline{N}}_{p\right)}\left(r,f\right),{\text{N}}_{\text{(}p}\text{(}r\text{,}f\text{)}$, and ${\overline{N}}_{\left(p}\left(r,f\right)$, respectively, if a = ∞.

Definition 1.2 Let f be a non-constant meromorphic function, and let a be any value in the extended complex plane, and let k be an arbitrary nonnegative integer. We define
${\delta }_{k}\left(a,f\right)=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{{N}_{k}\left(r,\frac{1}{f-a}\right)}{T\left(r,f\right)},$
(1)
where
${N}_{k}\left(r,\frac{1}{f-a}\right)=\overline{N}\left(r,\frac{1}{f-a}\right)+{\overline{N}}_{\left(2}\left(r,\frac{1}{f-a}\right)+\cdots +{\overline{N}}_{\left(k}\left(r,\frac{1}{f-a}\right).$
(2)

Remark 1.1. From (1) and (2), we have 0 ≤ δ k (a, f) ≤ δk- 1(a, f) ≤ δ1(a, f) ≤ Θ(a, f) ≤ 1.

Definition 1.3 Let f be a non-constant meromorphic function, and let a be any value in the extended complex plane, and let k be an arbitrary nonnegative integer.

We define
${\Theta }_{k\right)}\left(a,f\right)=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{{\overline{N}}_{k\right)}\left(r,\frac{1}{f-a}\right)}{T\left(r,f\right)}.$
(3)

Remark 1.2. From (3), we have 0 ≤ Θ(a, f) ≤ Θk)(a, f) ≤ Θk-1)(a, f) ≤ Θ1)(a, f) ≤ 1.

Definition 1.4 Let k be a positive integer. Let f and g be two non-constant meromorphic functions such that f and g share the value 1 IM. Let z0 be a 1-point of f with multiplicity p, and a 1-point of g with multiplicity q. We denote by ${\stackrel{̄}{N}}_{f>k}\left(r,\frac{1}{g-1}\right)$ the reduced counting function of those 1-points of f and g such that $p>q=k.\phantom{\rule{2.77695pt}{0ex}}{\stackrel{̄}{N}}_{g>k}\left(r,\frac{1}{f-1}\right)$ is defined analogously.

It is natural to ask the following question:

Question 1.1 What can be said about the relationship between two meromorphic functions f,g when two differential polynomials, generated by f and g, respectively, share certain values?

Regarding Question 1.1, we first recall the following result by Yang and Hua :

Theorem A. Let f(z) and g(z) be two non-constant meromorphic functions, n ≥ 11 an integer and a C - {0}. If f n f' and g n g' share the value a CM, then either f = tg for a constant t with tn+1= 1 or g(z) = c1e cz and f(z) = c2e -cz , where c, c1and c2 are constants satisfying (c1c2)n+1c2 = -a2.

Considering k th derivative instead of 1st derivative Fang  proved the following theorems.

Theorem B. Let f(z) and g(z) be two non-constant entire functions, and let n, k be two positive integers with n > 2k + 4. If [f n ](k)and [g n ](k)share 1 CM, then either f = tg for a constant t with t n = 1 or f(z) = c1e cz and g(z) = c2e-cz, where c, c1andc2 are constants satisfying ( -1) k (c1c2) n (nc)2k= 1.

Theorem C. Let f(z) and g(z) be two non-constant entire functions, and let n, k be two positive integers with n ≥ 2k + 8. If [f n (z)(f(z) - 1)](k)and [g n (z)(g(z) - 1)](k)share 1 CM, then f(z) ≡ g(z).

In 2008, Banerjee  proved the following theorem.

Theorem D. Let f and g be two transcendental meromorphic functions, and let n, k be two positive integers with n ≥ 9k + 14. Suppose that [f n ](k)and [g n ](k)share a non-zero constant b IM, then either f = tg for a constant t with t n = 1 or f(z) = c1e cz and g(z) = c2e -cz , where c, c1and c2 are constants satisfying ( -1) k (c1c2) n (nc)2k= b2.

Recently, Lahiri and Sahoo  proved the following theorem.

Theorem E. Let f and g be two non-constant meromorphic functions, and $\alpha \left(\not\equiv 0,\infty \right)$ be a small function of f and g. Let n and m(≥ 2) be two positive integers with n > max{4, 4m + 22 - 5Θ(∞, f) - 5Θ(∞, g) -min[Θ(∞, f), Θ(∞, g)]}. If f n (f m - a)f' and g n (g m - a)g' share α IM for a non-zero constant a, then either fg or f ≡ -g.

Also, the possibility f ≡ -g does not arise if n and m are both even, both odd or n is even and m is odd.

One may ask, what can be said about the relationship between f and g, if we relax the nature of sharing values of Theorem D and Theorem E ? In this paper, we prove:

Theorem 1.1. Let f(z) and g(z) be two non-constant meromorphic functions, and let n(≥ 1), k(≥ 1) and m(≥ 0) be three integers. Let [f n (f - 1) m ](k)and [g n (g - 1) m ](k)share the value 1 IM. Then, one of the following holds:
1. (i)

When m = 0 and n > 9k + 14, then either f(z) = c 1e cz and g(z) = c 2e-cz, where c,c 1 andc 2 are constants satisfying (-1) k (c 1 c 2) n (nc)2k= 1 or f = tg for a constant t with t n = 1.

2. (ii)

When m = 1, n > 9k + 18 and $\Theta \left(\infty ,f\right)>\frac{2}{n}$, then fg.

