© Wuhan University 2023

0 Introduction and Main Results

At the outset, it will be assumed that the standard definitions and terminologies of value distribution theory are known to the readers (see Ref.[1]). In the following, a meromorphic function $f\left(z\right)$ always means meromorphic in the whole complex plane. By $S(r,f)$, we denote any quantity satisfying $S(r,f)=o(T(r,f))$ as $r\to \mathrm{\infty }$ outside of an exceptional set $E$ with finite logarithmic measure. A meromorphic function $\alpha \left(z\right)$ is said to be a small function of $f$ if it satisfies $T(r,\alpha )=S(r,f)$. We say two nonconstant meromrophic functions $f$ and $g$ share the value$a$CM (IM) if $f-a$ and $g-a$ have the same zeros counting multiplicities (ignoring multiplicities). Let $k$ be a positive integer, we denote by $N_{k)}\left(r,\frac{\mathrm{1}}{f-a}\right)$ the counting function of a points of $f$ with multiplicity $\le k$ and by $N_{(k+\mathrm{1}}\left(r,\frac{\mathrm{1}}{f-a}\right)$ the counting function of a points of $f$ with multiplicity $>k$, where each a point is counted according to its multiplicity.

The following three theorems are classical and well-known results in the study of the uniqueness problems of meromorphic functions.

Theorem 1^[2] If two nonconstant meromorphic functions $f$ and $g$ share five distinct values $a_{\mathrm{1}},a_{\mathrm{2}},a_{\mathrm{3}},a_{\mathrm{4}},$$a_{\mathrm{5}}\in \mathbb{C}$$\cup \mathrm{\infty }$ IM, then $f\equiv g$.

Theorem 2^[2] If two nonconstant meromorphic functions $f$ and $g$ share four distinct values $a_{\mathrm{1}},a_{\mathrm{2}},a_{\mathrm{3}},$$a_{\mathrm{4}}\in \mathbb{C}\cup \mathrm{\infty }$ CM, then $f\equiv g$ or $f=T\circ g$, where $T$ is a Mobiüs transformation.

Theorem 3 ^[3] If two nonconstant meromorphic functions $f$ and $g$ share two values IM, and share two other values CM, then $f$ and $g$ share all values CM.

The assumption "4CM" in Theorem 2 has been improved to "2IM+2CM" by Gunderson ^[3]. However, Gunderson ^[4] gave an example to show that Theorem 2 is not true if the condition "4CM" is replaced by "4IM", and the question whether "4CM" in Theorem 2 can be replaced by "1CM+3IM" still remains open. Many researchers are devoted to studying the open problem. For partial progress on this, we refer the readers to survey ^[5]. Recently, Wang and Fang^[6] obtained some results from another direction.

Theorem 4 ^[6] Let $f$ and $g$ be nonconstant meromorphic functions, let$a$ be a finite nonzero value, and let $n$ be a positive integer satisfying $n\ge \mathrm{5}$. If $f^n\left(z\right)$ and $g^n\left(z\right)$ share $a$ CM, $f$ and $g$ share $\mathrm{\infty }$ CM, and $N_{\mathrm{1})}(r,f)=S(r,f)$, then either $f\left(z\right)\equiv tg\left(z\right)$ or $f\left(z\right)g\left(z\right)\equiv t$, where $t^n=\mathrm{1}$.

Naturally, a question arises:

Question 1 Can "2CM" be replaced by "1CM+1IM" in Theorem 4?

First, we give a partially answer to Question 1.

Theorem 5   Let $f$ and $g$ be nonconstant meromorphic functions, and let $n$ be a positive integer satisfying $n\ge \mathrm{3}$. Suppose that $f^n\left(z\right)$ and $g^n\left(z\right)$ share 1 CM, $f$ and $g$ share $\mathrm{\infty }$ IM, and $\overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,\frac{\mathrm{1}}{g})=S(r,f)+S(r,g)$, then either $f\left(z\right)\equiv tg\left(z\right)$ or $f\left(z\right)g\left(z\right)\equiv t$, where $t^n=\mathrm{1}$.

