A Malmquist–Steinmetz Theorem for Difference Equations

Abstract

It is shown that if the equation

$$\begin{aligned} f(z+1)^n=R(z,f), \end{aligned}$$

where R(z, f) is rational in both arguments and $\deg _f(R(z,f))\not =n$, has a transcendental meromorphic solution, then the equation above reduces into one out of several types of difference equations where the rational term R(z, f) takes particular forms. Solutions of these equations are presented in terms of Weierstrass or Jacobian elliptic functions, exponential type functions or functions which are solutions to a certain autonomous first-order difference equation having meromorphic solutions with preassigned asymptotic behavior. These results complement our previous work on the case $\deg _f(R(z,f))=n$ of the equation above and thus provide a complete difference analogue of Steinmetz’ generalization of Malmquist’s theorem.

1 Introduction

Nevanlinna theory (see, e.g., [10]) is a powerful tool when analyzing meromorphic solutions of complex differential equations. For example, by utilizing Nevanlinna theory, Yosida [28] and Laine [14] provided elegant alternate proofs of the classical Malmquist theorem on first-order differential equations [16], while Steinmetz [22], and Bank and Kaufman [3] gave a precise classification of the differential equation

$$\begin{aligned} (f')^n=R(z,f), \end{aligned}$$

(1.1)

where $n\in \mathbb {N}$ and R(z, f) is rational in both arguments. See also [15, Chapter 10] for Malmquist–Yosida–Steinmetz type theorems. In [13], we studied a natural difference analogue of Eq. (1.1), i.e., the first-order difference equation

$$\begin{aligned} f(z+1)^n=R(z,f), \end{aligned}$$

(1.2)

where $n\in \mathbb {N}$ and R(z, f) is rational in f with small functions of f as coefficients. Assuming that (1.2) has a meromorphic solution and $\deg _f(R(z,f))=n$, we showed that Eq. (1.2) can, by a bilinear transformation in f, be transformed into one in a list of twelve equations. In particular, we considered meromorphic solutions of hyper-order less than 1 of (1.2) and showed that if such a solution exists, then Eq. (1.2) with rational coefficients has to reduce into the difference linear or Riccati equation, or one in a list of five equations including Fermat type difference equations and a special case of the symmetric Quispel–Roberts–Thompson (QRT) map [19, 20]. We also showed that these five equations are explicitly solved in terms of Weierstrass or Jacobian elliptic functions, or of functions which are solutions of certain difference Riccati equations. These results provide a natural difference analogue of Steinmetz’ generalization of Malmquist’s theorem in the sense of Ablowitz, Halburd and Herbst [1], who proposed that the existence of sufficiently many finite-order meromorphic solutions of a difference equation is a good difference analogue of the Painlevé property. Their idea was refined and successfully implemented by Halburd and the second author [8] on the second-order difference equation

$$\begin{aligned} f(z+1)+f(z-1)=R(z,f), \end{aligned}$$

(1.3)

where R(z, f) is rational in f with small functions of f as coefficients, reducing equation (1.3) into a short list of canonical equations, including the difference Painléve I and II equations. The finite-order condition of the proposed difference Painléve property was relaxed into hyper-order strictly less than one by Halburd, the second author and Tohge [9], and recently into hyper-order equal to one with minimal hyper type by Zheng and the second author [29].

The purpose of this paper is to find out all transcendental meromorphic solutions for the case $\deg _f(R(z,f))\not =n$ of Eq. (1.2) without growth conditions and provide a complete difference analogue of Steinmetz’ generalization of Malmquist’s theorem. As is pointed out in [13], in this case all the transcendental meromorphic solutions of (1.2) are of hyper-order at least one. Our work is a continuation of many mathematicians’ research on first-order difference equations. For example, in [21], Shimomura showed that the difference equation

$$\begin{aligned} f(z+1) = P(f(z)), \end{aligned}$$

(1.4)

where P(f(z)) is a polynomial in f(z) with constant coefficients, always has a non-trivial entire solution; in [26], Yanagihara showed that the difference equation

$$\begin{aligned} f(z+1) = R(f(z)), \end{aligned}$$

(1.5)

where R(f(z)) is rational in f(z) having constant coefficients, has a non-trivial meromorphic solution with preassigned asymptotic behavior in a sector for all choices of $R\not \equiv 0$. Yanagihara [26] also showed that if (1.5), where R(f(z)) is replaced by R(z, f), which is rational in both arguments, has a transcendental meromorphic solution of hyper-order less than 1, then (1.5) must reduce into the difference Riccati equation. This can be viewed as a natural difference analogue of Malmquist’s theorem on first-order differential equations. For the higher-degree equation (1.2), the classification work in the special case where the right-hand side (RHS) of (1.2) is a polynomial in f(z) with constant coefficients has been done by Nakamura and Yanagihara [18] and Yanagihara [27]. The results obtained in the present paper, supplemented with results of the first part of this study [13], can be summarized as follows.

Theorem 1

If Eq. (1.2), where R(z, f) is rational in both arguments, has a transcendental meromorphic solution, then (1.2) can be reduced into the case $n=1$ or one out of 27 equations in Theorems 3–7 below and in [13, Theorem 2].

Here we have to point out that the list of equations in [13, Theorem 2] is not complete, since one equation in the case $n=2$ is left out in the proof there; see [30]. Equation (1.2) with $n=1$ actually includes 3 equations, namely the Eq. (2.4) below and two equations in [13, Theorem 2]. Autonomous versions of them are just (1.4) and (1.5). In other cases, we have counted each equation in Theorems 3–7 below once even when some of them have the same form but appear in two different theorems. Moreover, autonomous versions of all the 27 equations can be solved in terms of elliptic and elementary functions.

In this paper, we shall confine ourselves to considering equation (1.2) with rational coefficients. The main tools from Nevanlinna theory we use, both in our previous paper [13] and in the present one, are the generalizations of Nevanlinna’s second main theorem given by Yamanoi [24, 25]. We refer to [10, pp. 42–43] for the standard definitions of $\delta (a,f)$, $\theta (a,f)$ and $\varTheta (a,f)$, etc. Recall that a value $a\in \mathbb {C}\cup \{\infty \}$ is said to be a completely ramified value of f(z) when $f(z)-a=0$ has no simple roots. Denote the field of rational functions by $\mathcal {R}$ and set $\hat{\mathcal {R}}=\mathcal {R}\cup \{\infty \}$. Throughout the paper, we say that $c(z)\in \hat{\mathcal {R}}$ is a completely ramified rational function of a transcendental meromorphic function f(z) when the equation $f(z)=c(z)$ has at most finitely many simple roots and that c(z) is a Picard exceptional rational function of f(z) when $N(r,c,f)=O(\log r)$. We also say that c(z) has multiplicity at least m if all the roots of $f(z)=c(z)$ have multiplicity at least m with at most finitely many exceptions. As is mentioned in [13], the main theorem in [25] yields that the inequality

$$\begin{aligned} \sum _{i=1}^q\varTheta (c_i,f)\le 2 \end{aligned}$$

(1.6)

holds for any collection of $c_1,\cdots ,c_q\in \hat{\mathcal {R}}$ when f is transcendental. Moreover, we have

Theorem 2

A non-constant transcendental meromorphic function f(z) can have at most four completely ramified rational functions.

As in [13], when considering a meromorphic solution f(z) of (1.2), we will do a transformation to f using some algebraic functions and end up in a situation such that the considered functions are meromorphic on a finite-sheeted Riemann surface. Such functions are called algebroid functions (see, e.g., [11]). We will ignore considering the degree of these functions in this paper since in any case we are dealing with at most finitely many algebraic branch points and the inequality (1.6) and Theorem 2 hold true for such functions.

The remainder of this paper is organized in the following way. In Sect. 2, we will first set up some notation and build several lemmas concerning the roots of the numerator and the denominator of R(z, f) in (1.2). Our main results, i.e., Theorems 3–7, and the proofs for them are distributed in Sects. 3 and 4, respectively, where different cases of Eq. (1.2) are treated. Moreover, for all the equations we find we will present explicit solutions to them in the autonomous case. These solutions are presented in the corresponding theorems, however, for the elliptic solutions, in particular, we give a detailed discussion in Sect. 5 separately.

For simplicity, from now on we will use the suppressed notations: $f=f(z)$, $\underline{f}=f(z-1)$ and $\overline{f}=f(z+1)$ for a meromorphic, algebraic or algebroid, function f(z).

2 Lemmas on the Roots

We begin to consider the transcendental meromorphic solution f of the following difference equation:

$$\begin{aligned} \overline{f}^n=R(z,f), \end{aligned}$$

(2.1)

where $n\in \mathbb {N}$ and R(z, f) is rational in both arguments and $\deg _{f} (R(z,f))\not =n$. We first set up some notation for (2.1) that will be used throughout the proofs in the following sections. We denote

$$\begin{aligned} R(z,f) = \frac{P(z,f)}{Q(z,f)}, \end{aligned}$$

where

$$\begin{aligned} P(z,f) = a_pf^p + a_{p-1}f^{p-1}+\cdots +a_0 \end{aligned}$$

(2.2)

and

$$\begin{aligned} Q(z,f) = b_qf^q + b_{q-1}f^{q-1}+\cdots +b_0 \end{aligned}$$

(2.3)

are two polynomials in f having no common factors, $p,q\in \mathbb {N}$ and the coefficients $a_p$, $\cdots$, $a_0$, $b_q$, $\cdots$, $b_0$ are rational functions. We have $\deg _f (P(z,f))=p$ and $\deg _f (Q(z,f))=q$. Denote $d=\deg _{f}(R(z,f))$. Then $d=\max \{p,q\}\not =n$. If $n=1$, then Eq. (2.1) is just

$$\begin{aligned} \overline{f}=R(z,f). \end{aligned}$$

(2.4)

As mentioned in the introduction, in the autonomous case, equation (2.4) always has an entire or meromorphic solution independently of the degree of R. Therefore, we always assume that $n\ge 2$ in what follows. Under this assumption, if the degree of R(z, f) in f in (2.1) equals 1, then we have

$$\begin{aligned} \underline{f}=\frac{af^n+b}{cf^n+d}, \end{aligned}$$

where the coefficients are rational functions such that $ad-bc\not =0$. The equation above is included in (2.4) under the transformation $x=-z$. So we also always assume that $d\ge 2$. We write P(z, f) and Q(z, f) in the algebraic factorization form

$$\begin{aligned} P(z,f) = a_p\prod _{i=1}^{M}P_i(z,f)^{\mu _i}, \end{aligned}$$

(2.5)

and

$$\begin{aligned} Q(z,f) =b_q\prod _{j=1}^{N}Q_j(z,f)^{\nu _j}, \end{aligned}$$

(2.6)

where $P_i(z,f)$ and $Q_j(z,f)$ are irreducible polynomials in f of the form in (2.2) or (2.3), $\mu _i,\nu _j\in \mathbb {N}$ and $\mu _1+\cdots +\mu _M=p$ and $\nu _1+\cdots +\nu _N=q$. In this paper, we shall use the term ’a polynomial in f’ which means that the polynomial in f has rational or algebraic coefficients. Note that the inequality (1.6) also holds when f is replaced by an algebroid function with finitely many branch points and $c_1$, $\cdots$, $c_q$ are replaced by algebraic functions. For convenience, in the following we always use the terms ’completely ramified rational function’ and ’Picard exceptional rational function’ of f even though sometimes they actually refer to algebraic functions. We also write (2.5) and (2.6) as

$$\begin{aligned} P(z,f) = P_0(z,f)^n(f-\alpha _1)^{k_1}\cdots (f-\alpha _\mu )^{k_\mu } \end{aligned}$$

(2.7)

and

$$\begin{aligned} Q(z,f) =Q_0(z,f)^n(f-\beta _1)^{l_1}\cdots (f-\beta _\nu )^{l_\nu }, \end{aligned}$$

(2.8)

where $\alpha _1$, $\cdots$, $\alpha _\mu$, $\beta _1$, $\cdots$, $\beta _\nu$ are in general algebraic functions and $k_i$ and $l_j$ denote the orders of the roots $\alpha _i$ and $\beta _j$, respectively, and satisfy $n\not \mid k_i$ and $n\not \mid l_j$, and

$$\begin{aligned} P_0(z,f) =a_p^{1/n}\prod _{i=1,n\mid \mu _i}^{M}P_i(z,f)^{\mu _i/n}=a_p^{1/n}\prod _{i=1,n\mid \mu _i}^{M}(f-\alpha _{\mu _{i,1}})^{\mu _i/n}\cdots (f-\alpha _{\mu _{i,p_i}})^{\mu _i/n} \end{aligned}$$

and

$$\begin{aligned} Q_0(z,f) =b_q^{1/n}\prod _{j=1,n\mid \nu _j}^{N}Q_j(z,f)^{\nu _j/n}=b_p^{1/n}\prod _{j=1,n\mid \nu _j}^{N}(f-\beta _{\nu _{j,1}})^{\nu _j/n}\cdots (f-\beta _{\nu _{j,q_j}})^{\nu _j/n} \end{aligned}$$

are two polynomials in f of degrees $p_0$ and $q_0$, respectively, with $\alpha _{\mu _{i,1}}$, $\cdots$, $\alpha _{\mu _{i,p_i}}$, $\beta _{\nu _{j,1}}$, $\cdots$, $\beta _{\nu _{j,q_j}}$ being in general algebraic functions, and $\mu _i$ and $\nu _j$ denoting the orders of these roots of P(z, f) or Q(z, f), respectively. We have $p_0=\deg _f(P_0(z,f))$ and $q_0=\deg _f(Q_0(z,f))$. Note that $a_p^{1/n},b_p^{1/n}$ are in general algebraic functions. For convenience, when $q\ge 1$, we always suppose that $b_q=1$. Also note that $\alpha _i$ and $\beta _j$ are neither roots of $P_0(z,f)$ nor roots of $Q_0(z,f)$. Now, if $n\not \mid \mu _i$ for all $i=1,\cdots , M$ or $n\not \mid \nu _j$ for all $j=1,\cdots ,N$, then $P_0(z,f)^n=a_p$ or $Q_0(z,f)^n =b_q$. On the other hand, if there are no such $\mu _i$ or $\nu _j$ that $n\not \mid \mu _i$ and $n\not \mid \nu _j$, then after taking the nth root on both sides of (2.1) we get equation (2.4). So in the sequel we always suppose that there is at least one such $\mu _i$ or $\nu _j$. The combined number of $\alpha _i$ and $\beta _j$ in (2.7) and (2.8) is $\mu +\nu$ and, for convenience, we always denote $N_c:=\mu +\nu$ even when there is no such $\alpha _i$ or $\beta _j$. Moreover, we may suppose that the greatest common divisor of n, $k_1$, $\cdots$, $k_\mu$, $l_1$, $\cdots$, $l_\nu$, which is denoted by $k=(n,k_1,\cdots ,k_\mu ,l_1,\cdots ,l_\nu )$, is 1. Otherwise, after taking the kth root on both sides of (2.1), we get a new equation of the same form as (2.1) with the power of $\overline{f}$ being n/k.

With the notation above, we are now ready to construct lemmas regarding the roots of P(z, f) and Q(z, f) through elementary multiplicity analysis on f. For simplicity, in what follows, when considering the zeros, poles or $\alpha _i$-points of f, etc., we will omit giving the corresponding Taylor or Laurent series expansions for f. The first lemma below provides some basic upper bounds for $N_c$.

Lemma 1

Let f be a transcendental meromorphic solution of Eq. (2.1). Then $\alpha _i$ is either a Picard exceptional rational function of f or a completely ramified rational function of f with multiplicity $n/(n,k_i)$ and $\beta _j$ is either a Picard exceptional rational function of f or a completely ramified rational function of f with multiplicity $n/(n,l_j)$. Moreover, if $q=0$, then $\infty$ is a Picard exceptional rational function of f and $N_c\le 2$ and, in particular, if $n\ge 3$, then $N_c=1$; if $q\ge 1$, then $N_c\le 4$ and, in particular, if $n\ge 3$, then $N_c\le 3$; if $q\ge 1$ and $n\not \mid |p-q|$, then $\infty$ is either a Picard exceptional rational function of f or a completely ramified rational function of f with multiplicity $n/(n,|p-q|)$ and $N_c\le 3$.

Proof

By making use of the factorizations (2.7) and (2.8), it follows that $\alpha _1$, $\cdots$, $\alpha _\mu$ and $\beta _1$, $\cdots$, $\beta _\nu$ are roots of P(z, f) and Q(z, f), respectively. For each $\alpha _i$, if $\alpha _i$ is not a Picard exceptional rational function of f, then we let $z_0\in \mathbb {C}$ be such that $f(z_0)-\alpha _i(z_0)=0$ with multiplicity $m\in \mathbb {Z}^{+}$. We write $k_i=nk_{i1}+k_{i2}$, where $k_{i1},k_{i2}\in \mathbb {N}$ and $k_{i2}<n$. Note that $(n,k_{i2})=(n,k_i)$. Now, $n|mk_{i2}$ with at most finitely many exceptions since otherwise $z_0+1$ would be a branch point of f. Hence, $n\le mk_{i2}$ and so $m \ge n/(n,k_{i2})$, i.e., $m \ge n/(n,k_i)$. Therefore, we have

$$\begin{aligned} \overline{N}(r,\alpha _i,f) \le \frac{(n,k_i)}{n} N(r,\alpha _i,f)+O(\log r). \end{aligned}$$

In particular, we have $m\ge 2$ since $n/(n,k_i)>1$. Thus $\alpha _i$ is a completely ramified rational function of f with multiplicity at least $n/(n,k_i)$. The same analysis above applies for each $\beta _j$ by writing $l_j=nl_{j1}+l_{j2}$, where $l_{j1},l_{j2}\in \mathbb {N}$ and $l_{j2}<n$, as well. Therefore, if $\beta _j$ is not a Picard exceptional rational function of f, then we also have

$$\begin{aligned} \overline{N}(r,\beta _j,f) \le \frac{(n,l_j)}{n} N(r,\beta _j,f)+O(\log r), \end{aligned}$$

and thus $\beta _j$ is a completely ramified rational function of f with multiplicity at least $n/(n,l_j)$. Below we consider the two cases $q=0$ and $q\ge 1$, respectively.

When $q=0$, Eq. (2.1) takes the following form:

$$\begin{aligned} \overline{f}^n=P(z,f). \end{aligned}$$

(2.9)

We claim that f has at most finitely many poles. Otherwise, let $z_0\in \mathbb {C}$ be a pole of f with multiplicity $m\in \mathbb {Z}^{+}$. We may choose $z_0$ such that $|z_0|$ is large enough so that none of the coefficients of P(z, f) has poles or zeros outside of $\{z\in \mathbb {C}: |z|<|z_0|\}$. When $n>p$, from (2.9) we see that $z_0+1$ is a pole of f of order pm/n and by iterating along the pole sequence we have $z_0+s$ is a pole of f of order $p^sm/n^s$, $s\in \mathbb {N}$. By letting $s\rightarrow \infty$, it follows that there is necessarily a branch point of f at some $z_0+s_0$, $s_0\in \mathbb {N}$, a contradiction to our assumption that f is meromorphic. On the other hand, when $n<p$, from (2.9) we see that $z_0-s$ is a pole of f of order $n^sm/p^s$, $s\in \mathbb {N}$, and by letting $s\rightarrow \infty$ we still get the same contradiction as above. Therefore, f has at most finitely many poles, i.e., $\infty$ is a Picard exceptional rational function of f. Then the inequality (1.6) implies that $N_c\le 2$. In particular, when $n\ge 3$, $N_c=2$ is impossible; otherwise, $\alpha _i$ would have multiplicity at least n for at least one i under our assumptions, a contradiction to the inequality (1.6). Thus we have the assertions for the case $q=0$.

When $q\ge 1$, since $\alpha _i$, as well as $\beta _j$, is either a Picard exceptional rational function of f or a completely ramified rational function of f, then by Picard’s theorem and Theorem 2 we conclude that $N_c\le 4$. In particular, when $n\ge 3$, since $n/(n,k_i)>2$ or $n/(n,l_j)>2$ for at least one index i or j by our assumption, then such $\alpha _i$ or $\beta _j$ is either a Picard exceptional rational function of f or a completely ramified rational function with multiplicity at least 3 and so by the inequality (1.6) it follows that $N_c\le 3$; when $n\not \mid |p-q|$, by applying the above analysis to poles of f, it follows that $\infty$ is either a Picard exceptional rational function of f or a completely ramified rational function of f, then by the inequality (1.6) we also have $N_c\le 3$. Thus the assertions of the lemma for the case $q\ge 1$ follow and this also completes the proof. $\square$

By giving a more careful analysis on the roots of P(z, f) and Q(z, f), we have the four following Lemmas 2–5, which play key roles in reducing equation (2.1) into certain forms in the following sections.

Lemma 2

Let f be a transcendental meromorphic solution of Eq. (2.1). Suppose that some $\alpha _i$ in (2.7) is 0. Then 0 is a Picard exceptional rational function of f. Moreover, if $q=0$, or $q\ge 1$ and $n\not \mid |p-q|$, then $\infty$ is also a Picard exceptional rational function of f.

Proof

Suppose that f has infinitely many zeros. Let $z_0\in \mathbb {C}$ be a zero of f with multiplicity $m\in \mathbb {Z}^{+}$. We may choose $z_0$ such that $|z_0|$ is large enough so that none of the coefficients of P(z, f) and Q(z, f) has poles or zeros outside of $\{z\in \mathbb {C}: |z|<|z_0|\}$. Since some $\alpha _i$ is zero, from (2.1) we see that $z_0+1$ is a zero of f of order $k_im/n$ and by iterating along the zero sequence we have $z_0+s$ is a zero of f of order $k_i^sm/n^s$, $s\in \mathbb {N}$. By letting $s\rightarrow \infty$, since $n\not \mid k_i$, it follows that there is necessarily a branch point of f at some $z_0+s_0$, $s_0\in \mathbb {N}$, a contradiction to our assumption that f is meromorphic. Therefore, f has at most finitely many zeros, i.e., 0 is a Picard exceptional rational function of f.

Moreover, if $q=0$, or if $q\ge 1$ and $p<q$, then it follows immediately from the proof of Lemma 1 that $\infty$ is a Picard exceptional rational function of f. Consider the case when $p>q\ge 1$ and $n\not \mid (p-q)$. Suppose that f has infinitely many poles. Let $z_0\in \mathbb {C}$ be a pole of f with multiplicity $m\in \mathbb {Z}^{+}$. We may choose $z_0$ such that $|z_0|$ is large enough so that none of the coefficients of P(z, f) and Q(z, f) has poles or zeros outside of $\{z\in \mathbb {C}: |z|<|z_0|\}$. From Eq. (2.1) we see that f has a pole of order $(p-q)m/n$ at $z=z_0+1$ and by iterating along the pole sequence $z=z_0+s$ is a pole of f of order $(p-q)^{s}m/n^{s}$, $s\in \mathbb {N}$. Since $n\not \mid (p-q)$, then by letting $s\rightarrow \infty$, it follows that there is necessarily a branch point of f at $z_0+s_0$ for some $s_0\in \mathbb {N}$, a contradiction to our assumption that f is meromorphic. Therefore, when $p>q\ge 1$ and $n\not \mid (p-q)$, f has at most finitely many poles, i.e., $\infty$ is a Picard exceptional rational function of f. We complete the proof.

Lemma 3

Let f be a transcendental meromorphic solution of Eq. (2.1) and $\gamma \in \mathcal {R}\setminus \{0\}$ be a rational function. Then $\gamma$ cannot be a Picard exceptional rational function of f. Moreover, if $\gamma$ is a completely ramified function of f with multiplicity at least m, then $\omega \gamma$ is a completely ramified function of f with multiplicity at least m, where $\omega$ is the nth root of 1.

Proof

To prove the assertions of the lemma, we divide equation (2.1) into the following three cases:

• (1) at least one of $\alpha _i$ and $\beta _j$ in (2.7) and (2.8) is non-zero and $q=0$, or $q\ge 1$ and $n\not \mid |p-q|$;

• (2) at least one of $\alpha _i$ and $\beta _j$ in (2.7) and (2.8) is non-zero and $q\ge 1$ and $n\mid |p-q|$;

• (3) there is only one $\alpha _i$ or $\beta _j$ in (2.7) and (2.8) and this $\alpha _i$ or $\beta _j$ is zero.