3. (iii)

When m ≥ 2, n > 4m + 9k + 14, then fg or f and g satisfies the algebraic equation R(x, y) = x n (x - 1) m - y n (y - 1) m = 0.

Theorem 1.2. Let f(z) and g(z) be two non-constant meromorphic functions, and let m, n(≥ 2) and k be three positive integers such that n > 4m + 9k + 14. If [f n (f m - a)](k)and [g n (g m - a)](k)share the value 1 IM, where a(≠ 0) is a finite complex number, then either fg or f ≡ -g.

The possibility f ≡ -g does not arise if n and m are both odd or if n is even and m is odd or if n is odd and m is even.

Remark 1.3. If m = 0, m = 1, then the cases become Theorem 1.1 (i) (ii).

Theorem 1.3. Let f(z) and g(z) be two non-constant entire functions, and let n(≥ 1), k(≥ 1) and m(≥ 0) be three integers. Let [f n (f - 1) m ](k)and [g n (g - 1) m ](k)share the value 1 IM. Then, one of the following holds:
1. (i)

When m = 0 and n > 5k + 7, then either f(z) = c 1e cz and g(z) = c 2e-cz, where c,c 1 andc 2 are constants satisfying ( -1) k (c 1 c 2) n (nc)2k= 1 or f = tg for a constant t with t n = 1.

2. (ii)

When m ≥ 1, n > 4m + 5k + 7, then fg or f and g satisfies the algebraic equation R(x, y) = x n (x - 1) m - y n (y - 1) m = 0.

Theorem 1.4. Let f(z) and g(z) be two non-constant entire functions, and let m, n(≥ 1) and k be three positive integers such that n > 4m + 5k + 7. If [f n (f m - a)](k)and [g n (g m - a)](k)share the value 1 IM, where a(≠ 0) is a finite complex number, then either fg or f-g.

The possibility f-g does not arise if n and m are both odd or if n is even and m is odd or if n is odd and m is even.

Remark 1.4. If m = 0, then the cases becomes Theorem 1.3 (i).

## 2 Some lemmas

Lemma 2.1. (See [2, 3].) Let f(z) be a non-constant meromorphic function, k a positive integer and let c be a non-zero finite complex number. Then,
$\begin{array}{ll}\hfill T\left(r,f\right)& \le \stackrel{̄}{N}\left(r,f\right)+N\left(r,\frac{1}{f}\right)+N\left(r,\frac{1}{{f}^{\left(k\right)}-c}\right)-N\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le \stackrel{̄}{N}\left(r,f\right)+{N}_{k+1}\left(r,\frac{1}{f}\right)+\stackrel{̄}{N}\left(r,\frac{1}{{f}^{\left(k\right)}-c}\right)-{N}_{0}\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)+S\left(r,f\right).\phantom{\rule{2em}{0ex}}\end{array}$
(4)

where ${N}_{0}\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)$ is the counting function, which only counts those points such that f(k+1)= 0 but f(f(k)-c) ≠ 0

Lemma 2.2. (See .) Let f(z) be a non-constant meromorphic function, and let k be a positive integer.

Suppose that ${f}^{\left(k\right)}\not\equiv 0$, then
$N\left(r,\frac{1}{{f}^{\left(k\right)}}\right)\le N\left(r,\frac{1}{f}\right)+k\stackrel{̄}{N}\left(r,f\right)+S\left(r,f\right).$
Lemma 2.3. (See .) Let f(z) be a non-constant meromorphic function, s, k be two positive integers, then
${N}_{s}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)\le k\stackrel{̄}{N}\left(r,f\right)+{N}_{s+k}\left(r,\frac{1}{f}\right)+S\left(r,f\right).$

Clearly, $\stackrel{̄}{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)={N}_{1}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)$.

Lemma 2.4. (See .) Let f, g share (1,0). Then
1. (i)

${\stackrel{̄}{N}}_{f>1}\left(r,\frac{1}{g-1}\right)\le \stackrel{̄}{N}\left(r,\frac{1}{f}\right)+\stackrel{̄}{N}\left(r,f\right)-{N}_{0}\left(r,\frac{1}{{f}^{\prime }}\right)+S\left(r,f\right),$,

2. (ii)

${\stackrel{̄}{N}}_{g>1}\left(r,\frac{1}{f-1}\right)\le \stackrel{̄}{N}\left(r,\frac{1}{g}\right)+\stackrel{̄}{N}\left(r,g\right)-{N}_{0}\left(r,\frac{1}{{g}^{\prime }}\right)+S\left(r,g\right)$.

Lemma 2.5. Let f(z) and g(z) be two non-constant meromorphic functions such that f(k)and g(k)share 1 IM, where k be a positive integer. If
$\Delta =\left(2k+4\right)\Theta \left(\infty ,g\right)+\left(2k+3\right)\Theta \left(\infty ,f\right)+{\delta }_{k+2}\left(0,g\right)+{\delta }_{k+2}\left(0,f\right)+{\delta }_{k+1}\left(0,f\right)+2{\delta }_{k+1}\left(0,g\right)>4k+11$

then either f(k)g(k)≡ 1 or fg.

Proof. Let
$\Phi \left(z\right)=\frac{{f}^{\left(k+2\right)}}{{f}^{\left(k+1\right)}}-2\frac{{f}^{\left(k+1\right)}}{{f}^{\left(k\right)}-1}-\frac{{g}^{\left(k+2\right)}}{{g}^{\left(k+1\right)}}+2\frac{{g}^{\left(k+1\right)}}{{g}^{\left(k\right)}-1}.$
(5)

Clearly m(r, Φ) = S(r, f) + S(r, g). We consider the cases $\Phi \left(z\right)\not\equiv 0$ and Φ(z) ≡ 0.