Remark 1   Actually, in the proof of Theorem 5, if we replace $f^n$ and $g^n$ with $\frac{f^n}{a}$ and $\frac{g^n}{a}$, then the condition that $f^n\left(z\right)$ and $g^n\left(z\right)$ share 1 CM can be improved to $f^n\left(z\right)$ and $g^n\left(z\right)$ share $a$ CM, where $a$ is a nonzero finite value.

To reduce the number of shared values quickly, many authors began to consider the case that $f(z)$ and $g\left(z\right)$ have some special relationship. For example, when $g\left(z\right)$ is the shift of $f(z)$, Heittokangas et al ^[7] obtained some uniqueness results in 2009. The background for these considerations lies in the development of the difference version to the usual Nevanlinna theory, especially the difference-type logarithmic derivative lemma, which starts in the papers ^[8-10].

Theorem 6^[7] Let $f(z)$ be a meromorphic function of finite order and $c\in \mathbb{C}$, and let $a_{\mathrm{1}},a_{\mathrm{2}},a_{\mathrm{3}}\in S(f)\cup \mathrm{\infty }$be three distinct periodic functions with period $c$. If $f(z)$ and $f(z+c)$ share $a_{\mathrm{1}},a_{\mathrm{2}}$ CM and $a_{\mathrm{3}}$ IM, then $f(z)=f(z+c)$.

Following that, several researchers began to study shared values problems related to a meromorphic function $f(z)$ and its difference operators, see Refs.[11-19]. Here, we recall the following results from Qi ^[18] and Chen et al^[20].

Theorem 7 ^[1] Let $f(z)$ be a meromorphic function of finite order, let $a,c$ be two finite nonzero values, and let $n$ be a positive integer satisfying $n\ge \mathrm{9}$. If ${\left[f^{\text{'}}(z)\right]}^n$ and $f^n(z+c)$ share $a$ CM, $f^{\text{'}}(z)$ and $f(z+c)$ share $\mathrm{\infty }$ CM, then $f^{\text{'}}(z)\equiv tf(z+c)$, where $t^n=\mathrm{1}$.

Theorem 8 ^[20] Let $f$be a meromorphic function of finite order, $c\in \mathbb{C}$, and let $n$ be a positive integer satisfying $n\ge \mathrm{9}$. If $f^n\left(z\right)$and ${\left[{\mathrm{\Delta }}_{c}f\right]}^n$share $a$ CM, $f$ and ${\mathrm{\Delta }}_{c}f$share $\mathrm{\infty }$ CM, then $f(z)\equiv tf(z+c)$, where $t^n=\mathrm{1}$, $t\ne -\mathrm{1}$.

Theorem 7   and 8 both require that $f(z)$ have finite order. As a result, we pose two questions:

Question 2 What can happen if $f(z)$ is of infinite order in Theorem 7-8?

Question 3 Is it possible to widen the range of $n$ in Theorem 7-8?

In the following, as an application of Theorem 5, we give two uniqueness results about a meromorphic function related to its shifts and difference operators, which are answers to the two questions.

Theorem 9   Let $f$ be a meromorphic function, and let $n$ be a positive integer satisfying $n\ge \mathrm{3}$. If $f^n(z+c)$ and $(f{(z))}^n$ share 1 CM, $f(z+c)$ and $f^{\text{'}}(z)$ share $\mathrm{\infty }$ IM, and $\overline{N}(r,\frac{\mathrm{1}}{f^{\text{'}}})+\overline{N}(r,\frac{\mathrm{1}}{f\left(z+c\right)})=S(r,f)$, then either $f(z+c)\equiv$$t{f}^{\text{'}}(z)$, where $t^n=\mathrm{1}$.

Theorem 10   Let $f$ be a meromorphic function, and let $n$ be a positive integer satisfying $n\ge \mathrm{3}$ and $c\in \mathbb{C}\backslash \left\{\mathrm{0}\right\}$. If $f^n\left(z\right)$ and $({\mathrm{\Delta }}_c{f)}^n$share 1 CM, $f$ and ${\mathrm{\Delta }}_{c}f$ share $\mathrm{\infty }$IM, $\overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,\frac{\mathrm{1}}{{\mathrm{\Delta }}_{c}f})=S(r,f)$, then $f(z)\equiv tf(z+c)$, where $t^n=\mathrm{1}$.

