1 Introduction

Global existence of large classes of meromorphic solutions is a rare property for a differential equation to have. According to a classical result due to Malmquist [14], if the first-order differential equation

$$\begin{aligned} f'=R(z,f), \end{aligned}$$
(1.1)

where R(zf) is rational in both arguments, has a transcendental meromorphic solution, then (1.1) reduces to the Riccati equation

$$\begin{aligned} f'=a_2 f^2 + a_1 f + a_0 \end{aligned}$$
(1.2)

with rational coefficients \(a_0\), \(a_1\) and \(a_2\). Generalizations of Malmquist’s theorem for the equation

$$\begin{aligned} (f')^n=R(z,f),\qquad n\in {\mathbb {N}}, \end{aligned}$$
(1.3)

have been given by Yosida [31] and Laine [12]. Steinmetz [24] and Bank and Kaufman [3] proved that if (1.3) has rational coefficients and a transcendental meromorphic solution, then by a suitable Möbius transformation, (1.3) can be either mapped to (1.2) or to one of the equations in the following list:

$$\begin{aligned} \begin{aligned} (f')^2&= a(f-b)^2(f-\tau _1)(f-\tau _2),\\ (f')^2&= a(f-\tau _1)(f-\tau _2)(f-\tau _3)(f-\tau _4),\\ (f')^3&= a(f-\tau _1)^2(f-\tau _2)^2(f-\tau _3)^2,\\ (f')^4&= a(f-\tau _1)^2(f-\tau _2)^3(f-\tau _3)^3,\\ (f')^6&= a(f-\tau _1)^3(f-\tau _2)^4(f-\tau _3)^5,\\ \end{aligned} \end{aligned}$$

where a and b are rational functions and \(\tau _1,\ldots ,\tau _4\) are distinct constants.

The existence of globally meromorphic solutions is somewhat more common in the case of difference equations as compared to differential equations. It was shown by Shimomura [23] that the difference equation

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

where P(f(z)) is a polynomial in f(z) with constant coefficients, always has a nontrivial entire solution. On the other hand, Yanagihara [28] showed that the difference equation

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

where R(f(z)) is rational in f(z) having constant coefficients, has a nontrivial meromorphic solution no matter how R is chosen. Yanagihara [29] also considered higher order equations and showed, for instance, that the difference equation

$$\begin{aligned} \alpha _{n}f(z+n)+\alpha _{n-1}f(z+n-1)+\cdots +\alpha _{1}f(z+1)=R(f(z)), \qquad \alpha _1,\ldots ,\alpha _n\in {\mathbb {C}}, \end{aligned}$$

has a nontrivial meromorphic solution if the degree p of the numerator P(f(z)) of the rational function R(f(z)) satisfies \(p\ge q+2\), where q is the degree of the denominator Q(f(z)), and P(f(z)) and Q(f(z)) have no common factors.

Ablowitz, Halburd and Herbst [1] suggested that the existence of sufficiently many finite-order meromorphic solutions of a difference equation is a good difference analogue of the Painlevé property for differential equations. An ordinary differential equation is said to have the Painlevé property when all solutions are single-valued around all movable singularities. They showed, for instance, that if the difference equation

$$\begin{aligned} f(z+1)+f(z-1) = R(z,f(z)), \end{aligned}$$
(1.4)

where R(zf(z)) is rational in both arguments, has a transcendental meromorphic solution of finite order, then \(\deg _f(R(z,f(z)))\le 2\). Their results are consistent with Yanagihara’s work on the first-order equation [28] in which he proved that if

$$\begin{aligned} f(z+1)= R(z,f(z)), \end{aligned}$$
(1.5)

where R(zf(z)) is rational in both arguments, has a transcendental meromorphic solution of hyper-order strictly less than one, then \(\deg _f(R(z,f(z)))= 1\) and thus (1.5) reduces into the difference Riccati equation. This is a natural difference analogue of Malmquist’s 1913 result on differential equations. Halburd and the first author [7] showed that if (1.4), where the right-hand side has meromorphic coefficients, has an admissible meromorphic solution f of finite order, then either f satisfies a difference Riccati equation, or a linear transformation of (1.4) reduces it to one in a short list of difference equations that consists solely of difference Painlevé equations and equations related to them, linear equations, and linearizable equations. The finite-order condition was relaxed into hyper-order strictly less than one by Halburd, the first author, and Tohge [9].

The purpose of this paper is to present a natural difference analogue of Steinmetz’s generalization of Malmquist’s theorem. We will show that if the difference equation

$$\begin{aligned} f(z+1)^n=R(z,f), \end{aligned}$$
(1.6)

with rational coefficients has a transcendental meromorphic solution f of hyper-order \(<1\), then either f satisfies a difference linear or Riccati equation

$$\begin{aligned} f(z+1)= & {} a_1(z)f(z)+a_2(z), \end{aligned}$$
(1.7)
$$\begin{aligned} f(z+1)= & {} \frac{b_1(z)f(z)+b_2(z)(z)}{f(z)+b_3(z)}, \end{aligned}$$
(1.8)

where \(a_i(z)\) and \(b_j(z)\) are rational functions, or, by implementing a transformation \(f\rightarrow \alpha f\) or \(f\rightarrow 1/(\alpha f)\) with an algebraic function \(\alpha \) of degree at most 2, (1.6) reduces into one of the following equations:

$$\begin{aligned} f(z+1)^2= & {} 1-f(z)^2, \end{aligned}$$
(1.9)
$$\begin{aligned} f(z+1)^2= & {} 1-\left( \frac{\delta (z) f(z)-1}{f(z)-\delta (z)}\right) ^2, \end{aligned}$$
(1.10)
$$\begin{aligned} f(z+1)^{2}= & {} 1-\left( \frac{f(z)+3}{f(z)-1}\right) ^{2}, \end{aligned}$$
(1.11)
$$\begin{aligned} f(z+1)^{2}= & {} \frac{f(z)^2 - \kappa ^2}{f(z)^2-1}, \end{aligned}$$
(1.12)
$$\begin{aligned} f(z+1)^{3}= & {} 1-f(z)^{-3}, \end{aligned}$$
(1.13)

where \(\delta (z)\)\((\not \equiv \pm \,1)\) is an algebraic function of degree 2 at most and \(\kappa ^2\not =0,1\) is a constant. Equations (1.7) and (1.8) are the linear and the difference Riccati equation, respectively, Eqs. (1.9)–(1.11) and (1.13) are difference Fermat equations, while (1.12) is a special case of the symmetric Quispel–Roberts–Thompson (QRT) map. We will present finite-order meromorphic solutions to autonomous versions of all of these equations in terms of Weierstrass or Jacobi elliptic functions, or in terms of meromorphic functions that are solutions to a difference Riccati equation, in Sect. 3 below.

Under the condition that the meromorphic solution f of (1.6) is of hyper-order less than 1, it can actually be shown that \(\deg _f(R(z,f))=n\) by using an asymptotic relation between the Nevanlinna characteristics (see Sect. 2 below) \(T(r,f(z+1))\) and T(rf(z)) from [9], and the Valiron–Mohon’ko identity [15, 26] (see also [13]). By discarding the assumption that the meromorphic solution is of hyper-order \(<1\), and considering the more general case of admissible meromorphic solutions of (1.6) with meromorphic coefficients and such that \(\deg _f(R(z,f))=n\), it follows either that f satisfies (1.7) or (1.8), or (1.6) can be transformed into one of the Eqs. (1.9)–(1.13), but now with meromorphic coefficients \(a_i(z)\) and \(b_j(z)\), an algebroid function \(\delta (z)=\delta _2(z)\) of degree at most 2 and with \(\kappa ^2=\kappa _1(z)^2\) being a meromorphic periodic function of period 1, or (1.6) becomes one of the following equations:

$$\begin{aligned} f(z+1)^2= & {} \delta _1(z)(f(z)^2-1), \end{aligned}$$
(1.14)
$$\begin{aligned} f(z+1)^2= & {} \delta _3(z) (1-f(z)^{-2}), \end{aligned}$$
(1.15)
$$\begin{aligned} f(z+1)^2= & {} \frac{\kappa _2(z+1)^2f(z)^2-1}{f(z)^2-1}, \end{aligned}$$
(1.16)
$$\begin{aligned} f(z+1)^2= & {} \theta \frac{f(z)^2-\kappa _3(z)f(z)+1}{f(z)^2+\kappa _3(z)f(z)+1}, \end{aligned}$$
(1.17)
$$\begin{aligned} f(z+1)^3= & {} 1-f(z)^3, \end{aligned}$$
(1.18)

where \(\theta =\pm \,1\) and \(\delta _1(z)\), \(\delta _3(z)\), \(\kappa _2(z)^2\), \(\kappa _3(z)^2\) are meromorphic functions each of which satisfies a certain difference equation, see Sect. 3 below. In particular, if the coefficients of (1.6) are rational functions, then \(\theta =\pm \,1\) and \(\delta _1(z)\), \(\delta _3(z)\), \(\kappa _2(z)^2\), \(\kappa _3(z)^2\) are all constants. We will show that meromorphic solutions of autonomous versions of (1.14)–(1.18) can be characterized by Weierstrass or Jacobi elliptic functions composed with certain entire functions, but none of them is of hyper-order \(<1\).

The remainder of this paper is organized in the following way. Section 2 contains the necessary notation and preliminary results needed to prove our main theorem given in Sect. 3. Section 3 also contains a discussion of the Eqs. (1.9)–(1.18). In particular, we will show that either their autonomous versions have meromorphic solutions expressible in terms of Weierstrass or Jacobi elliptic functions, or solutions of these equations can be explicitly expressed in terms of meromorphic functions that are solutions of certain difference Riccati equations. The proof of our main result, Theorem 2 below, has been split into the three remaining Sects. 46.