Let $\Phi \left(z\right)\not\equiv 0$, then if z0 is a common simple 1-point of f(k)and g(k), substituting their Taylor series at z0 into (5), we see that z0 is a zero of Φ(z). Thus, we have
${N}_{11}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)={N}_{11}\left(r,\frac{1}{{g}^{\left(k\right)}-1}\right)\le \stackrel{̄}{N}\left(r,\frac{1}{\Phi }\right)\le T\left(r,\Phi \right)+O\left(1\right)\le N\left(r,\Phi \right)+S\left(r,f\right)+S\left(r,g\right).$
(6)
Our assumptions are that Φ(z) has poles, all simple only at zeros of f(k+1)and g(k+1)and poles of f and g, and 1-points of f whose multiplicities are not equal to the multiplicities of the corresponding 1-points of g. Thus, we deduce from (5) that
$\begin{array}{c}N\left(r,\Phi \right)\le \overline{N}\left(r,f\right)+\overline{N}\left(r,g\right)+{\overline{N}}_{\left(k+2}\left(r,\frac{1}{f}\right)+{\overline{N}}_{\left(k+2}\left(r,\frac{1}{g}\right)+{N}_{0}\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)\\ +{N}_{0}\left(r,\frac{1}{{g}^{\left(k+1\right)}}\right)+{\overline{N}}_{L}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)+{\overline{N}}_{L}\left(r,\frac{1}{{g}^{\left(k\right)}-1}\right).\end{array}$
(7)
here ${N}_{0}\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)$ has the same meaning as in Lemma 2.1. From Lemma 2.1, we have
$T\left(r,g\right)\le \stackrel{̄}{N}\left(r,g\right)+{N}_{k+1}\left(r,\frac{1}{g}\right)+\stackrel{̄}{N}\left(r,\frac{1}{{g}^{\left(k\right)}-1}\right)-{N}_{0}\left(r,\frac{1}{{g}^{\left(k+1\right)}}\right)+S\left(r,g\right).$
(8)
Since
$\overline{N}\left(r,\frac{1}{{g}^{\left(k\right)}-1}\right)={N}_{11}\left(\left(r,\frac{1}{{g}^{\left(k\right)}-1}\right)+{\overline{N}}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)+{\overline{N}}_{{g}^{\left(k\right)}>1}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right).$
(9)
Thus, we deduce from (6)-(9) that
$\begin{array}{c}T\left(r,g\right)\le 2\overline{N}\left(r,g\right)+\overline{N}\left(r,f\right)+{N}_{k+1}\left(r,\frac{1}{g}\right)+{\overline{N}}_{\left(k+2}\left(r,\frac{1}{f}\right)+{\overline{N}}_{\left(k+2}\left(r,\frac{1}{g}\right)+{N}_{0}\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)+{\overline{N}}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)\\ +{\overline{N}}_{L}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)+{\overline{N}}_{L}\left(r,\frac{1}{{g}^{\left(k\right)}-1}\right)+{\overline{N}}_{{g}^{\left(k\right)}>1}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)+S\left(r,f\right)+S\left(r,g\right).\end{array}$
(10)
From the definition of ${N}_{0}\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)$, we see that
${N}_{0}\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)+{\overline{N}}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)+{N}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)-{\overline{N}}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)\le N\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right).$
The above inequality and Lemma 2.2 give
$\begin{array}{c}{N}_{0}\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)+{\overline{N}}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)\le N\left(r,\frac{1}{{f}^{\left(k+1\right)}}\right)-{N}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+{\overline{N}}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)\\ \le N\left(r,\frac{1}{{f}^{\left(k\right)}}\right)-{N}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+{\overline{N}}_{\left(2}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+\overline{N}\left(r,f\right)+S\left(r,f\right)\\ \le \overline{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+\overline{N}\left(r,f\right)+S\left(r,f\right).\end{array}$
(11)
Substituting (11) in (10), we get
$\begin{array}{c}T\left(r,g\right)\le 2\overline{N}\left(r,g\right)+\overline{N}\left(r,f\right)+{N}_{k+1}\left(r,\frac{1}{g}\right)+{\overline{N}}_{\left(k+2}\left(r,\frac{1}{f}\right)+{\overline{N}}_{\left(k+2}\left(r,\frac{1}{g}\right)+\overline{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+\overline{N}\left(r,f\right)\\ +{\overline{N}}_{L}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)+{\overline{N}}_{L}\left(r,\frac{1}{{g}^{\left(k\right)}-1}\right)+{\overline{N}}_{{g}^{\left(k\right)}>1}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)+S\left(r,f\right)+S\left(r,g\right)\\ \le 2\overline{N}\left(r,g\right)+2\overline{N}\left(r,f\right)+{N}_{k+2}\left(r,\frac{1}{g}\right)+{\overline{N}}_{\left(k+2}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+{\overline{N}}_{L}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)\\ +{\overline{N}}_{L}\left(r,\frac{1}{{g}^{\left(k\right)}-1}\right)+{\overline{N}}_{{g}^{\left(k\right)}>1}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)+S\left(r,f\right)+S\left(r,g\right).\end{array}$
(12)
According to Lemma 2.3,
$\stackrel{̄}{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)={N}_{1}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)\le {N}_{k+1}\left(r,\frac{1}{f}\right)+k\stackrel{̄}{N}\left(r,f\right)+S\left(r,f\right).$
(13)
Therefore,
$\begin{array}{ll}\hfill {\stackrel{̄}{N}}_{L}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)& \le N\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)-\stackrel{̄}{N}\left(r,\frac{1}{{f}^{\left(k\right)}-1}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\le N\left(r,\frac{{f}^{\left(k\right)}}{{f}^{\left(k+1\right)}}\right)\le N\left(r,\frac{{f}^{\left(k+1\right)}}{{f}^{\left(k\right)}}\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\le \stackrel{̄}{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+\stackrel{̄}{N}\left(r,f\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\le {N}_{k+1}\left(r,\frac{1}{f}\right)+\left(k+1\right)\stackrel{̄}{N}\left(r,f\right)+S\left(r,f\right).\phantom{\rule{2em}{0ex}}\end{array}$
similarly,
${\stackrel{̄}{N}}_{L}\left(r,\frac{1}{{g}^{\left(k\right)}-1}\right)\le {N}_{k+1}\left(r,\frac{1}{g}\right)+\left(k+1\right)\stackrel{̄}{N}\left(r,g\right)+S\left(r,g\right).$
Combining the above inequality, Lemma 2.4 and (12), we obtain
$\begin{array}{ll}\hfill T\left(r,g\right)& \le \left(2k+4\right)\stackrel{̄}{N}\left(r,g\right)+\left(2k+3\right)\stackrel{̄}{N}\left(r,f\right)+{N}_{k+2}\left(r,\frac{1}{g}\right)+{N}_{k+2}\left(r,\frac{1}{f}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+{N}_{k+1}\left(r,\frac{1}{f}\right)+2{N}_{k+1}\left(r,\frac{1}{g}\right)-{N}_{0}\left(r,\frac{1}{{g}^{\left(k+1\right)}}\right)+S\left(r,f\right)+S\left(r,g\right)\phantom{\rule{2em}{0ex}}\\ \le \left(2k+4\right)\stackrel{̄}{N}\left(r,g\right)+\left(2k+3\right)\stackrel{̄}{N}\left(r,f\right)+{N}_{k+2}\left(r,\frac{1}{g}\right)+{N}_{k+2}\left(r,\frac{1}{f}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+{N}_{k+1}\left(r,\frac{1}{f}\right)+2{N}_{k+1}\left(r,\frac{1}{g}\right)+S\left(r,f\right)+S\left(r,g\right).\phantom{\rule{2em}{0ex}}\end{array}$
Without loss of generality, we suppose that there exists a set I with infinite measure such that T(r, f) ≤ T(r, g) for r I. Hence,
$\begin{array}{c}T\left(r,g\right)\le \left\{\left(2k+4\right)\left[1-\Theta \left(\infty ,g\right)\right]+\left(2k+3\right)\left[1-\Theta \left(\infty ,f\right)\right]+\left[1-{\delta }_{k+2}\left(0,g\right)\right]+\left[1-{\delta }_{k+2}\left(0,f\right)\right]\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}+\left[1-{\delta }_{k+1}\left(0,f\right)\right]+2\left[1-{\delta }_{k+1}\left(0,g\right)\right]+\epsilon \right\}T\left(r,g\right)+S\left(r,g\right).\end{array}$