1 Preliminary Lemmas

To prove our results of this paper, the following lemmas are required.

Lemma 1 ^[21] Let $f(z)$ be a nonconstant meromorphic function in the complex plane. Let $q\ge \mathrm{3}$ and $a_{\mathrm{1}}\cdots a_q$ be $q$ distinct finite complex numbers. Then

$(q-\mathrm{2})T(r,f)\le {\displaystyle \sum _{j=\mathrm{1}}^q}\overline{N}(r,\frac{\mathrm{1}}{f-a_q})-N_{\mathrm{1}}(r)+S(r,f)$

where $N_{\mathrm{1}}(r)=(\mathrm{2}N(r,f)-N(r,f^{\text{'}}))+N(r,\frac{\mathrm{1}}{f^{\text{'}}})$.

Lemma 2   Let $f$ and $g$ be nonconstant meromorphic functions with $\overline{N}(r,f)=S(r,f)$ and $\overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,\frac{\mathrm{1}}{g})=S(r,f)+S(r,g)$. Suppose that $f^n$ and $g^n$ share 1 CM, and $f$ and $g$ share $\mathrm{\infty }$ IM. If

$N_{(\mathrm{2}}(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\ne S(r,f)+S(r,g)$(1)

Then $f^n=g^n$.

Proof   Set

$\omega (z)=\frac{{\left(f^n\right)}^{\mathrm{\text{'}}}}{f^n}-\frac{{\left(g^n\right)}^{\mathrm{\text{'}}}}{g^n}$(2)

From (2) and the lemma of logarithmic derivative, we have $m(r,\omega )=S(r,f)+S(r,g)$. If $\omega (z)\cancel{\equiv }\mathrm{0}$, then we see that the possible poles of $\omega (z)$ can occur at the poles of $f^n$ and $g^n$, and zeros of $f^n$ and $g^n$. Therefore, we get

$\begin{array}{l}N(r,\omega )\le \overline{N}(r,f^n)+\overline{N}(r,g^n)+\overline{N}(r,\frac{\mathrm{1}}{f^n})\\ +\overline{N}(r,\frac{\mathrm{1}}{g^n})+S(r,f)+S(r,g)\end{array}$(3)

Since $\overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,\frac{\mathrm{1}}{g})=S(r,f)+S(r,g)$, we get

$N(r,\omega )\le \overline{N}(r,f)+\overline{N}(r,g)+S(r,f)+S(r,g)$

Then from the assumption that $f$ and $g$ share $\mathrm{\infty }$ IM and $\overline{N}(r,f)=S(r,f)$, we have

$N(r,\omega )\le \mathrm{2}\overline{N}(r,f)+S(r,f)+S(r,g)\le S(r,f)+S(r,g)$

If $z_{\mathrm{0}}$ is a 1 point of $f^n$ with multiplicities $\ge \mathrm{2}$, then by a short caculation with laurent series and (2) we see that $z_{\mathrm{0}}$ is a zero of $\omega \left(z\right)$. Since $f^n$and $g^n$ share 1 CM, we know the 1 points of $f^n$and $g^n$are not poles of $\omega \left(z\right)$. From (2) and Nevanlinna's first fundamental theorem, it is easy to deduce that

$\begin{array}{l}{N}_{(\mathrm{2}}(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\le N(r,\frac{\mathrm{1}}{\omega })\le T(r,\omega )\\ \le N(r,\omega )+S(r,f)+S(r,g)\le S(r,f)+S(r,g)\end{array}$

It is a contradiction.

Thus $\omega (z)\equiv \mathrm{0}$. From (2) we have

$f^n=c{g}^n$(4)

where $c$ is a nonconstant. By the assumption that $N_{(\mathrm{2}}(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\ne S(r,f)+S(r,g)$, there exists $z_{\mathrm{0}}$ such that $f(z_{\mathrm{0}})=g(z_{\mathrm{0}})=\mathrm{1}$, so $c=\mathrm{1}$.