2 Preliminaries

In this section, we introduce some preliminaries for the proof of our main results. In the following, a meromorphic function is always meromorphic on the whole complex plane \({\mathbb {C}}\). Let \({\mathcal {M}}\) denote the field of meromorphic functions, and let \(f(z)\in {\mathcal {M}}\). We assume that the readers are familiar with the basic notation and fundamental results of Nevanlinna theory [10], such as m(rf), N(rf), T(rf), and Nevanlinna’s main theorems. Moreover, we use the notation \(\sigma (f)\) to denote the order of growth of f(z), which is defined to be

$$\begin{aligned} \sigma (f)=\limsup _{r\rightarrow \infty }\frac{\log T(r,f)}{\log {r}}, \end{aligned}$$

and, if f(z) is of order \(\sigma (f)=\infty \), we use the notation \(\sigma _2(f)\) to denote the hyper-order of f(z), which is defined to be

$$\begin{aligned} \sigma _2(f)=\limsup _{r\rightarrow \infty }\frac{\log \log T(r,f)}{\log {r}}. \end{aligned}$$

Let S(rf) denote any quantity that satisfies the condition \(S(r,f)=o(1)T(r,f)\), \(r\rightarrow \infty \), outside of a possible exceptional set of finite logarithmic measure. For a meromorphic function \(c(z)\in {\mathcal {M}}\), if c(z) satisfies \(T(r,c(z))=S(r,f)\), then c(z) is said to be small compared to f(z). For example, all rational functions are small with respect to any transcendental meromorphic function. Denote by \({\mathcal {S}}(f)\) the field of all small functions of f(z), i.e.,

$$\begin{aligned} {\mathcal {S}}(f)=\{c(z)\in {\mathcal {M}}: T(r,c(z))=S(r,f)\}. \end{aligned}$$

Set \(\hat{{\mathcal {S}}}(f)={\mathcal {S}}(f)\cup \{\infty \}\). A meromorphic solution f(z) of a differential equation is called admissible if all coefficients of the equation are in \({\mathcal {S}}(f)\) [13]. This definition is very natural for differential equations, but for difference equations to be studied in this paper we need to be slightly more careful due to the possible effect of the shift operation on the growth of the solution and of the coefficients. We define the following set for small functions of a meromorphic function f(z):

$$\begin{aligned} {\mathcal {S}}'(f)=\{c(z)\in {\mathcal {M}}: T(r,c(z+n))=S(r,f(z+n)), n\in {\mathbb {N}}\}, \end{aligned}$$

and in what follows we say that a meromorphic solution f(z) of a difference equation is admissible if all coefficients of the equation are in \({\mathcal {S}}'(f)\). If the hyper-order of f(z) is less than one, then by [9, Lemma 8.3] it follows that

$$\begin{aligned} T(r,f(z+1)) = T(r,f) + S(r,f), \end{aligned}$$
(2.1)

and so in this case shifting a difference equation does not affect the admissibility of solutions.

The fundamental results of Nevanlinna’s theory are the First Main Theorem and the Second Main Theorem. Nevanlinna’s Second Main Theorem can be simplified into the following form:

$$\begin{aligned} (q-2)T(r,f)\le \sum _{i=1}^q{\overline{N}}(r,a_i,f)+S(r,f), \end{aligned}$$
(2.2)

where \(q\ge 3\), \(a_i\in {\mathbb {C}}\cup \{\infty \}\), and \({\overline{N}}(r,a_i,f)\) is the truncated counting function for \(a_i\)-points (poles) of f(z), and the error term \(S(r,f)=O(\log r)\), \(r\rightarrow \infty \), when f(z) is of finite order and \(S(r,f)=O(\log rT(r,f))\), \(r\rightarrow \infty \), outside a possible set of finite linear measure when f(z) is of infinite order. Yamanoi [27, Corollary 1] generalized Nevanlinna’s Second Main Theorem by proving that, for any collection of functions \(c_i(z)\in \hat{{\mathcal {S}}}(f)\), \(i=1,\ldots ,q\), of f(z), the inequality

$$\begin{aligned} (q-2-\varepsilon )T(r,f)\le \sum _{i=1}^q{\overline{N}}(r,c_i,f), \quad \text {for all} \quad \varepsilon >0, \end{aligned}$$
(2.3)

holds outside an exceptional set \(E=E(\varepsilon )\subset (0,\infty )\) satisfying \(\int _{E}d\log \log r<\infty \). The inequality (2.3) implies a generalization of the defect relation for small functions of f(z) [27, Corollary 1] and that

$$\begin{aligned} \sum _{i=1}^q\varTheta (c_i,f)\le 2, \end{aligned}$$
(2.4)

where the quantity \(\varTheta (c,f)\), \(c=c(z)\in \hat{{\mathcal {S}}}(f)\), is defined as

$$\begin{aligned} \varTheta (c,f)=1-\limsup _{r\rightarrow \infty }\frac{{\overline{N}}(r,c,f)}{T(r,f)}. \end{aligned}$$

Recall that a value a is said to be a completely ramified value of f(z) when \(f(z)-a=0\) has no simple roots. In what follows, we say that \(c(z)\in \hat{{\mathcal {S}}}(f)\) is a completely ramified small function of f(z) when \(f(z)-c(z)=0\) has at most S(rf) many simple roots and that \(c(z)\in \hat{{\mathcal {S}}}(f)\) is a Picard exceptional small function of f(z) when \(N(r,c,f)=S(r,f)\). A nonconstant meromorphic function f(z) can have at most two Picard exceptional small functions. Moreover, by (2.4) and the First Main Theorem, we have the following result.

Theorem 1

A nonconstant meromorphic function f(z) can have at most four completely ramified small functions.

In what follows, when considering a meromorphic solution f of (1.6), we will do a transformation to f using some algebroid functions and end up in a situation such that the considered function has some finite-sheeted branching. The classical version of Nevanlinna theory introduced above cannot be used to handle this situation. In this case, we need the algebroid version of Nevanlinna theory introduced by Selberg [20,21,22], Ullrich [25] and Valiron [26] (see also [11]), which studies meromorphic functions on a finitely sheeted Riemann surface. All algebroid functions we need to consider in this paper are either small functions with respect to an admissible meromorphic solution f of (1.6), or can be obtained from it by a Möbius transformation with small algebroid coefficients. Such functions could be described as “almost meromorphic” in the sense of Nevanlinna theory, since the presence of branch points actually only affects the small error term S(rf) in any of the estimates involving Nevanlinna functions. Correspondingly, T(rf) and N(rf) will denote the characteristic and counting functions of a finite-sheeted algebroid function f, and similarly with the rest of the Nevanlinna functions involving f.

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

3 Extension of the Difference Malmquist Theorem

As mentioned in the introduction, from Eq. (1.5) Yanagihara [28] obtained a difference analogue of Malmquist’s theorem. Let \(n\in {\mathbb {N}}\). Nakamura and Yanagihara [16], and Yanagihara [30], considered meromorphic solutions of a more general case,

$$\begin{aligned} {\overline{f}}^n=P(z,f), \end{aligned}$$
(3.1)

where P(zf) is a polynomial in f with constants as coefficients, and, in particular, Yanagihara [30] showed that if (3.1) has a meromorphic solution of finite order, then either \({\overline{f}}=Af+B\) for some constants AB or (3.1) reduces to \({\overline{f}}^2=1-f^2\). We consider the difference equation

$$\begin{aligned} {\overline{f}}^n=R(z,f), \end{aligned}$$
(3.2)

where R(zf) is rational in f having meromorphic coefficients. If Eq. (3.2) has an admissible meromorphic solution f such that the hyper-order \(\sigma _2(f)\) of f satisfies \(\sigma _2(f)<1\), then, by (3.2), it follows from (2.1) and the Valiron–Mohon’ko identity [15, 26] (see also [13]) that

$$\begin{aligned} \begin{aligned} \deg _f(R(z,f))T(r,f)&= T(r,R(z,f)) + S(r,f)\\&= T(r,{\overline{f}}^n) + S(r,f) \\&= nT(r,{\overline{f}}) + S(r,f) \\&= nT(r,f) + S(r,f), \end{aligned} \end{aligned}$$

which implies that \(\deg _f(R(z,f))=n\). Assume that \(\deg _f(R(z,f))=n\) and that all the coefficients of (3.2) are in \({\mathcal {S}}'(f)\). Then the characteristic function of f(z) satisfies \(T(r,f(z+j))=T(r,f(z))+S(r,f)\), \(j\in {\mathbb {N}}\). Under these conditions, but discarding the assumption \(\sigma _2(f)<1\), we now use Nevanlinna theory to simplify Eq. (3.2). Our main result, which is an extension of the result on (3.1) of Yanagihara [30], is stated as follows.

Theorem 2

Let \(n\in {\mathbb {N}}\). If the difference equation (3.2) with \(\deg _f(R(z,f))=n\) has an admissible meromorphic solution, then either f satisfies a difference linear or Riccati equation:

$$\begin{aligned} {\overline{f}}= & {} a_1f+a_2, \end{aligned}$$
(3.3)
$$\begin{aligned} {\overline{f}}= & {} \frac{b_1f+b_2}{f+b_3}, \end{aligned}$$
(3.4)

where \(a_i,b_j\) are small meromorphic functions; or, by a transformation \(f\rightarrow \alpha f\) or \(f\rightarrow 1/(\alpha f)\) with a small algebroid function \(\alpha \) of degree at most 3, (3.2) reduces to one of the following equations:

$$\begin{aligned} {\overline{f}}^2= & {} 1-f^2, \end{aligned}$$
(3.5)
$$\begin{aligned} {\overline{f}}^2= & {} \delta _1(f^2-1), \end{aligned}$$
(3.6)
$$\begin{aligned} {\overline{f}}^2= & {} 1-\left( \frac{\delta _2 f-1}{f-\delta _2}\right) ^2, \end{aligned}$$
(3.7)
$$\begin{aligned} {\overline{f}}^2= & {} \delta _3(1-f^{-2}), \end{aligned}$$
(3.8)
$$\begin{aligned} {\overline{f}}^{2}= & {} 1-\left( \frac{f+3}{f-1}\right) ^{2}, \end{aligned}$$
(3.9)
$$\begin{aligned} {\overline{f}}^2= & {} \frac{f^2-\kappa _1^2}{f^2-1}, \end{aligned}$$
(3.10)
$$\begin{aligned} {\overline{f}}^2= & {} \frac{{\overline{\kappa }}_2^2f^2-1}{f^2-1}, \end{aligned}$$
(3.11)
$$\begin{aligned} {\overline{f}}^2= & {} \theta \frac{f^2-\kappa _3f+1}{f^2+\kappa _3f+1}, \end{aligned}$$
(3.12)
$$\begin{aligned} {\overline{f}}^{3}= & {} 1-f^3, \end{aligned}$$
(3.13)
$$\begin{aligned} {\overline{f}}^{3}= & {} 1-f^{-3}, \end{aligned}$$
(3.14)

where \(\theta =\pm \,1\), \(\delta _2\not \equiv \pm \,1\) is a small algebroid function of degree at most 2 and \(\delta _1\), \(\delta _3\), \(\kappa _1^2\), \(\kappa _2^2\), \(\kappa _3^2\) are all small meromorphic functions satisfying \({\overline{\delta }}_1(\delta _1+1)+1=0\), \({\overline{\delta }}_3\delta _3={\overline{\delta }}_3+\delta _3\), \({\overline{\kappa }}_1^2=\kappa _1^2\), \({\overline{\kappa }}_2^2\kappa _2^2=1\), and \({\overline{\kappa }}_3^2(\kappa _3^2-4)=2(1-\theta )\kappa _3^2-8(1+\theta )\), respectively.