for I and 0 < ε < Δ - (4k +11)

Therefore, we can get T(r, g) ≤ S(r, g),r I, by the condition, a contradiction.

Hence, we get Φ(z) ≡ 0. Then, by (5), we have
$\frac{{f}^{\left(k+2\right)}}{{f}^{\left(k+1\right)}}-\frac{2{f}^{\left(k+1\right)}}{{f}^{\left(k\right)}-1}\equiv \frac{{g}^{\left(k+2\right)}}{{g}^{\left(k+1\right)}}-\frac{2{g}^{\left(k+1\right)}}{{g}^{\left(k\right)}-1}.$
By integrating two sides of the above equality, we obtain
$\frac{1}{{f}^{\left(k\right)}-1}=\frac{b{g}^{\left(k\right)}+a-b}{{g}^{\left(k\right)}-1}.$
(14)

where a(≠ 0) and b are constants. We consider the following three cases:

Case 1. b ≠ 0 and a = b
1. (i)

If b = -1, then from (14), we obtain that f (k) g (k)≡ 1.

2. (ii)
If b ≠ -1, then from (14), we get
${f}^{\left(k\right)}=\frac{\left(1+b\right){g}^{\left(k\right)}-1}{b{g}^{\left(k\right)}}.$
(15)

From (15), we get
$\stackrel{̄}{N}\left(r,\frac{1}{{g}^{\left(k\right)}-1∕\left(1+b\right)}\right)=\stackrel{̄}{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right).$
(16)
Combing (13) (16) and Lemma 2.1, we have
$\begin{array}{ll}\hfill T\left(r,g\right)& \le \stackrel{̄}{N}\left(r,g\right)+{N}_{k+1}\left(r,\frac{1}{g}\right)+\stackrel{̄}{N}\left(r,\frac{1}{{g}^{\left(k\right)}-1∕\left(b+1\right)}\right)-{N}_{0}\left(r,\frac{1}{{g}^{\left(k+1\right)}}\right)+S\left(r,g\right)\phantom{\rule{2em}{0ex}}\\ \le \stackrel{̄}{N}\left(r,g\right)+{N}_{k+1}\left(r,\frac{1}{g}\right)+k\stackrel{̄}{N}\left(r,f\right)+{N}_{k+1}\left(r,\frac{1}{f}\right)+S\left(r,f\right)+S\left(r,g\right).\phantom{\rule{2em}{0ex}}\end{array}$
(17)
From (17), we get
$\Theta \left(\infty ,g\right)+k\Theta \left(\infty ,f\right)+{\delta }_{k+1}\left(0,g\right)+{\delta }_{k+1}\left(0,f\right)\le k+2.$

By the condition, we get a contradiction.