We first suppose that $\gamma$ is a Picard exceptional rational function of f. Under this assumption, below we show that each of the above three cases will lead to contradictions.

In the first case, we let $\beta$ be such that $\beta =\alpha _i$ or $\beta =\beta _j$ for some $\alpha _i$ or $\beta _j$ in (2.7) and (2.8) and $\beta \not =0$. Denote the order of this root $\alpha _i$ or $\beta _j$ by $t_1$. Put

$$\begin{aligned} u=\frac{\overline{f}}{f-\beta }, \quad v=\frac{1}{f-\beta }. \end{aligned}$$

(2.10)

Then u and v are two algebroid functions with at most finitely many branch points and we have

$$\begin{aligned} \overline{f}=\frac{u}{v}, \quad f=\frac{1}{v}+\beta , \end{aligned}$$

and it follows that (2.1) becomes

$$\begin{aligned} u^n=\frac{P_1(z,v)}{Q_1(z,v)}v^{n_1}, \end{aligned}$$

(2.11)

where $n_1\in \mathbb {Z}$, $P_1(z,v)$ and $Q_1(z,v)$ are two polynomials in v having no common factors and none of the roots of $P_1(z,v)$ or $Q_1(z,v)$ is zero. Denote by $p_1=\deg _{v}(P_1(z,v))$ the degree of $P_1(z,v)$ in v and by $q_1=\deg _{v}(Q_1(z,v))$ the degree of $Q_1(z,v)$ in v, respectively. Note that $q=0$, or $q\ge 1$ and $n\not \mid |p-q|$. By elementary calculations, when $q=0$ we get $n_1=n-p$, $p_1=p-t_1$ and $q_1=0$; when $q\ge 1$ and $\beta =\alpha _i$ we get $n_1=n+q-p$, $p_1=p-t_1$ and $q_1=q$; when $q\ge 1$ and $\beta =\beta _j$ we get $n_1=n+q-p$, $p_1=p$ and $q_1=q-t_1$. Therefore, we always have $n_1\not =0$, $n_1\not =n$ and $p_1-q_1+n_1\not =n$. We consider

$$\begin{aligned} \overline{f}-\overline{\gamma }=\frac{u}{v}-\overline{\gamma }=\frac{u-\overline{\gamma }v}{v}. \end{aligned}$$

(2.12)

Let $(u_0,v_0)$ be any pair of non-zero functions satisfying the following system of equations:

$$\begin{aligned} u_0^n=\frac{P_1(z,v_0)}{Q_1(z,v_0)}v_0^{n_1}, \quad u_0-\overline{\gamma }v_0=0. \end{aligned}$$

(2.13)

By assumption, the equation $\overline{f}-\overline{\gamma }=0$ has at most finitely many roots. Let $z_0\in \mathbb {C}$ be such that $u(z_0)=u_0(z_0)$, $v(z_0)=v_0(z_0)$ and $u_0=\overline{\gamma }v_0$ hold simultaneously. Then from (2.12) we see that the equation $u-\overline{\gamma }v=0$ can have at most finitely many such roots $z_0$. Now we consider the equation $\overline{f}-\omega \overline{\gamma }=0$, where $\omega$ is the nth root of 1. We have

$$\begin{aligned} \overline{f}-\omega \overline{\gamma }=\frac{u}{v}-\omega \overline{\gamma }=\frac{u-\omega \overline{\gamma }v}{v}. \end{aligned}$$

(2.14)

Let $\hat{z}_0\in \mathbb {C}$ be such that $v(\hat{z}_0)=0$ or $v(\hat{z}_0)=\infty$. Since $n_1\not =0$, $n_1\not =n$ and $p_1-q_1+n_1\not =n$, we see from (2.11) that the multiplicity of $\hat{z}_0$ for $v(\hat{z}_0)=0$ or $v(\hat{z}_0)=\infty$ equals the multiplicity of $\hat{z}_0$ for $u(\hat{z}_0)=0$ or $u(\hat{z}_0)=\infty$ for at most finitely many such $\hat{z}_0$. Then it follows from (2.14) that the two equations $v=0$ and $\overline{f}-\omega \overline{\gamma }=0$, as well as the two equations $v=\infty$ and $\overline{f}-\omega \overline{\gamma }=0$, cannot have infinitely many common roots; otherwise, $\overline{\gamma }$ would be identically equal to 0 or $\infty$, a contradiction to our assumption. Let $(u_1,v_1)$ be any pair of non-zero functions satisfying the following system of equations:

$$\begin{aligned} u_1^n=\frac{P_1(z,v_1)}{Q_1(z,v_1)}v_1^{n_1}, \quad u_1-\omega \overline{\gamma }v_1=0. \end{aligned}$$

(2.15)

Recall that $\omega$ is the nth root of 1. Then the two systems of equations in (2.13) and (2.15) yield identically the same algebraic equations for $v_0$ and $v_1$. Since $u_0^n$ and $u_1^n$ equals the same rational term in $v_0$ or $v_1$, we see that the two systems of equations in (2.13) and (2.15) have the same pair of non-zero solutions, apart from permutations. By summarizing the above results, we conclude that the equation $\overline{f}-\omega \overline{\gamma }{=}0$ can have at most finitely many roots, i.e., $\omega \gamma$ is a Picard exceptional rational function of f.

Now, since $n\ge 2$, we must have $n=2$; otherwise, f would have three or more distinct Picard exceptional rational functions, a contradiction to Picard’s theorem. However, when $q=0$, from Lemma 1 we know $\infty$ is also a Picard exceptional rational function of f, a contradiction to Picard’s theorem; when $q\ge 1$, since $n\not \mid |p-q|$, from Lemma 1 we know $\infty$ is either a Picard exceptional rational function of f or a completely ramified rational function of f, a contradiction to Picard’s theorem or the inequality (1.6).

In the second case, we must have $N_c\ge 2$ since $n\mid |p-q|$. We let $\beta$ be such that $\beta =\alpha _i$ or $\beta =\beta _j$ for some $\alpha _i$ or $\beta _j$ in (2.7) and (2.8) and $\beta \not =0$, and $\alpha$ be such that $\alpha =\alpha _i$ or $\alpha =\beta _j$ for another $\alpha _i$ or $\beta _j$ in (2.7) and (2.8) distinct from $\beta$. Put

$$\begin{aligned} u=\frac{\overline{f}(f-\alpha )}{f-\beta }, \quad v=\frac{f-\alpha }{f-\beta }. \end{aligned}$$

(2.16)

Then u and v are two algebroid functions with at most finitely many branch points and we have

$$\begin{aligned} \overline{f}=\frac{u}{v}, \quad f=\frac{\beta v-\alpha }{v-1}, \end{aligned}$$

and it follows that (2.1) becomes

$$\begin{aligned} u^n=\frac{P_2(z,v)}{Q_2(z,v)}v^{n_2}, \end{aligned}$$

where $n_2\in \mathbb {Z}$, $P_2(z,v)$ and $Q_2(z,v)$ are two polynomials in v having no common factors and none of the roots of $P_2(z,v)$ or $Q_2(z,v)$ is zero. Denote by $p_2=\deg _{v}(P_2(z,v))$ the degree of $P_2(z,v)$ in v and by $q_2=\deg _{v}(Q_2(z,v))$ the degree of $Q_2(z,v)$ in v, respectively. Note that $q\ge 1$ and $n\mid |p-q|$. As in the previous case we can show that $n_2\not =0$, $n_2\not =n$ and $p_2-q_2+n_2\not =n$ by elementary calculations. Then we consider the roots of the equation $\overline{f}-\overline{\gamma }=0$ and of the equation $\overline{f}-\omega \overline{\gamma }=0$, respectively, and by the same arguments as in the previous case we get the same conclusion as there.

We have $n=2$ in both cases (1) and (2). If some $\alpha _i$ or $\beta _j$ in (2.7) and (2.8) is distinct from $\pm \gamma$, then from Lemma 1 it follows that this $\alpha _i$ or $\beta _j$ is either a Picard exceptional rational function of f or a completely ramified rational function of f, a contradiction to Picard’s theorem or the inequality (1.6) since $\pm \gamma$ are both Picard exceptional rational functions of f. Therefore, $N_c=2$ and the two roots $\alpha _i$ or $\beta _j$ in (2.7) and (2.8) are equal to $\pm \gamma$. We consider

$$\begin{aligned} \overline{f}^2-\overline{\gamma }^2=\frac{P(z,f)-\overline{\gamma }^2Q(z,f)}{Q(z,f)} =\frac{a_{p_{\tau }}(f-\gamma _1)^{t_1}\cdots (f-\gamma _{\tau })^{t_{\tau }}}{Q(z,f)}, \end{aligned}$$

(2.17)

where $\gamma _1$, $\cdots$, $\gamma _{\tau }$ are in general algebraic functions distinct from each other, $t_1,\cdots ,t_{\tau }\in \mathbb {N}$ denote the orders of the roots $\gamma _1$, $\cdots$, $\gamma _{\tau }$, respectively, and $t_1+\cdots +t_{\tau }=p_{\tau }\in \mathbb {N}$. Note that the equation $\overline{f}^2-\overline{\gamma }^2=0$ can have at most finitely many roots. If $p_{\tau }<q$, then we obtain from (2.17) that $\infty$ is a Picard exceptional rational function of f, a contradiction to Picard’s theorem. Therefore, $p_{\tau }\ge q$. From the previous discussions we see that none of $\gamma _1$, $\cdots$, $\gamma _{\tau }$ is equal to $\pm \gamma$. But then we have $\gamma _1$, $\cdots$, $\gamma _{\tau }$ are all Picard exceptional rational functions of f by analyzing the roots of the equations $f-\gamma _{i}=0$, $i=1,\cdots , \tau$, again a contradiction to Picard’s theorem since $p_{\tau }\ge 1$.

In the third case, it follows from the assumptions that $n\not \mid |p-q|$. Moreover, we have $q\ge 1$ and this $\alpha _i$ or $\beta _j$ is $\beta _1$ and $\beta _1\equiv 0$; otherwise, this $\alpha _i$ or $\beta _j$ is $\alpha _1$ and $\alpha _1=0$, but it follows by Lemma 2 that 0 and $\infty$ are also both Picard exceptional rational functions of f, a contradiction to Picard’s theorem. Put

$$\begin{aligned} u=\frac{\overline{f}}{f}, \quad v=\frac{1}{f}. \end{aligned}$$

(2.18)

Then u and v are two algebroid functions with at most finitely branch points and we have

$$\begin{aligned} \overline{f}=\frac{u}{v}, \quad f=\frac{1}{v}, \end{aligned}$$

and it follows that (2.1) becomes

$$\begin{aligned} u^n=\frac{P_3(z,v)}{Q_3(z,v)}v^{n_3}, \end{aligned}$$

where $n_3\in \mathbb {Z}$, $P_3(z,v)$ and $Q_3(z,v)$ are two polynomials in v having no common factors and none of the roots of $P_3(z,v)$ or $Q_3(z,v)$ is zero. Denote by $p_3=\deg _{v}(P_3(z,v))$ the degree of $P_3(z,v)$ in v and by $q_3=\deg _{v}(Q_3(z,v))$ the degree of $Q_3(z,v)$ in v, respectively. As in case (1) we can show that $n_3\not =0$, $n_3\not =n$ and $p_3-q_3+n_3\not =n$ by elementary calculations. Then we consider the roots of the equation $\overline{f}-\overline{\gamma }=0$ and of the equation $\overline{f}-\omega \overline{\gamma }=0$, respectively, and by the same arguments as in case (1) we get that $\omega \gamma$ is a Picard exceptional rational function of f. But this is impossible by the inequality (1.6) since from Lemma 1 it follows that $\infty$ is either a Picard exceptional rational function of f or a completely ramified rational function of f.

From the above reasoning, we conclude that $\gamma$ cannot be a Picard exceptional rational function of f. This gives our first assertion of the lemma.

Next, we suppose that $\gamma$ is a completely ramified rational function of f with multiplicity at least m. We also consider the three cases in the beginning of the proof. In the first case, we do the transformations in (2.10) and get the equation in (2.11). By assumption, all roots of the equation $\overline{f}-\overline{\gamma }=0$ with at most finitely many exceptions have multiplicities at least m. Therefore, for any pair of non-zero functions $(u_0,v_0)$ such that the system of equations in (2.13) holds, if we let $z_0\in \mathbb {C}$ be such that $u(z_0)=u_0(z_0)$, $v(z_0)=v_0(z_0)$ and $u_0=\overline{\gamma }v_0$ hold simultaneously, then from (2.12) we see that the equation $u-\overline{\gamma }v=0$ has at most finitely many such roots $z_0$ with multiplicity less than m. Moreover, letting $\hat{z}_0\in \mathbb {C}$ be such that $f(\hat{z}_0+1)-\omega \gamma (\hat{z}_0+1)=0$, from previous discussions we know that $v(\hat{z}_0)=0$ or $v(\hat{z}_0)=\infty$ for at most finitely many such $\hat{z}_0$. Also, we know that the two systems of equations in (2.13) and (2.15) have the same pair of solutions, apart from permutations. Then we conclude from the above reasoning that the equation $\overline{f}-\omega \overline{\gamma }=0$ can have at most finitely many roots with multiplicities less than m. The second and the third cases can be discussed in an analogous way as above after doing the transformations in (2.16) and (2.18), respectively. We omit those details. Thus we have the second assertion of the lemma and also the complete proof. $\square$

Lemma 4

Let f be a transcendental meromorphic solution of Eq. (2.1). Suppose that one of the following cases occurs:

• (1) $q=0$ and P(z, f) has a root $\alpha$ of order k such that $2\le k<n$ and $k\not \mid n$, or $k\ge n+1$;

• (2) $q\ge 1$ and P(z, f) has a root $\alpha$ of order k such that $2\le k<n$ and $k\not \mid n$, or $k\ge n+1$;

• (3) $q\ge 1$ and Q(z, f) has a root $\beta$ of order l such that $2\le l<n$ and $l\not \mid n$, or $l\ge n+1$;

• (4) $q\ge 1$ and p and q satisfy $2\le |p-q|<n$ and $|p-q|\not \mid n$, or $|p-q|\ge n+1$.

Then if $q=0$, then f cannot have 2 completely ramified rational functions in $\mathcal {R}$; if $q\ge 1$, then f cannot have 4 completely ramified rational functions in $\mathcal {R}$ and cannot have 3 non-zero completely ramified rational functions $\gamma _1,\gamma _2,\gamma _3\in \mathcal {R}$ such that $\sum _{i=1}^3\varTheta (\gamma _i,f)=2$.

Proof

First, when $q=0$, by Lemma 1 we know that $\infty$ is a Picard exceptional rational function of f and thus $\varTheta (\infty ,f)=1$. Suppose that $\gamma _1$ and $\gamma _2$ are both completely ramified rational functions of f. By the inequality (1.6) it follows that $\gamma _1$ and $\gamma _2$ both have multiplicities 2 and that $\varTheta (\gamma _1,f)+\varTheta (\gamma _2,f)=1$. Further, by the Second Main Theorem of Yamanoi for rational functions as targets [25], letting $\gamma$ be any rational function distinct from $\gamma _i$, we must have $\overline{N}(r,\gamma ,f)=T(r,f)+o(T(r,f))$, $N(r,\gamma _i,f)=T(r,f)+o(T(r,f))$ and $\overline{N}(r,\gamma _i,f)=\frac{1}{2}T(r,f)+o(T(r,f))$, where $r\rightarrow \infty$ outside an exceptional set E with finite linear measure. If either $\gamma _1=0$ or $\gamma _2=0$, say $\gamma _1=0$, then by Lemma 3 it follows that $\omega \gamma _2$ is a completely ramified rational function of f, where $\omega$ is the nth root of 1, a contradiction to the inequality (1.6). Therefore, 0 is not a completely ramified rational function of f. Let $z_0\in \mathbb {C}$ be such that $f(z_0)-\alpha (z_0)=0$ with multiplicity $m\in \mathbb {Z}^{+}$. Since $\alpha$ is a root of P(z, f) of order k such that $2\le k<n$ and $k\not \mid n$, or $k\ge n+1$, then from (2.1) we see that $z_0$ is a root of $f(z+1)=0$ with multiplicity $m_0\in \mathbb {Z}^{+}$ such that $m_0=mk/n\ge 2$ with at most finitely many exceptions. By the Valiron–Mohon’ko identity [17, 23] (see also [15]), we have from (2.1) that $nT(r,\overline{f})=dT(r,f)+O(\log r)$. Now there are at least $T(r,f)+o(T(r,f))$ many points $z_0$ such that $f(z_0+1)=0$ with multiplicity $m_0\ge 2$. This implies that $n<d$. Denote by S the set of zeros of $\overline{f}$ in the finite disk $D=\{z\in \mathbb {C}: |z|< t\}$, where $t>0$, and by $S_1$ the set of zeros with multiplicity $\ge 2$ of $\overline{f}$ in D. Denote by $n_{S_1}(t,1/\overline{f})$ and $\overline{n}_{S_1}(t,1/\overline{f})$ the number of zeros of $\overline{f}$ in $S_1$, counting or ignoring multiplicities, respectively; denote by $n_{S\setminus S_1}(t,1/\overline{f})$ and $\overline{n}_{S\setminus S_1}(t,1/\overline{f})$ the number of zeros of $\overline{f}$ for the complement of $S_1$, counting or ignoring multiplicities, respectively. We may suppose that $f(1)\not =0$. By the definition of the truncated counting function $\overline{N}(r,1/\overline{f})$, we deduce that

$$\begin{aligned} \begin{aligned} \overline{N}(r,1/\overline{f})&=\int _{0}^{r}\overline{n}_{S\setminus S_1}(t,1/\overline{f})\frac{dt}{t}+\int _{0}^{r}\overline{n}_{S_1}(t,1/\overline{f})\frac{dt}{t}\\&\le \int _{0}^{r}n_{S\setminus S_1}(t,1/\overline{f}) \frac{dt}{t}+\frac{1}{2}\int _{0}^{r}n_{S_1}(t,1/\overline{f})\frac{dt}{t}\\&=\int _{0}^{r}n_{S}(t,1/\overline{f})\frac{dt}{t}-\frac{1}{2}\int _{0}^{r}n_{S_1}(t,1/\overline{f})\frac{dt}{t}\\&\le T(r,1/\overline{f})-\frac{1}{2}[T(r,f)+o(T(r,f))]. \end{aligned} \end{aligned}$$

By the First Main Theorem of Nevanlinna we have $T(r,1/\overline{f})=T(r,\overline{f})+O(1)$ and it follows that $T(r,1/\overline{f})=\frac{d}{n}T(r,f)+O(\log r)$. Then by combining the above results together we get

$$\begin{aligned} \varTheta (0,\overline{f})=1-\limsup _{r\rightarrow \infty }\frac{\overline{N}(r,1/\overline{f})}{T(r,\overline{f})}\ge \frac{n}{2d}. \end{aligned}$$

In general, the quantity $\varTheta (\gamma _i,f)$ may not be shift-invariant [12], but under our assumptions we already have $\sum _{i=1}^2\varTheta (\overline{\gamma }_i,\overline{f})=1$, and thus the inequality above is impossible. Therefore, if case (1) occurs, then f cannot have 2 completely ramified rational functions in $\mathcal {R}$. This is the first assertion of the lemma.

Second, we consider the case where $q\ge 1$. We suppose that f has four completely ramified rational functions $\gamma _i$, $i=1,2,3,4$, in $\mathcal {R}$. By Theorem 2, $\infty$ is not a completely ramified rational function of f. Moreover, none of $\gamma _1$, $\gamma _2$, $\gamma _3$ and $\gamma _4$ is zero; otherwise, say $\gamma _1=0$, by Lemma 3 it follows that $\omega \gamma _2$, $\omega \gamma _3$ and $\omega \gamma _4$ are all completely ramified rational functions of f, where $\omega$ is the cubic root of 1, and thus by Theorem 2 we must have $n\ge 3$. However, by Lemma 1 it follows that at least one of $\alpha _i$ in (2.7) and $\beta _j$ in (2.8) is a completely ramified rational function of f with multiplicity $\ge 3$, a contradiction to the inequality (1.6). Therefore, 0 is not a completely ramified rational function of f. Below we consider the three cases (2), (3) and (4), respectively.

If case (2) occurs, then $\alpha$ is not a Picard exceptional rational function of f. As in case (1), for the point $z_0\in \mathbb {C}$ such that $f(z_0)-\alpha (z_0)=0$ with multiplicity $m\in \mathbb {Z}^{+}$, we have that $z_0$ is a root of $f(z+1)=0$ with multiplicity $m_0\in \mathbb {Z}^{+}$ such that $m_0=mk/n\ge 2$ with at most finitely many exceptions. Since f has four completely ramified rational functions $\gamma _i$, $i=1,2,3,4$, then by the inequality (1.6) we know that $\gamma _i$, $i=1,2,3,4$, all have multiplicities 2. Further, by the Second Main Theorem of Yamanoi for rational functions as targets [25], letting $\gamma$ be any rational function distinct from $\gamma _i$, we must have $\overline{N}(r,\gamma ,f)=T(r,f)+o(T(r,f))$, $N(r,\gamma _i,f)=T(r,f)+o(T(r,f))$ and $\overline{N}(r,\gamma _i,f)=\frac{1}{2}T(r,f)+o(T(r,f))$, where $r\rightarrow \infty$ outside an exceptional set E with finite linear measure. Then, similarly as in case (1), we can obtain a contradiction by computing $\varTheta (0,\overline{f})$. Therefore, if case (2) occurs, then f cannot have 4 completely ramified rational functions.

If case (3) occurs, then for the point $z_0\in \mathbb {C}$ such that $f(z_0)-\beta (z_0)=0$ we have that $z_0$ is a root of $f(z+1)=\infty$ with multiplicity $m_0\in \mathbb {Z}^{+}$ such that $m_0=ml/n\ge 2$ with at most finitely many exceptions, and then we can obtain a contradiction by computing $\varTheta (\infty ,\overline{f})$. Therefore, we still have the same conclusion as in case (2).

If case (4) occurs, then we let $z_0$ be a pole of f with multiplicity $m\in \mathbb {Z}^{+}$ and it follows that $z_0$ is a root of $f(z+1)=0$ or $f(z+1)=\infty$ with multiplicity $m_0\in \mathbb {Z}^{+}$ such that $m_0=m|p-q|/n\ge 2$ with at most finitely many exceptions. If $p<q$, then we get a contradiction by computing $\varTheta (0,f)$; if $p>q$, then we get a contradiction by computing $\varTheta (\infty ,f)$. Therefore, we still have the same conclusion as in case (2).

Last, we suppose that f has 3 non-zero completely ramified rational functions $\gamma _1,\gamma _2,\gamma _3\in \mathcal {R}$ such that $\sum _{i=1}^3\varTheta (\gamma _i,f)=2$. From [10, p. 46] we know that the possible multiplicity sets $(m_1,m_2,m_3)$ corresponding to $\gamma _1$, $\gamma _2$, $\gamma _3$ are (2, 4, 4), (2, 3, 6) or (3, 3, 3), apart from permutations. Also, by the Second Main Theorem of Yamanoi for rational functions as targets [25], we have $\overline{N}(r,\gamma _i,f)=\frac{1}{m_i}T(r,f)+o(T(r,f))$, where $r\rightarrow \infty$ outside an exceptional set E with finite linear measure. Note that $\infty$ is not a completely ramified rational function of f. If one of the three cases (2), (3) or (4) occurs, then we can use the same arguments as above to compute $\varTheta (0,\overline{f})$ or $\varTheta (\infty ,\overline{f})$ and obtain similar contradictions. Thus our assertion follows. $\square$

In Lemma 4, we did not deal with Eq. (2.1) for the case when f has three non-zero completely ramified functions $\gamma _i$ such that $\sum _{i=1}^3\varTheta (\gamma _i,f)=2$ and one $\gamma _i$ is $\infty$. We will exclude out this possibility in the proof of Theorem 6 in Sect. 4 with applications of the analysis in the proof of Lemma 4.