Assume that in Theorem 2 all the coefficients of (3.2) are rational functions. Then it is seen that the coefficients \(\delta _1,\delta _3,\kappa _1^2,\kappa _2^2,\kappa _3^2\) are all constants. In this case, it is easy to show by iteration and substitution that solutions to (3.5), (3.6), (3.8), (3.10), (3.11), (3.13), and (3.14) are all periodic functions, which implies that for these equations the coefficient \(\alpha \) in the transformation \(f\rightarrow \alpha f\) or \(f\rightarrow 1/(\alpha f)\) has no branch points. Also, from the proof of Theorem 2 we see that when obtaining (3.9) and (3.12), the coefficient \(\alpha \) in the transformation \(f\rightarrow \alpha f\) or \(f\rightarrow 1/(\alpha f)\) is rational. Thus the algebraic case of \(\alpha \) can only occur when obtaining Eq. (3.7).

In the rational coefficients case of (3.2), solutions to (3.13) and (3.14) can be characterized by Weierstrass elliptic functions, solutions to Eqs. (3.6), (3.8), (3.10)–(3.12) can be characterized by Jacobi elliptic functions, and Eqs. (3.5), (3.7) and (3.9) can be explicitly solved in terms of functions that are solutions of certain difference Riccati equations, as is shown below.

Equations (3.5), (3.7), (3.9), (3.13), and (3.14) are so-called Fermat difference equations. In general, a Fermat equation is a function analogue of the Fermat diophantine equation \(x^n+y^n=1\); i.e., \(h(z)^n+g(z)^n=1\), where \(n\ge 2\) is an integer. Meromorphic solutions to Fermat equations have been clearly characterized, see [2, 5, 6], for example. In particular, when \(n=3\), all meromorphic solutions 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

$$\begin{aligned} H(z)=\frac{1+\wp '(z)/\sqrt{3}}{2\wp (z)}, \quad G(z)=\frac{1-\wp '(z)/\sqrt{3}}{2\wp (z)}, \end{aligned}$$
(3.15)

is a pair of solutions of the Fermat equation with \(\wp (z)\) being the particular Weierstrass elliptic function satisfying \(\wp '(z)^2=4\wp (z)^3-1\). Now, for Eq. (3.13) the solution takes the form \(f=H(\varphi )\), \({\overline{f}}=\eta G(\varphi )=H(-\eta ^2\varphi )\). It follows that \({\overline{\varphi }}=-\eta ^2\varphi +\omega \), where \(\omega \) is a period of \(\wp (\varphi )\). This implies that \(\varphi =\upsilon (-\eta ^2)^z+\omega /(1+\eta ^2)\), where \(\upsilon \) is an arbitrary periodic entire function of period 1, and thus \(\varphi \) is a transcendental entire function of order at least 1, of mean type. From (3.13) we have \(T(r,{\overline{f}})=T(r,f)+O(1)\). Moreover, we have \(\eta f+{\overline{f}}=\eta /\wp (\varphi )\), which yields \(T(r,\wp (\varphi ))\le 2T(r,f)+O(1)\). By taking the derivatives on both sides of \(\eta f+{\overline{f}}=\eta /\wp (\varphi )\) and combining it with the resulting equation, we get

$$\begin{aligned} \varphi '^2=\left( \frac{(\eta f+{\overline{f}})'}{\eta f+{\overline{f}}}\right) ^2\cdot \frac{\wp (\varphi )^2}{4\wp (\varphi )^3-1}. \end{aligned}$$
(3.16)

Since \(\wp (\varphi )\) satisfies \(\wp '(\varphi )^2=4\wp (\varphi )^3-1\), we see that \(\wp (\varphi )\), as a function of z, is twofold ramified over poles and each of \(\vartheta _i\), \(i=1,2,3\), where \(\vartheta _i\) are the three cubic roots of 1 / 4. Then the First Main Theorem and the Second Main Theorem (2.2) yield that \(m(r,\wp (\varphi ))+\sum ^3_{i=1}m(r,\frac{1}{\wp (\varphi )-\vartheta _i})\le O(\log rT(r,\wp (\varphi )))\). Thus, by taking the proximity functions on both sides of (3.16), we get

$$\begin{aligned} m(r,\varphi '^2)\le & {} O(\log rT(r,f))+2m(r,\wp (\varphi ))\\&+\sum ^3_{i=1}m\left( r, \frac{1}{\wp (\varphi )-\vartheta _i}\right) +O(1)=O(\log rT(r,f))+O(1). \end{aligned}$$

Then, \(T(r,\varphi '^2)=m(r,\varphi '^2)\le O(\log rT(r,f))+O(1)\), which together with [13, Lemma 1.1.1] yields that \(\sigma _2(f)\ge 1\), and so Eq. (3.13) cannot have meromorphic solutions of hyper-order less than 1.

We now show that the autonomous versions of (3.6), (3.8), (3.11), and (3.12) cannot admit meromorphic solutions of hyper-order \(<1\) either. We only need to consider Eqs. (3.6) and (3.11) since (3.8) and (3.12) can be transformed into (3.6) or (3.11) in the following way. For Eq. (3.8), \(\delta _3=2\), and we see that \(f^2-1\) is written as \(f^2-1=-(i{\overline{f}}f/\sqrt{2})^2=-f_0^2\). Then we have \(T(r,f)=T(r,f_0)+O(1)\), and by substituting \(f^2=-f_0^2+1\) into (3.8) we get the autonomous case of Eq. (3.11). For Eq. (3.12), if we let \(w=f+1/f\), then w satisfies \(T(r,w)=2T(r,f)+O(1)\) by the Valiron–Mohon’ko identity [15, 26] (see also [13]), and it follows that

$$\begin{aligned} {\overline{f}}^2=\theta \frac{w-\kappa _3}{w+\kappa _3}, \quad \frac{1}{{\overline{f}}^2}=\theta \frac{w+\kappa _3}{w-\kappa _3}. \end{aligned}$$

The above two equations yield

$$\begin{aligned} {\overline{w}}^2=\frac{2(\theta +1)w^2+2(\theta -1) \kappa _3^2}{w^2-\kappa _3^2}. \end{aligned}$$

If \(\theta =1\), then we have \({\overline{\kappa }}_3^2(\kappa _3^2-4)=-16\), and by doing a transformation \(w\rightarrow \kappa _3/w\) we get \({\overline{w}}^2=d_1(w^2-1)\), which is the Eq. (3.6) since \(d_1=-{\overline{\kappa }}_3^2/4\) satisfies \({\overline{d}}_1(d_1+1)+1=0\); if \(\theta =-1\), then we have \({\overline{\kappa }}_3^2(\kappa _3^2-4)=4\kappa _3^2\), and by doing a transformation \(w\rightarrow \kappa _3/w\) we get \({\overline{w}}^2=d_2(1-w^{-2})\), which is the Eq. (3.8) since \(d_2={\overline{\kappa }}_3^2/4\) satisfies \({\overline{d}}_2d_2={\overline{d}}_2+d_2\).

Recall that the Jacobi elliptic function \(\text {sn}(z,k)\) with the elliptic modulus \(k\in (0,1)\) satisfies the differential equation \(\text {sn}'(z)^2=(1-\text {sn}(z)^2)(1-k^2\text {sn}(z)^2)\). For Eq. (3.6), \(\delta _1\) is now a cubic root of 1, and we see that f is twofold ramified over \(\pm \,1\). We write \(\delta _1=\eta ^2\), where \(\eta \) is the cubic root of 1 satisfying \(\eta ^2(\eta ^2+1)+1=\eta ^2+\eta +1=0\). Let \(z_0\) be such that \(f(z_0)=\pm \, i\eta \), by (3.6) we have \(f(z_0+1)=\pm \,1\), and so f is also twofold ramified at \(\pm \, i\eta \)-points. Let \(\varrho _1=\exp (7i\pi /12)\), and set \(\tau _1=(1-\varrho _1)/(1+\varrho _1)\). Then

$$\begin{aligned} g_1=\tau _1\frac{f+\varrho _1}{f-\varrho _1} \end{aligned}$$

is twofold ramified over each of \(\pm \,1,\pm \tau _1^2\), where \(-\tau _1^2=\tan (7\pi /24)^2>1\). Let \(\text {sn}(\phi )=\text {sn}(\phi ,-1/\tau _1^2)\) and \(\phi _0\) be such that \(\text {sn}(\phi _0)\not =\pm \,1,\pm \tau _1^2\). Letting \(z_0\) be such that \(g_1(z_0)=\text {sn}(\phi _0)\), it follows that there is a neighborhood U of \(z_0\) such that \(\phi _1(z)=\text {sn}^{-1}(g_1(z))\) is defined and holomorphic in U. By following the reasoning in the proof of [30, Lemma 4.1], we know that \(\phi _1(z)\) can be continued analytically throughout the complex plane to an entire function. Therefore, \(g_1\) is written as \(g_1(z)=\text {sn}(\phi _1(z))\), where \(\text {sn}(\phi _1)=\text {sn}(\phi _1,-1/\tau _1^2)\) satisfies the differential equation \(\text {sn}'(\phi _1)^2=(1-\text {sn}(\phi _1)^2)(1-\text {sn} (\phi _1)^2/\tau _1^4)\). By taking the derivative of \(g_1\) and combining it with the resulting equation, we get

$$\begin{aligned} \phi _1'^2=\frac{\text {sn}(\phi _1)^2}{\text {sn}'(\phi _1)^2} \cdot \left( \frac{g_1'}{g_1}\right) ^2=\frac{g_1^2}{(1-g_1^2) (1-g_1^2/\tau _1^4)}\cdot \left( \frac{g_1'}{g_1}\right) ^2. \end{aligned}$$
(3.17)