Case 2. b ≠ 0 and ab.
1. (i)
If b = -1, then a ≠ 0, from (14) we obtain
${f}^{\left(k\right)}=\frac{a}{a+1-{g}^{\left(k\right)}}.$
(18)

From (18), we get
$\stackrel{̄}{N}\left(r,\frac{1}{{g}^{\left(k\right)}-\left(a+1\right)}\right)=\stackrel{̄}{N}\left(r,f\right).$
(19)
From (19) and Lemma 2.1 and in the same manner as in the proof of (17), we get
$\begin{array}{ll}\hfill T\left(r,g\right)& \le \stackrel{̄}{N}\left(r,g\right)+{N}_{k+1}\left(r,\frac{1}{g}\right)+\stackrel{̄}{N}\left(r,\frac{1}{{g}^{\left(k\right)}-\left(a+1\right)}\right)+S\left(r,g\right)\phantom{\rule{2em}{0ex}}\\ \le \stackrel{̄}{N}\left(r,g\right)+{N}_{k+1}\left(r,\frac{1}{g}\right)+\stackrel{̄}{N}\left(r,f\right)+S\left(r,g\right).\phantom{\rule{2em}{0ex}}\end{array}$
Using the argument as in case 1, we get a contradiction.
1. (ii)
If b ≠ -1, then from (14), we get
${f}^{\left(k\right)}-\left(1+\frac{1}{b}\right)=\frac{-a}{{b}^{2}\left[{g}^{\left(k\right)}+\frac{a-b}{b}\right]}.$
(20)

From (20), we get
$\stackrel{̄}{N}\left[r,\frac{1}{{g}^{\left(k\right)}+\left(\frac{a-b}{b}\right)}\right]=\stackrel{̄}{N}\left[{f}^{\left(k\right)}-\left(1+\frac{1}{b}\right)\right]=\stackrel{̄}{N}\left(r,{f}^{\left(k\right)}\right)=\stackrel{̄}{N}\left(r,f\right).$
(21)

Using the argument as in case 1, we get a contradiction.

Case 3. b = 0. From (14), we obtain
${f}^{\left(k\right)}=\frac{1}{a}{g}^{\left(k\right)}+1-\frac{1}{a},$
(22)
$f=\frac{1}{a}g+p\left(z\right).$
(23)
where p(z) is a polynomial with its degree ≤ k. If $p\left(z\right)\not\equiv 0$, then by second fundamental theorem for small functions, we have
$\begin{array}{ll}\hfill T\left(r,g\right)& \le \stackrel{̄}{N}\left(r,g\right)+\stackrel{̄}{N}\left(r,\frac{1}{g}\right)+\stackrel{̄}{N}\left(r,\frac{1}{g+ap\left(z\right)}\right)+S\left(r,g\right)\phantom{\rule{2em}{0ex}}\\ \le \stackrel{̄}{N}\left(r,g\right)+\stackrel{̄}{N}\left(r,\frac{1}{g}\right)+\stackrel{̄}{N}\left(r,\frac{1}{f}\right)+S\left(r,g\right).\phantom{\rule{2em}{0ex}}\end{array}$
(24)

Using the argument as in Case 1, we get a contradiction. Therefore, p(z) ≡ 0. So from (22) and (23), we obtain a = 1 and so fg. This proves the lemma.

Lemma 2.6. Let f(z) and g(z) be two non-constant entire functions such that f(k)and g(k)share 1 IM, where k be a positive integer. If
$\Delta ={\delta }_{k+2}\left(0,g\right)+{\delta }_{k+2}\left(0,f\right)+{\delta }_{k+1}\left(0,f\right)+2{\delta }_{k+1}\left(0,g\right)>4$

then either f(k)g(k)≡ 1 or fg.

Proof. Since f and g are entire functions, we have $\stackrel{̄}{N}\left(r,f\right)=0$ and $\stackrel{̄}{N}\left(r,g\right)=0$. Proceeding as in the proof of Lemma 2.5, we obtain conclusion of Lemma 2.6.

Lemma 2.7. (See .) Let f(z) be a non-constant entire function, and let k(≥ 2) be a positive integer. If f f(k)≠ 0, then f = eaz+b,where a ≠ 0, b are constants.

Lemma 2.8. (See .) Let f(z) be a non-constant meromorphic function. Let k be a positive integer, and let c be a non-zero finite complex number. Then,
$T\left(r,{a}_{n}{f}^{n}+{a}_{n-1}{f}^{n-1}+\cdots +{a}_{0}\right)=nT\left(r,f\right)+S\left(r,f\right).$

## 3 Proof of theorems

### 3.1 Proof of Theorem 1.1

Let F = f n (f - 1) m and G = g n (g - 1) m .