Finally, we consider the equation $\overline{f}^n-\overline{\gamma }^n=0$ further, where $\gamma \in \mathcal {R}{\setminus }\{0\}$ is a completely ramified rational function of f with multiplicity at least $m\ge 2$. By Lemma 3, $\omega \gamma$ is a completely ramified rational function of f with multiplicity at least m, where $\omega$ is the nth root of 1. By (2.1), when $q=0$ we have

$$\begin{aligned} \overline{f}^n-\overline{\gamma }^n=P(z,f)-\overline{\gamma }^n=a_p(f-\gamma _1)^{t_1}\cdots (f-\gamma _{\tau })^{t_{\tau }}, \end{aligned}$$

(2.19)

or, when $q\ge 1$, we have

$$\begin{aligned} \overline{f}^n-\overline{\gamma }^n=\frac{P(z,f) -\overline{\gamma }^nQ(z,f)}{Q(z,f)}=\frac{a_{p_{\tau }}(f-\gamma _1)^{t_1}\cdots (f-\gamma _{\tau })^{t_{\tau }}}{Q(z,f)}, \end{aligned}$$

(2.20)

where $\gamma _1$, $\cdots$, $\gamma _{\tau }$ are in general algebraic functions distinct from each other and $t_1,\cdots ,t_{\tau }\in \mathbb {N}$ denote the orders of the roots $\gamma _1$, $\cdots$, $\gamma _{\tau }$, respectively, and $t_1+\cdots +t_{\tau }=p_{\tau }\in \mathbb {N}$. We apply the analysis in the proof of Lemma 4 to Eqs. (2.19) and (2.20), respectively, and get the following

Lemma 5

Let f be a transcendental meromorphic solution of Eq. (2.1) and $\gamma \in \mathcal {R}\setminus \{0\}$ be a completely ramified rational function of f with multiplicity $m\ge 2$. Suppose that $\zeta _i$, $\cdots$, $\zeta _t$ are Picard exceptional rational functions of f or completely ramified rational functions of f such that $\sum _{i=1}^t\varTheta (\zeta _i,f)=2$. For each $\gamma _i$ in (2.19) or (2.20), if $\gamma _i$ is not a completely ramified rational function of f, then $t_i=m$; if $\gamma _i$ is a completely ramified rational function of f with multiplicity $m_i\ge 2$, then $t_im_i=m$. In particular, for (2.20), when $1\le p_{\tau }<q$, if $\infty$ is not a completely ramified rational function of f, then $q-p_{\tau }=m$; if $\infty$ is a completely ramified rational function of f with multiplicity $m_{\infty }\ge 2$, then $(q-p_{\tau })m_{\infty }=m$.

Proof

By the assumption $\sum _{i=1}^t\varTheta (\zeta _i,f)=2$, we know from the proof of Lemma 4 that $t=3$ or $t=4$. Moreover, for each $\gamma _i$ in (2.19) or (2.20) we have $\overline{N}(r,\gamma _i,f)=T(r,f)+o(T(r,f))$ when $\gamma _i$ is not a completely ramified rational function of f and $\overline{N}(r,\gamma _i,f)=\frac{1}{m_i}T(r,f)+o(T(r,f))$ when $\gamma _i$ is a completely ramified rational function of f with multiplicity $m_i$, where $r\rightarrow \infty$ outside an exceptional set E with finite linear measure. In particular, we have $N(r,\infty ,f)=O(\log r)$ when $\infty$ is a Picard exceptional rational function of f and, otherwise, we have $N(r,\infty ,f)=T(r,f)+o(T(r,f))$ when $\infty$ is not a completely ramified rational function of f and $\overline{N}(r,\infty ,f)=\frac{1}{m_{\infty }}T(r,f)+o(T(r,f))$ when $\infty$ is a completely ramified rational function of f with multiplicity $m_{\infty }$, where again $r\rightarrow \infty$ outside an exceptional set E with finite linear measure. Note that for Eq. (2.20), when $p_{\tau }\ge 1$, from the proof of Lemma 3 we have that $\infty$ cannot be a Picard exceptional rational function of f. We also have $\overline{N}(r,\gamma ,f)=\frac{1}{m}T(r,f)+o(T(r,f))$, where $r\rightarrow \infty$ outside an exceptional set E with finite linear measure. Now there are $T(r,f)+o(T(r,f))$ many points $z_0\in \mathbb {C}$ such that $f(z_0)-\gamma _i(z_0)=0$ (or $f(z_0)=\infty$ when $p_{\tau }<q$) and from (2.19) or (2.20) it follows that $f(z_0+1)^n-\gamma (z_0+1)^n=0$. For such $z_0$, by comparing the multiplicities on both sides of the Eq. (2.19) or (2.20), we obtain the desired results. We omit those details. $\square$

3 Equation (2.1) with $q=0$

3.1 Equation (2.1) with $q=0$ and $n>p\ge 2$

For the case $q=0$ and $n>p\ge 2$ of Eq. (2.1), we prove the following

Theorem 3

Suppose that $q=0$ and $n>p\ge 2$. Let f be a transcendental meromorphic solution of Eq. (2.1). Then there exists a rational function $\alpha$ such that the linear transformation $f\rightarrow \alpha f$ reduces (2.1) into

$$\begin{aligned} \overline{f}^n=cf^p, \end{aligned}$$

(3.1)

where c is a non-zero constant. Moreover, solutions of Eq. (3.1) are represented as

$$\begin{aligned} f=c^{\frac{1}{n-p}}\exp [\pi (z)(p/n)^z], \end{aligned}$$

(3.2)

where $\pi (z)$ is an arbitrary entire periodic function with period 1.

Proof

From Lemma 1 we know that $\infty$ is a Picard exceptional rational function of f and $N_c=1$. Therefore, we have

$$\begin{aligned} \overline{f}^n=a_p(f-\alpha _1)^{k_1}, \end{aligned}$$

where $k_1=p$ and $(n,k_1)=1$. If $\alpha _1\not =0$, then by Lemmas 1 and 3 it follows that $\omega \alpha _1$ is a completely ramified rational function of f with multiplicity at least n, where $\omega$ is the nth root of 1, a contradiction to the inequality (1.6) since $n\ge 3$. Therefore, $\alpha _1=0$ and thus we have

$$\begin{aligned} \overline{f}^n=a_pf^p. \end{aligned}$$

(3.3)

By Lemma 2, it follows that 0 is also a Picard exceptional rational function of f. Then there is a non-zero rational function $\alpha$ such that $\alpha f$ is a zero-free entire function. Since f satisfies (3.3), it follows that

$$\begin{aligned} (\overline{\alpha }\overline{f})^n=\overline{\alpha }^n\overline{f}^n =\overline{\alpha }^na_pf^p=a_p\frac{\overline{\alpha }^n}{\alpha ^p}(\alpha f)^p. \end{aligned}$$

By redefining $\alpha f$ as f, we have

$$\begin{aligned} \overline{f}^n=cf^p, \end{aligned}$$

(3.4)

where $c=a_p\frac{\overline{\alpha }^n}{\alpha ^p}$ is a non-zero constant. By taking the logarithm on both sides of (3.4), then $g=\log f$ is entire and satisfies

$$\begin{aligned} n\overline{g}=\log c+pg. \end{aligned}$$

Therefore, we can solve f as

$$\begin{aligned} f=c^{\frac{1}{n-p}}\exp [\pi (z)(p/n)^z], \end{aligned}$$

where $\pi (z)$ is an arbitrary entire periodic function with period 1. This completes the proof. $\square$

3.2 Equation (2.1) with $q=0$ and $2\le n<p$

For the case $q=0$ and $2\le n<p$ of Eq. (2.1), we prove the following

Theorem 4

Suppose that $q=0$ and $2\le n< p$. Let f be a transcendental meromorphic solution of Eq. (2.1). Then there exists a rational function $\alpha$ such that by doing a linear transformation $f\rightarrow \alpha f$, we have either

• (1) equation (2.1) reduces into

  $$\begin{aligned} \overline{f}^n=cf^p, \end{aligned}$$

  (3.5)

  where c is a non-zero constant; solutions of (3.5) are represented as

  $$\begin{aligned} f(z)=c^{\frac{1}{n-p}}\exp [\pi (z)(p/n)^z], \end{aligned}$$

  (3.6)

  where $\pi (z)$ is an arbitrary entire periodic function with period 1; or

• (2) when $n=2$ and $p=2p_0+1$, $p_0\ge 1$, Eq. (2.1) reduces into

  $$\begin{aligned} \overline{f}^2=P_0(f)^2(f-1), \end{aligned}$$

  (3.7)

  and $P_0(f)$ is a polynomial in f such that

  $$\begin{aligned} P_0(f)=\frac{\pm i}{2^{1/2}}[U_{p_0}(f)+U_{p_0-1}(f)] \end{aligned}$$

  with the Chebyshev polynomials $U_{p_0}$ and $U_{p_0-1}$ of the second kind, i.e.,

  $$\begin{aligned} U_{p_0}(\cos x)=\frac{\sin (p_0+1)x}{\sin x}, \end{aligned}$$

  i.e.,

  $$\begin{aligned} U_{p_0}(f)=\sum _{t=0}^{[p_0/2]}\frac{(-1)^t(p_0-t)!}{t!(p_0-2t)!}(2f)^{p_0-2t}; \end{aligned}$$

  therefore, if we write

  $$\begin{aligned} P_0(f)=\sum _{t=0}^{p_0}A_{p_0}f^{p_0}, \end{aligned}$$

  then

  $$\begin{aligned} \begin{aligned} A_{p_0-2t}&=(-1)^t2^{p_0-2t}\frac{(p_0-t)\cdots (p_0-2t+1)}{t!},\\ A_{p_0-2t-1}&=(-1)^t2^{p_0-2t-1}\frac{(p_0-t-1)\cdots (p_0-2t)}{t!}; \end{aligned} \end{aligned}$$

  solutions of (3.7) are represented as

  $$\begin{aligned} f(z)=\frac{1}{2}(\delta ^2+\delta ^{-2}), \end{aligned}$$

  (3.8)

  where $\delta$ is given by

  $$\begin{aligned} \delta =(\pm i)^{\frac{1}{1-2p_0}}\exp [\pi (z)(p_0+1/2)^z], \end{aligned}$$

  or

  $$\begin{aligned} \delta =(\pm i)^{\frac{1}{3+2p_0}}\exp [\pi (z)(-p_0-1/2)^z], \end{aligned}$$

  where $\pi (z)$ is an arbitrary entire periodic function with period 1; or

• (3) when $n=2$ and $p=2p_0+2$, $p_0\ge 1$, Eq. (2.1) reduces into

  $$\begin{aligned} \overline{f}^2=P_0(f)^2(f^2-1), \end{aligned}$$

  (3.9)

  and $P_0(f)$ is a polynomial in f such that

  $$\begin{aligned} P_0(f)=\pm i\sum _{l=0}^{[p_0/2]}\left( {\begin{array}{c}p_0+1\\ 2l+1\end{array}}\right) f^{p_0-2l}(f^2-1)^{l}, \end{aligned}$$

  (3.10)

  where $[p_0/2]$ denotes the greatest integer not exceeding $p_0/2$; solutions of (3.9) are represented as

  $$\begin{aligned} f=\frac{1}{2}(\lambda +\lambda ^{-1}), \end{aligned}$$

  (3.11)

  where $\lambda$ is given by

  $$\begin{aligned} \lambda =(\pm i)^{-\frac{1}{p_0}}\exp [\pi (z)(p_0+1)^z], \end{aligned}$$

  or

  $$\begin{aligned} \begin{aligned} \lambda =(\pm i)^{\frac{1}{2+p_0}}\exp [\pi (z)(-p_0-1)^z], \end{aligned} \end{aligned}$$

  where $\pi (z)$ is an arbitrary entire periodic function with period 1.

We remark that the solutions with the form (3.8) of Eq. (3.7) are not given in [18] where an existence theorem for entire solutions of (3.7) is stated instead; we also remark that the polynomial in (3.10) has different form from the one in [18, Theorem 4(a)] since we have chosen different form of the solutions (3.11).

Equations (3.1) and (3.5), as well as their solutions (3.2) and (3.6), are apparently of the same form. We note that in Theorem 4, when $n\ge 3$, we only have equation (3.5). In fact, when $n\ge 3$, if some $\alpha _i$ in (2.7) is non-zero, then by Lemmas 1 and 3 we have a contradiction to the inequality (1.6) since $\omega \alpha _i$ is a completely ramified rational function of f, where $\omega$ is the nth root of 1, and $\infty$ is a Picard exceptional rational function of f. Then (3.5) follows by applying Lemma 2 and Picard’s theorem.

Proof of Theorem 4

From Lemma 1 we know that $\infty$ is a Picard exceptional rational function of f and $N_c\le 2$. If some $\alpha _i$ in (2.7) is zero, then by Lemma 2 it follows that 0 is also a Picard exceptional rational function of f. Then by Picard’s theorem we conclude that P(z, f) cannot have any non-zero root and thus $P_0(z,f)^n=a_p$, i.e., we have the following equation:

$$\begin{aligned} \overline{f}^n=a_pf^p. \end{aligned}$$

(3.12)

Similarly as in the proof of Theorem 3, we choose a rational function $\alpha$ such that $\alpha f$ has no zeros and poles and then write equation (3.12) as

$$\begin{aligned} \overline{f}^n=cf^p, \end{aligned}$$

(3.13)

where $c=a_p\frac{\overline{\alpha }^n}{\alpha ^p}$ is a non-zero constant and f above is a zero-free entire function. Moreover, solutions of (3.13) can be solved as

$$\begin{aligned} f=c^{\frac{1}{n-p}}\exp [\pi (z)(p/n)^z], \end{aligned}$$

where $\pi (z)$ is an arbitrary entire periodic function with period 1. This is the first part of Theorem 4.

From now on, we suppose that none of $\alpha _i$ in (2.7) is zero. By Lemmas 1 and 3 it follows that $\omega \alpha _i$ is a completely ramified rational function of f, where $\omega$ is the nth root of 1. Since $\infty$ is a Picard exceptional rational function of f, then by the inequality (1.6) we must have $n=2$ and $N_c\le 2$. Below we consider the two cases where $N_c=1$ and $N_c=2$ separately.

Case 1: $N_c=1$.

In this case, $\infty$ is a Picard exceptional rational function of f and p is odd. Therefore, we have

$$\begin{aligned} \overline{f}^2=P_0(z,f)^2(f-\alpha _1)^{k_1}, \end{aligned}$$

(3.14)

where $\alpha _1\not =0$ and $k_1$ is an odd integer. By Lemmas 1 and 3 it follows that $\pm \alpha _1$ are both completely ramified rational functions of f with multiplicities 2. Then by Lemma 4, we must have $k_1=1$ and $P_0(z,f)$ is a polynomial in f with simple roots only. We may let $\alpha _1=1$ by doing a linear transformation $f\rightarrow \alpha _1f$. We consider

$$\begin{aligned} \overline{f}^2-1=P_0(z,f)^2(f-1)-1. \end{aligned}$$

(3.15)

The RHS of (3.15) is a polynomial in f with odd degree and thus has at least one root, say $\gamma _1$, of odd order. Since f has no other completely ramified rational functions besides $\pm 1$, then by Lemma 5 we conclude that $\gamma _1$ must be $-1$ and there is only one such $\gamma _1$; moreover, the RHS of (3.15) is of the form $P_1(z,f)^2(f+1)$ for some polynomial $P_1(z,f)$ in f with simple roots only. Now we have

$$\begin{aligned} \overline{f}^2=P_0(z,f)^2(f-1), \end{aligned}$$

(3.16)

and further that

$$\begin{aligned} \overline{f}^2-1=P_1(z,f)^2(f+1). \end{aligned}$$

(3.17)

From (3.16) and (3.17), we see that the degree of $P_1(z,f)$ is $p_0$, at least 1. Put

$$\begin{aligned} f=\frac{1}{2}(\lambda +\lambda ^{-1}). \end{aligned}$$

(3.18)

Since both $\pm 1$ have multiplicities 2, we may write $f+1=g^2$ with an algebroid function g and g has at most finitely many algebraic branch points. It follows that the RHS of equation (3.17) becomes $[P_1(z,g^2-1)g]^2$, which implies that $\lambda$ is an algebroid function with at most finitely many algebraic branch points. Moreover, 0 and $\infty$ are both Picard exceptional rational functions of $\lambda$. Put

$$\begin{aligned} \lambda =\delta ^2. \end{aligned}$$

(3.19)

Then $\delta$ is an algebroid function with at most finitely many algebraic branch points. Now it follows from Eq. (3.17) that

$$\begin{aligned} \frac{1}{2}(\overline{\delta }^2-\overline{\delta }^{-2})=P_1(z,f)\frac{\delta ^2+1}{2^{1/2}\delta }. \end{aligned}$$

(3.20)

By solving equation (3.20) together with (3.16) and (3.17), we get

$$\begin{aligned} \overline{\delta }^2=\frac{P_1\left( z,\frac{\delta ^4+1}{2\delta ^2}\right) (\delta ^2+1)\pm P_0\left( z,\frac{\delta ^4+1}{2\delta ^2}\right) (\delta ^2-1)}{2^{1/2}\delta }:=\frac{P_{11}(z,\delta )}{2^{1/2}\delta (2\delta ^2)^{p_0}}, \end{aligned}$$

(3.21)

where $P_{11}(z,\delta )$ is a polynomial in $\delta$ of degree at most $4p_0+2$. Since 0 is a Picard exceptional rational function of $\lambda$, then by Picard’s theorem we see from Eq. (3.21) that $P_{11}(z,\delta )$ cannot have any non-zero root. By the Valiron–Mohon’ko identity [17, 23] (see also [15]), we have from (3.16), (3.18) and (3.19) that $4T(r,\overline{\delta })=(4p_0+2)T(r,\delta )+O(\log r)$. Therefore, we have either

$$\begin{aligned} \begin{aligned} \overline{\delta }^2=T_0\delta ^{2p_0+1}, \end{aligned} \end{aligned}$$

(3.22)

or

$$\begin{aligned} \begin{aligned} \overline{\delta }^2=T_0\delta ^{-2p_0-1}, \end{aligned} \end{aligned}$$

(3.23)

where $T_0$ is an algebraic function. We write

$$\begin{aligned} P_1(z,f) = c_{p_0}f^{p_0} + c_{p_0-1}f^{p_0-1}+\cdots +c_0, \end{aligned}$$

(3.24)

where $c_{p_0}$, $\cdots$, $c_0$ are algebraic functions and $c_{p_0}\not =0$. If we have (3.22), then by substituting (3.18), (3.19) and (3.22) into (3.20) and then comparing the terms on both sides of the resulting equation together with (3.24), we get

$$\begin{aligned} \overline{\delta }^2=2^{1/2}2^{-p_0}c_{p_0}\delta ^{2p_0+1}, \end{aligned}$$

(3.25)

and

$$\begin{aligned} \overline{\delta }^2=-2^{-1/2}2^{p_0}(1/c_{p_0})\delta ^{2p_0+1}. \end{aligned}$$

(3.26)

On the other hand, if we have (3.23), then similarly as above from (3.18), (3.19), (3.20) and (3.24) we get

$$\begin{aligned} \overline{\delta }^2=2^{1/2}2^{-p_0}c_{p_0}\delta ^{-2p_0-1}, \end{aligned}$$

(3.27)

and

$$\begin{aligned} \overline{\delta }^2=-2^{-1/2}2^{p_0}(1/c_{p_0})\delta ^{-2p_0-1}. \end{aligned}$$

(3.28)

We obtain from (3.25) and (3.26), as well as from (3.27) and (3.28), that $c_{p_0}=\pm i2^{p_0-\frac{1}{2}}$ and it follows that $T_0=\pm i$. Thus the solution f of (3.16) is represented by (3.18) and (3.19) with $\delta$ such that

$$\begin{aligned} \overline{\delta }^2=\pm i\delta ^{2p_0+1}. \end{aligned}$$

(3.29)

or

$$\begin{aligned} \overline{\delta }^2=\pm i\delta ^{-2p_0-1}. \end{aligned}$$

(3.30)

Note that $\delta$ has at most finitely many zeros, poles and branch points. Then we can solve $\delta$ from (3.29) and (3.30) as

$$\begin{aligned} \delta =(\pm i)^{\frac{1}{1-2p_0}}\exp [\pi (z)(p_0+1/2)^z], \end{aligned}$$

(3.31)

or

$$\begin{aligned} \delta =(\pm i)^{\frac{1}{3+2p_0}}\exp [\pi (z)(-p_0-1/2)^z], \end{aligned}$$

(3.32)

respectively, where $\pi (z)$ is an arbitrary entire periodic function with period 1. We conclude that solutions of Eq. (3.16) are given by (3.18), (3.19) with (3.31) or (3.32).

Now we determine the polynomial $P_0(z,f)$ in f in (3.16) by using (3.18), (3.19), (3.29) and (3.30). From (3.16) and (3.29) we have

$$\begin{aligned} \frac{1}{2}(\pm i)(\delta ^{2p_0+1}-\delta ^{-2p_0-1})=P_0(z,f)\frac{\delta ^2-1}{2^{1/2}\delta }. \end{aligned}$$

(3.33)

Moreover, by (3.18) and (3.19) we have

$$\begin{aligned} \begin{aligned} \delta ^2&=f\pm (f^2-1)^{1/2},\\ \delta ^{-2}&=f\mp (f^2-1)^{1/2}, \end{aligned} \end{aligned}$$

(3.34)

and also that

$$\begin{aligned} \begin{aligned} \delta ^{2p_0+1}-\delta ^{-2p_0-1}=\frac{\delta ^{4p_0+2}-1}{\delta ^{2p_0+1}}. \end{aligned} \end{aligned}$$

(3.35)

From (3.33), (3.34) and (3.35) we see that $P_0(z,f)=P_0(f)$ is a polynomial in f with constant coefficients and

$$\begin{aligned} P_0(f)=\frac{\pm i}{2^{1/2}}\left\{ \frac{\delta ^{2p_0+2}-\delta ^{-2p_0}}{\delta ^2-1}\right\} =\frac{\pm i}{2^{1/2}}\left\{ \frac{[f\pm (f^2-1)^{1/2}]^{p_0+1}-[f\mp (f^2-1)^{1/2}]^{p_0}}{f\pm (f^2-1)^{1/2}-1}\right\} .\nonumber \\ \end{aligned}$$

(3.36)

Now the polynomial $P_0(f)$ with constant coefficients takes the form in [18, Theorem 4(b)]. But the proof there is rather complicated and here we give a simple one. Note that f has no finite Picard exceptional values. It suffices to take the value $f=\cos x$, where $x\in (-\pi /2,\pi /2)$ is real. By substituting $f=\cos x$ into (3.36), we get

$$\begin{aligned} P_0(\cos x)= & {} \frac{\pm i}{2^{1/2}}\left\{ \frac{[\cos x\pm i\sin x]^{p_0+1}-[\cos x\mp i\sin x]^{p_0}}{\cos x\pm i\sin x-1}\right\} \nonumber \\{} & {} =\frac{\pm i}{2^{1/2}}\left\{ \frac{[\cos (x/2)\pm i\sin (x/2)]^{2(p_0+1)}-[\cos (x/2)\mp i\sin (x/2)]^{2p_0}}{[\cos (x/2)\pm i\sin (x/2)]^2-1}\right\} .\nonumber \\ \end{aligned}$$

(3.37)

By the well-known de Moivre’s formula and the basic formula $\sin (x+y)+\sin (x-y)=2\sin x\cos y$ on real trigonometric functions, we deduce from (3.37) that

$$\begin{aligned} \begin{aligned} P_0(\cos x)&=\frac{\pm i}{2^{1/2}}\left\{ \frac{e^{\pm i(p_0+1)x}-e^{\mp ip_0x}}{e^{\pm ix}-1}\right\} =\frac{\pm i}{2^{1/2}}\left\{ \frac{e^{i(p_0+1/2)x}-e^{-i(p_0+1/2)x}}{e^{ix/2}-e^{-ix/2}}\right\} \\&=\frac{\pm i}{2^{1/2}}\left\{ \frac{\sin (p_0+1/2)x}{\sin (x/2)}\right\} =\frac{\pm i}{2^{1/2}}\left\{ \frac{2\sin (p_0+1/2)x\cos (x/2)}{2\sin (x/2)\cos (x/2)}\right\} \\&=\frac{\pm i}{2^{1/2}}\left\{ \frac{\sin (p_0+1)x+\sin p_0x}{\sin x}\right\} =\frac{\pm i}{2^{1/2}}\left\{ \frac{\sin (p_0+1)x}{\sin x}+\frac{\sin p_0x}{\sin x}\right\} . \end{aligned} \end{aligned}$$

Note that in the second step the equation holds for both choices of the signs ± in the exponential function. Denote $U_{p_0}(\cos x):=\frac{\sin (p_0+1)x}{\sin x}$. Then $U_{p_0}$ is the Chebyshev polynomial of the second kind [5, p. 184]. Thus the coefficients of $P_0(f)$ are independent of the choice of x. We conclude that

$$\begin{aligned} \begin{aligned} P_0(f)=\frac{\pm i}{2^{1/2}}[U_{p_0}(f)+U_{p_0-1}(f)], \end{aligned} \end{aligned}$$

where $U_{p_0}$ and $U_{p_0-1}$ are the Chebyshev polynomials of the second kind. This corresponds to the second part of Theorem 4.