Lemma 3   Let $f$ and $g$ be nonconstant meromorphic functions with $\overline{N}(r,f)=S(r,f)$ and $\overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,\frac{\mathrm{1}}{g})=S(r,f)+S(r,g)$. Suppose that $f^n$and $g^n$ share 1 CM, and $f$ and $g$ share $\mathrm{\infty }$ IM. If $f^n$ is not a Mobiüs transformation of $g^n$, then

$N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\le N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})+N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(g^n\right)}^{\mathrm{\text{'}}}})+S(r,f)+S(r,g)$(5)

where $N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})$ denotes the zeros of ${\left(f^n\right)}^{\mathrm{\text{'}}}$ which are not zeros of $f^n\left(f^n-\mathrm{1}\right)$.

Proof   Set

$\phi (z)=\frac{{\left(f^n\right)}^{\mathrm{\text{'}}\mathrm{\text{'}}}}{{\left(f^n\right)}^{\mathrm{\text{'}}}}-\mathrm{2}\frac{{\left(f^n\right)}^{\mathrm{\text{'}}}}{f^n-\mathrm{1}}-\left(\frac{{\left(g^n\right)}^{\mathrm{\text{'}}\mathrm{\text{'}}}}{{\left(g^n\right)}^{\mathrm{\text{'}}}}-\mathrm{2}\frac{{\left(g^n\right)}^{\mathrm{\text{'}}}}{g^n-\mathrm{1}}\right)$(6)

If $\phi \left(z\right)\equiv \mathrm{0}$, then from (6), we have $f^n$ is a Mobiüs transformation of $g^n$, a contradiction. Thus, $\phi \left(z\right)\cancel{\equiv }\mathrm{0}$, and $m(r,\phi )=S(r,f)+S(r,g)$. First, we know

$N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\le N_{\mathrm{1})}(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})+N_{(\mathrm{2}}(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})+S(r,f)$(7)

From Lemma 2, we deduce that $N_{(\mathrm{2}}(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})=S(r,f)+S(r,g)$. If not, then $f^n$ is a transformation of $g^n$, a contradiction. Moreover, by a short calculation with Laurent series, it is obvious that the simple 1 points of $f^n$ must be zeros of $\phi (z)$.Therefore, we can deduce from (7) that

$\begin{array}{l}N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\le N(r,\frac{\mathrm{1}}{\phi })+S(r,f)\le T(r,\phi )+S(r,f)\\ =N(r,\phi )+S(r,f)+S(r,g)\end{array}$

Next, we need to estimate the poles of $\phi$. By a calculation with Laurent series, $f^n$ and $g^n$ share 1 CM, and $f$ and $g$ share ∞IM, we get that the poles of $\phi$ can only occur at the poles of $f^n$ and $g^n$, the zeros of $f^n$ and $g^n$, the zeros of ${\left(f^n\right)}^{\mathrm{\text{'}}}$ which are not zeros of $f^n\left(f^n-\mathrm{1}\right)$, and the zeros of of ${\left(g^n\right)}^{\mathrm{\text{'}}}$ which are not zeros of $g^n\left(g^n-\mathrm{1}\right)$. Hence, we can get

$\begin{array}{l}N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\le \overline{N}(r,f^n)+\overline{N}(r,g^n)+\overline{N}(r,\frac{\mathrm{1}}{f^n})+\overline{N}(r,\frac{\mathrm{1}}{g^n})\\ +N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})+N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(g^n\right)}^{\mathrm{\text{'}}}})+S(r,f)+S(r,g)\\ \le \mathrm{2}\overline{N}(r,f)+\overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,\frac{\mathrm{1}}{g})+N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})\\ +N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(g^n\right)}^{\mathrm{\text{'}}}})+S(r,f)+S(r,g)\end{array}$

From the assumption that $f$ and $g$ share $\mathrm{\infty }$ IM, $\overline{N}(r,f)=S(r,f)$, and $\overline{N}(r,\frac{\mathrm{1}}{f})+$$\overline{N}(r,\frac{\mathrm{1}}{g})=S(r,f)+S(r,g)$, so we have

$N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\le N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})+N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(g^n\right)}^{\mathrm{\text{'}}}})+S(r,f)+S(r,g)$

This completes the proof of Lemma 3.