By Lemma 2.8, we have
$\begin{array}{c}\Theta \left(\infty ,F\right)=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{\overline{N}\left(r,F\right)}{T\left(r,F\right)}=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{\overline{N}\left(r,{f}^{n}{\left(f-1\right)}^{m}\right)}{\left(m+n\right)T\left(r,f\right)}\\ \ge 1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{T\left(r,f\right)}{\left(m+n\right)T\left(r,f\right)}\ge \frac{n+m-1}{m+n},\end{array}$
$\begin{array}{c}{\delta }_{k+1}\left(0,F\right)=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{{N}_{k+1}\left(r,\frac{1}{F}\right)}{T\left(r,F\right)}=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{{N}_{k+1}\left(r,\frac{1}{{f}^{n}{\left(f-1\right)}^{m}}\right)}{\left(m+n\right)T\left(r,f\right)}\\ \ge 1-\frac{\left(k+m+1\right)T\left(r,f\right)}{\left(m+n\right)T\left(r,f\right)}\ge \frac{n-k-1}{m+n},\end{array}$

Similarly, $\Theta \left(\infty ,G\right)\ge \frac{n+m-1}{m+n},{\delta }_{k+1}\left(0,G\right)\ge \frac{n-k-1}{m+n},\phantom{\rule{0.3em}{0ex}}{\delta }_{k+2}\left(0,F\right)\ge \frac{n-k-2}{m+n},{\delta }_{k+2}\left(0,G\right)\ge \frac{n-k-2}{m+n}$.

Therefore,
$\begin{array}{ll}\hfill \Delta & =\left(2k+4\right)\Theta \left(\infty ,G\right)+\left(2k+3\right)\Theta \left(\infty ,F\right)+{\delta }_{k+2}\left(0,G\right)+{\delta }_{k+2}\left(0,F\right)+{\delta }_{k+1}\left(0,F\right)+2{\delta }_{k+1}\left(0,G\right)\phantom{\rule{2em}{0ex}}\\ \ge \left(2k+4\right)\cdot \frac{m+n-1}{m+n}+\left(2k+3\right)\cdot \frac{m+n-1}{m+n}+\frac{n-k-2}{m+n}+\frac{n-k-2}{m+n}+\frac{n-k-1}{m+n}+2\cdot \frac{n-k-1}{m+n}\phantom{\rule{2em}{0ex}}\end{array}$

If n > 4m + 9k + 14, we obtain Δ > 4k + 11.

So by Lemma 2.5, we get either F(k)G(k)≡ 1 or FG.

Case 1. F(k)G(k)≡ 1, that is,
${\left({f}^{n}{\left(f-1\right)}^{m}\right)}^{\left(k\right)}{\left({g}^{n}{\left(g-1\right)}^{m}\right)}^{\left(k\right)}\equiv 1.$
(25)
Case 1.1 when m = 0, that is,
${\left({f}^{n}\right)}^{\left(k\right)}{\left({g}^{n}\right)}^{\left(k\right)}\equiv 1.$
(26)

Next, we prove f ≠ 0, ∞ and g ≠ 0, ∞.

Suppose that f has a zero z0 of order p, then z0 is a pole of g of order q. By (26), we get np - k = nq + k, i.e., n(p - q) = 2k, which is impossible since n > 9k + 14.

Therefore, we conclude that f ≠ 0 and g ≠ 0.

Similarly, Suppose that f has a pole ${z}_{0}^{\prime }$ of order p', then ${z}_{0}^{\prime }$ is a zero of g of order q'. By (26), we get np' + k = nq' - k, i.e., n(q' - p') = 2k, which is impossible since n > 9k + 14.

Therefore, we conclude that f ≠ ∞ oo and g ≠ ∞.

From (26), we get
${\left({f}^{n}\right)}^{\left(k\right)}\ne 0\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}{\left({g}^{n}\right)}^{\left(k\right)}\ne 0.$
(27)

From (26)-(27) and Lemma 2.7, we get that f(z) = c1e cz and g(z) = c2e -cz , where c, c1 and c2 are three constants satisfying ( -1) k (c1c2) n (nc)2k= 1.

Case 1.2 when m ≥ 1

Let f has a zero z1 of order p1. From (25), we get z1 is a pole of g. Suppose that z1 is a pole of g of order q1. Again by (25), we obtain np1 - k = nq1 + mq1 + k, i.e., n(p1 - q1) = mq1 + 2k, which implies that p1q1 + 1 and mq1 + 2kn. From n > 4m + 9k + 14, we can deduce p1 ≥ 6.

Let f - 1 has a zero z2 of order p2, then z2 is a zero of [f n (f - 1) m ](k)of order mp2 - k. Therefore from (25), we obtain z2 is a pole of g of order q2. Again by (25), we obtain mp2 - k = (n + m)q2 + k, i.e., mp2 = (n + m)q2 + 2k, i.e., ${p}_{2}\ge \frac{m+n}{m}+\frac{2k}{m}$.

Let z3 be a zero of f' of order p3 that not a zero of f(f - 1), as above, we obtain from (25), p3 - (k - 1) = (n + m)q3 + k, i.e., p3n + m + 2k - 1.

Moreover, in the same manner as above, we have similar results for the zeros of [g n (g-1) m ](k).

On the other hand, Suppose z4 is a pole of f, from (25), we get z4 is a zero of [g n (g - 1) m ](k).