Since \(g_1\) is twofold ramified over each of \(\pm \,1,\pm \tau _1^2\), the First Main Theorem and the Second Main Theorem (2.2) yield that \(2m(r,g_1)+\sum _{i=1}^{4}m(r,\frac{1}{g_1-\iota _i})\le O(\log rT(r,g_1))\), \(\iota _i=\pm \,1,\pm \tau _1^2\). Note that \(T(r,f)=T(r,g_1)+O(1)\). Now taking the proximity function on both sides of (3.17) gives \(T(r,\phi _1'^2)=m(r,\phi _1'^2)\le O(\log rT(r,f))+O(1)\), which together with [13, Lemma 1.1.1] yields \(\sigma (\phi _1)\le \sigma _2(f)\). Suppose that \(\sigma _2(f)<1\). By iterating (3.6) we obtain \(f(z+3)^2=f(z)^2\), and so \(f(z+6)=f(z)\) and it follows that \(\text {sn}(\phi _1(z+6))=\text {sn}(\phi _1(z))\) giving \(\phi _1(z+6)=\phi _1(z)+K_1\) with a period \(K_1\) of \(\text {sn}(\phi _1)\), which is possible only when \(\phi _1\) is a polynomial of degree 1 since \(\phi _1'(z+6)=\phi _1'(z)\). This implies that f is of finite order. But according to [30, Lemma 9.1], (3.6) cannot admit any meromorphic solutions of finite order when \(\delta _1\) is a constant. Therefore, (3.6) cannot admit any meromorphic solutions of hyper-order strictly less than 1 when \(\delta _1\) is a constant.

For Eq. (3.11), if \(\kappa _2\) is a constant, then \(\kappa _2^2=-1\) and we see that f is twofold ramified over each of \(\pm \, i,\pm \,1\). Let \(\varrho _2=\exp (i\pi /4)\), and set \(\tau _2=(1+\varrho _2)/(1-\varrho _2)\). Then

$$\begin{aligned} g_2=\tau _2\frac{f-\varrho _2}{f+\varrho _2} \end{aligned}$$

is twofold ramified over each of \(\pm \,\tau _2^2,\pm \,1\), where \(-\tau _2^2=(\sqrt{2}+1)^2>1\). By similar arguments as above, there exists an entire function \(\phi _2(z)\) such that \(g_2\) is written as \(g_2(z)=\text {sn}(\phi _2(z))\), where \(\text {sn}(\phi _2)=\text {sn}(\phi _2,-1/\tau _2^2)\). Thus

$$\begin{aligned} f=-\varrho _2\frac{\text {sn}(\phi _2)+\tau _2}{\text {sn}(\phi _2)-\tau _2}. \end{aligned}$$

Suppose that \(\sigma _2(f)<1\). Then we have \(\sigma (\phi _2')\le \sigma _2(f)<1\). Iterating (3.11) gives \(f(z+4)^2=f(z)^2\). Therefore, we have \(f(z+8)=f(z)\) and then \(\text {sn}(\phi _2(z+8))=\text {sn}(\phi _2(z))\), giving \(\phi _2(z+8)=\phi _2(z)+K_2\) with a period \(K_2\) of \(\text {sn}(\phi _2)\), which is possible only when \(\phi _2\) is a polynomial of degree 1 since \(\phi _2'(z+8)=\phi _2'(z)\). Write \(\phi _2(z)=Cz+D\) for two constants CD. Supposing that \(\text {sn}(\phi _2(z_0))=1\); we then have \(f(z_0)=1\). By (3.11) we have \(f(z_0+1)=\infty \) and it follows that \(\text {sn}(\phi _2(z_0+1))=\tau _2\). On the other hand, if \(\text {sn}(\phi _2(z_1))=\tau _2^2\), then \(f(z_1)=-1\) and it follows from (3.11) that \(f(z_1+1)=\infty \) and hence \(\text {sn}(\phi _2(z_1+1))=\tau _2\). Thus, \(\text {sn}(\phi _2(z_0+1))=\text {sn}(\phi _2(z_1+1))\), which implies that \(\phi _2(z_0)-\phi _2(z_1)=C(z_0-z_1)=\phi _2(z_0+1)-\phi _2(z_1+1)=K_2\) with a period \(K_2\), but \(\text {sn}(\phi _2(z_0))\not =\text {sn}(\phi _2(z_1))\), a contradiction. Therefore, (3.11) cannot admit any meromorphic solutions of hyper-order strictly less than 1 when \(\kappa _2\) is a constant.

From the above discussions, we obtain the following corollary, which is a natural difference analogue of Steinmetz’s generalization [24] of Malmquist’s 1913 result on differential equations.

Corollary 1

Let \(n\in {\mathbb {N}}\). If the difference equation (3.2) with rational coefficients has a transcendental meromorphic solution of hyper-order strictly less than 1, then either f satisfies (3.3) or (3.4) with rational coefficients or, by a transformation \(f\rightarrow \alpha f\) or \(f\rightarrow 1/(\alpha f)\) with an algebraic function \(\alpha \) of degree at most 2, (3.2) reduces to one of the following equations:

$$\begin{aligned} {\overline{f}}^2= & {} 1-f^2, \end{aligned}$$
(3.18)
$$\begin{aligned} {\overline{f}}^2= & {} 1-\left( \frac{\delta f-1}{f-\delta }\right) ^2, \end{aligned}$$
(3.19)
$$\begin{aligned} {\overline{f}}^{2}= & {} 1-\left( \frac{f+3}{f-1}\right) ^{2}, \end{aligned}$$
(3.20)
$$\begin{aligned} {\overline{f}}^2= & {} \frac{f^2-\kappa ^2}{f^2-1}, \end{aligned}$$
(3.21)
$$\begin{aligned} {\overline{f}}^{3}= & {} 1-f^{-3}, \end{aligned}$$
(3.22)

where \(\delta \not \equiv \pm \,1\) is an algebraic function of degree 2 at most and \(\kappa ^2\not =0,1\) is a constant.

Below we will show that Eqs. (3.18)– (3.22) can indeed have meromorphic solutions of finite order.

Let’s look at the second-degree Fermat difference Eqs. (3.18), (3.19), and (3.20) first. For Eq. (3.18), we know from [30, Theorem 2] that the solution f is represented as \(f=(\beta +\beta ^{-1})/2\), where \(\beta \) satisfies \({\overline{\beta }}=i\beta ^{\pm \,1}\). For Eq. (3.19), if we put \(f=(\gamma +\gamma ^{-1})/2\), then we have

$$\begin{aligned} \frac{1}{4}\left( {\overline{\gamma }}-\frac{1}{{\overline{\gamma }}} \right) ^2=-\left( \frac{\delta f-1}{f-\delta }\right) ^2. \end{aligned}$$

It follows that

$$\begin{aligned} {\overline{\gamma }}^2-2i\frac{\delta \gamma ^2-2\gamma +\delta }{\gamma ^2-2\delta \gamma +1}{\overline{\gamma }}-1=0. \end{aligned}$$

Solving the above equation, we get the difference Riccati equation

$$\begin{aligned} {\overline{\gamma }}=\left\{ -\theta \frac{(i\delta -\sqrt{1-\delta ^2}) \gamma +i}{\gamma -\delta +i\sqrt{1-\delta ^2}}\right\} ^{\theta }, \quad \theta =\pm \,1. \end{aligned}$$

For Eq. (3.20), we have

$$\begin{aligned} {\overline{f}}^{2}=-\frac{8(f+1)}{(f-1)^{2}}. \end{aligned}$$

If we put

$$\begin{aligned} \sqrt{-8}u=\frac{{\overline{f}}(f-1)}{f+1}, \quad v=\frac{1}{f+1}, \end{aligned}$$

then we have

$$\begin{aligned} u^{2}=v, \quad f =\frac{1}{v}-1, \quad {\overline{f}}=\frac{\sqrt{-8}u}{1-2v}, \end{aligned}$$

and further

$$\begin{aligned} (\sqrt{2}{\overline{u}})^{2}=2{\overline{v}}=\frac{2}{{\overline{f}} +1}=\frac{2(2u^{2}-1)}{2u^{2}-\sqrt{-8}u-1}=1-\left( \frac{ \sqrt{2}iu-1}{\sqrt{2}u-i}\right) ^2. \end{aligned}$$

Putting \(\sqrt{2}u=(\lambda +\lambda ^{-1})/2\), then we have

$$\begin{aligned} f=\frac{1-u^2}{u^2}=\frac{8\lambda ^2-(\lambda ^2+1)^2}{(\lambda ^2+1)^2}, \quad {\overline{\lambda }}=\left\{ -\theta \frac{-(1+\sqrt{2})\lambda +i}{\lambda -i+i\sqrt{2}}\right\} ^{\theta }, \quad \theta =\pm \,1. \end{aligned}$$

Since the autonomous version of the difference Riccati equation is solvable explicitly in terms of exponential functions, so are Eqs. (3.18), (3.19), and (3.20) when \(\delta \) is a constant.

Equation (3.21) can be rewritten as \({\overline{f}}^2f^2-({\overline{f}}^2+f^2)+\kappa ^2=0\), which is a special case of symmetric QRT map [17, 18]. By doing a suitable Möbius transformation [4, 19], for example, \(f\rightarrow a(f+1)/(f-1)\), where a (\(|a|>1\)) is a constant satisfying \(2a^4-2a^2-1=0\) and \(\kappa ^2=a^4\), we get

$$\begin{aligned} {\overline{f}}^2f^2+{\overline{f}}^2+f^2+4(1+4a^2){\overline{f}}f+1=0, \end{aligned}$$

which is solvable in terms of Jacobi elliptic functions with finite order of growth [8].