2 Proof of Theorem 5

Proof   Set

$\varphi (z)=\frac{{\left(f^n\right)}^{\mathrm{\text{'}}}}{f^n\left(f^n-\mathrm{1}\right)}-\frac{{\left(g^n\right)}^{\mathrm{\text{'}}}}{g^n\left(g^n-\mathrm{1}\right)}$(8)

Now we consider the following cases.

Case 1 $\varphi (z)\equiv \mathrm{0}$. By (8) we get

$\frac{f^n-\mathrm{1}}{f^n}=C\frac{g^n-\mathrm{1}}{g^n}$(9)

where $C\ne \mathrm{0}$ is a constant.

If $C=\mathrm{1}$, from (9) we get $f^n={\mathrm{g}}^n$, and thus $f\left(z\right)\equiv tg\left(z\right)$, where $t^n=\mathrm{1}$.

If $C\ne \mathrm{1}$, by (9) we obtain

$\frac{\mathrm{1}}{f^n}-\frac{C}{g^n}=\mathrm{1}-C$(10)

Clearly, this leads to

$T(r,f)=T(r,g)+S(r,f),S(r,f)=S(r,g)$(11)

By Lemma 1, (10) and the assumption that $f,g$ share $\mathrm{\infty }$ IM, we get

$nT(r,f)=T(r,f^n)$

$\begin{array}{l}\le \overline{N}(r,f^n)+\overline{N}(r,\frac{\mathrm{1}}{f^n})+\overline{N}(r,\frac{\mathrm{1}}{f^n-\frac{\mathrm{1}}{\mathrm{1}-C}})+S(r,f)\\ \le \overline{N}(r,f)+\mathrm{2}\overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,g)+S(r,f)\\ \le \mathrm{2}\overline{N}(r,f)+\mathrm{2}\overline{N}(r,\frac{\mathrm{1}}{f})+S(r,f)\\ \le \mathrm{2}T(r,f)+S(r,f)+S(r,g)\end{array}$

Similarly, we have

$nT(r,g)=T(r,g^n)$

$\begin{array}{l}\le \overline{N}(r,g^n)+\overline{N}(r,\frac{\mathrm{1}}{g^n})+\overline{N}(r,\frac{\mathrm{1}}{g^n-\frac{C}{\mathrm{1}-C}})+S(r,g)\\ \le \overline{N}(r,g)+\mathrm{2}\overline{N}(r,\frac{\mathrm{1}}{g})+\overline{N}(r,f)+S(r,g)\\ \le \mathrm{2}\overline{N}(r,g)+\mathrm{2}\overline{N}(r,\frac{\mathrm{1}}{g})+S(r,g)\\ \le \mathrm{2}T(r,g)+S(r,f)+S(r,g)\end{array}$

Combining the above two inequalities, we have

$n(T(r,g)+T(r,f))\le \mathrm{2}(T(r,g)+T(r,f))+S(r,f)+S(r,g)$

which contradicts the condition $n>\mathrm{3}$.

Case 2 $\varphi (z)\ne \mathrm{0}$. If $f$ has a pole $z_{\mathrm{0}}$ with multiplicity $p(\ge \mathrm{1})$, $z_{\mathrm{0}}$ is a pole of $g$ with multiplicity $q(\ge \mathrm{1})$, then $z_{\mathrm{0}}$ is a pole of $f^n$ with multiplicity $np$, and it is also a pole of $g^n$ with multiplicity $nq$. By a short calculation, we deduce that $z_{\mathrm{0}}$ is a zero of $\varphi (z)$. Since $f$ and $g$ share $\mathrm{\infty }$ IM, we have

$\begin{array}{l}\overline{N}(r,f)=\overline{N}(r,g)\le \overline{N}(r,\frac{\mathrm{1}}{\varphi })+S(r,f)+S(r,g)\\ \le T(r,\varphi )+S(r,f)+S(r,g)\\ \le m(r,\varphi )+N(r,\varphi )+S(r,f)+S(r,g)\end{array}$(12)