Case 2: $N_c=2$.

In this case, $\infty$ is a Picard exceptional rational function of f and p is even. Moreover, by the inequality (1.6) it follows that all $\alpha _i$ are completely ramified rational functions of f with multiplicities 2. Therefore, by Lemma 4 we have

$$\begin{aligned} \overline{f}^2=P_0(z,f)^2(f-\alpha _1)(f-\alpha _2), \end{aligned}$$

(3.38)

where $\alpha _1$ and $\alpha _2$ are in general both non-zero algebraic functions and distinct from each other, and $P_0(z,f)$ is a polynomial in f with simple roots only. By Lemmas 1 and 3, it follows that $\pm \alpha _1$ and $\pm \alpha _2$ are all completely ramified rational functions of f and so by Theorem 2 we must have $\alpha _1+\alpha _2=0$. We may let $\alpha _1=1$ and $\alpha _2=-1$ by doing a linear transformation $f\rightarrow \alpha _1f$. We consider

$$\begin{aligned} \overline{f}^2-1=P_0(z,f)^2(f^2-1)-1. \end{aligned}$$

(3.39)

Since $\pm 1$ are both completely ramified rational functions of f with multiplicities 2 and f has no other completely ramified rational functions, then by Lemma 5 we conclude that the RHS of (3.39) is of the form $P_1(z,f)^2$ for some polynomial $P_1(z,f)$ in f with simple roots only. Now we have

$$\begin{aligned} \overline{f}^2=P_0(z,f)^2(f^2-1), \end{aligned}$$

(3.40)

and further that

$$\begin{aligned} \overline{f}^2-1=P_1(z,f)^2. \end{aligned}$$

(3.41)

Put

$$\begin{aligned} f=\frac{1}{2}(\lambda +\lambda ^{-1}). \end{aligned}$$

(3.42)

Then from (3.41) we see that $\lambda$ is an algebroid function with at most finitely many algebraic branch points. Moreover, 0 and $\infty$ are both Picard exceptional rational functions of $\lambda$. It follows from (3.41) that

$$\begin{aligned} \frac{1}{2}(\overline{\lambda }-\overline{\lambda }^{-1})=P_1(z,f). \end{aligned}$$

(3.43)

From (3.40) and (3.41), we see that the degree of $P_1(z,f)$ in f is $p_0+1$, which is greater than or equal to 2. By solving equation (3.43) together with (3.40) and (3.41), we get

$$\begin{aligned} \overline{\lambda }=P_1(z,f)\pm P_0(z,f)(f^2-1)^{1/2}= & {} \frac{P_1\left( z,\frac{\lambda ^2+1}{2\lambda }\right) (2\lambda )\pm P_0\left( z,\frac{\lambda ^2+1}{2\lambda }\right) (\lambda ^2-1)}{2\lambda }\nonumber \\:= & {} \frac{P_{11}(z,\lambda )}{(2\lambda )^{p_0+2}}, \end{aligned}$$

(3.44)

where $P_{11}(z,\lambda )$ is a polynomial in $\lambda$ of degree at most $2p_0+3$. Since 0 is a Picard exceptional rational function of $\lambda$, then by Picard’s theorem we see from Eq. (3.44) that $P_{11}(z,\lambda )$ cannot have any non-zero root. Moreover, by the Valiron–Mohon’ko identity [17, 23] (see also [15]), we have from (3.43) that $2T(r,\overline{\lambda })=(p_0+1)T(r,f)+O(\log r)=2(p_0+1)T(r,\lambda )+O(\log r)$. Therefore, similarly as in previous case, we may show that the solution f of Eq. (3.40) is represented by (3.42) with $\lambda$ such that

$$\begin{aligned} \overline{\lambda }= \pm i\lambda ^{p_0+1}, \end{aligned}$$

(3.45)

or

$$\begin{aligned} \overline{\lambda }= \pm i\lambda ^{-p_0-1}. \end{aligned}$$

(3.46)

Note that $\lambda$ has at most finitely many zeros, poles and branch points. Then we can solve $\lambda$ from Eqs. (3.45) and (3.46) as

$$\begin{aligned} \lambda =(\pm i)^{-\frac{1}{p_0}}\exp [\pi (z)(p_0+1)^z], \end{aligned}$$

(3.47)

or

$$\begin{aligned} \lambda =(\pm i)^{\frac{1}{2+p_0}}\exp [\pi (z)(-p_0-1)^z], \end{aligned}$$

(3.48)

respectively, where $\pi (z)$ is an arbitrary entire periodic function with period 1. We conclude that solutions of Eq. (3.40) are given by (3.42) with (3.47) or (3.48).

Now we determine the polynomial $P_0(z,f)$ in f in (3.40) by using (3.42) with (3.45) or with (3.46). From (3.40) and (3.45), or (3.40) and (3.46), we have

$$\begin{aligned} \frac{\pm i}{2}(\lambda ^{p_0+1}-\lambda ^{-p_0-1})=\frac{1}{2}(\lambda -\lambda ^{-1})P_0(z,f). \end{aligned}$$

(3.49)

By (3.42), we have

$$\begin{aligned} \begin{aligned} \lambda&=f\pm (f^2-1)^{1/2},\\ \lambda ^{-1}&=f\mp (f^2-1)^{1/2}, \end{aligned} \end{aligned}$$

(3.50)

and it follows that

$$\begin{aligned} \begin{aligned} \lambda ^{p_0+1}-\lambda ^{-p_0-1}&=\sum _{k=0}^{p_0+1}\left( {\begin{array}{c}p_0+1\\ k\end{array}}\right) f^{p_0+1-k}\left\{ [\pm (f^2-1)^{1/2}]^{k}-[\mp (f^2-1)^{1/2}]^{k}\right\} ,\\&=2\sum _{l=0}^{[p_0/2]}\left( {\begin{array}{c}p_0+1\\ 2l+1\end{array}}\right) f^{p_0-2l}[\pm (f^2-1)^{1/2}]^{2l+1}. \end{aligned} \end{aligned}$$

(3.51)

From (3.49), (3.50) and (3.51) we see that $P_0(z,f)=P_0(f)$ is a polynomial in f with constant coefficients and

$$\begin{aligned} P_0(f)=\pm i\sum _{l=0}^{[p_0/2]}\left( {\begin{array}{c}p_0+1\\ 2l+1\end{array}}\right) f^{p_0-2l}(f^2-1)^{l}, \end{aligned}$$

where $[p_0/2]$ denotes the greatest integer not exceeding $p_0/2$. This corresponds to the third part of Theorem 4 and also completes the proof. $\square$

4 Equation (2.1) with $q\ge 1$ and $d=\max \{p,q\}\ge 2$

4.1 Equation (2.1) with $q\ge 1$ and $n>d\ge 2$

For the case $q\ge 1$ and $n>d\ge 2$ of Eq. (2.1), we prove the following

Theorem 5

Suppose that $q\ge 1$ and $n>d\ge 2$. Let f be a transcendental meromorphic solution of Eq. (2.1). Then there exists a rational function $\alpha$ such that the linear transformation $f\rightarrow \alpha f$ reduces (2.1) into

$$\begin{aligned} \overline{f}^n=cf^{-q}, \end{aligned}$$

(4.1)

where c is a non-zero constant. Moreover, solutions of Eq. (4.1) are represented as

$$\begin{aligned} f=c^{\frac{1}{n+q}}\exp [\pi (z)(-q/n)^z], \end{aligned}$$

(4.2)

where $\pi (z)$ is an arbitrary entire periodic function with period 1.

Proof of Theorem 5

Suppose that P(z, f) has a non-zero root, say $\alpha _1$. Since $n\ge 3$, then by Lemmas 1 and 3 it follows that $\omega \alpha _1$ is a completely ramified rational function of f, where $\omega$ is the nth root of 1. Then by the inequality (1.6) we conclude that $n=3$ or $n=4$. In particular, when $n=3$ we see that $\eta \alpha _1$ has multiplicity 3 since the order $k_1$ of the root $\alpha _1$ equals 1 or 2 under the assumption that $n>d$, where $\eta$ is the cubic root of 1. Thus by the inequality (1.6), when $n=3$ we have $\sum _{i=1}^3\varTheta (\eta _i\alpha _1,f)=2$, where $\eta _i$ are the three numbers such that $\eta _i^3=1$. On the other hand, when $n=4$ we have $\sum _{i=1}^4\varTheta (\omega _i\alpha _1,f)=2$, where $\omega _i$ are the four numbers such that $\omega _i^4=1$. By Lemma 1 and the inequality (1.6) we see that 0 cannot be a root of P(z, f) or Q(z, f). Now, if $p\not =q$, then by Lemma 1 it follow that $\infty$ is either a Picard exceptional rational function of f or a completely ramified function of f, a contradiction to the inequality (1.6). Therefore, $p=q$. From the above reasoning, when $n=3$, we have $p=q=2$ and by Lemma 4 we see that each of the roots of P(z, f) and Q(z, f) is simple since 0 is not a completely ramified rational function of f, i.e., we have $N_c=4$, a contradiction to Lemma 1. When $n=4$, we have $p=q=2$ or $p=q=3$, and by Lemma 4 we see that each of the roots of P(z, f) and Q(z, f) has double order since 0 is not a completely ramified rational function of f, but it follows that the case where $p=q=3$ is impossible and when $p=q=2$ we have a contradiction to our assumption that at least one of $\alpha _i$ and $\beta _j$ in (2.7) and (2.8) has order that is not divided by n. We conclude that P(z, f) does not have any non-zero root. Similarly, Q(z, f) cannot have any non-zero root either. Since $q\ge 1$, then by assumption we must have $P(z,f)=a_p$, i.e., we have the following equation:

$$\begin{aligned} \overline{f}^n= a_pf^{-q}. \end{aligned}$$

(4.3)

We claim that f has at most finitely many poles. Suppose on the contrary that f has infinitely many poles. Let $z_0\in \mathbb {C}$ be a pole of f with multiplicity $m\in \mathbb {Z}^{+}$. We may choose $z_0$ such that $|z_0|$ is large enough so that $a_p$ has no poles or zeros outside of $\{z\in \mathbb {C}: |z|<|z_0|\}$. Then from (4.3) we see that $z_0+1$ is a zero of f of order qm/n and it follows that $z_0+2$ is a pole of f of order $q^2m/n^2$. By iteration we have that f has a pole of order $q^{2s}m/n^{2s}$ at the point $z_0+2s$, $s\in \mathbb {N}$. Since $(n,q)=1$, then by letting $s\rightarrow \infty$, it follows that there is necessarily a branch point at some $z_0+2s_0$, $s_0\in \mathbb {N}$, a contradiction to our assumption that f is meromorphic. Therefore, f has at most finitely many poles. From (4.3) we also have that f has at most finitely many zeros. Similarly as in the proof of Theorem 3, we choose a rational function $\alpha$ such that $\alpha f$ has no zeros and poles and then write equation (4.3) as

$$\begin{aligned} \overline{f}^n= cf^{-q}, \end{aligned}$$

(4.4)

where $c=a_p\overline{\alpha }^n\alpha ^q$ is a non-zero constant and f above is a zero-free entire function. By taking the logarithm on both sides of (4.4), then $g=\log f$ is entire and satisfies

$$\begin{aligned} n\overline{g}=\log c-qg. \end{aligned}$$

Therefore, we can solve f from (4.4) as

$$\begin{aligned} f=c^{\frac{1}{n+q}}\exp [\pi (z)(-q/n)^z], \end{aligned}$$

where $\pi (z)$ is an arbitrary entire periodic function with period 1. This completes the proof. $\square$

4.2 Equation (2.1) with $q\ge 1$ and $2\le n<d$

In this section, we consider the two cases $n\not \mid |p-q|$ and $n\mid |p-q|$ of Eq. (2.1) separately. For the case $n\not \mid |p-q|$, we actually have $N_c=1$; we prove the following

Theorem 6

Suppose that $q\ge 1$ and $2\le n<d$ and $n\not \mid |p-q|$. Let f be a transcendental meromorphic solution of Eq. (2.1). Then there exists a rational function $\alpha$ such that by doing a linear transformation $f\rightarrow \alpha f$, we have either

• (1) equation (2.1) reduces into

  $$\begin{aligned} \overline{f}^n = cf^{-q}, \end{aligned}$$

  (4.5)

  and solutions of Eq. (4.5) are represented as

  $$\begin{aligned} f=c^{\frac{1}{n+q}}\exp [\pi (z)(-q/n)^z], \end{aligned}$$

  (4.6)

  where $\pi (z)$ is an arbitrary entire periodic function with period 1; or

• (2) q is even and $q=2q_0$, $q_0\ge 1$, and Eq. (2.1) reduces into

  $$\begin{aligned} \overline{f}^2 = \frac{P_0(z,f)^2}{Q_0(z,f)^2}(f-1)^{k_1}, \end{aligned}$$

  (4.7)

  where $k_1$ is an odd integer and $2p_0+k_1< 2q_0$; moreover, we have

  $$\begin{aligned} \begin{aligned} Q_0(z,f)=Q_1(z,f)^2(f+1)^{l_0}=\frac{1}{2i}[P_{011}(z,f)^2-P_{012}(z,f)^2(f-1)^{k_1}], \end{aligned} \end{aligned}$$

  (4.8)

  where $l_0\in \mathbb {N}$ is zero or an odd integer, $Q_1(z,f)$ is a polynomial in f and $P_{011}(z,f)$ and $P_{012}(z,f)$ are two polynomials in f with no common roots such that $P_{011}(z,f)P_{012}(z,f)=P_0(z,f)$; solutions of (4.7) are represented as

  $$\begin{aligned} f=\frac{1}{2}(\delta ^2+\delta ^{-2}), \end{aligned}$$

  and $\delta$ is a function such that

  $$\begin{aligned} \overline{\delta }=\pm i^{1/2}\frac{P_{021}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) +\theta P_{022}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \left( \frac{\delta ^2-1}{2^{1/2}\delta }\right) ^{k_1}}{P_{021}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) -\theta P_{022}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \left( \frac{\delta ^2-1}{2^{1/2}\delta }\right) ^{k_1}}, \quad \theta =\pm 1, \end{aligned}$$

  (4.9)

  when $l_0=0$, where $P_{021}(z,f)$ and $P_{022}(z,f)$ are two polynomials in f with no common roots such that $P_{021}(z,f)P_{022}(z,f)=P_{012}(z,f)$, or such that

  $$\begin{aligned} \overline{\delta }=\pm (-i)^{1/2}\frac{P_{023}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \delta ^{t_1}+\theta P_{024}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) }{P_{023}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \delta ^{t_1}-\theta P_{024}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) }, \quad \theta =\pm 1, \end{aligned}$$

  (4.10)

  when $l_0>0$, where $t_1\in \mathbb {Z}\setminus \{0\}$ is an odd integer, $P_{023}(z,f)$ and $P_{024}(z,f)$ are two polynomials in f with no common roots such that $P_{023}(z,f)P_{024}(z,f)=P_{011}(z,f)$.

Equations (4.1) and (4.5), as well as their solutions (4.2) and (4.6), are apparently of the same form. We note that in Theorem 6, when $n\ge 3$, we only have equation (4.5). In fact, when $n\ge 3$, if some $\alpha _i$ in (2.7) or $\beta _j$ in (2.8) is non-zero, say $\alpha _i\not =0$ for some i, then by Lemmas 1 and 3 we have a contradiction to the inequality (1.6) since $\omega \alpha _i$ is a completely ramified rational function of f, where $\omega$ is the nth root of 1, and $\infty$ is either a Picard exceptional rational function or a completely ramified rational function of f with multiplicity at least n. Then the reasoning in the proof of Theorem 5 yields equation (4.5).

In the autonomous case, the RHS of Eq. (4.9) or (4.10) becomes a rational term $R_0(\delta )$ in $\delta$ after multiplying $(2^{1/2}\delta )^{q_0}$ to both the numerator and the denominator and thus $\overline{\delta }=R_0(\delta )$ always has a meromorphic solution for any given $P_0(z,f)$ and $Q_1(z,f)$ such that the relation in (4.8) holds, as mentioned in the introduction. We note that when $l_0>0$, the polynomials $P_0(z,f)$ and $Q_0(z,f)$ satisfying the relation in (4.8) exist. For example, for the two polynomials $P_0(f)$ and $P_1(f)$ satisfying (3.16) and (3.17) in the proof of Theorem 4, we have $iP_0(f)^2(f-1)-iP_1(f)^2(f+1)=i$. In the simplest case $p_0=k_1=l_0=1$, we have $2i(f-1/2)^2(f+1)-2i(f+1/2)^2(f-1)=i$.

Proof of Theorem 6

First, we show that under the assumptions of Theorem 6 the case $p>q$ cannot occur. Since $n\not \mid (p-q)$, then from the proof of Lemma 2, we see that $\infty$ is a Picard exceptional rational function of f no matter whether or not some $\alpha _i$ in (2.7) is zero. Let $\beta$ be any root of Q(z, f). Then from (2.1) we see that the equation $f-\beta =0$ has at most finitely many roots and so $\beta$ is a Picard exceptional rational function of f. By Picard’s theorem we see that there is only one such $\beta$. Then by Lemma 1 and the inequality (1.6) we conclude that (2.1) takes the following form:

$$\begin{aligned} \overline{f}^n=\frac{P_0(z,f)^n}{(f-\beta )^q}, \end{aligned}$$

(4.11)

where $\beta$ is a rational function. Moreover, since $p>q$, we see that $\beta \not \equiv 0$ since otherwise all the roots of $P_0(z,f)$ are Picard exceptional rational functions of f, a contradiction to Picard’s theorem. Denote $g=(f-\beta )^{1/n}$. Then g is an algebroid function with at most finitely many algebraic branch points and it follows that $f=g^n+\beta$. Then we can rewrite equation (4.11) as follows

$$\begin{aligned} \overline{g}^{n}g^{q}=P_0(z,g^n+\beta )-\overline{\beta }g^{q}. \end{aligned}$$

(4.12)

Note that 0 and $\infty$ are both Picard exceptional rational functions of g. Let $u_0$ be a function such that

$$\begin{aligned} P_0(z,u_0^n+\beta )-\overline{\beta }u_0^{q}=0. \end{aligned}$$

Then $u_0$ is an algebraic function. Since $q<p$ and $P_0(z,f)$ has at least one root distinct from $\beta$, we see that the equation above has at least one non-zero root and from (4.12) we see that for the non-zero $u_0$ we have that $g-u_0=0$ has at most finitely many roots, i.e., $u_0$ is a Picard exceptional rational function of g, a contradiction to Picard’s theorem. Therefore, the case where $p>q\ge 1$ and $n\not \mid (q-p)$ cannot occur. In particular, since $p\not =q$, the above analysis also implies that the case where 0 is a root of P(z, f) of order $k_0$ such that $n\not \mid k_0$ cannot occur; otherwise, by doing a bilinear transformation $f\rightarrow 1/f$ to (2.1), we always get

$$\begin{aligned} \overline{f}^n=\frac{P_1(z,f)}{Q_1(z,f)}, \end{aligned}$$

(4.13)

where $P_1(z,f)$ is a polynomial in f of degree d and $Q_1(z,f)$ is a polynomial in f of degree $d-k_0$, which is impossible from previous discussions.

Second, we show that $N_c=1$. Suppose on the contrary that $N_c\ge 2$. Then at least one of $\alpha _i$ and $\beta _j$ is non-zero, say $\alpha _i$. By Lemmas 1 and 3, $\omega \alpha _i$ is a completely ramified rational function of f, where $\omega$ is the nth root of 1. Moreover, since $p<q$ and $n\not \mid |p-q|$, then by Lemma 1 it follows that $\infty$ is also a completely ramified rational function of f with multiplicity at least n. By the inequality (1.6), we must have $n=2$. In this case, we see that $N_c$ is an odd integer. If $N_c\ge 3$, then none of $\alpha _i$ and $\beta _j$ is zero; otherwise, 0 has multiplicity at least 2 and it follows that $\infty$ has multiplicity at least 4, a contradiction to the inequality (1.6). For convenience, we denote these $\alpha _i$ and $\beta _j$ by $\gamma _j$, $j=1,2,\cdots ,k$. By Lemmas 1 and 3, $\pm \gamma _j$ are all completely ramified rational functions of f. Then by Theorem 2 we conclude that $\gamma _j^2$ must be equal to each other for all j, i.e., $\gamma _1^2=\gamma _2^2=\cdots =\gamma _k^2$ holds for $k\ge 3$, which is impossible. Therefore, we must have $N_c=1$.

Third, we show that 0 cannot be a root of Q(z, f) of order $l_0<q$ such that $n\not \mid l_0$. Note that now we have $0\le p<q$ under the assumption $p\not =q$. Otherwise, (2.1) can be written as

$$\begin{aligned} \overline{f}^n=\frac{P(z,f)}{f^{l_0}Q_2(z,f)}, \end{aligned}$$

(4.14)

where $1\le l_0\le q-1$ satisfies $n\not \mid l_0$, and $Q_2(z,f)$ is a polynomial in f of degree $q-l_0$. Now we must have $n\mid (q-l_0)$ and $n\mid p$ by the fact that $N_c=1$ and it follows that $q-p=k_0n+l_0$ for some integer $k_0$. Suppose that f has infinitely many zeros and let $z_0\in \mathbb {C}$ be a zero of f with multiplicity $m\in \mathbb {Z}^{+}$. We may choose $z_0$ such that $|z_0|$ is large enough so that none of the coefficients of P(z, f) and Q(z, f) has poles or zeros outside of $\{z\in \mathbb {C}: |z|<|z_0|\}$. By (4.14), $z_0+1$ is a pole of f of order $l_0m/n$. It follows that $z=z_0+2$ is a zero of f of order $l_0(k_0n+l_0)m/n^2$. Then, by iteration it follows that $z=z_0+2s$, $s\in \mathbb {N}$, is a zero of f of order $(k_0n+l_0)^sl_0^{s}m/n^{2s}$. Since $n^2\not \mid (k_0n+l_0)l_0$, then by letting $s\rightarrow \infty$, it follows that there is necessarily a branch point of f at $z_0+2s_0$ for some $s_0\in \mathbb {N}$, a contradiction to our assumption that f is meromorphic. Therefore, f has at most finitely many zeros, i.e., 0 is a Picard exceptional rational function of f. Also, from (4.14) we see that f has at most finitely many poles since $p<q$ and then, since $l_0\le q-1$, it follows that there exists another non-zero $\beta$ such that $\beta$ is a root of Q(z, f) and $f-\beta =0$ has at most finitely many roots, that is to say, f has at least 3 Picard exceptional rational functions, a contradiction to Picard’s theorem. Therefore, 0 cannot be a root of Q(z, f) of order $l_0<q$ such that $n\not \mid l_0$ when assuming $n \not \mid (q-p)$.

By combining all the above results together, we see that we only need to consider two cases of (2.1) under our assumption: (1) the case where 0 is the only root of Q(z, f); or (2) the case where 0 is not a root of Q(z, f) of order $0<l_0<q$ such that $n\not \mid l_0$. In particular, in the latter case if 0 is not a root of $Q_0(z,f)$, then from the previous discussions we can assume that 0 is not a root of P(z, f) of order $k_0$ such that $n\not \mid k_0$.