For the third-degree Fermat difference equation (3.22), the solution f satisfies \(f^{-1}=H(\varphi )\), \({\overline{f}}=\eta G(\varphi )\), where HG are defined in (3.15) and \(\varphi \) is an entire function. Choose \(\eta =1\). It follows that

$$\begin{aligned} \frac{1+\wp '({\overline{\varphi }})/\sqrt{3}}{2\wp ({\overline{\varphi }})}\cdot \frac{1-\wp '(\varphi )/\sqrt{3}}{2\wp (\varphi )}=1. \end{aligned}$$

Using the addition law of the Weierstrass elliptic function together with the relation \(\wp '^2=4\wp ^3-1\), it can be shown that this equation is solved by a polynomial of degree 1 satisfying \({\overline{\varphi }}=\varphi +a\), where a is a constant such that \(\wp '(a)=-\sqrt{3}\) and \(\wp (a)=1\). It follows that the order of growth of f is 2.

The following three sections contain the proof of Theorem 2. In Sect. 4, we will first find some restrictions on the roots and degrees of the numerator and denominator of R(zf). This allows us to only consider two cases of Eq. (3.2) where \(p=n,q=0\) or \(p=q=n\) after a possible bilinear transformation to f. These two cases will be discussed in Sects. 5 and 6, respectively. The results obtained in the case \(p=n,q=0\) in Sect. 5 are the admissible counterparts of the results on (3.1) by Yanagihara [30].

4 Restrictions on the Roots and Degrees of (3.2)

We define

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

where

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

and

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

are polynomials in f having no common factors and \(p,q\in {\mathbb {N}}\). Then we have \(\deg _f (P(z,f))=p\), \(\deg _f (Q(z,f))=q\), and by assumption \(\deg _f (R(z,f))=\max \{p,q\}=n\). In what follows we also write

$$\begin{aligned} P(z,f) = a_p(f-\alpha _1)^{k_1}\cdots (f-\alpha _\mu )^{k_\mu } \end{aligned}$$
(4.1)

and

$$\begin{aligned} Q(z,f) = b_q(f-\beta _1)^{l_1}\cdots (f-\beta _\nu )^{l_\nu }, \end{aligned}$$
(4.2)

where the coefficients \(\alpha _1,\ldots ,\alpha _\mu \) and \(\beta _1,\ldots ,\beta _\nu \) are in general algebroid functions and \(k_i\) and \(l_j\) denote the order of the roots \(\alpha _i\) and \(\beta _j\), respectively. Without losing generality, we may suppose that the greatest common divisor of \(k_1,\ldots ,k_\mu ,l_1,\ldots ,l_\mu \), which is denoted by \(k=(k_1,\ldots ,k_\mu ,l_1,\ldots ,l_\mu )\), is 1. Otherwise, after taking the k-th root on both sides of (3.2), Eq. (3.2) reduces to a difference equation of degree n / k with meromorphic coefficients. Note that under this assumption if \(n\ge 2\), then either P(zf) or Q(zf) has at least two distinct roots when \(p=0\) or \(q=0\) or \(p=q=n\).

Let now f be an admissible meromorphic solution of (3.2). For the simple case \(n=1\), (3.2) is the linear difference equation (3.3) or the difference Riccati equation (3.4). From now on, we take n to be \(\ge 2\). By making use of the factorizations (4.1) and (4.2), it follows that the roots of P(zf) and Q(zf) are \(\alpha _1,\ldots ,\alpha _\mu \) and \(\beta _1,\ldots ,\beta _\nu \), respectively. Suppose that \(z_0\in {\mathbb {C}}\) is such that

$$\begin{aligned} f(z_0)-\alpha _i(z_0)=0 \end{aligned}$$

with multiplicity \(m\in {\mathbb {Z}}^+\). Now, \(n|mk_i\) since otherwise \(z_0+1\) would be an algebraic branch point of f. Hence, \(n\le mk_i\) and so \(m \ge n/k_i\). The same inequality holds for the roots \(\beta _j\), as well. Therefore,

$$\begin{aligned} {\overline{N}}\left( r,\frac{1}{f-\alpha _i}\right) \le \frac{k_i}{n} N\left( r,\frac{1}{f-\alpha _i}\right) \end{aligned}$$
(4.3)

for all \(i=1,\ldots ,\mu \) and

$$\begin{aligned} {\overline{N}}\left( r,\frac{1}{f-\beta _j}\right) \le \frac{l_j}{n} N\left( r,\frac{1}{f-\beta _j}\right) \end{aligned}$$
(4.4)

for all \(i=1,\ldots ,\nu \). This is trivially true if there is no such \(z_0\). Note that the above two inequalities (4.3) and (4.4) still hold true for \({\overline{f}}-{\overline{\alpha }}_i\) and \({\overline{f}}-{\overline{\beta }}_j\), respectively. Now,

$$\begin{aligned} k_1+\cdots +k_\mu \le n, \quad l_1+\cdots +l_\nu \le n, \end{aligned}$$

and so by applying the Second Main Theorem (2.3) together with (4.3) and (4.4), we have, for any \(\varepsilon > 0\),

$$\begin{aligned} \begin{aligned} (\mu +\nu -2-\varepsilon ) T(r,f)&\le \sum _{i=1}^\mu {\overline{N}}\left( r,\frac{1}{f-\alpha _i}\right) + \sum _{j=1}^\nu {\overline{N}}\left( r,\frac{1}{f-\beta _j}\right) \\&\le \sum _{i=1}^\mu \frac{k_i}{n} N\left( r,\frac{1}{f-\alpha _i}\right) + \sum _{j=1}^\nu \frac{l_j}{n} N\left( r,\frac{1}{f-\beta _j}\right) \\&\le \frac{1}{n}\left( \sum _{i=1}^\mu k_i + \sum _{j=1}^\nu l_j\right) T(r,f) \\&\le 2T(r,f), \end{aligned} \end{aligned}$$

which implies that \(\mu +\nu \le 4\) and therefore the combined number of distinct roots of P(zf) and Q(zf) is at most 4. In particular, if \(k_1+\cdots +k_\mu < n\) or \(l_1+\cdots +l_\nu <n\), then \(\mu +\nu \le 3\).

Consider first the case \(p=n\) and \(1\le q \le n-1\). We show that this case cannot occur. Suppose that \(N(r,f)\not =S(r,f)\). Then there are more than S(rf) points \(z_0\in {\mathbb {C}}\) such that

$$\begin{aligned} f(z)= C(z-z_0)^{-m} + O\left( (z-z_0)^{-m+1}\right) ,\quad C\not =0, \quad m\in {\mathbb {Z}}^+, \end{aligned}$$

in a neighborhood of \(z_0\). Let these poles be our starting points for iteration. Note that \(T(r,f(z+j))=T(r,f)+S(r,f)\), \(j\in {\mathbb {N}}\), and all the coefficients of (3.2) are in \({\mathcal {S}}'(f)\). It follows by [7, Lemma 3.1] that, for an arbitrarily small \(\varepsilon \ge 0\), there are more than S(rf) points \(z_0\in {\mathbb {C}}\) at which \(f(z+1)^n\) has a pole of order at most \((1+\varepsilon )(n-q)m\) and so there are more than S(rf) poles of f of order at most \((1+\varepsilon )(n-q)m/n\) at \(z=z_0+1\). By continuing the above iteration it follows that there are more than S(rf) poles of f of order at most \((1+\varepsilon )^2(n-q)^2m/n^2\) at \(z=z_0+2\), and more than S(rf) poles of f of order at most \((1+\varepsilon )^s(n-q)^{s}m/n^{s}\) at \(z=z_0+s\), \(s\in {\mathbb {N}}\). By letting \(s\rightarrow \infty \), it follows by induction 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, we have \(N(r,f)=S(r,f)\). Also, from (3.2) we see that \(N(r,1/(f-\beta _j))=S(r,f)\). Thus \(\infty ,\beta _j\), \(j=1,\ldots ,\nu \), are all Picard exceptional small functions of f. By the First Main Theorem, we conclude from (4.3) and (2.4) that \(\mu =\nu =1\). It follows that (3.2) assumes the following form:

$$\begin{aligned} {\overline{f}}^n=\frac{c(f-\alpha _1)^n}{(f-\beta _1)^q}, \end{aligned}$$
(4.5)

where \(\alpha _1,\beta _1\) are meromorphic functions. Moreover, \(\beta _1\not \equiv 0\) since otherwise \(\alpha _1\) is also a Picard exceptional small function of f, which is impossible. Set \(c_1=c^{1/n}\) and \(g=(f-\beta _1)^{1/n}\). Then \(c^{1/n},g\) are both algebroid functions of degree at most n, and from the above reasoning we have \(N(r,g)=S(r,f)\) and \(N(r,1/g)=S(r,f)\). We take the n-th root on both sides of (4.5) and then rewrite the resulting equation as follows:

$$\begin{aligned} {\overline{g}}^ng^q=c_1g^n-{\overline{\beta }}_1g^q+c_1(\beta _1-\alpha _1). \end{aligned}$$
(4.6)

Let u be a function such that

$$\begin{aligned} c_1u^n-{\overline{\beta }}_1u^{q}+c_1(\beta _1-\alpha _1)=0. \end{aligned}$$

Then u is an algebroid function of degree at most \(n^2\). Since \(q\le n-1\) and \(\beta _1\not \equiv \alpha _1\), the above equation has at least one nonzero root \(u_0\) and from (4.6) we see that \(N(r,1/(g-u_0))=S(r,g)\). Since \(T(r,f)=nT(r,g)+S(r,f)\), then \(u_0^n+\beta _1\) is also a Picard exceptional small function of f, which is impossible. Therefore, the case where \(p=n\) and \(1\le q \le n-1\) cannot occur. This also implies that the case where 0 is a root of P(zf) of order less than n cannot occur since otherwise by doing a bilinear transformation \(f\rightarrow 1/f\) to (3.2), we get

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

where \(P_1(z,f)\) is a polynomial in f of degree n and \(Q_1(z,f)\) is a polynomial in f of degree less than n, which is impossible. From the above reasoning, we conclude that \(q=0\) or \(q=n\) when \(p=n\) and that 0 is not a root of P(zf) of order less than n.