Thus,
$\begin{array}{ll}\hfill \stackrel{̄}{N}\left(r,f\right)& \le \stackrel{̄}{N}\left(r,\frac{1}{g}\right)+\stackrel{̄}{N}\left(r,\frac{1}{g-1}\right)+\stackrel{̄}{N}\left(r,\frac{1}{{g}^{\prime }}\right)\phantom{\rule{2em}{0ex}}\\ \le \frac{1}{6}N\left(r,\frac{1}{g}\right)+\frac{m}{m+n+2k}N\left(r,\frac{1}{g-1}\right)+\frac{1}{n+m+2k-1}N\left(r,\frac{1}{{g}^{\prime }}\right).\phantom{\rule{2em}{0ex}}\end{array}$
We get
$\stackrel{̄}{N}\left(r,f\right)\le \left(\frac{1}{6}+\frac{m}{m+n+2k}+\frac{1}{n+m+2k-1}\right)T\left(r,g\right)+S\left(r,g\right).$
From this and the second fundamental theorem, we obtain
$\begin{array}{ll}\hfill T\left(r,f\right)& \le \stackrel{̄}{N}\left(r,f\right)+\stackrel{̄}{N}\left(r,\frac{1}{f-1}\right)+\stackrel{̄}{N}\left(r,\frac{1}{f}\right)+S\left(r,f\right)\phantom{\rule{2em}{0ex}}\\ \le \left(\frac{1}{6}+\frac{m}{m+n+2k}+\frac{1}{n+m+2k-1}\right)T\left(r,g\right)+\left(\frac{1}{6}+\frac{m}{m+n+2k}\right)T\left(r,f\right)+S\left(r,f\right)+S\left(r,g\right).\phantom{\rule{2em}{0ex}}\end{array}$
Similarly, we have
$T\left(r,g\right)\le \left(\frac{1}{6}+\frac{m}{m+n+2k}+\frac{1}{n+m+2k-1}\right)T\left(r,f\right)+\left(\frac{1}{6}+\frac{m}{m+n+2k}\right)T\left(r,g\right)+S\left(r,f\right)+S\left(r,g\right).$
We can deduce from above
$T\left(r,f\right)+T\left(r,g\right)\le \left(\frac{1}{3}+\frac{2m}{m+n+2k}+\frac{1}{n+m+2k-1}\right)\left[T\left(r,f\right)+T\left(r,g\right)\right]+S\left(r,f\right)+S\left(r,g\right).$
Since n > 4m + 9k + 14, we obtain
$T\left(r,f\right)+T\left(r,g\right)\le \left(\frac{1}{3}+\frac{2}{31}+\frac{1}{30}\right)\left[T\left(r,f\right)+T\left(r,g\right)\right]+S\left(r,f\right)+S\left(r,g\right).$

i.e., 0.57[T(r, f) + T(r, g)] ≤ S(r, f) + S(r, g),

Case 2. FG, i.e.,
${f}^{n}{\left(f-1\right)}^{m}\equiv {g}^{n}{\left(g-1\right)}^{m}.$
(28)

Now we consider following three cases.

Case 2.1 when m = 0, then from (28), we get f = tg for a constant t such that t n = 1.

Case 2.2 when m = 1, then from (28), we have
${f}^{n}\left(f-1\right)\equiv {g}^{n}\left(g-1\right).$
(29)

Suppose $f\not\equiv g$. Let $h=\frac{f}{g}$ be a constant. Then from (29), it follows that h ≠ 1, h n ≠ 1, hn+1≠ 1 and $g=\frac{1-{h}^{n}}{1-{h}^{n+1}}=\mathsf{\text{constant}}$, a contradiction. So we suppose h is not a constant. Since $f\not\equiv g$, we have $h\not\equiv 1$.

From (29), we obtain $g=\frac{1-{h}^{n}}{1-{h}^{n+1}}$ and $f=\frac{h\left(1-{h}^{n}\right)}{1-{h}^{n+1}}$. Hence, it follows that T(r, f) = nT(r, h) + S(r, f).

By the second fundamental theorem, we have
$\stackrel{̄}{N}\left(r,f\right)=\sum _{i=1}^{n}\stackrel{̄}{N}\left(r,\frac{1}{h-{a}_{i}}\right)\ge \left(n-2\right)T\left(r,h\right)+S\left(r,f\right)$

where a i (≠ 1) (i = 1, 2,..., n) are distinct roots of the equation hn+1= 1.

So we obtain
$\Theta \left(\infty ,f\right)=1-\frac{\stackrel{̄}{N}\left(r,f\right)}{T\left(r,f\right)}\le \frac{2}{n},$

which contradicts the assumption $\Theta \left(\infty ,f\right)>\frac{2}{n}$, thus fg.

Case 2.3 when m ≥ 2, then from (28), we obtain
${f}^{n}\left[{f}^{m}+\cdots +{\left(-1\right)}^{i}{C}_{m}^{m-i}{f}^{m-i}+\cdots +{\left(-1\right)}^{m}\right]\equiv {g}^{n}\left[{g}^{m}+\cdots +{\left(-1\right)}^{i}{C}_{m}^{m-i}{g}^{m-i}+\cdots +{\left(-1\right)}^{m}\right].$
(30)
Let $h=\frac{f}{g}$, if h is a constant, then substituting f = gh into (30), we deduce
${g}^{n+m}\left({h}^{n+m}-1\right)\cdots +{\left(-1\right)}^{i}{C}_{m}^{m-i}{g}^{m+n-i}\left({h}^{n+m-i}-1\right)+\cdots +{\left(-1\right)}^{m}{g}^{n}\left({h}^{n}-1\right)=0,$

which implies h = 1. Thus, f(z) ≡ g(z). If h is not a constant, then we know by (30) that f and g satisfies the algebraic equation R(f, g) ≡ 0, where $R\left({w}_{1},{w}_{2}\right)={w}_{1}^{n}{\left({w}_{1}-1\right)}^{m}-{w}_{2}^{n}{\left({w}_{2}-1\right)}^{m}$.

This completes the proof of Theorem 1.1.

### 3.2 Proof of Theorem 1.2

Consider F = f n (f m - a), G = g n (g m - a), then F(k)and G(k)share 1 IM.