Now, if 0 is the only root of Q(z, f) and $n\not \mid q$, from the above reasoning we must have

$$\begin{aligned} \overline{f}^n = \frac{P_0(z,f)^n}{f^{q}}. \end{aligned}$$

In this case, since $p<q$ and $n\not \mid (q-p)$, then by applying exactly the same analysis as in the previous discussions on the case where 0 is a root of Q(z, f) of order $l_0<q$ and $n\not \mid l_0$, we can show that f has at most finitely many zeros and poles, i.e., 0 and $\infty$ are both Picard exceptional rational functions of f. By Picard’s theorem, we see that $P_0(z,f)$ has no non-zero roots and thus $P_0(z,f)^n=a_p$. Then by doing a linear transformation $f\rightarrow \alpha f$ with a suitably chosen rational function $\alpha$, we get

$$\begin{aligned} \overline{f}^n = cf^{-q}, \end{aligned}$$

(4.15)

where $c=\frac{a_p}{\overline{\alpha }^n\alpha ^q}$ is a non-zero constant. Also, as in Theorem 5, solutions of (4.15) are represented as

$$\begin{aligned} f=c^{\frac{1}{n+q}}\exp [\pi (z)(-q/n)^z], \end{aligned}$$

where $\pi (z)$ is an arbitrary entire periodic function with period 1. Otherwise, we have that 0 is the only root of Q(z, f) with $n\mid q$, or 0 is not the only root of Q(z, f). Recalling that we have excluded out the two possibilities that 0 is a root of P(z, f) of order $k_0$ such that $n\not \mid k_0$ and that 0 is a root of Q(z, f) of order $l_0$ such that $l_0<q$ and $n\not \mid l_0$, we see that in either case the only $\alpha _i$ in P(z, f) or $\beta _j$ in (2.8) is non-zero. Therefore, we only need to consider the following two equations:

$$\begin{aligned} \overline{f}^2= & {} \frac{P_0(z,f)^2}{Q_0(z,f)^2}(f-\alpha _1)^{k_1}, \end{aligned}$$

(4.16)

$$\begin{aligned} \overline{f}^2= & {} \frac{P_0(z,f)^2}{Q_0(z,f)^2}\frac{1}{(f-\beta _1)^{l_1}}, \end{aligned}$$

(4.17)

where $\alpha _1\not =0$ and $\beta _1\not =0$, $k_1,l_1\in \mathbb {N}$ are odd integers, and in Eq. (4.16) we have $2p_0+k_1< 2q_0$ and in Eq. (4.17) we have $2p_0< 2q_0+l_1$. Below we discuss the two equations above separately.

Subcase 1: Equation (4.16).

From the previous discussions we see that $\pm \alpha _1$ and $\infty$ are all completely ramified rational functions of f. In fact, $\infty$ cannot be a Picard exceptional rational function of f; otherwise, the roots of Q(z, f) are all Picard’s exceptional rational functions of f, which is impossible by the inequality (1.6). Also, from (4.16) we see that all roots of $f-\alpha _1=0$ with at most finitely many exceptions are of even multiplicities. If $Q_0(z,f)$ has a root, say $\beta$, of odd order, then by applying the same analysis as in the proof of Lemma 1 and considering the multiplicities of the roots of $f-\beta =0$ together with the fact that $\infty$ is a completely ramified rational function of f, we obtain that $\beta$ is a completely ramified rational function of f. Moreover, $\beta \not =0$ since otherwise from the proof of Lemma 1 we have that $\pm \alpha _1$ are completely ramified rational functions of f with multiplicity 4, a contradiction to the inequality (1.6). By Lemma 3 it follows that $\pm \beta$ are both completely ramified rational functions of f. Then by Theorem 2 we must have $\beta =-\alpha _1$ and there is only one such $\beta$. We conclude that $Q_0(z,f)$ is of the form $Q_0(z,f)=Q_1(z,f)^2(f+\alpha _1)^{l_0}$ for some polynomial $Q_1(z,f)$ in f and $l_0\in \mathbb {N}$ is 0 or an odd integer. We may let $\alpha _1=-1$ by doing a linear transformation $f\rightarrow \alpha _1f$. We consider

$$\begin{aligned} \overline{f}^2-1= \frac{P_0(z,f)^2(f-1)^{k_1}-Q_0(z,f)^2}{Q_0(z,f)^2}. \end{aligned}$$

(4.18)

Note that the numerator of the RHS of (4.18) is a polynomial in f with degree $2q_0$. If it has two distinct roots, say $\gamma _1$ and $\gamma _2$, of odd orders, then by considering the multiplicities of the roots of $f-\gamma _1=0$ and $f-\gamma _2=0$, respectively, together with the fact that $\pm 1$ are both completely ramified rational functions of f and that the roots of $f\pm 1=0$ have even multiplicities with at most finitely many exceptions, we obtain that $\gamma _1$ and $\gamma _2$ are both completely ramified rational functions of f. By Lemma 3, it follows that $\pm \gamma _1$ and $\pm \gamma _2$ are all completely ramified rational functions of f; this yields a contradiction to Theorem 2 even when $\gamma _1+\gamma _2=0$. Therefore, the numerator of the RHS of (4.18) must be of the form $P_1(z,f)^2$ for some polynomial $P_1(z,f)$ in f. Now we have

$$\begin{aligned} \overline{f}^2= \frac{P_0(z,f)^2(f-1)^{k_1}}{Q_0(z,f)^2}, \end{aligned}$$

(4.19)

and further that

$$\begin{aligned} \overline{f}^2-1 =\frac{P_1(z,f)^2}{Q_0(z,f)^2}. \end{aligned}$$

(4.20)

It follows that

$$\begin{aligned} \begin{aligned} P_0(z,f)^2(f-1)^{k_1}&=P_1(z,f)^2+Q_0(z,f)^2\\&=[P_1(z,f)+iQ_0(z,f)][P_1(z,f)-iQ_0(z,f)], \end{aligned} \end{aligned}$$

and so

$$\begin{aligned} \begin{aligned} P_1(z,f)+iQ_0(z,f)&=P_{01}(z,f),\\ P_1(z,f)-iQ_0(z,f)&=P_{02}(z,f), \end{aligned} \end{aligned}$$

(4.21)

where $P_{01}(z,f)$ and $P_{02}(z,f)$ are two polynomials in f such that $P_{01}(z,f)P_{02}(z,f)=P_0(z,f)^2(f-1)^{k_1}$. Since $P_{01}(z,f)$ and $P_{02}(z,f)$ have no common roots, without loss of generality, we may write

$$\begin{aligned} \begin{aligned} P_{01}(z,f)&=P_{011}(z,f)^2,\\ P_{02}(z,f)&=P_{012}(z,f)^2(f-1)^{k_1}, \end{aligned} \end{aligned}$$

where $P_{011}(z,f)$ and $P_{012}(z,f)$ are two polynomials in f with no common roots and $P_{011}(z,f)P_{012}(z,f)=P_0(z,f)$. From equations in (4.21) together with previous discussions we have

$$\begin{aligned} \begin{aligned} Q_0(z,f)&=\frac{1}{2i}[P_{011}(z,f)^2-P_{012}(z,f)^2(f-1)^{k_1}]=Q_1(z,f)^2(f+1)^{l_0},\\ P_1(z,f)&=\frac{1}{2}[P_{011}(z,f)^2+P_{012}(z,f)^2(f-1)^{k_1}]. \end{aligned} \end{aligned}$$

(4.22)

With the above two expressions for $Q_0(z,f)$ and $P_1(z,f)$, solutions of Eq. (4.19) can be obtained in the following way. Put

$$\begin{aligned} f=\frac{1}{2}(\lambda +\lambda ^{-1}). \end{aligned}$$

(4.23)

Note that the leading coefficient of the polynomial $P_1(z,f)^2$ is $-1$. From (4.20) we see that $\lambda$ is a meromorphic function and it follows that

$$\begin{aligned} \frac{1}{2}\frac{\overline{\lambda }^2-1}{\overline{\lambda }}= \frac{P_1(z,f)}{Q_0(z,f)}. \end{aligned}$$

(4.24)

Since $\infty$ is a completely ramified rational function of f, then we see that 0 and $\infty$ are both completely ramified rational functions of $\lambda$. Moreover, since all zeros of $f-1$ have multiplicities at least 2, from (4.19) we see that the leading coefficient of the polynomial $P_0(z,f)^2(f-1)^{k_1}$ is a square of some rational function and it follows that all poles of f have even multiplicities. Put

$$\begin{aligned} \lambda =\delta ^2. \end{aligned}$$

(4.25)

Then $\delta$ is a meromorphic function. By solving equation (4.24) together with (4.19), (4.20) and (4.25), we get

$$\begin{aligned} \overline{\delta }^2= \frac{P_1(z,f)}{Q_0(z,f)}\pm \frac{P_0(z,f)(f-1)^{k_1/2}}{Q_0(z,f)}= \frac{P_1\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \pm P_0\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \left( \frac{\delta ^2-1}{2^{1/2}\delta }\right) ^{k_1}}{Q_0\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) }. \end{aligned}$$

Note that $P_0(z,f)=P_{011}(z,f)P_{012}(z,f)$. By combining the equation above and the equations in (4.22), we have

$$\begin{aligned} \overline{\delta }^2=i\frac{P_{011}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) +\theta P_{012}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \left( \frac{\delta ^2-1}{2^{1/2}\delta }\right) ^{k_1}}{P_{011}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) -\theta P_{012}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \left( \frac{\delta ^2-1}{2^{1/2}\delta }\right) ^{k_1}}, \quad \theta =\pm 1. \end{aligned}$$

(4.26)

Denote the degrees of the polynomials $Q_1(z,f)$, $P_{011}(z,f)$ and $P_{012}(z,f)$ by $s_0$, $s_1$, $s_2$, respectively. By comparing the degrees of the polynomials in the first equation of (4.22) on both sides, we see that if $l_0=0$, then $s_0=s_1$ and if $l_0>0$, then $2s_0+l_0=2s_2+k_1$. We discuss these two cases below further.

When $l_0=0$, we have from the first equation in (4.22) that

$$\begin{aligned} \begin{aligned} {[}P_{011}(z,f)+(2i)^{1/2}Q_1(z,f)][P_{011}(z,f)-(2i)^{1/2}Q_1(z,f)]=P_{012}(z,f)^2(f-1)^{k_1}. \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \begin{aligned} P_{011}(z,f)+(2i)^{1/2}Q_1(z,f)&=P_{013}(z,f),\\ P_{011}(z,f)-(2i)^{1/2}Q_1(z,f)&=P_{014}(z,f), \end{aligned} \end{aligned}$$

(4.27)

where $P_{013}(z,f)$ and $P_{014}(z,f)$ are two polynomials in f such that $P_{013}(z,f)P_{014}(z,f)=P_{012}(z,f)^2(f-1)^{k_1}$. Since $P_{013}(z,f)$ and $P_{014}(z,f)$ have no common roots, without loss of generality, we may write

$$\begin{aligned} \begin{aligned} P_{013}(z,f)&=P_{021}(z,f)^2,\\ P_{014}(z,f)&=P_{022}(z,f)^2(f-1)^{k_1}, \end{aligned} \end{aligned}$$

(4.28)

where $P_{021}(z,f)$ and $P_{022}(z,f)$ are two polynomials in f with no common roots and $P_{021}(z,f)P_{022}(z,f)=P_{012}(z,f)$. Then we have from equations in (4.27) that

$$\begin{aligned} \begin{aligned} P_{011}(z,f)&=\frac{1}{2}[P_{013}(z,f)+P_{014}(z,f)]. \end{aligned} \end{aligned}$$

(4.29)

By combining equations in (4.28) and (4.29), we have from Eq. (4.26) that

$$\begin{aligned} \overline{\delta }=\pm i^{1/2}\frac{P_{021}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) +\theta P_{022}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \left( \frac{\delta ^2-1}{2^{1/2}\delta }\right) ^{k_1}}{P_{021}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) -\theta P_{022}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \left( \frac{\delta ^2-1}{2^{1/2}\delta }\right) ^{k_1}}, \quad \theta =\pm 1. \end{aligned}$$

(4.30)

When $l_0>0$, we let h and g be such that $h^2+1=f$ and $g^2-1=f$, respectively. Then we have from the first equation in (4.22) that

$$\begin{aligned} \begin{aligned} {[}P_{012}(z,f)h^{k_1}+(-2i)^{1/2}Q_1(z,f)g^{l_0}][P_{012}(z,f)h^{k_1}-(-2i)^{1/2}Q_1(z,f)g^{l_0}]=P_{011}(z,f)^2. \end{aligned} \end{aligned}$$

By (4.23) and (4.25), we may write h and g as $h=\frac{\delta ^2-1}{2^{1/2}\delta }$ and $g=\frac{\delta ^2+1}{2^{1/2}\delta }$, respectively, and it follows that

$$\begin{aligned} \begin{aligned} P_{012}(z,f)h^{k_1}&=P_{012}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \left( \frac{\delta ^2-1}{2^{1/2}\delta }\right) ^{k_1}:=\frac{P_{0121}(z,\delta ^2)}{(2^{1/2})^{2s_2+k_1}\delta ^{2s_2+k_1}},\\ Q_1(z,f)g^{l_0}&=Q_1\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \left( \frac{\delta ^2+1}{2^{1/2}\delta }\right) ^{l_0}:=\frac{Q_{11}(z,\delta ^2)}{(2^{1/2})^{2s_0+l_0}\delta ^{2s_0+l_0}}, \end{aligned} \end{aligned}$$

where $P_{0121}(z,\delta ^2)$ and $Q_{11}(z,\delta ^2)$ are two polynomials in $\delta ^2$. Here, none of the roots of $P_{0121}(z,\delta ^2)$ and $Q_{11}(z,\delta ^2)$ in $\delta ^2$ is equal to $\pm 1$. Note that the leading coefficients of the two polynomials $P_{012}(z,f)^2(f-1)^{k_1}$ and $-2iQ_1(z,f)^2(f+1)^{l_0}$ are equal. Recalling that $2s_0+l_0=2s_2+k_1$, we see that one of the two polynomials $P_{0121}(z,\delta ^2)+(-2i)^{1/2}Q_{11}(z,\delta ^2)$ and $P_{0121}(z,\delta ^2)-(-2i)^{1/2}Q_{11}(z,\delta ^2)$ in $\delta ^2$ has some zero roots. Consider the pair of equations

$$\begin{aligned} \begin{aligned} P_{012}(z,f)h^{k_1}+(-2i)^{1/2}Q_1(z,f)g^{l_0}&=0,\\ P_{012}(z,f)h^{k_1}-(-2i)^{1/2}Q_1(z,f)g^{l_0}&=0. \end{aligned} \end{aligned}$$

Since the two polynomials $P_{012}(z,f)$ and $Q_1(f)$ have no common roots, then together with the relations $h^2+1=f$ and $g^2-1=f$ we see that the above two equations with respect to f cannot have any common solution and thus each root of the polynomial $P_{011}(z,f)^2$ satisfy only one of them. This implies the following two results: First, the two polynomials $P_{0121}(z,\delta ^2)+(-2i)^{1/2}Q_{11}(z,\delta ^2)$ and $P_{0121}(z,\delta ^2)-(-2i)^{1/2}Q_{11}(z,\delta ^2)$ in $\delta ^2$ have no common non-zero roots; second, for each root $\gamma$ of $P_{011}(z,f)$, the two distinct non-zero solutions of the equation $\delta ^4-2\gamma \delta ^2+1=0$ with respect to $\delta ^2$ must be either both roots of the polynomial $P_{0121}(z,\delta ^2)+(-2i)^{1/2}Q_{11}(z,\delta ^2)$ in $\delta ^2$ or the polynomial $P_{0121}(z,\delta ^2)-(-2i)^{1/2}Q_{11}(z,\delta ^2)$ in $\delta ^2$. Since $2s_0+l_0$ is an odd integer and since $P_{011}(z,f)=P_{011}\left( z,\frac{1}{2}(\delta ^2+\delta ^{-2})\right)$ is a rational function in $\delta ^2$ whose denominator has only zero root, then from the above discussions we see that there must be an odd integer $t_1\in \mathbb {Z}\setminus \{0\}$ such that

$$\begin{aligned} \begin{aligned} P_{012}(z,f)h^{k_1}+(-2i)^{1/2}Q_1(z,f)g^{l_0}&=P_{015}(z,f)\delta ^{t_1},\\ P_{012}(z,f)h^{k_1}-(-2i)^{1/2}Q_1(z,f)g^{l_0}&=P_{016}(z,f)\delta ^{-t_1}, \end{aligned} \end{aligned}$$

(4.31)

where $P_{015}(z,f)$ and $P_{016}(z,f)$ are two polynomials in f with no common roots and $P_{015}(z,f)P_{016}(z,f)=P_{011}(z,f)^2$. We may write

$$\begin{aligned} \begin{aligned} P_{015}(z,f)&=P_{023}(z,f)^2,\\ P_{016}(z,f)&=P_{024}(z,f)^2, \end{aligned} \end{aligned}$$

(4.32)

where $P_{023}(z,f)$ and $P_{024}(z,f)$ are two polynomials in f with no common roots and $P_{023}(z,f)P_{024}(z,f)=P_{011}(z,f)$. Then we have from equations in (4.31) that

$$\begin{aligned} \begin{aligned} P_{012}(z,f)h^{k_1}&=\frac{1}{2}[P_{015}(z,f)\delta ^{t_1}+P_{016}(z,f)\delta ^{-t_1}]. \end{aligned} \end{aligned}$$

(4.33)

By combining equations (4.32) and (4.33), we have from (4.26) that

$$\begin{aligned} \overline{\delta }=\pm (-i)^{1/2}\frac{P_{023}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \delta ^{t_1}+\theta P_{024}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) }{P_{023}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) \delta ^{t_1}-\theta P_{024}\left( z,\frac{\delta ^4+1}{2\delta ^{2}}\right) }, \quad \theta =\pm 1. \end{aligned}$$

(4.34)

We conclude that solutions of (4.19) are represented by (4.23) and (4.25), i.e., $f=\frac{1}{2}(\delta ^2+\delta ^{-2})$ with $\delta$ being a solution of (4.30) or (4.34).

Subcase 2: Equation (4.17).

Since $\beta _1\not =0$, then by similar arguments as in the previous case we know that $\pm \beta _1$ and $\infty$ are all completely ramified rational functions of f. We may let $\beta _1=1$ by doing a linear transformation $f\rightarrow \beta _1f$. Further, by considering the multiplicities of the roots of $f-1=0$ together with the fact that $\infty$ has multiplicity at least 2 and by Lemma 3, we obtain from (4.17) that $\pm 1$ both have multiplicities 4 and it follows that $\varTheta (\infty ,f)+\varTheta (1,f)+\varTheta (-1,f)=2$. Then by applying the analysis in the proof of Lemma 4 to the roots of $P_0(z,f)$ and poles of f and comparing the multiplicities of the zeros on both sides of (4.17) and to the roots of $Q_0(z,f)^2(f-1)^{l_1}$ and comparing the multiplicities of the poles on both sides of (4.17), respectively, we obtain that $l_1=1$, $p_0=q_0$, $P_0(z,f)$ has simple roots only, $Q_0(z,f)=Q_1(z,f)^2$ for some polynomial $Q_1(z,f)$ in f with simple roots only and none of these roots equals $\pm 1$. We consider

$$\begin{aligned} \overline{f}^2-1 = \frac{P_0(z,f)^2-Q_0(z,f)^2(f-1)}{Q_0(z,f)^2(f-1)}. \end{aligned}$$

(4.35)

The numerator of the RHS of (4.35) is a polynomial in f with degree $2q_0+1$ and thus has at least one root, say $\gamma _1$, of odd order. By applying the same analysis as in the proof of Lemma 1 and considering the multiplicities of the roots of $f-\gamma _1=0$ together with the fact that both $\pm 1$ have multiplicities 4, we obtain that $\gamma _1$ is also a completely ramified rational function of f. By Lemma 3 and Theorem 2 we must have $\gamma _1=-1$. Then by Lemma 5 we conclude that the order of $\gamma _1$ equals 1. Also, by Theorem 2 we see that there is only one such $\gamma _1$ that has odd order. Now we have

$$\begin{aligned} \overline{f}^2= \frac{P_0(z,f)^2}{Q_0(z,f)^2(f-1)}, \end{aligned}$$

(4.36)

and also that

$$\begin{aligned} \overline{f}^2-1 = \frac{P_1(z,f)^2(f+1)}{Q_0(z,f)^2(f-1)}. \end{aligned}$$

(4.37)

Moreover, by Lemma 5 it follows that $P_1(z,f)=P_2(z,f)^2$ for some polynomials $P_2(z,f)$ with simple roots only. We let $(f+1)/(f-1)=g^2$. Then g is an algebroid function with at most finitely many algebraic branch points and it follows that the RHS of (4.37) becomes $[P_1(z,f)g/Q_0(z,f)]^2$. Put

$$\begin{aligned} f=\frac{1}{2}(\lambda +\lambda ^{-1}). \end{aligned}$$

(4.38)

With the setting $(f+1)/(f-1)=g^2$ we see from (4.37) that $\lambda$ is an algebroid function with at most finitely many algebraic branch points. Recall that $\pm 1$ both have multiplicities 4 and $\infty$ has multiplicity 2. By a simple multiplicity analysis on the zeros, poles and $\pm 1$-points of $\lambda$, we obtain from the definition in (4.38) that 0, $\infty$ and $\pm 1$ are all completely ramified rational functions of $\lambda$ with multiplicities 2. By substituting (4.38) into (4.37) we get

$$\begin{aligned} \frac{1}{2}\frac{\overline{\lambda }^2-1}{\overline{\lambda }}= \frac{P_1(z,f)}{Q_0(z,f)}\frac{\lambda +1}{\lambda -1}. \end{aligned}$$

(4.39)

We put

$$\begin{aligned} \lambda =\delta ^2. \end{aligned}$$

(4.40)

Then $\delta$ is an algebroid function with at most finitely many algebraic branch points. Moreover, by the definition of $\lambda$ we see that $\pm 1$ and $\pm i$ are all completely ramified rational functions of $\delta$ with multiplicity 2. By solving equation (4.39) together with Eqs. (4.36), (4.37) and (4.40), we get

$$\begin{aligned} \overline{\delta }^2= & {} \frac{P_1(z,f)}{Q_0(z,f)}\frac{\lambda +1}{\lambda -1}\pm \frac{P_0(z,f)}{Q_0(z,f)(f-1)^{1/2}}\nonumber \\= & {} \frac{P_1\left( z,\frac{\delta ^4+1}{2\delta ^2}\right) (\delta ^2+1)\pm P_0\left( z,\frac{\delta ^4+1}{2\delta ^2}\right) (2^{1/2}\delta )}{Q_0\left( z,\frac{\delta ^4+1}{2\delta ^2}\right) (\delta ^2-1)}. \end{aligned}$$

(4.41)

Recall that $p_0=q_0$ from the beginning of this subcase. Also, from previous discussions we see that the degree of the polynomial $P_1(z,f)$ in (4.37), denoted by $p_1$, satisfies $p_1=p_0$. By multiplying $(2\delta ^2)^{p_0}$ to both of the numerator and the denominator of the RHS of (4.41), we have

$$\begin{aligned} \overline{\delta }^2= \frac{P_{10}(z,\delta ^2)(\delta ^2+1)\pm P_{00}(z,\delta ^2)(2^{1/2}\delta )}{Q_{00}(z,\delta ^2)(\delta ^2-1)}, \end{aligned}$$

(4.42)

where $P_{10}(z,\delta ^2)$, $P_{00}(z,\delta ^2)$ and $Q_{00}(z,\delta ^2)$ are polynomials in $\delta$ of the same degrees $4q_0$. Note that $\delta$ cannot have any other completely ramified rational function besides $\pm 1$ and $\pm i$. Since $\pm 1$ are not roots of $P_0(z,f)$, $Q_0(z,f)$ or $P_1(z,f)$, then from the above reasoning we see that $\pm 1$ and $\pm i$ are not roots of the polynomials $P_{10}(z,\delta ^2)$, $P_{00}(z,\delta ^2)$ or $Q_{00}(z,\delta ^2)$. By applying Lemma 4 to Eq. (4.42), we conclude that the numerator of the RHS of (4.42) must be a square of some polynomial in $\delta$ with simple roots only and none of these roots is equal to $\pm 1$ or $\pm i$. Moreover, since $p<q$ and we have assumed $b_q=1$, we see from the above reasoning that the leading coefficient of the numerator of the RHS of (4.42) is $\pm i$. Therefore, we may write

$$\begin{aligned} P_{10}(z,\delta ^2)^2(\delta ^2+1)\pm P_{00}(z,\delta ^2)2^{1/2}\delta =\pm i(\delta -\epsilon _1)^2\cdots (\delta -\epsilon _{2q_{0}+1})^2:=P_{11}(z,\delta ), \end{aligned}$$

(4.43)

where the roots $\epsilon _1$, $\cdots$, $\epsilon _{2q_{0}+1}$ are in general algebraic functions, distinct from each other and from $\pm 1$ and $\pm i$. Since $P_{11}(z,\delta )$ is a polynomial in $\delta$ of degree $4q_0+2$, we may denote by $\varrho _{4q_{0}+2}$, $\cdots$, $\varrho _1$ the roots of $P_{11}(z,\delta )$. Then from the equations in (4.41) and (4.42) we see that 0 is not a root of the polynomial $P_{11}(z,\delta )$ and that $\varrho _{4q_{0}+2}^{-1}$, $\cdots$, $\varrho _1^{-1}$ are also roots of $P_{11}(z,\delta )$. However, since $\epsilon _1$, $\cdots$, $\epsilon _{2q_{0}+1}$ are distinct from each other, we see that $\varrho _i=\varrho _i^{-1}$ for at least one $i\in \{4q_0+2,\cdots ,1\}$ and thus $\varrho _i=\pm 1$, a contradiction to our previous observations. This implies that Eq. (4.36) cannot have any meromorphic solution, which completes the proof. $\square$

For the case $q\ge 1$ and $2\le n<d$ and $n\mid |p-q|$ of Eq. (2.1), we prove the following Theorem 7. In this theorem, we will do a bilinear transformation to f in (2.1); for simplicity, we always assume that the leading coefficient of the resulting $Q_0(z,f)^n$ is 1 and denote the leading coefficient of the resulting $P_0(z,f)^n$ by A.