Consider now the case \(q=n\) and \(0\le p\le n-1\). Below we show that in this case 0 cannot be a root of Q(zf) of order less than n. Otherwise, (3.2) can be written as

$$\begin{aligned} {\overline{f}}^n=\frac{P(z,f)}{f^{l_0}{\hat{Q}}(z,f)}, \end{aligned}$$
(4.7)

where \(1\le l_0\le n-1\), and \({\hat{Q}}(z,f)\) is a polynomial in f of degree \(n-l_0\). If \(N(r,1/f)\not =S(r,f)\), then there are more than S(rf) points \(z_0\in {\mathbb {C}}\) such that

$$\begin{aligned} f(z)= C(z-z_0)^{m} + O\left( (z-z_0)^{m+1}\right) ,\quad C\not =0, \quad m\in {\mathbb {N}}^+ \end{aligned}$$

in a neighborhood of \(z_0\). We now iterate (3.2) with these points as our starting points. Recall that \(T(r,f(z+j))=T(r,f)+S(r,f)\) for all \(j\in {\mathbb {N}}\) and all the coefficients of (3.2) are in \({\mathcal {S}}'(f)\). According to [7, Lemma 3.1], for an arbitrarily small \(\varepsilon \ge 0\), by (4.7), there are more than S(rf) points \(z_0\in {\mathbb {C}}\) at which \(f(z+1)^n\) has a pole of order at most \((1+\varepsilon )l_0m\) and so there are more than S(rf) poles of f of order at most \((1+\varepsilon )l_0m/n\) at \(z=z_0+1\). It follows that there are more than S(rf) points \(z_0\in {\mathbb {C}}\) at which \(f(z+2)^n\) has a zero of order at most \((1+\varepsilon )l_0(n-p)m\), that is, there are more than S(rf) zeros of f of order at most \((1+\varepsilon )l_0(n-p)m/n\) at \(z=z_0+2\). Then, by continuing the iteration it follows that there are more than S(rf) zeros of f of order \((1+\varepsilon )^s(n-p)^sl_0^{s}m/n^{s}\) at \(z=z_0+2s\), \(s\in {\mathbb {N}}\). By letting \(s\rightarrow \infty \), it follows by induction 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, \(N(r,1/f)=S(r,f)\). Also, from (4.7) we see that \(N(r,f)=S(r,f)\) since \(p<q\) and then, since \(l_0\le n-1\), it follows that there exists another nonzero \(\beta _j\) such that \(N(r,1/(f-\beta _j)=S(r,f)\), that is to say, f has at least 3 Picard exceptional small functions, which is impossible. Therefore, 0 cannot be a root of Q(zf) of order \(<n\) when \(0\le p\le n-1\). Now, since 0 is not a root of P(zf) either, by doing a bilinear transformation \(f\rightarrow 1/f\) to (3.2), we get

$$\begin{aligned} {\overline{f}}^n=\frac{P_2(z,f)}{Q_2(z,f)}, \end{aligned}$$

where \(P_2(z,f)\) and \(Q_2(z,f)\) are two polynomials in f, both of them having degree equal to n. In particular, \(Q_2(z,f)=f^n\) when \(p=0\).

From the above discussions, we conclude that we only need to consider (3.2) for two cases where \(p=n\), \(q=0\), or \(p=q=n\). Moreover, if P(zf) has two or more distinct roots, then none of them vanishes identically.

5 Equation (3.2) with \(p=n\) and \(q=0\)

In Sect. 4, we have shown that the combined number of distinct roots of P(zf) is at most 3 when \(p=n\) and \(q=0\) and assuming that P(zf) has at least 2 distinct roots. Hence we have the following two possibilities:

$$\begin{aligned} {\overline{f}}^n= & {} c(f-\alpha _1)^{\kappa _1}(f-\alpha _2)^{\kappa _2}, \quad \alpha _1\alpha _2\not \equiv 0, \end{aligned}$$
(5.1)
$$\begin{aligned} {\overline{f}}^n= & {} c(f-\alpha _1)^{\mu _1}(f-\alpha _2)^{\mu _2}(f-\alpha _3)^{\mu _3}, \quad \alpha _1\alpha _2\alpha _3\not \equiv 0, \end{aligned}$$
(5.2)

where \(\kappa _i\), \(\mu _j\) are positive integers satisfying \(\kappa _1+\kappa _2=n\), \(\mu _1+\mu _2+\mu _3=n\).

We first look at Eq. (5.1). Note that \(\alpha _1,\alpha _2\) are both meromorphic functions when \(n\ge 3\) and are algebroid functions of degree at most 2 when \(n=2\). Suppose that \(\kappa _1+\kappa _2=n\ge 3\). Since \((\kappa _1,\kappa _2)=(n,\kappa _2)=(n,\kappa _2)=1\), from (5.1) we see that the multiplicity of zero \(z_0\) of \(f-\alpha _1\) (also of \(f-\alpha _2\)) is at least n since otherwise \(z_0\) would be an algebraic branch point of f. Thus we have

$$\begin{aligned} {\overline{N}}\left( r,\frac{1}{f-\alpha _i}\right) \le \frac{1}{n} N\left( r,\frac{1}{f-\alpha _i}\right) , \quad i=1,2. \end{aligned}$$

If \(n=3\) or \(n=4\), then obviously \(\min \{\kappa _1,\kappa _2\}=1\); if \(n\ge 5\) and \(\min \{\kappa _1,\kappa _2\}\ge 2\), then from (5.1) we see that f is also completely ramified at zeros, which, with the First Main Theorem and (2.4), yields that \(2-2/n+1/2\le 2\), a contradiction. Hence we must have \(\min \{\kappa _1,\kappa _2\}=1\), and it follows that (5.1) is of the form

$$\begin{aligned} {\overline{f}}^{n} = c(f-\alpha _1)^{n-1}(f-\alpha _2). \end{aligned}$$
(5.3)

Moreover, since \(n\ge 3\), we have \(N(r,1/(f-\alpha _2))\not =S(r,f)\), for otherwise f is n-fold ramified at \(\alpha _1\)-points and \((n-1)\)-fold ramified at zeros, which yields a contradiction to (2.4). Write \(c_1=[c(\alpha _2-\alpha _1)]^{1/n}\) and \(c_2=1/(\alpha _2-\alpha _1)\). Then \(c_1\) is an algebroid function of degree at most n. Put

$$\begin{aligned} c_1u=\frac{{\overline{f}}}{f-\alpha _1}, \quad v=\frac{1}{f-\alpha _1}, \end{aligned}$$
(5.4)

where u is an algebroid function with at most S(rf) many branch points. By the Valiron–Mohon’ko identity [15, 26] (see also [13]) we have \(T(r,v)=T(r,f)+S(r,f)\), and thus all the coefficients of (5.3) are small functions of v. Then we have

$$\begin{aligned} f=\frac{1}{v}+\alpha _1, \quad {\overline{f}}=\frac{c_1u}{v}, \end{aligned}$$

and from (5.3) and (5.4) that

$$\begin{aligned} u^{n}=c_2-v. \end{aligned}$$
(5.5)

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\alpha }}_2v}{v}={\overline{f}}-{\overline{\alpha }}_2. \end{aligned}$$
(5.6)

Since zeros of \({\overline{f}}-{\overline{\alpha }}_2\) are n-fold ramified, we see that zeros of \(F=c_1u-{\overline{\alpha }}_2v\) are also n-fold ramified. Let \(u_0\) and \(v_0\) be two functions such that

$$\begin{aligned} u_0^{n}=c_2-v_0, \quad c_1u_0-{\overline{\alpha }}_2v_0=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives

$$\begin{aligned} {\overline{\alpha }}_2^{n}v_0^{n}+c_1^{n}v_0-c_1^{n}c_2=0, \end{aligned}$$
(5.7)

from which we see that \(v_0\) is an algebroid function of degree at most n. Moreover, it is easy to see that \(c_2\) is not a root of (5.7). If (5.7) has only one root, then, since \(n\ge 3\), by Vieta’s formulas we see that \(v_0=0\) and it follows from (5.7) that \(c_1^{n}c_2\equiv 0\), a contradiction. Therefore, (5.7) has at least two distinct roots, say, \(v_1\) and \(v_2\). Rewrite (5.6) as

$$\begin{aligned} -\frac{{\overline{\alpha }}_2^{n}v^n+c_1^nv-c_1^nc_2}{v^n}= {\overline{f}}^n-{\overline{\alpha }}_2^{n}. \end{aligned}$$
(5.8)

Denote the order of the roots \(v_1\) and \(v_2\) by \(l_1\) and \(l_2\), respectively. Then \(1\le l_1,l_2\le n-1\), and we have

$$\begin{aligned} {\overline{\alpha }}_2^{n}v^n+c_1^nv-c_1^nc_2= {\overline{\alpha }}_2^{n}(v-v_1)^{l_1}(v-v_2)^{l_2}P(z,v), \end{aligned}$$

where P(zv) is a polynomial in v of degree \(n-l_1-l_2\) with small functions of v as coefficients. Suppose that \(z_0\in {\mathbb {C}}\) is such that

$$\begin{aligned} f(z_0+1)-\alpha _2(z_0+1)=0 \end{aligned}$$

with multiplicity \(n_0\ge {n}\). If \(v(z_0)-v_1(z_0)=0\) with multiplicity \(m\in {{\mathbb {Z}}^+}\), then from (5.8), we have \(n_0|ml_1\). Hence, \(n_0\le ml_1\) and so \(m \ge 2\). Therefore,

$$\begin{aligned} {\overline{N}}\left( r,\frac{1}{v-v_1}\right) \le \frac{1}{2} N\left( r,\frac{1}{v-v_1}\right) . \end{aligned}$$

The same inequality holds for \(v_2\), as well. Thus v is completely ramified at \(v_1\)-points and \(v_2\)-points. But from (5.4) we see that v is n-fold ramified at poles and \(c_2\)-points, which yields a contradiction to (2.4) since \(n\ge 3\). Hence \(n=2\) and we therefore have

$$\begin{aligned} {\overline{f}}^{2} = c(f-\alpha _1)(f-\alpha _2). \end{aligned}$$
(5.9)

In what follows, we will still use the expressions in (5.4) to transform (5.9) into (5.5) for the case \(n=2\). Moreover, from the above reasoning we know that

$$\begin{aligned} {\overline{\alpha }}_2^{2}v_0^{2}+c_1^{2}v_0-c_1^{2}c_2=0, \end{aligned}$$

which has two distinct roots if

$$\begin{aligned} (4{\overline{\alpha }}_2^2c_2+c_1^2)c_1^2\not =0. \end{aligned}$$
(5.10)