By Lemma 2.8, we have
$\begin{array}{c}\Theta \left(\infty ,F\right)=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{\overline{N}\left(r,F\right)}{T\left(r,F\right)}=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{\overline{N}\left(r,{f}^{n}\left({f}^{m}-a\right)\right)}{\left(m+n\right)T\left(r,f\right)}\\ \ge 1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{T\left(r,f\right)}{\left(m+n\right)T\left(r,f\right)}\ge \frac{m+n-1}{m+n},\end{array}$
and
$\begin{array}{c}{\delta }_{k+1}\left(0,F\right)=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{{N}_{k+1}\left(r,\frac{1}{F}\right)}{T\left(r,F\right)}=1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{{N}_{k+1}\left(r,\frac{1}{{f}^{n}\left({f}^{m}-a\right)}\right)}{\left(m+n\right)T\left(r,f\right)}\\ \ge 1-\underset{n\to \infty }{\overline{\mathrm{lim}}}\frac{\left(k+m+1\right)T\left(r,f\right)}{\left(m+n\right)T\left(r,f\right)}\ge \frac{n-k-1}{m+n}.\end{array}$

Similarly, $\Theta \left(\infty ,G\right)\ge \frac{m+n-1}{m+n},{\delta }_{k+1}\left(0,G\right)\ge \frac{n-k-1}{m+n},{\delta }_{k+2}\left(0,F\right)\ge \frac{n-k-2}{m+n},{\delta }_{k+2}\left(0,G\right)\ge \frac{n-k-2}{m+n}$.

Therefore,
$\begin{array}{ll}\hfill \Delta & =\left(2k+4\right)\Theta \left(\infty ,G\right)+\left(2k+3\right)\Theta \left(\infty ,F\right)+{\delta }_{k+2}\left(0,G\right)+{\delta }_{k+2}\left(0,F\right)+{\delta }_{k+1}\left(0,F\right)+2{\delta }_{k+1}\left(0,G\right)\phantom{\rule{2em}{0ex}}\\ \ge \left(2k+4\right)\cdot \frac{m+n-1}{m+n}+\left(2k+3\right)\cdot \frac{m+n-1}{m+n}+\frac{n-k-2}{m+n}+\frac{n-k-2}{m+n}+\frac{n-k-1}{m+n}+2\cdot \frac{n-k-1}{m+n}\phantom{\rule{2em}{0ex}}\end{array}$

Since n > 4m + 9k + 14, we get Δ > 4k + 11, then by Lemma 2.5, we obtain either F(k)G(k)≡ 1 or FG.

Let F(k)G(k)≡ 1, i.e.,
${\left[{f}^{n}\left({f}^{m}-a\right)\right]}^{\left(k\right)}{\left[{g}^{n}\left({g}^{m}-a\right)\right]}^{\left(k\right)}\equiv 1,$
(31)
We can rewrite (31) as
${\left[{f}^{n}\left(f-{a}_{1}\right)\cdots \left(f-{a}_{m}\right)\right]}^{\left(k\right)}{\left[{g}^{n}\left(g-{a}_{1}\right)\cdots \left(g-{a}_{m}\right)\right]}^{\left(k\right)}\equiv 1,$
(32)

where a1, a2,..., a m are roots of w m - a = 0.

By the similar argument for (32) of case 1.2 of Theorem 1.1, the case F(k)G(k)≡ 1 does not arise.

Let FG, i.e.,
${f}^{n}\left({f}^{m}-a\right)\equiv {g}^{n}\left({g}^{m}-a\right).$
(33)

Obviously, if m and n are both odd or if m is odd and n is even or if m is even and n is odd, then f ≡ - g contradicts F ≡ G. Let $f\not\equiv g$ and $f\not\equiv -g$. We put $h=\frac{f}{g}$, then $h\not\equiv 1$ and $h\not\equiv -1$. So from (33), we get ${g}^{m}=\frac{a\left(1-{h}^{n}\right)}{1-{h}^{n+m}}$.

Since g is non-constant, we see that h is not a constant. Again since g m has no simple pole, h - h k has no simple zero, where ${h}_{k}=\mathsf{\text{exp}}\left(\frac{2\pi ki}{n+m}\right)$ and k = 1, 2,..., n + m - 1. Hence, $\Theta \left({h}_{k},h\right)\ge \frac{1}{2}$ for k = 1,2,...,n + m - 1, which is impossible.

Therefore either fg or f ≡ - g.

This completes the proof of Theorem 1.2.

### 3.3 Proof of Theorem 1.3

Since f and g are entire functions, we have N(r, f) = N(r, g) = 0. Proceeding as in the proof of Theorem 1.1 and applying Lemma 2.6, we obtain that Theorem 1.3 holds.

### 3.4 Proof of Theorem 1.4

Since f and g are entire functions, we have N(r, f) = N(r, g) = 0. Proceeding as in the proof of Theorem 1.2 and applying Lemma 2.6, we can easily prove Theorem 1.4.

## Declarations

### Acknowledgements

The author want to thanks the referee for his/her thorough review and valuable suggestions toward improved of the paper. This work is supported in part by NSF of China (11071266), in part by NSF project of CQ CSTC (2010BB9218) and in part by ST project of CQEC (KJ110609).

## Authors’ Affiliations

(1)
College of Mathematics and Statistics, Chongqing University, Chongqing, 401331, People's Republic of China
(2)
College of Mathematics Science, Chongqing Normal University, Chongqing, 401331, People's Republic of China

## References 