Theorem 7

Suppose that $q\ge 1$ and $2\le n<d$ and $n\mid |p-q|$. Let f be a transcendental meromorphic solution of Eq. (2.1). Then there exists a rational or algebraic function $\alpha$ such that by doing a linear transformation $f\rightarrow \alpha f$ or $f\rightarrow 1/(\alpha f)$, (2.1) becomes (3.7), (3.9) or (4.7), or one of the following equations:

• (1) the first equation is

  $$\begin{aligned} \overline{f}^2= \frac{P_0(z,f)^2}{Q_0(z,f)^2}(f-1)(f-\kappa ), \end{aligned}$$

  (4.44)

  where $\kappa \not =0,\pm 1$, $P_0(z,f)$ and $Q_0(z,f)$ are two polynomials in f with simple roots only and $p_0-q_0\in \{-2,-1,0\}$; moreover,

  $$\begin{aligned} \begin{aligned} P_0(z,f)^2(f-1)(f-\kappa )-Q_0(z,f)^2&=P_1(z,f)^2(f+1)(f+\kappa ),\\ P_0(z,f)^2(f-1)(f-\kappa )-\overline{\kappa }^2 Q_0(z,f)^2&=P_2(z,f)^2, \end{aligned} \end{aligned}$$

  (4.45)

  where $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f with simple roots only and of degrees $p_1$ and $p_2$ such that $2p_1+2\in \{d,d-2\}$ and $2p_2\in \{d,d-2\}$;

• (2) the second equation is

  $$\begin{aligned} \overline{f}^2= \frac{P_0(z,f)^2}{Q_0(z,f)^2}(f-1)^{k_1}(f+1)^{k_2}, \end{aligned}$$

  (4.46)

  where $k_1,k_2$ are two positive odd integers such that $P_0(z,f)^2(f-1)^{k_1}(f+1)^{k_2}-Q_0(z,f)^2=P_1(z,f)^2$ for some polynomial $P_1(z,f)$ in f;

• (3) the third equation is

  $$\begin{aligned} \overline{f}^2= \frac{P_0(z,f)^2}{Q_0(z,f)^2}(f^2-1), \end{aligned}$$

  (4.47)

  where $P_0(z,f)$ and $Q_0(z,f)$ are two polynomials in f with simple roots only such that $p_0-q_0\in \{-2,-1,0\}$; moreover,

  $$\begin{aligned} \begin{aligned} P_0(z,f)^2(f^2-1)-Q_0(z,f)^2&=P_1(z,f)^2(f^2-\gamma ^2),\\ P_0(z,f)^2(f^2-1)-\overline{\gamma }^2 Q_0(z,f)^2&=P_2(z,f)^2, \end{aligned} \end{aligned}$$

  (4.48)

  where $\gamma \not =0,\pm 1$, $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f with simple roots only and of degrees $p_1$ and $p_2$ such that $2p_1+2\in \{d,d-2\}$ and $2p_2\in \{d,d-2\}$;

• (4) the fourth equation is

  $$\begin{aligned} \overline{f}^2= \frac{P_0(z,f)^2(f-\kappa )}{Q_0(z,f)^2(f-1)}, \end{aligned}$$

  (4.49)

  where $\kappa \not =0,1$, $P_0(z,f)$ and $Q_0(z,f)$ are two polynomials in f with simple roots only and $p_0-q_0\in \{-1,0,1\}$; moreover, when $\kappa =-1$, we have

  $$\begin{aligned} \begin{aligned} P_0(z,f)^2(f+1)-Q_0(z,f)^2(f-1)&=P_1(z,f)^2(f-\gamma ),\\ P_0(z,f)^2(f-1)-\overline{\gamma }^2 Q_0(z,f)^2(f-1)&=P_2(z,f)^2(f+\gamma ), \end{aligned} \end{aligned}$$

  (4.50)

  where $\gamma \not =0,\pm 1$, $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f with simple roots only and of degrees $p_1$ and $p_2$ such that $2p_1+1\in \{d,d-2\}$ and $2p_2+1\in \{d,d-2\}$; or when $\kappa \not =-1$, we have

  $$\begin{aligned} \begin{aligned} P_0(z,f)^2(f-\kappa )-Q_0(z,f)^2(f-1)&=P_1(z,f)^2(f+\kappa ),\\ P_0(z,f)^2(f-\kappa )-\overline{\kappa }^2 Q_0(z,f)^2(f-1)&=P_2(z,f)^2(f+1), \end{aligned} \end{aligned}$$

  (4.51)

  where $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f with simple roots only and of degrees $p_1$ and $p_2$ such that $2p_1+1\in \{d,d-2\}$ and $2p_2+1\in \{d,d-2\}$;

• (5) the fifth equation is

  $$\begin{aligned} \overline{f}^3= \frac{P_0(z,f)^3(f-1)}{Q_0(z,f)^3(f-\eta )}, \end{aligned}$$

  (4.52)

  where $\eta$ is a cubic root of 1 such that $\eta ^2+\eta +1=0$, $P_0(z,f)$ and $Q_0(z,f)$ are two polynomials in f with simple roots only and $p_0-q_0\in \{-1,0,1\}$, and $P_0(z,f)^3(f-1)-Q_0(z,f)^3(f-\eta )=P_1(z,f)^3(f-\eta ^2)$ for some polynomial $P_1(z,f)$ in f with simple roots only and of degree $p_1$ such that $3p_1 + 1\in \{d,d-3\}$;

• (6) the sixth equation is

  $$\begin{aligned} \overline{f}^3= \frac{P_0(z,f)^3}{Q_0(z,f)^3}(f^3-1), \end{aligned}$$

  (4.53)

  where $P_0(z,f)$ and $Q_0(z,f)$ are two polynomials in f with simple roots only such that $p_0-q_0\in \{-2,-1,0\}$, and $P_0(z,f)^3(f^3-1)-Q_0(z,f)^3=P_1(z,f)^3$ for some polynomial $P_1(z,f)$ in f with simple roots only and of degree $p_1$ such that $3p_1\in \{d,d-3\}$;

• (7) the seventh equation is

  $$\begin{aligned} \overline{f}^2= \frac{P_0(z,f)^2(f^2-\kappa ^2)}{Q_0(z,f)^2(f^2-1)}, \end{aligned}$$

  (4.54)

  where $\kappa \not =0,\pm 1$, $P_0(z,f)$ and $Q_0(z,f)$ are two polynomials in f with simple roots only such that $p_0-q_0\in \{-1,0,1\}$; moreover,

  $$\begin{aligned} \begin{aligned} P_0(z,f)^2(f^2-\kappa ^2)-Q_0(z,f)^2(f^2-1)&=P_1(z,f)^2,\\ P_0(z,f)^2(f^2-\kappa ^2)-\overline{\kappa }^2Q_0(z,f)^2(f^2-1)&=P_2(z,f)^2, \end{aligned} \end{aligned}$$

  (4.55)

  where $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f with simple roots only and of degrees $p_1$ and $p_2$ such that $2p_1\in \{d,d-2\}$ and $2p_2\in \{d,d-2\}$;

• (8) the eighth equation is

  $$\begin{aligned} \overline{f}^2= \frac{P_0(z,f)^2(f-\kappa )(f-1)}{Q_0(z,f)^2(f+\kappa )(f+1)}, \end{aligned}$$

  (4.56)

  where $\kappa \not =0,\pm 1$, $P_0(z,f)$ and $Q_0(z,f)$ are two polynomials in f with simple roots only such that $p_0-q_0\in \{-1,0,1\}$; moreover,

  $$\begin{aligned} \begin{aligned} P_0(z,f)^2(f-\kappa )(f-1)-Q_0(z,f)^2(f+\kappa )(f+1)&=P_1(z,f)^2,\\ P_0(z,f)^2(f-\kappa )(f-1)-\overline{\kappa }^2Q_0(z,f)^2(f+\kappa )(f+1)&=P_2(z,f)^2, \end{aligned} \end{aligned}$$

  (4.57)

  where $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f with simple roots only and of degrees $p_1$ and $p_2$ such that $2p_1\in \{d,d-2\}$ and $2p_2\in \{d,d-2\}$;

• (9) the ninth equation is

  $$\begin{aligned} \overline{f}^2= \frac{P_0(z,f)^2}{Q_0(z,f)^2}(f^2-\kappa ^2)(f^2-1), \end{aligned}$$

  (4.58)

  where $\kappa \not =0,\pm 1$, $P_0(z,f)$ and $Q_0(z,f)$ are two polynomials in f with simple roots only such that $p_0-q_0\in \{-3,-2,-1\}$; moreover,

  $$\begin{aligned} \begin{aligned} P_0(z,f)^2(f^2-\kappa ^2)(f^2-1)-Q_0(z,f)^2&=P_1(z,f)^2,\\ P_0(z,f)^2(f^2-\kappa ^2)(f^2-1)-\overline{\kappa }^2Q_0(z,f)^2&=P_2(z,f)^2, \end{aligned} \end{aligned}$$

  (4.59)

  where $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f with simple roots only and of degrees $p_1$ and $p_2$ such that $2p_1\in \{d,d-2\}$ and $2p_2\in \{d,d-2\}$;

Before getting into the proof, we make some remarks on how to solve the nine equations in Theorem 7. Unlike in the case of Eq. (2.1) with $n=d$, it seems impossible to give explicit expressions for the two polynomials $P_0(z,f)$ and $Q_0(z,f)$ in the general case. However, when the degrees $p_0$ and $q_0$ are given, it is possible to do this via some basic computations.

First, we show how to solve the second equation (4.46). Since $P_0(z,f)^2(f-1)^{k_1}(f+1)^{k_2}-Q_0(z,f)^2=P_1(z,f)^2$ for some polynomial $P_1(z,f)$ in f, we have the equation

$$\begin{aligned} \overline{f}^2-1= \frac{P_1(z,f)^2}{Q_0(z,f)^2}, \end{aligned}$$

(4.60)

and may write

$$\begin{aligned} \begin{aligned} P_1(z,f)+iQ_0(z,f)&=P_{01}(z,f),\\ P_1(z,f)-iQ_0(z,f)&=P_{02}(z,f), \end{aligned} \end{aligned}$$

(4.61)

where $P_{01}(z,f)$ and $P_{02}(z,f)$ are two polynomials in f such that $P_{01}(z,f)P_{02}(z,f)=P_0(z,f)^2(f-1)^{k_1}(f+1)^{k_2}$. Here, $P_{01}(z,f)$ and $P_{02}(z,f)$ have no common roots. Without loss of generality, we may write

$$\begin{aligned} \begin{aligned} P_{01}(z,f)&=P_{011}(z,f)^2(f-1)^{k_1}(f+1)^{k_2},\\ P_{02}(z,f)&=P_{012}(z,f)^2, \end{aligned} \end{aligned}$$

(4.62)

or

$$\begin{aligned} \begin{aligned} P_{01}(z,f)&=P_{011}(z,f)^2(f-1)^{k_1},\\ P_{02}(z,f)&=P_{012}(z,f)^2(f+1)^{k_2}, \end{aligned} \end{aligned}$$

(4.63)

where $P_{011}(z,f)$ and $P_{012}(z,f)$ are two polynomials in f such that $P_0(z,f)=P_{011}(z,f)P_{012}(z,f)$. Now, $P_{011}(z,f)$ and $P_{012}(z,f)$ have no common roots. From equations in (4.61) we get

$$\begin{aligned} \begin{aligned} Q_0(z,f)&=\frac{1}{2i}[P_{01}(z,f)-P_{02}(z,f)],\\ P_1(z,f)&=\frac{1}{2}[P_{01}(z,f)+P_{02}(z,f)]. \end{aligned} \end{aligned}$$

(4.64)

Put $f=\frac{1}{2}(\lambda +\lambda ^{-1})$. Then from (4.60) we see that $\lambda$ is an algebroid function with at most finitely many algebraic branch points and it follows from (4.60) that

$$\begin{aligned} \frac{1}{2}(\overline{\lambda }-\overline{\lambda }^{-1})=\frac{P_1(z,f)}{Q_0(z,f)}. \end{aligned}$$

(4.65)

By solving equation (4.65) together with Eq. (4.60) and equations in (4.64), we get

$$\begin{aligned} \overline{\lambda }=\frac{\frac{1}{2}\left[ P_{01}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) +P_{02}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) \right] \pm P_0\left( z,\frac{\lambda ^2+1}{2\lambda }\right) \frac{(\lambda -1)^{k_1}(\lambda +1)^{k_2}}{(2\lambda )^{(k_1+k_2)/2}}}{\frac{1}{2i}\left[ P_{01}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) -P_{02}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) \right] }. \end{aligned}$$

Therefore, when equations in (4.62) hold, we have

$$\begin{aligned} \overline{\lambda }=i\frac{P_{011}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) \frac{(\lambda -1)^{k_1}(\lambda +1)^{k_2}}{(2\lambda )^{(k_1+k_2)/2}}+\theta P_{012}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) }{P_{011}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) \frac{(\lambda -1)^{k_1}(\lambda +1)^{k_2}}{(2\lambda )^{(k_1+k_2)/2}}-\theta P_{012}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) }, \quad \theta =\pm 1, \end{aligned}$$

(4.66)

or, when equations in (4.63) hold, we have

$$\begin{aligned} \overline{\lambda }=i\frac{P_{011}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) \frac{(\lambda -1)^{k_1}}{(2\lambda )^{k_1/2}}+\theta P_{012}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) \frac{(\lambda +1)^{k_2}}{(2\lambda )^{k_2/2}}}{P_{011}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) \frac{(\lambda -1)^{k_1}}{(2\lambda )^{k_1/2}}-\theta P_{012}\left( z,\frac{\lambda ^2+1}{2\lambda }\right) \frac{(\lambda +1)^{k_2}}{(2\lambda )^{k_2/2}}}, \quad \theta =\pm 1. \end{aligned}$$

(4.67)

We conclude that solutions of Eq. (4.60) are represented by $f=\frac{1}{2}(\lambda +\lambda ^{-1})$ with $\lambda$ being a solution of Eq. (4.66) or (4.67). In the autonomous case, the RHS of Eq. (4.66) or (4.67) becomes a rational term $R_1(\lambda )$ in $\lambda$ after multiplying $(2\lambda )^{d}$ to both of the numerator and the denominator; thus the equation $\overline{\lambda }=R_1(\lambda )$ always has a meromorphic solution for any given $P_{01}(f)$ and $P_{02}(f)$ in (4.62) or in (4.63) and any given $Q_0(f)$ such that (4.64) holds, as mentioned in the introduction.

Second, we show how to determine the polynomials $P_0(z,f)$ and $Q_0(z,f)$ from the seven pairs of polynomial equations in (4.45), (4.48), (4.50), (4.51), (4.55), (4.57) and (4.59). Take the pair of equations in (4.45) as an example. When $p_0=q_0-1$, we write

$$\begin{aligned} \begin{aligned} P_0(z,f)^2&=A[(f-a_{p_0})(f-a_{p_0-1})\cdots (f-a_1)]^2,\\ Q_0(z,f)^2&=[(f-b_{q_0})(f-b_{q_0-1})\cdots (f-b_1)]^2, \end{aligned} \end{aligned}$$

(4.68)

where $a_i$ and $b_j$ are in general algebraic functions, distinct from each other. If $A\not =1,\overline{\kappa }^2$, we also write

$$\begin{aligned} \begin{aligned} P_1(z,f)^2&=(A-1)[(f-c_{p_0})(f-c_{p_0-1})\cdots (f-c_1)]^2,\\ P_2(z,f)^2&=(A-\overline{\kappa }^2)[(f-d_{q_0})(f-d_{q_0-1})\cdots (f-d_1)]^2, \end{aligned} \end{aligned}$$

(4.69)

where $c_k$ and $d_l$ are in general algebraic functions, distinct from each other. By comparing the coefficients on both sides of the two polynomial equations in (4.45) we obtain $4p_0+4$ polynomial equations with respect to the unknowns A, $\kappa$, $a_i$, $b_j$, $c_k$ and $d_l$, whose combined number is $4p_0+4$. This implies that $\overline{\kappa }$ and $\kappa$ satisfy a polynomial equation $U(\overline{\kappa },\kappa )=0$ with respect to $\overline{\kappa }$ and $\kappa$. In general, it is difficult to determine whether or not $\kappa$ is a constant, but this can be done once the degrees $p_0$ and $q_0$ are given. The case when $A=1$ or $A=\overline{\kappa }^2$ is similar. It seems that the pair of polynomial equations in (4.45) is solvable only when $p_0=q_0-1$ since we would obtain $4p_0+4$ polynomial equations with respect to $4p_0+3$ unknowns when $p_0=q_0$ and we would obtain $4p_0+8$ polynomial equations with respect to $4p_0+7$ unknowns when $p_0=q_0-2$. The polynomial equations in (4.48), (4.50), (4.51), (4.52), (4.55), (4.57) and (4.59) can be given similar discussions. Also, for the three pairs of polynomial equations in (4.50), (4.51) and (4.57), it seems that they are solvable only when $p_0=q_0$. For given small $p_0$ and $q_0$, one may readily obtain some explicit examples for each of the seven pairs of polynomial equations by doing some basic computations. In particular, in all the three cases $p=q$, $p=q+2$ and $p=q-2$ one may obtain some examples for each of the three equations (4.47), (4.56) and (4.58), where $P_0(z,f)$ and $Q_0(z,f)$ are in fact both polynomials in $f^2$. It is not known if this is always true for any given $p_0$ and $q_0$. For the fifth equation (4.52), we note that the polynomial equation $P_0(z,f)^3(f-1)-Q_0(z,f)^3(f-\eta )=P_1(z,f)^3(f-\eta ^2)$ is solvable when $p_0=q_0$ by the same arguments as above. In fact, we may even show that all coefficients of $P_0(z,f)^3(f^3-1)$ and $Q_0(z,f)^3$ are constants since we do not need to deal with the shift of a function.

Third, for the sixth equation (4.53), below we show that the polynomial equation $P_0(z,f)^3(f^3-1)-Q_0(z,f)^3=P_1(z,f)^3$ is also solvable. We may write

$$\begin{aligned} {[}P_1(z,f)+Q_0(z,f)][P_1(z,f)+\eta Q_0(z,f)][P_1(z,f)+\eta ^2Q_0(z,f)]=P_0(z,f)^3(f^3-1). \end{aligned}$$

Since any two of the polynomials $P_1(z,f)+Q_0(z,f)$, $P_1(z,f)+\eta Q_0(z,f)$ and $P_1(z,f)+\eta ^2Q_0(z,f)$ have no common roots, we may write

$$\begin{aligned} \begin{aligned} P_1(z,f)+Q_0(z,f)&=P_{01}(z,f)^3(f-1)^{\theta _{11}}(f-\eta )^{\theta _{12}}(f-\eta ^2)^{\theta _{13}},\\ P_1(z,f)+\eta Q_0(z,f)&=P_{02}(z,f)^3(f-1)^{\theta _{21}}(f-\eta )^{\theta _{22}}(f-\eta ^2)^{\theta _{23}},\\ P_1(z,f)+\eta ^2Q_0(z,f)&=P_{03}(z,f)^3(f-1)^{\theta _{31}}(f-\eta )^{\theta _{32}}(f-\eta ^2)^{\theta _{33}}, \end{aligned} \end{aligned}$$

(4.70)

where $\theta _{ij}\in \{0,1\}$ and $\theta _{1j}+\theta _{2j}+\theta _{3j}=1$, and $P_{01}(z,f)$, $P_{02}(z,f)$ and $P_{03}(z,f)$ are three polynomials in f such that $P_{01}(z,f)P_{02}(z,f)P_{03}(z,f)=P_0(z,f)$ and any two of them have no common roots. Denote the degrees of the three polynomials on the RHS of Eq. (4.70) by $p_1$, $p_2$ and $p_3$, respectively. Consider the case where $p_1\le p_2\le p_3$. By eliminating $P_1(z,f)$ from the first two equations in (4.70) and then from the second and the third equations, respectively, we can obtain two expressions for the polynomial $Q_0(z,f)$. By comparing the degrees of these two polynomials, we see that the cases where $p_1< p_2< p_3$ and $p_1= p_2< p_3$ cannot occur. Therefore, we have $p_1<p_2=p_3$ or $p_1=p_2=p_3$. Moreover, if the case $p_1<p_2=p_3$ occurs, then we must have that the two polynomials $P_1(z,f)$ and $Q_0(z,f)$ have the same degrees and also that the leading coefficients of them have opposite signs; in this case, we must have $p<q$ and we see from equations in (4.70) that $p_1+p_2+p_3=p=q-3$. It follows that $p_1-p_2+3p_2=3q_0-3=3p_2-3$, which gives $p_2-p_1=3$. We conclude that the three integers $p_1$, $p_2$ and $p_3$ are equal to each other or one is less by 3 than the other two. Recall that $P_0(z,f)$ and $Q_0(z,f)$ can have simple roots only. From the above discussions, we have only the following two possibilities:

$$\begin{aligned} \begin{aligned} P_1(z,f)+Q_0(z,f)&=P_{01}(z,f)^3(f-\eta _1),\\ P_1(z,f)+\eta Q_0(z,f)&=P_{02}(z,f)^3(f-\eta _2),\\ P_1(z,f)+\eta ^2Q_0(z,f)&=P_{03}(z,f)^3(f-\eta _3), \end{aligned} \end{aligned}$$

(4.71)

where $\eta _1$, $\eta _2$ and $\eta _3$ are the three distinct roots of 1, or

$$\begin{aligned} \begin{aligned} P_1(z,f)+Q_0(z,f)&=P_{01}(z,f)^3(f^3-1),\\ P_1(z,f)+\eta Q_0(z,f)&=P_{02}(z,f)^3,\\ P_1(z,f)+\eta ^2Q_0(z,f)&=P_{03}(z,f)^3. \end{aligned} \end{aligned}$$

(4.72)

Denote the degrees of the three polynomials $P_{01}(z,f)$, $P_{02}(z,f)$ and $P_{03}(z,f)$ by $s_1$, $s_2$ and $s_3$, respectively, and denote the leading coefficients of the three polynomials $P_{01}(z,f)^3$, $P_{02}(z,f)^3$ and $P_{03}(z,f)^3$ by $a_{11}$, $a_{12}$ and $a_{13}$, respectively.