If (5.10) holds, then v must be completely ramified at poles, \(c_2\)-points, \(v_1\)-points, and \(v_2\)-points; if not, then we have

$$\begin{aligned} c_1^2+4{\overline{\alpha }}_2^2c_2=\frac{4{\overline{\alpha }}_2^2 +c(\alpha _2-\alpha _1)^2}{\alpha _2-\alpha _1}=0. \end{aligned}$$
(5.11)

Rewrite (5.9) as

$$\begin{aligned} {\overline{f}}^{2} = c(g^2-c_4^2), \end{aligned}$$

where \(g=f-c_3\), \(c_3=(\alpha _1+\alpha _2)/2\) are both meromorphic functions and \(c_4=(\alpha _2-\alpha _1)/2=1/(2c_2)\) is an algebroid function of degree at most 2. Note that \(c_4^2=[(\alpha _2+\alpha _1)^2-4\alpha _1\alpha _2]/4\) is a meromorphic function. Put

$$\begin{aligned} g=\frac{c_4}{2}\left( \beta +\frac{1}{\beta }\right) . \end{aligned}$$

Then substitution yields

$$\begin{aligned} {\overline{f}}^{2} =\frac{cc_4^2}{4}\left( \beta -\frac{1}{\beta }\right) ^2, \end{aligned}$$

and, without loss of generality, we may suppose that

$$\begin{aligned} \beta ^2-\frac{2}{c^{1/2}c_4}\left( {\overline{\alpha }}_1 +\frac{1}{{\overline{v}}}\right) \beta -1=0. \end{aligned}$$
(5.12)

Note that \(4c_2^2c_4^2=1\) and \(c=c_1^2c_2\). We have \(4cc_4^2=c_1^2/c_2\). The discriminant of the quadratic equation (5.12) with respect to \(\beta \),

$$\begin{aligned} \frac{4({\overline{\alpha }}_1{\overline{v}}+1)^2+4cc_4^2 {\overline{v}}^2}{cc_4^2{\overline{v}}^2}=\frac{4c_2 ({\overline{\alpha }}_1{\overline{v}}+1)^2+c_1^2 {\overline{v}}^2}{c_2cc_4^2{\overline{v}}^2}= \frac{4{\overline{f}}^2}{cc_4^2}+4=\frac{4(cc_4^2+ {\overline{f}}^2)}{cc_4^2}=\frac{4g^2}{c_4^2}, \end{aligned}$$

implies that \(\beta \) is meromorphic apart from at most S(rf) many points. Moreover, roots of the equation

$$\begin{aligned} (4{\overline{\alpha }}_1^2c_2+c_1^2){\overline{v}}^2+8c_2 {\overline{\alpha }}_1{\overline{v}}+4c_2=0 \end{aligned}$$
(5.13)

with respect to \({\overline{v}}\) must be either \(c_2\) or coincide with \(v_1,v_2\).

If \(c_2\) is a root of (5.13), then we have

$$\begin{aligned} \frac{4{\overline{\alpha }}_1^2c_2+c_1^2}{({\overline{\alpha }}_2 -{\overline{\alpha }}_1)^2}+\frac{8c_2{\overline{\alpha }}_1}{{\overline{\alpha }}_2-{\overline{\alpha }}_1}+4c_2=0, \end{aligned}$$

which yields (5.11). On the other hand, by exchanging the roles of \(\alpha _1\) and \(\alpha _2\), we also have

$$\begin{aligned} c_1^2+4{\overline{\alpha }}_1^2c_2=0. \end{aligned}$$
(5.14)

From (5.11) and (5.14) we see that \(\alpha _2=-\alpha _1\), and it follows that \(c=-{\overline{\alpha }}_1^2/\alpha _1^2\). By doing the transformation \(f\rightarrow \alpha _1f\), (5.9) becomes

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

which is the Eq. (3.5).

If \(c_2\) is not a root of (5.13), then roots of (5.13) must coincide with \(v_1,v_2\), which means that

$$\begin{aligned} \frac{\overline{{\overline{\alpha }}}_2^{2}}{4{\overline{\alpha }}_1^2c_2 +c_1^2}=\frac{{\overline{c}}_1^{2}}{8c_2{\overline{\alpha }}_1} =\frac{-{\overline{c}}_1^{2}{\overline{c}}_2}{4c_2}. \end{aligned}$$
(5.15)

Recall that \(c_1=[c(\alpha _2-\alpha _1)]^{1/2}\) and \(c_2=1/(\alpha _2-\alpha _1)\). The second equation of (5.15) gives \(-2{\overline{c}}_2{\overline{\alpha }}_1=1\), i.e., \({\overline{\alpha }}_2=-{\overline{\alpha }}_1\). It follows from the first equation of (5.15) that

$$\begin{aligned} \alpha _1^2{\overline{c}}c+{\overline{\alpha }}_1^2{\overline{c}} +\overline{{\overline{\alpha }}}_1^2=0. \end{aligned}$$

Set \(d=c\alpha _1^2/{\overline{\alpha }}_1^2\). Then d is a meromorphic function satisfying \({\overline{d}}(d+1)+1=0\). By doing the transformation \(f\rightarrow \alpha _1f\), (5.9) becomes

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

which is the Eq. (3.6).

We now look at Eq. (5.2). Set \(k_1=(n,\mu _1)\), \(k_2=(n,\mu _2)\), \(k_3=(n,\mu _3)\), and let \(K_1=n/k_1\), \(K_2=n/k_2\), \(K_3=n/k_3\). Obviously, \(K_i\ge 2\), \(i=1,2,3\), and \(1/K_1+1/K_2+1/K_3=(k_1+k_2+k_3)/n\le 1\). Now f is \(K_i\)-fold ramified at \(\alpha _i\)-points, \(i=1,2,3\), which with (2.4) implies that

$$\begin{aligned} 1-\frac{1}{K_1}+1-\frac{1}{K_2}+1-\frac{1}{K_3}\le 2, \end{aligned}$$

and so \(1/K_1+1/K_2+1/K_3=(k_1+k_2+k_3)/n=1\). Thus we have the following three possibilities:

$$\begin{aligned} \begin{aligned} 1, \quad&K_1=2, K_2=3, K_3=6;\\ 2, \quad&K_1=2, K_2=4, K_3=4;\\ 3, \quad&K_1=3, K_2=3, K_3=3; \end{aligned} \end{aligned}$$

which implies that (5.2) assumes one of the following three forms:

$$\begin{aligned} {\overline{f}}^3= & {} c(f-\alpha _1)(f-\alpha _2)(f-\alpha _3), \quad \alpha _1\alpha _2\alpha _3\not \equiv 0, \end{aligned}$$
(5.16)
$$\begin{aligned} {\overline{f}}^4= & {} c(f-\alpha _1)^{2}(f-\alpha _2)(f-\alpha _3), \quad \alpha _1\alpha _2\alpha _3\not \equiv 0, \end{aligned}$$
(5.17)
$$\begin{aligned} {\overline{f}}^6= & {} c(f-\alpha _1)^{3}(f-\alpha _2)^{2}(f-\alpha _3), \quad \alpha _1\alpha _2\alpha _3\not \equiv 0. \end{aligned}$$
(5.18)

Note that \(\alpha _1\) in (5.17) and \(\alpha _1,\alpha _2,\alpha _3\) in (5.18) are all meromorphic functions.

Let’s first consider Eq. (5.16). Put

$$\begin{aligned} \frac{{\overline{f}}}{f-\alpha _1}=c_1u, \quad \frac{1}{f-\alpha _1}=b_2v+b_1, \end{aligned}$$
(5.19)

where u and v are two algebroid functions with at most S(rf) many branch points and

$$\begin{aligned} \begin{aligned} c_1&= [4cb_2^2(\alpha _2-\alpha _1)(\alpha _3-\alpha _1)]^{1/3},\\ b_1&=\frac{1}{2}\left( \frac{1}{\alpha _2-\alpha _1}+\frac{1}{\alpha _3-\alpha _1}\right) ,\\ b_2&=\frac{1}{2}\left( \frac{1}{\alpha _2-\alpha _1}-\frac{1}{\alpha _3-\alpha _1}\right) \end{aligned} \end{aligned}$$
(5.20)

are algebroid functions of degrees at most 9, 3, 3, respectively. By writing v in terms of f with small coefficients and the Valiron–Mohon’ko identity [15, 26] (see also [13]), we have \(T(r,v)=T(r,f)+S(r,f)\) and thus all the coefficients of (5.16) are small functions of v. Then we have

$$\begin{aligned} f=\alpha _1+\frac{1}{b_2v+b_1}, \quad {\overline{f}}=\frac{c_1u}{b_2v+b_1}, \end{aligned}$$

and from (5.16) and (5.19) that

$$\begin{aligned} v^2=4u^3+1. \end{aligned}$$
(5.21)

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\alpha }}_2(b_2v+b_1)}{b_2v+b_1} ={\overline{f}}-{\overline{\alpha }}_2. \end{aligned}$$
(5.22)

Since zeros of \({\overline{f}}-{\overline{\alpha }}_2\) are threefold ramified, we see that zeros of \(F=c_1u-{\overline{\alpha }}_2(b_2v+b_1)\) are also threefold ramified. Let \(u_0\) and \(v_0\) be two functions such that

$$\begin{aligned} v_0^2=4u_0^3+1, \quad c_1u_0-{\overline{\alpha }}_2(b_2v_0+b_1)=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives

$$\begin{aligned} 4{\overline{\alpha }}_2^3(b_2v_0+b_1)^3-c_1^3(v_0^2-1)=0, \end{aligned}$$
(5.23)

from which we see that \(v_0\) is an algebroid function of degree at most 27. Moreover, it is seen that \(\pm \,1\) are not roots of (5.23), for otherwise we have \(b_2{\pm \,}b_1=0\), a contradiction. Rewrite (5.22) as

$$\begin{aligned} -\frac{4{\overline{\alpha }}_2^3(b_2v+b_1)^3-c_1^3(v^2-1)}{4(b_2v+b_1)^3}={\overline{f}}^3-{\overline{\alpha }}_2^3. \end{aligned}$$
(5.24)