Clearly, from (8), we see that $m(r,\varphi )=S(r,f)+S(r,g)$. On the other hand, since $f^n$ and $g^n$ share 1 CM, by a calculation with Laurent series, we know the 1 points of $f^n$ and $g^n$ are not the poles of $\varphi$. Thus, the poles of $\varphi$ can only occur at the zeros of $f$ and $g$. Therefore, from (12) and the hypothesis that $\overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,\frac{\mathrm{1}}{g})=S(r,f)+S(r,g)$, we deduce

$\begin{array}{l}\overline{N}(r,f)=\overline{N}(r,g)\le \overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,\frac{\mathrm{1}}{g})+S(r,f)+S(r,g)\\ =S(r,f)+S(r,g)\end{array}$(13)

Next, we still discuss two cases.

Case 2.1 If $f^n$ is not a Mobiüs transformation of $g^n$, then from Lemma 3, we get

$\begin{array}{l}N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\le N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})+N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(g^n\right)}^{\mathrm{\text{'}}}})\\ +S(r,f)+S(r,g)\end{array}$(14)

and

$\begin{array}{l}N(r,\frac{\mathrm{1}}{g^n-\mathrm{1}})\le N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})+N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(g^n\right)}^{\mathrm{\text{'}}}})\\ +S(r,f)+S(r,g)\end{array}$(15)

Thus, by Lemma 1, we obtain

$\begin{array}{l}T(r,f^n)\le \overline{N}(r,f^n)+\overline{N}(r,\frac{\mathrm{1}}{f^n})+N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\\ -N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})+S(r,f)\\ \le \overline{N}(r,f)+\overline{N}(r,\frac{\mathrm{1}}{f})+N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})\\ -N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})+S(r,f)\end{array}$(16)

Similarly,

$\begin{array}{l}T(r,g^n)\le \overline{N}(r,g)+\overline{N}(r,\frac{\mathrm{1}}{g})+N(r,\frac{\mathrm{1}}{g^n-\mathrm{1}})\\ -N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(g^n\right)}^{\mathrm{\text{'}}}})+S(r,g)\end{array}$(17)

By (13)-(17), we get

$\begin{array}{l}T(r,f^n)+T(r,g^n)\\ \le \overline{N}(r,f)+\overline{N}(r,g)+\overline{N}(r,\frac{\mathrm{1}}{f})+\overline{N}(r,\frac{\mathrm{1}}{g})\\ +N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})+N(r,\frac{\mathrm{1}}{g^n-\mathrm{1}})-N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(f^n\right)}^{\mathrm{\text{'}}}})\end{array}$

$\begin{array}{l}-N_{\mathrm{0}}(r,\frac{\mathrm{1}}{{\left(g^n\right)}^{\mathrm{\text{'}}}})+S(r,f)+S(r,g)\\ \le \frac{\mathrm{1}}{\mathrm{2}}\left(N(r,\frac{\mathrm{1}}{f^n-\mathrm{1}})+N(r,\frac{\mathrm{1}}{g^n-\mathrm{1}})\right)+S(r,f)+S(r,g)\\ \le \frac{\mathrm{1}}{\mathrm{2}}\left\{T(r,f^n)+T(r,g^n)\right\}+S(r,f)+S(r,g)\end{array}$

It is a contradiction.

Case 2.2 If $f^n$ is a Mobi$\ddot{\mathrm{u}}$'s transformation of $g^n$, that is

$f^n=\frac{a{g}^n+b}{c{g}^n+d}$(18)

where $a,b,c,d$ are constants, and $ad-bc\ne \mathrm{0}$.

Next we discuss following two cases.

Case 2.2.1 $c=\mathrm{0}$. Thus $ad\ne \mathrm{0}$. From (18), we have $f^n=\frac{a}{d}{g}^n+\frac{b}{d}$. If $b\ne \mathrm{0}$, $d{f}^n=a\left(g^n+\frac{b}{a}\right)$, then $\overline{N}(r,\frac{\mathrm{1}}{f^n})=\overline{N}(r,\frac{\mathrm{1}}{g^n+\frac{b}{a}})$. By Nevanlinna's second fundamental theorem and (13), we get

$\begin{array}{l}T(r,g^n)\le \overline{N}(r,g^n)+\overline{N}(r,\frac{\mathrm{1}}{g^n})+\overline{N}(r,\frac{\mathrm{1}}{g^n+\frac{b}{a}})+S(r,g)\\ \le \overline{N}(r,g)+\overline{N}(r,\frac{\mathrm{1}}{g})+\overline{N}(r,\frac{\mathrm{1}}{f})+S(r,g)\\ \le S(r,f)+S(r,g)\end{array}$

It is a contradiction.