For the three equations in (4.71), when $p=q+3$, we consider the case when $\eta _2=\eta \eta _1$ and $\eta _3=\eta ^2\eta _1$ and also that $P_{01}(z,f)^3=P_{02}(z,\eta f)^3=P_{03}(z,\eta ^2 f)^3$, i.e., $P_{02}(z,f)^3=P_{01}(z,\eta ^2f)^3$ and $P_{03}(z,f)^3=P_{01}(z,\eta f)^3$. In this case, if $P_1(z,f)$ is of the form $P_1(z,f)=P_{11}(z,f^3)f$ for some polynomial $P_{11}(z,f^3)$ in $f^3$ and $Q_0(z,f)$ is of the form $Q_0(z,f)=Q_{11}(z,f^3)$ for some polynomial $Q_{11}(z,f^3)$ in $f^3$, then by doing the transformation $f\rightarrow \eta f$ for the second equation in (4.71) and dividing by $\eta$ both sides of the resulting equation and applying the transformation $f\rightarrow \eta ^2f$ for the third equation in (4.71) and dividing by $\eta ^2$ both sides of the resulting equation, respectively, we get exactly the first equation in (4.71). With $P_1(z,f)$ and $Q_0(z,f)$ above, the first equation in (4.71) is solvable by the same arguments as in the discussions previously. We note that when $p=q$ and $A=1$, we can deal with (4.71) in exactly the same way as above by just changing the positions of $P_1(z,f)$ and $Q_0(z,f)$. On the other hand, for the case when $p=q-3$, we consider equations in (4.72). We have

$$\begin{aligned} \begin{aligned} P_{01}(z,f)^3(f^3-1)+\eta P_{02}(z,f)^3+\eta ^2P_{03}(z,f)^3=0. \end{aligned} \end{aligned}$$

(4.73)

When $p=q-3$, we may suppose that the leading coefficient of $P_1(z,f)$ is $-\eta$. Then if we write

$$\begin{aligned} \begin{aligned} P_{01}(z,f)^3&=a_{11}[(f^3-a_{s_1})(f^3-a_{s_1-1})\cdots (f^3-a_1)]^3,\\ P_{02}(z,f)^3&=a_{12}[(f^3-b_{s_1})(f^3-b_{s_1-1})\cdots (f^3-b_1)]^3,\\ P_{03}(z,f)^3&=a_{13}[(f^3-c_{s_1})(f^3-c_{s_1-1})\cdots (f^3-c_1)]^3f^3,\\ \end{aligned} \end{aligned}$$

(4.74)

where $a_i$, $b_j$ and $c_k$ are in general algebraic functions and $a_{11}=-\eta +1$, $a_{13}=-\eta +\eta ^2$ and $a_{11}a_{12}a_{13}=A$. Denoting $g=f^3$, then Eq. (4.73) becomes a polynomial equation with respect to g. By comparing the coefficients on both sides of this equation we obtain $3s_1+1$ polynomial equations with respect to the unknowns A, $a_i$, $b_j$ and $c_k$, whose combined number is $3s_1+1$. For example, when $s_1=0$, we get $P_{01}(z,f)^3=1-\eta$, $P_{02}(z,f)^3=\eta ^2-1$ and $P_{03}(z,f)^3=(\eta ^2-\eta )f^3$. It follows that $P_0(z,f)^3=3(\eta -\eta ^2)f^3(f^3-1)$, $Q_0(z,f)=f^3+\eta$ and $P_1(z,f)=-\eta (f^3+\eta ^2)$. By looking at the above examples, one may ask if $P_0(z,f)$ and $Q_0(z,f)$ are both polynomials in $f^3$ for any given $p_0$ and $q_0$.

Solutions of the eight equations (4.44), (4.47), (4.49), (4.52), (4.53), (4.54), (4.56) and (4.58) in the autonomous case are elliptic functions. We will discuss them further in Sect. 5. Below we begin to prove Theorem 7.

Proof of Theorem 7

Since $n\mid |p-q|$, then we have $2\le N_c\le 4$. Below we consider the three cases where $N_c= 2$, $N_c=3$ and $N_c=4$, respectively.

Case 1: $N_c= 2$.

Since $n\mid |p-q|$, then we have the following three possibilities:

$$\begin{aligned} \overline{f}^n= & {} \frac{P_0(z,f)^n}{Q_0(z,f)^n}(f-\alpha _1)^{k_1}(f-\alpha _2)^{k_2}, \end{aligned}$$

(4.75)

$$\begin{aligned} \overline{f}^n= & {} \frac{P_0(z,f)^n}{Q_0(z,f)^n}\frac{1}{(f-\beta _1)^{l_1}(f-\beta _2)^{l_2}}, \end{aligned}$$

(4.76)

$$\begin{aligned} \overline{f}^n= & {} \frac{P_0(z,f)^n}{Q_0(z,f)^n}\frac{(f-\alpha _1)^{k_1}}{(f-\beta _1)^{l_1}}, \end{aligned}$$

(4.77)

where in (4.75) we have $n\mid (k_1+k_2)$, in (4.76) we have $n\mid (l_1+l_2)$ and in (4.77) we have $n\mid |k_1-l_1|$. Since $n\mid |p-q|$, we see that $n\not \mid k_i$ and $n\not \mid l_j$. For convenience, we denote $\alpha _i$ or $\beta _j$ in each of the above equations by $\gamma _1$ and $\gamma _2$. By Lemmas 1 and 3 it follows that if $\gamma _i\not \equiv 0$, then $\omega \gamma _i$ is a completely ramified rational function of f with multiplicity at least n, where $\omega$ is the nth root of 1. Therefore, if one of $\gamma _1$ and $\gamma _2$ is zero, we must have $n=2$; otherwise, say $\gamma _1=0$, if $n\ge 3$, then 0 and $\omega \gamma _2$ are completely ramified rational functions of f with multiplicity at least 3, a contradiction to the inequality (1.6).

For Eq. (4.76), we divide the following two cases: (1) $\beta _1$ and $\beta _2$ are both non-zero; or (2) at least one of $\beta _1$ and $\beta _2$ is zero. In the first case, if $p_0\ge 2$ and $P_0(z,f)^n=a_pf^{np_0}$ particularly, then the analysis in the proof of Lemma 2 applies and 0 is a Picard exceptional rational function of f; also, we have $p\ge q$ for otherwise $\infty$ is also a Picard exceptional rational function of f by analyzing on the poles of f and it follows that $\beta _1$ and $\beta _2$ are both Picard exceptional rational functions of f, a contradiction to Picard’s theorem. Moreover, by the inequality (1.6), we must have $n=2$ and $\beta _1+\beta _2=0$ and it follows by Lemma 4 that $l_1=l_2=1$. Then by doing a bilinear transformation $f\rightarrow 1/f$, we get equation (3.38), which leads to Eq. (3.9) in Theorem 4. Otherwise, by doing a bilinear transformation $f\rightarrow 1/f$, we get equation (4.75). In the second case, since $n=2$, then by the bilinear transformation $f\rightarrow 1/f$, we get equation (4.16), which leads to Eq. (4.7) in Theorem 6. Therefore, in this section we only need to consider equations (4.75) and (4.77).

Further, for Eq. (4.75), we may suppose that $\alpha _1$ and $\alpha _2$ are both non-zero; otherwise, say $\alpha _1=0$, by doing the transformation $f\rightarrow 1/f$ we get equation (4.13) with $P_1(z,f)$ being a polynomial in f of degree d and $Q_1(z,f)$ being a polynomial in f of degree $d-k_1$, which cannot have any meromorphic solution as shown in the proof of Theorem 6. For Eq. (4.77), when $\alpha _1=0$, by Lemma 2 it follows that 0 is a Picard exceptional rational function of f. Similarly as in the previous paragraph, it follows that $p\ge q$. Moreover, we have $P_0(z,f)^n=a_p$; otherwise, the roots of $P_0(z,f)$ are also Picard’s exceptional rational functions, a contradiction to Picard’s theorem. Since $n=2$, then by doing a bilinear transformation $f\rightarrow 1/f$, we get equation (3.14), which leads to Eq. (3.7) in Theorem 4. When $\beta _1=0$, since $n=2$, by doing a bilinear transformation $f\rightarrow 1/f$, we get equation (4.17) since $2\mid |k_1-l_1|$, which cannot have any meromorphic solution as shown in the proof of Theorem 6. Therefore, in this section we only need to consider equations (4.75) and (4.77) for the case where $\alpha _i$ and $\beta _j$ are both non-zero. Under this assumptions, below we consider these two equations separately.

Subcase 1: Equation (4.75) with $\alpha _1\alpha _2\not \equiv 0$.

In this case, we claim that $n=2$. Suppose that $n\ge 3$. By Lemma 3 it follows that $\omega \alpha _1$ and $\omega \alpha _2$ are completely ramified rational functions of f with multiplicity at least 3, where $\omega$ is the nth root of 1. By the inequality (1.6) we must have $n=3$ and also that $\alpha _1^3=\alpha _2^3$. However, since $3\mid (k_1+k_2)$, we have $k_1\not \mid 3$ or $k_2\not \mid 3$ and in either case we will get a contradiction to Lemma 4. Therefore, $n=2$.

When $\alpha _1+\alpha _2\not =0$, $\pm \alpha _1$ and $\pm \alpha _2$ are all completely ramified rational functions of f with multiplicities 2. By Lemma 4 it follows that $P_0(z,f)$ and $Q_0(z,f)$ both have simple roots only and also that $k_1=k_2=1$ and $p-q\in \{-2,0,2\}$. We consider

$$\begin{aligned} \overline{f}^2-\overline{\alpha }_1^2= \frac{P_0(z,f)^2(f-\alpha _1)(f-\alpha _2)-\overline{\alpha }_1^2Q_0(z,f)^2}{Q_0(z,f)^2}. \end{aligned}$$

(4.78)

By analyzing the multiplicities of poles of f as in the proof Lemma 5 we get that the numerator of the RHS of (4.78) is a polynomial in f with even degree; moreover, if some root of this polynomial is not equal to $\pm \alpha _1$ or $\pm \alpha _2$, then this root has order two. Therefore, $-\alpha _1$ and $-\alpha _2$ are either both simple roots of the numerator of the RHS of (4.78), or neither of them are. In the first case, we then consider $\overline{f}^2-\overline{\alpha }_2^2$ and conclude by Lemma 5 that the polynomial $P_0(z,f)^2(f-\alpha _1)(f-\alpha _2)-\overline{\alpha }_2^2Q_0(z,f)^2$ is a square of some polynomial in f. In the latter case, we claim that $-\alpha _1$ and $-\alpha _2$ are both simple roots of the polynomial $P_0(z,f)^2(f-\alpha _1)(f-\alpha _2)-\overline{\alpha }_2^2Q_0(z,f)^2$ when considering $\overline{f}^2-\overline{\alpha }_2^2$. Otherwise, we may suppose that $R(z,-\alpha _1)=\gamma ^2$ for some algebraic function $\gamma ^2$ which is distinct from $\overline{\alpha }_1^2$ and $\overline{\alpha }_2^2$. Let $z_0\in \mathbb {C}$ be such that $f(z_0)+\alpha _1(z_0)=0$. Then we have $f(z_0+1)^2-\gamma (z_0)^2=0$ and by applying the analysis in the proof of Lemma 4 we get that $z_0$ is a root of the equation $f(z_0+1)-\gamma (z_0)=0$ or $f(z_0+1)+\gamma (z_0)=0$ with multiplicity 2 and there are $T(r,f)+o(T(r,f))$ many such points, where $r\rightarrow \infty$ outside an exceptional set of finite linear measure. But we then get a contradiction to the inequality (1.6) by computing the quantity $\varTheta (\gamma ,\overline{f})$ or $\varTheta (-\gamma ,\overline{f})$ as in the proof of Lemma 4 since at least one of these two quantities is strictly positive. Without loss of generality, we may suppose the following two equations:

$$\begin{aligned} \overline{f}^2-\overline{\alpha }_1^2= \frac{P_1(z,f)^2(f+\alpha _1)(f+\alpha _2)}{Q_0(z,f)^2}, \end{aligned}$$

(4.79)

and

$$\begin{aligned} \overline{f}^2-\overline{\alpha }_2^2= \frac{P_2(z,f)^2}{Q_0(z,f)^2}, \end{aligned}$$

(4.80)

where $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f. By Lemma 5, $P_1(z,f)$ and $P_2(z,f)$ both have simple roots only and none of the roots of $P_1(z,f)$ and $P_2(z,f)$ is equal to $\pm \alpha _1$ or $\pm \alpha _2$. Moreover, when $p=q$, if $a_p=\overline{\alpha }_1^2$ or $a_p=\overline{\alpha }_2^2$, then the degree of the numerator in (4.79) or in (4.80) decreases since the terms with the highest degrees in $P_0(z,f)^2$ and $Q_0(z,f)^2$ cancel out when considering $\overline{f}^2-\overline{\alpha }_1^2$ or $\overline{f}^2-\overline{\alpha }_2^2$; by Lemma 5 we see that it decreases by 2. By doing the transformation $f\rightarrow \alpha _1 f$, we get the first equation of Theorem 7.

When $\alpha _1+\alpha _2=0$, we may let $\alpha _1=1$ and $\alpha _2=-1$ by doing a linear transformation $f\rightarrow \alpha _1f$. We consider

$$\begin{aligned} \overline{f}^2-1= \frac{P_0(z,f)^2(f-1)^{k_1}(f+1)^{k_2}-Q_0(z,f)^2}{Q_0(z,f)^2}. \end{aligned}$$

(4.81)

If the numerator of the RHS of (4.81) has at least three distinct roots, say $\gamma _i$, $i=1,2,3$, of odd order, then by applying the analysis in the proof of Lemma 1 together with the fact that $\pm 1$ are completely ramified rational functions of f and that the roots of $f\pm 1=0$ have even multiplicities, we obtain that $\gamma _1,\gamma _2,\gamma _3$ are all completely ramified rational functions of f, a contradiction to Theorem 2 since $\gamma _i$ are all distinct from $\pm 1$. Therefore, the numerator of the RHS of (4.81) can have at most two distinct roots of odd order. Suppose that there is only one such root, say $\gamma _1$. Since p and q are both even integers, then we must have $p=q$ and $A=1$ in which case the terms with the highest degrees in the polynomials $P_0(z,f)^2(f-1)^{k_1}(f+1)^{k_2}$ and $Q_0(z,f)^2$ cancel out when considering (4.81) so that the degree of the polynomial $P_0(z,f)^2(f-1)^{k_1}(f+1)^{k_2}-Q_0(z,f)^2$ decreases to be an odd integer. Then by considering the multiplicity of the poles of f as in the proof of Lemma 1, we see from (4.81) that $\infty$ is also a completely ramified rational function of f. If $\gamma _1=0$, then 0 is a completely ramified rational function of f. Then by considering the multiplicities of the roots of $f\pm 1=0$ as in the proof of Lemma 1 and then by Lemma 3, we obtain from (4.75) that $\pm 1$ both have multiplicities at least 4, a contradiction to the inequality (1.6). On the other hand, if $\gamma _1\not \equiv 0$, then by Lemma 3 it follows that $\pm \gamma _1$ are both completely ramified rational functions of f, a contradiction to Theorem 2 since $\infty$ is also a completely ramified rational function of f. Therefore, the numerator of the RHS of (4.81) has no roots of odd order, or has two distinct roots of odd order.

If the numerator of the RHS of (4.81) has no roots of odd order, then we have $P_0(z,f)^2(f-1)^{k_1}(f+1)^{k_2}-1=P_1(z,f)^2$, where $P_1(z,f)$ is a polynomial in f. This gives the second equation of Theorem 7.

If the numerator of the RHS of (4.81) has two distinct roots of odd order, then we have $P_0(z,f)^2(f-1)^{k_1}(f+1)^{k_2}-Q_0(z,f)^2=P_1(z,f)^2(f-\gamma _1)^{t_1}(f-\gamma _2)^{t_2}$ for some polynomial $P_1(z,f)$ in f, and $\gamma _1$ and $\gamma _2$ are distinct from each other. In this case, $\gamma _1$ and $\gamma _2$ are also both completely ramified rational functions of f and by Lemma 3 it follows that if $\gamma _i\not \equiv 0$, then $\pm \gamma _i$ are both completely ramified rational functions of f. By Theorem 2 we must have $\gamma _1+\gamma _2=0$. Moreover, by Lemmas 4 and 5, we have $k_1=k_2=t_1=t_2=1$. Therefore, by denoting $\gamma =\gamma _1$, we have

$$\begin{aligned} \overline{f}^2-1= \frac{P_1(z,f)^2}{Q_0(z,f)^2}(f^2-\gamma ^2). \end{aligned}$$

(4.82)

Now $\pm 1$ and $\pm \gamma$ are all completely ramified rational functions of f with multiplicities 2 and f has no other completely ramified rational functions and has no Picard exceptional rational functions. Thus by applying the analysis on (4.81) to $\overline{f}^2-\overline{\gamma }^2$ we conclude that the polynomial $P_0(z,f)^2-\overline{\gamma }^2Q_0(z,f)^2$ cannot have any root of odd order; otherwise, this root is distinct from $\pm 1$ and $\pm \gamma$ and is also a completely ramified rational function of f, a contradiction to Theorem 2. Therefore, we have the following

$$\begin{aligned} \overline{f}^2-\overline{\gamma }^2= \frac{P_2(z,f)^2}{Q_0(z,f)^2}, \end{aligned}$$

(4.83)

where $P_2(z,f)$ is a polynomial in f. By Lemmas 4 and 5, we have $P_0(z,f)$, $Q_0(z,f)$, $P_1(z,f)$ and $P_2(z,f)$ all have simple roots only and none of the roots of $P_1(z,f)$ and $P_2(z,f)$ is equal to $\pm 1$ or $\pm \gamma$, and the degrees p and q satisfy $p-q\in \{-2,0,2\}$. As for Eqs. (4.79) and (4.80), when $p=q$, if $A=1$ or $A=\overline{\gamma }^2$, the degree of the numerator in (4.82) or in (4.83) decreases and by Lemma 5 it decreases by 2. This gives the third equation of Theorem 7.

Subcase 2: Equation (4.77) with $\alpha _1\beta _1\not \equiv 0$.

In this case, we discuss the two cases $n=2$ and $n\ge 3$ separately.

When $n=2$, by Lemmas 1 and 3 it follows that $\pm \alpha _1$, as well as $\pm \beta _1$, are completely ramified rational functions of f. We may let $\alpha _1=\kappa$ and $\beta _1=1$ by doing a linear transformation $f\rightarrow \beta _1f$. We consider

$$\begin{aligned} \overline{f}^2-1= \frac{P_0(z,f)^2(f-\kappa )^{k_1}-Q_0(z,f)^2(f-1)^{l_1}}{Q_0(z,f)^2(f-1)^{l_1}}. \end{aligned}$$

(4.84)

Recall that the leading coefficient of the polynomial $P_0(z,f)^2$ is denoted by A and that the polynomial $Q_0(z,f)^2$ is monic. When $p=q$ and $A=1$, the degree of the numerator of the RHS of (4.84) decreases due to the cancellation of the terms with the highest degrees in $P_0(z,f)^2(f-\kappa )^{k_1}$ and $Q_0(z,f)^2(f-1)^{l_1}$. Suppose that the degree of the polynomial $P_0(z,f)^2(f-\kappa )^{k_1}-Q_0(z,f)^2(f-1)^{l_1}$ in f decreases to be an even integer. By considering the multiplicities of the poles of f together with the fact that $\pm 1$ are both completely ramified rational functions of f and that the roots of $f\pm 1=0$ have even multiplicities with at most finitely many exceptions, we get that $\infty$ is also a completely ramified rational function of f. Further, by considering the multiplicities of the roots of $f-1=0$ as in the proof of Lemma 3, we obtain from (4.77) that $\pm 1$ both have multiplicities at least 4 and it follows that $\infty$ is also a completely ramified function of f with multiplicity at least 4, a contradiction to the inequality (1.6). This implies that the numerator of the RHS of Eq. (4.84) always has a root, say $\gamma$, of odd order. Then by applying the same analysis as in the proof of Lemma 1 together with the fact that $\pm 1$ are completely ramified rational functions of f we obtain that $\gamma$ is a completely ramified rational function of f. If $\kappa \not =-1$, then by Lemma 3, $\pm 1$ and $\pm \kappa$ are all completely ramified rational functions of f and thus by Theorem 2 we have $\gamma =-1$ or $\gamma =-\kappa$ and thus $\gamma \not =0$. If $\kappa =-1$ and $\gamma =0$, then by considering the multiplicities of the roots of $f+1=0$ as in the proof of Lemma 1 and then by Lemma 3, we obtain from (4.77) that $\pm 1$ both have multiplicities at least 4. But it follows by repeating the analysis after (4.84) that 0 is a completely ramified rational function of f with multiplicity at least 4, a contradiction to the inequality (1.6). Therefore, when $\kappa =-1$, we also have $\gamma \not =0$. Now, by Lemma 3 it follows that $\pm \gamma$ are both completely ramified rational functions of f. From the above reasoning, we see that f has four completely ramified rational functions $\pm 1$ and $\pm \kappa$ (or $\pm \gamma$), all of which have multiplicities 2. By Lemma 4 we must have $k_1=l_1=1$ and all the roots of $P_0(z,f)$ and $Q_0(z,f)$ are simple and also that $p_0-q_0\in \{-1,0,1\}$. As for Eqs. (4.79) and (4.80), when $p=q$, if the degree of the numerator in (4.84) decreases, then by Lemma 5 it decreases by 2. Now we have

$$\begin{aligned} \overline{f}^2 = \frac{P_0(z,f)^2}{Q_0(z,f)^2}\frac{(f-\kappa )}{(f-1)}. \end{aligned}$$

(4.85)

When $\kappa =-1$, from the above discussions we know that $P_0(z,f)^2(f+1)-Q_0(z,f)^2(f-1)$ is of the form $P_1(z,f)^2(f-\gamma )$ for some non-zero algebraic function $\gamma$ and a polynomial $P_1(z,f)$ in f. Since $\pm 1$ and $\pm \gamma$ are all completely ramified rational functions of f, then by Theorem 2 and Lemma 5 and considering $\overline{f}^2-\overline{\gamma }^2$, we see that $P_0(z,f)^2(f+1)-\overline{\gamma }^2Q_0(z,f)^2(f-1)$ must be of the form $P_2(z,f)^2(f+\gamma )$ for some polynomial $P_2(z,f)$ in f; moreover, both $P_1(z,f)$ and $P_2(z,f)$ have simple roots only and none of these roots equals $\pm 1$ or $\pm \gamma$. Therefore, we have

$$\begin{aligned} \overline{f}^2-1= \frac{P_1(z,f)^2}{Q_0(z,f)^2}\frac{(f-\gamma )}{(f-1)}, \end{aligned}$$

(4.86)

and

$$\begin{aligned} \overline{f}^2-\overline{\gamma }^2= \frac{P_2(z,f)^2}{Q_0(z,f)^2}\frac{(f+\gamma )}{(f-1)}. \end{aligned}$$

(4.87)

Note that, when $p=q$, if the degree of the numerator in (4.86) or in (4.87) decreases, then by Lemma 5 it decreases by 2. On the other hand, when $\kappa \not =-1$, by Theorem 2 and Lemma 5 and considering $\overline{f}^2-1$ and $\overline{f}^2-\overline{\kappa }^2$, respectively, we see that $P_0(z,f)^2(f-\kappa )-Q_0(z,f)^2(f-1)$ (and also $P_0(z,f)^2(f-\kappa )-\overline{\kappa }^2Q_0(z,f)^2(f-1)$) must be of the form $P_1(z,f)^2(f+1)$ or $P_2(z,f)^2(f+\kappa )$ for some polynomials $P_1(z,f)$ and $P_2(z,f)$ in f with simple roots only and none of these roots equals $\pm \kappa$ and $\pm 1$. Without loss of generality, we may consider the following two equations:

$$\begin{aligned} \overline{f}^2-1= \frac{P_1(z,f)^2}{Q_0(z,f)^2}\frac{(f+\kappa )}{(f-1)}, \end{aligned}$$

(4.88)

and

$$\begin{aligned} \overline{f}^2-\overline{\kappa }^2= \frac{P_2(z,f)^2}{Q_0(z,f)^2}\frac{(f+1)}{(f-1)}. \end{aligned}$$

(4.89)

Note that, when $p=q$, if the degree of the numerator in (4.88) or in (4.89) decreases, then by Lemma 5 it decreases by 2. This gives the fourth equation of Theorem 7.

Now we consider the case when $n\ge 3$. Since $\omega \alpha _1$ and $\omega \beta _1$ are all completely ramified rational functions of f with multiplicities at least 3, where $\omega$ is the nth root of 1, then by the inequality (1.6) we must have $n=3$ and $\alpha _1^3=\beta _1^3$. By Lemma 4 we conclude that $k_1=l_1=1$. We fix one $\eta$ such that $\eta ^2+\eta +1=0$ and choose without loss of generality that $\beta _1=\eta \alpha _1$. We may let $\alpha _1=1$ by doing a linear transformation $f\rightarrow \alpha _1f$. Then we have

$$\begin{aligned} \overline{f}^3= \frac{P_0(z,f)^3(f-1)}{Q_0(z,f)^3(f-\eta )}. \end{aligned}$$

(4.90)

Also, by Lemma 4, we conclude that $P_0(z,f)$ and $Q_0(z,f)$ can have simple roots only and $p_0-q_0\in \{-1,0,1\}$. We consider

$$\begin{aligned} \overline{f}^3-1= \frac{P_0(z,f)^3(f-1)-Q_0(z,f)^3(f-\eta )}{Q_0(z,f)^3(f-\eta )}. \end{aligned}$$

(4.91)

Note that f has three completely ramified rational functions with multiplicities 3. As for Eqs. (4.79) and (4.80), when $p_0=q_0$ and the leading coefficient of the polynomial $P_0(z,f)^3$ satisfies $A=1$, the degree of the numerator of the RHS of (4.91) decreases and by Lemma 5 it decreases by 3. Therefore, the numerator of the RHS of the equation above always has one root of order $l_1$ such that $3\not \mid l_1$ and by Lemma 5 we must have $l_1=1$ and then by the inequality (1.6) we see that the root must be $\eta ^2$. Therefore, the numerator of the RHS of (4.91) is of the form $P_1(z,f)^3(f-\eta ^2)$ for a polynomial $P_1(z,f)$ in f with simple roots only, i.e.,

$$\begin{aligned} \overline{f}^3-1= \frac{P_1(z,f)^3(f-\eta ^2)}{Q_0(z,f)^3(f-\eta )}. \end{aligned}$$

(4.92)

This gives the fifth equation of Theorem 7.

Case 2: $N_c=3$.

Since $n\mid |p-q|$, we must have $n\ge 3$. For convenience, we denote the three roots by $\gamma _1$, $\gamma _2$ and $\gamma _3$ and their orders by $t_1$, $t_2$ and $t_3$, respectively. Without loss of generality, we may suppose that $\gamma _1\gamma _2\not =0$. Since $n\ge 3$, then by Lemmas 1 and 3 it follows that $\omega \gamma _1$ is a completely ramified rational function of f, where $\omega$ is the nth root of 1, and so by the inequality (1.6) we must have $n=3$ or 4. However, when $n=4$, $t_1$ and $t_2$ must be both even integers; otherwise, $\omega \gamma _1$ (or $\omega \gamma _2$) would have multiplicity at least 4, where $\omega$ is the fourth root of 1, which is impossible. But since $n\mid |p-q|$, we see that $t_3$ is also an even integer, a contradiction to our assumption that at least one of $\alpha _i$ and $\beta _j$ in (2.7) and (2.8) has no common factors with n. Therefore, we must have $n=3$. We see that $\eta \gamma _1$ has multiplicity 3 since we must have $(n,t_1)=1$, where $\eta$ is the cubic root of 1. Moreover, by the inequality (1.6) we have none of $\gamma _1$, $\gamma _2$ and $\gamma _3$ is zero and by Lemma 4 we also have $t_1=t_2=t_3=1$. By noting that $n\mid |p-q|$, when $n=3$ we have only the following two possibilities:

$$\begin{aligned} \overline{f}^3= & {} \frac{P_0(z,f)^3}{Q_0(z,f)^3}(f-\alpha _1)(f-\alpha _2)(f-\alpha _3), \end{aligned}$$

(4.93)

$$\begin{aligned} \overline{f}^3= & {} \frac{P_0(z,f)^3}{Q_0(z,f)^3}\frac{1}{(f-\beta _1)(f-\beta _2)(f-\beta _3)}. \end{aligned}$$

(4.94)

For each of the above two equations, by Lemma 4 we have that all the roots of $P_0(z,f)$ and $Q_0(z,f)$ are simple and also that $p-q\in \{-3,0,3\}$. Now, for Eq. (4.94), if 0 is the only root of $P_0(z,f)$, then $p=3$ and it follows that $q=6$ under our assumption that $n<d$; in this case we have $p_0=q_0=1$ and that $Q_0(z,f)$ has a non-zero root. Since none of $\alpha _i$ and $\beta _j$ is zero, then by doing a bilinear transformation $f\rightarrow 1/f$, both of the above two cases of Eq. (4.94) become (4.93). Thus we only need to consider equation (4.93).

Since $\eta \alpha _i$ has multiplicity 3, where $\eta$ is the cubic root of 1, then by the inequality (1.6) we must have $\alpha _1^3=\alpha _2^3=\alpha _3^3$. We may let $\alpha _1=1$ by doing a linear transformation $f\rightarrow \alpha _1f$. We consider

$$\begin{aligned} \overline{f}^3-1=\frac{P_0(z,f)^3(f^3-1)-Q_0(z,f)^3}{Q_0(z,f)^3}. \end{aligned}$$

(4.95)

Let $\eta$ a fixed cubic root of 1 such that $\eta ^2+\eta +1=0$. Since 1, $\eta$ and $\eta ^2$ all have multiplicities 3, then by Lemma 5 we conclude that the numerator of the RHS of Eq. (4.95) is of the form $P_1(z,f)^3$ for some polynomial $P_1(z,f)$ in f with simple roots only and these roots are distinct from 1, $\eta$ and $\eta ^2$. As for Eqs. (4.79) and (4.80) when $p=q$ and $A=1$, the degree of the numerator in (4.95) decreases and by Lemma 5 it decreases by 3. This gives the sixth equation of Theorem 7.

Case 3: $N_c= 4$.

In this case, by Lemma 1 we know that $\alpha _i$ and $\beta _j$ are all completely ramified rational functions of f. Then by the inequality (1.6) we must have $n=2$. By noting that $2\mid |p-q|$ and Lemma 4 we have the following possibilities:

$$\begin{aligned} \overline{f}^2= & {} \frac{P_0(z,f)^2}{Q_0(z,f)^2}\frac{(f-\alpha _1)(f-\alpha _2)}{(f-\beta _1)(f-\beta _2)}, \end{aligned}$$

(4.96)

$$\begin{aligned} \overline{f}^2= & {} \frac{P_0(z,f)^2}{Q_0(z,f)^2}(f-\alpha _1)(f-\alpha _2)(f-\alpha _3)(f-\alpha _4), \end{aligned}$$

(4.97)

$$\begin{aligned} \overline{f}^2= & {} \frac{P_0(z,f)^2}{Q_0(z,f)^2}\frac{1}{(f-\beta _1)(f-\beta _2)(f-\beta _3)(f-\beta _4)}, \end{aligned}$$

(4.98)

$$\begin{aligned} \overline{f}^2= & {} \frac{P_0(z,f)^2}{Q_0(z,f)^2}\frac{(f-\alpha _1)(f-\alpha _2)(f-\alpha _3)}{(f-\beta _1)}, \end{aligned}$$

(4.99)

$$\begin{aligned} \overline{f}^2= & {} \frac{P_0(z,f)^2}{Q_0(z,f)^2}\frac{(f-\alpha _1)}{(f-\beta _1)(f-\beta _2)(f-\beta _3)}. \end{aligned}$$

(4.100)

For convenience, we denote the four roots $\alpha _i$ and $\beta _j$ in each of the above equations by $\gamma _1$, $\gamma _2$, $\gamma _3$ and $\gamma _4$. If $\gamma _i\not \equiv 0$ for some i, then by Lemmas 1 and 3 it follows that $\pm \gamma _i$ are both completely ramified rational functions of f with multiplicities 2. This implies that none of $\gamma _1$, $\gamma _2$, $\gamma _3$ and $\gamma _4$ is zero for otherwise f would have at least five completely ramified rational functions, a contradiction to Theorem 2. Moreover, by the inequality (1.6) we must have $\gamma _1^2=\gamma _2^2$ and $\gamma _3^2=\gamma _4^2$, apart from permutations. Also, by Lemma 4 we know that in each of the above equations all the roots of $P_0(z,f)$ and $Q_0(z,f)$ are simple and the degrees of P(z, f) and Q(z, f) satisfy $p-q\in \{-2,0,2\}$. In particular, for Eq. (4.98), we see that if 0 is the only root of $P_0(z,f)$ then we must have $p=2$ and $q=4$ under the the assumption that $n<d$. Therefore, by doing a linear transformation $f\rightarrow 1/f$, Eqs. (4.98) and (4.100) become (4.97) and (4.99), respectively. From the above discussions, we conclude that we only need to consider the three equations (4.96), (4.97) and (4.99).

Further, Eq. (4.99) cannot have any meromorphic solution, as is shown below. From the previous discussions, we may suppose $\alpha _1+\alpha _2=0$ and $\alpha _3+\beta _1=0$. We consider

$$\begin{aligned} \overline{f}^2-\overline{\alpha }_1^2= \frac{P_0(z,f)^2(f^2-\alpha _1^2)(f+\beta _1)-\overline{\alpha }_1^2Q_0(z,f)^2(f-\beta _1)}{Q_0(z,f)^2(f-\beta _1)}. \end{aligned}$$

(4.101)

Since $\alpha _1$, $\alpha _2$, $\alpha _3$ and $\beta _1$ are four completely ramified rational functions of f, then by Lemma 5 we conclude that the numerator of the RHS of Eq. (4.101) is of the form $P_1(z,f)^2$ for some polynomial $P_1(z,f)$ in f with simple roots only and none of these roots is equal to $\alpha _1$, $\alpha _2$, $\alpha _3$ or $\beta _1$. Note that $p-q\in \{-2,0,2\}$. Since the degrees p and q are both odd integers, this is possible only when $p=q$ and the leading coefficient $a_p$ of the numerator P(z, f) satisfies $a_p=\overline{\alpha }_1^2$ so that the terms with the highest degree in the two polynomials $P_0(z,f)^2(f^2-\alpha _1^2)(f+\beta _1)$ and $\overline{\alpha }_1^2Q_0(z,f)^2(f-\beta _1)$ cancel out. It follows by these arguments that $a_p=\overline{\alpha }_1^2=\overline{\alpha }_2^2=\overline{\alpha }_3^2=\overline{\beta }_2^2$, which is impossible. Therefore, we only need to consider the two equations (4.96) and (4.97). Below we discuss them, respectively.

Subcase 1: Equation (4.96).

From the previous discussions, we have two cases to consider: (1) $\alpha _1+\alpha _2=0$ and $\beta _1+\beta _2=0$; or (2) $\alpha _1+\beta _1=0$ and $\alpha _2+\beta _2=0$.

When $\alpha _1+\alpha _2=0$ and $\beta _1+\beta _2=0$, we may let $\alpha _1=\kappa$ and $\beta _1=1$ by doing a linear transformation $f\rightarrow \beta _1 f$. We consider

$$\begin{aligned} \overline{f}^2-1= \frac{P_0(z,f)^2(f^2-\kappa ^2)-\overline{\kappa }^2Q_0(z,f)^2(f^2-1)}{Q_0(z,f)^2(f^2-1)}. \end{aligned}$$

(4.102)

Since $\pm 1$ and $\pm \kappa$ are four completely ramified rational functions of f, then by Lemma 5 we conclude that the numerator of the RHS of Eq. (4.102) is of the form $P_1(z,f)^2$ for some polynomial $P_1(z,f)$ in f with simple roots only and none of these roots is $\pm 1$ or $\pm \kappa$. Similarly, by considering $\overline{f}^2-1$, we also have $P_0(z,f)^2(f^2-\kappa ^2)-\overline{\kappa }^2Q_0(z,f)^2(f^2-1)=P_2(z,f)^2$ for some polynomial $P_2(z,f)$ in f with simple roots only and none of these roots is $\pm 1$ or $\pm \kappa$. Now we have

$$\begin{aligned} \overline{f}^2-1= \frac{P_1(z,f)^2}{Q_0(z,f)^2(f^2-1)}, \end{aligned}$$

(4.103)

and

$$\begin{aligned} \overline{f}^2-\overline{\kappa }^2= \frac{P_2(z,f)^2}{Q_0(z,f)^2(f^2-1)}. \end{aligned}$$

(4.104)

As for equations (4.79) and (4.80), when $p=q$, if $A=1$ or $A=\overline{\kappa }^2$, the degree of the numerator in (4.103) or in (4.104) decreases and by Lemma 5 it decreases by 2. This gives the seventh equation of Theorem 7.

When $\alpha _1+\beta _1=0$ and $\alpha _2+\beta _2=0$, we may let $\alpha _1=\kappa$ and $\alpha _2=1$ by doing a linear transformation $f\rightarrow \alpha _2f$. By considering $\overline{f}^2-1$ and $\overline{f}^2-\overline{\kappa }^2$ similarly as in the first case, respectively, we have

$$\begin{aligned} \overline{f}^2-1= \frac{P_1(z,f)^2}{Q_0(z,f)^2(f+\kappa )(f+1)}, \end{aligned}$$

(4.105)

and

$$\begin{aligned} \overline{f}^2-\overline{\kappa }^2= \frac{P_2(z,f)^2}{Q_0(z,f)^2(f+\kappa )(f+1)}, \end{aligned}$$

(4.106)

where $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f with simple roots only and none of these roots is equal to $\pm \kappa$ or $\pm 1$. As for Eqs. (4.79) and (4.80), when $p=q$, if $A=1$ or $A=\overline{\kappa }^2$, the degree of the numerator in (4.105) or in (4.106) decreases and by Lemma 5 it decreases by 2. This gives the eighth equation of Theorem 7.

Subcase 2: Equation (4.97).

We have $\alpha _1+\alpha _2=0$ and $\alpha _3+\alpha _4=0$. We may suppose that $\alpha _1=1$ and $\alpha _3=\kappa$ by doing a linear transformation $f\rightarrow \alpha _1f$. Now we have

$$\begin{aligned} \overline{f}^2 = \frac{P_0(z,f)^2}{Q_0(z,f)^2}(f^2-1)(f^2-\kappa ^2), \end{aligned}$$

(4.107)

and, further, by applying the analysis after equation (4.102) to $\overline{f}^2-1$ and $\overline{f}^2-\overline{\kappa }^2$, respectively, we have

$$\begin{aligned} \overline{f}^2-1= \frac{P_1(z,f)^2}{Q_0(z,f)^2}, \end{aligned}$$

(4.108)

and

$$\begin{aligned} \overline{f}^2-\overline{\kappa }^2= \frac{P_2(z,f)^2}{Q_0(z,f)^2}, \end{aligned}$$

(4.109)

where $P_1(z,f)$ and $P_2(z,f)$ are two polynomials in f with simple roots only and none of these roots is equal to $\pm \kappa$ or $\pm 1$. As for Eqs. (4.79) and (4.80), when $p=q$, if $A=1$ or $A=\overline{\kappa }^2$, the degree of the numerator in (4.108) or in (4.109) decreases and by Lemma 5 it decreases by 2. This gives the ninth equation of Theorem 7 and also completes the proof. $\square$

5 Discussion

In Sects. 3 and 4, we gave a classification of Eq. (2.1) under the assumptions that Eq. (2.1) has a transcendental meromorphic solution and the degree of R(z, f) in f satisfies $d\not =n$. These results together with the main theorem in our previous paper [13], where the case $d=n$ of Eq. (2.1) was considered, provide a complete difference analogue of Steinmetz’ generalization of Malmquist’s theorem. The classification in Sects. 3 and 4 is according to the number $N_c$ of the roots $\alpha _i$ in (2.7) and $\beta _j$ in (2.8) and whether some of these roots is zero. We did this by mainly using five lemmas, i.e., Lemmas 1–5 in Sect. 2. From their proofs, we see that with some simple adjustments these lemmas also apply to the case $d=n$ of Eq. (2.1). In [30], we have shown how to simplify the proof of the main theorem in [13].

We have shown that if Eq. (2.1) with $d\not =n$ has a transcendental meromorphic solution, then (2.1) reduces into one in a list of 17 equations. In the beginning of Sect. 2, we point out that Eq. (2.1) may reduce into (2.4) in some special cases. In Sect. 3, we consider the case where $q=0$; from the results in Theorems 3 and 4, we see that the polynomial term P(z, f) takes particular form and the solutions f are expressed in terms of exponential type functions explicitly. In Sect. 4, we consider the case where $q\ge 1$. In this case, if $n>d$ or $3\le n<d$, then from Theorems 5 and 6 we see that solutions of (2.1) are also expressed in terms of exponential type functions. But for the case $n=2$ and $n<d$, Eq. (2.1), as well as its solutions, becomes much more complicated. When $q\ge 1$, $n=2$ and $n\not \mid |p-q|$, the polynomials $P_0(z,f)$ and $Q_0(z,f)$ are determined and the solutions are clear, as seen in Theorem 6. When $q\ge 1$, $n=2$ or $n=3$ and $n\mid |p-q|$, we obtain nine equations in Theorem 7. Below, we discuss the eight equations (4.44), (4.47), (4.49), (4.52), (4.53), (4.54), (4.56) and (4.58) in the autonomous case.

Solutions to the Eqs. (4.52) and (4.53) are Weierstrass elliptic functions, composed with entire functions. Below we show their relations with the Fermat type equation $h(z)^3+g(z)^3=1$; see [2, 6, 7]. All meromorphic solutions of the Fermat type equation $h^3+g^3=1$ can be represented as: $h=H(\varphi )$, $g=\eta G(\varphi )=\eta H(-\varphi )=H(-\eta ^2\varphi )$, where $\varphi =\varphi (z)$ is an entire function and $\eta$ is a cubic root of 1, and

$$\begin{aligned} H(z)=\frac{1+\wp '(z)/\sqrt{3}}{2\wp (z)}, \quad G(z)=\frac{1-\wp '(z)/\sqrt{3}}{2\wp (z)} \end{aligned}$$

(5.1)

is a pair of solutions of the Fermat equation $H^3+G^3=1$ with $\wp (z)$ being the particular Weierstrass elliptic function such that $\wp '(z)^2=4\wp (z)^3-1$. For Eq. (4.53), we have

$$\begin{aligned} \overline{f}^3 +\left[ -\frac{P_1(f)}{Q_0(f)}\right] ^3=1. \end{aligned}$$

Therefore, we have $\overline{f}=H(\phi _1)$ and $P_1(f)/Q_0(f)=-\eta G(\phi _1)$, where $\phi _1=\phi _1(z)$ is an entire function, and H(z) and G(z) are defined as in Eq. (5.1). Moreover, there exist two constants $A_1\not =0$ and $B_1$ dependent on the coefficients of P(f) and Q(f) such that $\overline{\phi }_1=A_1\phi _1+B_1$. On the other hand, Eq. (4.52) can also be transformed into the Fermat type equation in the following way: Recall from the proof that we have equation (4.92) and $1,\eta _1,\eta _1^2$, where $\eta _1$ is a cubic root of 1 such that $\eta _1^2+\eta _1+1=0$, are completely ramified values of f with multiplicities 3. We let $(f-\eta _1^2)/(f-\eta _1)=g^3$. Then g is a meromorphic function and it follows that $f=(\eta _1 g^3-\eta _1^2)/(g^3-1)$. By substituting this equation into (4.92), we get

$$\begin{aligned} \overline{f}^3+\left[ -\frac{P_{01}(g^3)g}{Q_{01}(g^3)}\right] ^3=1, \end{aligned}$$

where $P_{01}(g^3)$ and $Q_{01}(g^3)$ are two polynomials in g with no common roots. Then we have $\overline{f}=H(\phi _1)$ and $P_{01}(g^3)g/Q_{01}(g^3)=-\eta G(\phi _2)$, where $\phi _2=\phi _2(z)$ is an entire function, and H(z) and G(z) are defined as in (5.1). Moreover, there exist two constants $A_2\not =0$ and $B_2$ dependent on the coefficients of $P_0(f)$ and $Q_0(f)$ such that $\overline{\phi }_2=A_2\phi _2+B_2$.

Solutions to the six equations (4.44), (4.47), (4.49), (4.54), (4.56) and (4.58) are Jacobian elliptic functions, composed with entire functions. Below we show their relations with the symmetric biquadratic equation of the form $x^2y^2-(x+y)+c^2=0$; see [4, p. 471]). For Eqs. (4.54), (4.56), (4.58), from the two equations (4.103) and (4.104), or from the two equations (4.105) and (4.106), or from the two equations (4.108) and (4.109), we get an equation of the following form:

$$\begin{aligned} \frac{\overline{f}^2-1}{\overline{f}^2-\kappa ^2}=\frac{P_1(f)^2}{P_2(f)^2}, \end{aligned}$$

where $P_1(f)$ and $P_2(f)$ are two polynomials in f with simple roots only and with no common roots. Denoting $R_1(f)=P_1(f)/P_2(f)$, it follows that

$$\begin{aligned} \overline{f}^2R_1(f)^2-[\overline{f}^2+R_1(f)^2]+\kappa ^2=0, \end{aligned}$$

(5.2)

which is a symmetric biquadratic equation in $\overline{f}$ and $R_1$. The equation above can be solved as $\overline{f}=k_1^{1/2}\text {sn}(\varphi _1(z)\pm \tau _1)$ and $R_1(f)=k_1^{1/2}\text {sn}(\varphi _1(z))$, where $k_1$ and $\tau _1$ are two parameters dependent on the constant $\kappa ^2$, $\text {sn}(\varphi _1)$ is the Jacobian elliptic function with modulus $k_1$ and $\varphi _1=\varphi _1(z)$ is an entire function. Then there are two constants $C_1\not =0$ and $D_1$ such that $\overline{\varphi }_1=C_1\varphi _1+D_1$. On the other hand, Eqs. (4.44), (4.47) and (4.49) can also be transformed into symmetric biquadratic equations similar to (5.2) in the following way: For Eq. (4.44), we let $(f+\kappa )/(f+1)=g^2$; for Eq. (4.47) we let $(f+\gamma )/(f-\gamma )=g^2$; for Eq. (4.47) we let $(f+\gamma )/(f-\gamma )=g^2$ when $\kappa =-1$ and let $(f+\kappa )/(f+1)=g^2$ when $\kappa \not =-1$. From the proof in Theorem 7, we see that each g in the above expressions is a meromorphic function. By writing f in terms of $g^2$, then from the two equations (4.79) and (4.80), or from the two equations (4.82) and (4.83), or from the two equations (4.86) and (4.87), or from the two equations (4.88) and (4.89), we get an equation of the following form:

$$\begin{aligned} \frac{\overline{f}^2-\gamma ^2}{\overline{f}^2-1}=\frac{P_{01}(g)^2}{P_{02}(g)^2}, \end{aligned}$$

where $P_{01}(g)$ and $P_{02}(g)$ are two polynomials in g with simple roots only and with no common roots. Denoting $R_2(g)=P_{01}(g)/P_{02}(g)$, it follows that

$$\begin{aligned} \overline{f}^2R_2(g)^2-[\overline{f}^2+R_2(g)^2]+\gamma ^2=0, \end{aligned}$$

which is a symmetric biquadratic equation in $\overline{f}$ and $R_2$. Then we have $\overline{f}=k_2^{1/2}\text {sn}(\varphi _2(z)\pm \tau _2)$ and $R_2(g)=k_2^{1/2}\text {sn}(\varphi _2(z))$, where $k_2$ and $\tau _2$ are two parameters dependent on the constant $\gamma ^2$, $\text {sn}(\varphi _2)$ is the Jacobian elliptic function with modulus $k_2$ and $\varphi _2=\varphi _2(z)$ is an entire function. Moreover, there are two constants $C_2\not =0$ and $D_2$ such that $\overline{\varphi }_2=C_2\varphi _2+D_2$.