Hence $b=\mathrm{0}$, so $f^n=\frac{a}{d}{g}^n$. If $f^n$ does not have 1 point, it is easy to get a contradiction by Nevanlinna's second fundamental theorem. So there exists $z_{\mathrm{0}}$ such that $f^n\left(z_{\mathrm{0}}\right)=g^n\left(z_{\mathrm{0}}\right)=\mathrm{1}$. Thus we get $\frac{a}{d}=\mathrm{1}$, that is $f^n=g^n$, $f=tg$, where $t^n=\mathrm{1}$.

Case 2.2.2 $c\ne \mathrm{0}$. If $d\ne \mathrm{0}$, then from (18) and $f,g$ share $\mathrm{\infty }$ IM, we obtain $f\ne \mathrm{\infty },g\ne \mathrm{\infty }$, $g^n\ne -\frac{d}{c}$. By Nevanlinna's second fundamental theorem, we get a contradiction.

So $d=\mathrm{0}$, then $b\ne \mathrm{0}$. From (18), we have $f^n=\frac{a{g}^n+b}{c{g}^n}$, $c{f}^n{g}^n=a{g}^n+b$. It is easy to get $f\ne \mathrm{\infty }$ and $g\ne \mathrm{\infty }$. If $a\ne \mathrm{0}$, we get $g^n\ne -\frac{b}{a}$, which contradicts with Nevanlinna's second fundamental theorem. So $a=\mathrm{0}$, then $f^n{g}^n=\frac{b}{c}$. By similar reasoning as in Case 2.2.1, we can get $\frac{b}{c}=\mathrm{1}$. Thus we get $f^n{g}^n=\mathrm{1}$, $fg=t$, where $t^n=\mathrm{1}$. This completes the proof of Theorem 5.

3 Proof of Theorem 9

By Theorem 5, we get $f(z+c)f^{\text{'}}(z)=t$ or $f(z+c)=t{f}^{\text{'}}(z)$, where $t^n=\mathrm{1}$. If $f(z+c)f^{\text{'}}(z)=t$,$t^n=\mathrm{1}$. From the assumption that $f(z+c)$, $f^{\text{'}}(z)$ share $\mathrm{\infty }$ IM, we obtain $f(z)\ne \mathrm{0},\mathrm{\infty }$ and $f^{\text{'}}(z)\ne \mathrm{0},\mathrm{\infty }$. Otherwise, if $f(z)=\mathrm{0}$ or $\mathrm{\infty }$, then there exists $z_{\mathrm{0}}$ such that $f(z_{\mathrm{0}})=\mathrm{0}$ or $\mathrm{\infty }$, then $f(z_{\mathrm{0}})f^{\text{'}}(z_{\mathrm{0}}-c)=\mathrm{0}$ or $\mathrm{\infty }$. This is a contradiction. Similarly, we have $f^{\text{'}}(z)\ne \mathrm{0},f^{\text{'}}(z)\ne \mathrm{\infty }$. Thus $f(z)={\mathrm{e}}^{\alpha \left(z\right)}$, where $\alpha \left(z\right)$ be a nonconstant entire function, and hence $f^{\text{'}}(z)={\alpha }^{\text{'}}\left(z\right){\mathrm{e}}^{\alpha \left(z\right)}$. Since $f^{\text{'}}(z)\ne \mathrm{0},\mathrm{\infty }$, we can deduce that ${\alpha }^{\text{'}}\left(z\right)={\mathrm{e}}^{\beta \left(z\right)}$, where $\beta \left(z\right)$ is a nonconstant entire function. Hence, we have ${\alpha }^{\text{'}}\left(z\right){\mathrm{e}}^{\alpha \left(z\right)+\alpha \left(z+c\right)}=t$.