From Eq. (5.21), we see that v is threefold ramified over \(\pm \,1,\infty \). This implies that (5.23) can admit only one root, for otherwise by a similar analysis as to (5.8) we obtain from (5.24) that (5.23) has at least two roots, say, \(v_1\) and \(v_2\), distinct from \(\pm \,1,\infty \) such that v is completely ramified at \(v_1\)-points and \(v_2\)-points, which is impossible by Theorem 1. Expanding (5.23) gives

$$\begin{aligned} 4{\overline{\alpha }}_2^3b_2^3v_0^3+(12{\overline{\alpha }}_2^3b_2^2b_1 -c_1^3)v_0^2+12{\overline{\alpha }}_2^3b_2b_1^2v_0 +4{\overline{\alpha }}_2^3b_1^3+c_1^3=0, \end{aligned}$$

and so by Vieta’s formulas, we get

$$\begin{aligned} \frac{12{\overline{\alpha }}_2^3b_2^2b_1-c_1^3}{ 4{\overline{\alpha }}_2^3b_2^3}=-3v_0, \quad \frac{3b_1^2}{b_2^2}=3v_0^2. \end{aligned}$$
(5.25)

From the second equation of (5.25) we get \(v_0=\pm \, b_1/b_2\). If \(v_0=-b_1/b_2\), then substitution into (5.23) yields \(b_1^2=b_2^2\), a contradiction. Hence \(v_0=b_1/b_2\). Combining this equation with the first equation of (5.25), we get

$$\begin{aligned} c_1^3=24{\overline{\alpha }}_2^3b_2^2b_1. \end{aligned}$$

Similarly, the above equation also holds when replacing \({\overline{\alpha }}_2\) in (5.22) with \({\overline{\alpha }}_1\) and \({\overline{\alpha }}_3\), which implies that \(\alpha _1^3=\alpha _2^3=\alpha _3^3\). Thus, \(\alpha _2=\eta \alpha _1\), \(\alpha _3=\eta ^2\alpha _1\), where \(\eta \) is the cubic root of 1 satisfying \(\eta ^2+\eta +1=0\). Moreover, we have

$$\begin{aligned} c_1^3=4cb_2^2(\alpha _2-\alpha _1)(\alpha _3-\alpha _1) =24{\overline{\alpha }}_1^3b_2^2b_1, \end{aligned}$$

which yields \(c=-{\overline{\alpha }}_1^3/\alpha _1^3\). By doing the transformation \(f\rightarrow \alpha _1f\), (5.16) becomes

$$\begin{aligned} {\overline{f}}^3=1-f^3, \end{aligned}$$

which is the Eq. (3.13).

We now consider Eq. (5.17). Put

$$\begin{aligned} \frac{{\overline{f}}}{f-\alpha _1}=c_1u, \quad \frac{1}{f-\alpha _1}=b_2v+b_1, \end{aligned}$$
(5.26)

where u and v are two algebroid functions with at most S(rf) many branch points and

$$\begin{aligned} \begin{aligned} c_1&=[-b_2^2c(\alpha _2-\alpha _1)(\alpha _3-\alpha _1)]^{1/4},\\ b_1&=\frac{1}{2}\left( \frac{1}{\alpha _2-\alpha _1}+\frac{1}{\alpha _3-\alpha _1}\right) ,\\ b_2&=\frac{1}{2}\left( \frac{1}{\alpha _2-\alpha _1}-\frac{1}{\alpha _3-\alpha _1}\right) \end{aligned} \end{aligned}$$
(5.27)

are algebroid functions of degrees at most 8, 2, 2, respectively. By writing v in terms of f with small coefficients and the Valiron–Mohon’ko identity [15, 26] (see also [13]), we have \(T(r,v)=T(r,f)+S(r,f)\) and thus all the coefficients of (5.17) are small functions of v. Then we have

$$\begin{aligned} f=\frac{1}{b_2v+b_1}+\alpha _1, \quad {\overline{f}}=\frac{c_1u}{b_2v+b_1} \end{aligned}$$

and from (5.17) and (5.26) that

$$\begin{aligned} v^2=1-u^4. \end{aligned}$$
(5.28)

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\alpha }}_2(b_2v+b_1)}{b_2v+b_1} ={\overline{f}}-{\overline{\alpha }}_2. \end{aligned}$$
(5.29)

Since zeros of \({\overline{f}}-{\overline{\alpha }}_2\) are fourfold ramified, we see that zeros of \(F=c_1u-{\overline{\alpha }}_2(b_2v+b_1)\) are also fourfold ramified. Let \(u_0\) and \(v_0\) be two functions such that

$$\begin{aligned} v_0^2=1-u_0^4, \quad c_1u_0-{\overline{\alpha }}_2b_2v_0-{\overline{\alpha }}_2b_1=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives

$$\begin{aligned} {\overline{\alpha }}_2^4(b_2v_0+b_1)^4-c_1^4(1-v_0^2)=0, \end{aligned}$$
(5.30)

from which we see that \(v_0\) is an algebroid function of degree at most 16. Moreover, it is seen that \(\pm \,1\) are not roots of (5.30), for otherwise we have \(b_2{\pm }b_1=0\), a contradiction. Rewrite (5.29) as

$$\begin{aligned} \frac{c_1^4(1-v^2)-{\overline{\alpha }}_2^4(b_2v+b_1)^4}{(b_2v+b_1)^4}={\overline{f}}^4-{\overline{\alpha }}_2^4. \end{aligned}$$
(5.31)

From Eq. (5.28), we see that v is completely ramified over \(\pm \,1,\infty \). This implies that (5.30) can admit only one root, for otherwise by a similar analysis as to (5.8) we obtain from (5.31) that (5.30) has at least two roots, say, \(v_1\) and \(v_2\), distinct from \(\pm \,1,\infty \) such that v is completely ramified at \(v_1\)-points and \(v_2\)-points, which is impossible by Theorem 1. Expanding (5.30) gives

$$\begin{aligned} {\overline{\alpha }}_2^4(b_2^4v_0^4+4b_2^3b_1v_0^3+\ldots +b_1^4) -c_1^4(1-v_0^2)=0, \end{aligned}$$

and so by Vieta’s formulas, we obtain \(4b_1/b_2=-4v_0\), which gives \(b_2v_0+b_1=0\) and it follows from (5.30) that \(v_0^2=1\), a contradiction. Therefore, (5.17) cannot admit any meromorphic solutions.

Finally, let’s consider Eq. (5.18). Put

$$\begin{aligned} \frac{{\overline{f}}}{f-\alpha _1}=c_1u, \quad \frac{1}{f-\alpha _1}=b_2v-b_1, \end{aligned}$$
(5.32)

where u is an algebroid function with at most S(rf) many branch points and

$$\begin{aligned} \begin{aligned} c_1&=[-cb_2^3(\alpha _2-\alpha _1)^2(\alpha _3-\alpha _1)]^{1/6},\\ b_1&=-\frac{1}{3}\left( \frac{2}{\alpha _2-\alpha _1} +\frac{1}{\alpha _3-\alpha _1}\right) ,\\ b_2&=\frac{1}{\alpha _2-\alpha _1}+b_1=\frac{1}{3} \left( \frac{1}{\alpha _2-\alpha _1}-\frac{1}{\alpha _3-\alpha _1}\right) , \end{aligned} \end{aligned}$$
(5.33)

and \(c_1\) is an algebroid function of degree at most 6. By writing v in terms of f with small coefficients and the Valiron–Mohon’ko identity [15, 26] (see also [13]), we have \(T(r,v)=T(r,f)+S(r,f)\), and thus all the coefficients of (5.18) are small functions of v. Then we have

$$\begin{aligned} f=\frac{1}{b_2v-b_1}+\alpha _1, \quad {\overline{f}}=\frac{c_1u}{b_2v-b_1} \end{aligned}$$

and from (5.18) and (5.32) that

$$\begin{aligned} u^6=(v-1)^2(v+2)=v^3-3v+2. \end{aligned}$$
(5.34)

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\alpha }}_3(b_2v-b_1)}{b_2v-b_1} ={\overline{f}}-{\overline{\alpha }}_3. \end{aligned}$$
(5.35)

Since zeros of \({\overline{f}}-{\overline{\alpha }}_3\) are sixfold ramified, we see that zeros of \(F=c_1u-{\overline{\alpha }}_3(b_2v-b_1)\) are also sixfold ramified. Let \(u_0\) and \(v_0\) be two functions such that

$$\begin{aligned} u_0^6=v_0^3-3v_0+2, \quad c_1u_0-{\overline{\alpha }}_3b_2v_0+{\overline{\alpha }}_3b_1=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives

$$\begin{aligned} {\overline{\alpha }}_2^6(b_2v_0-b_1)^6-c_1^6(v_0^3-3v_0+2)=0, \end{aligned}$$
(5.36)

from which we see that \(v_0\) is an algebroid function of degree at most 6. Moreover, it is seen that 1 and \(-2\) are not roots of (5.36), for otherwise we have \(b_2-b_1=0\) or \(2b_2+b_1=0\), a contradiction. Rewrite (5.35) as

$$\begin{aligned} \frac{c_1^6(v^3-3v+2)-{\overline{\alpha }}_3^6(b_2v-b_1)^6}{(b_2v-b_1)^6}={\overline{f}}^6-{\overline{\alpha }}_3^6. \end{aligned}$$
(5.37)

From Eq. (5.34), we see that v is completely ramified over \(1,-2,\infty \). This implies that (5.36) can admit only one root, for otherwise by a similar analysis as to (5.8) we obtain from (5.37) that (5.36) has at least two roots, say, \(v_1\) and \(v_2\), distinct from \(1,-2,\infty \) such that v is completely ramified at \(v_1\)-points and \(v_2\)-points, which is impossible by Theorem 1. Expanding (5.36) gives

$$\begin{aligned} {\overline{\alpha }}_2^6(b_2^6v_0^6-6b_2^5b_1v_0^5+\cdots +b_1^6) -c_1^6(v_0^3-3v_0+2)=0, \end{aligned}$$

and so by Vieta’s formulas, we obtain \(-6b_1/b_2=-6v_0\), which gives \(b_2v_0-b_1=0\) and it follows from (5.36) that \(v_0^3-3v_0+2=0\), a contradiction. Therefore, (5.18) cannot admit any meromorphic solutions.

6 Equation (3.2) with \(p=q=n\)

In this section, we consider the cases where Q(zf) has one, two, or three roots.

6.1 Q(zf) has Only One Root

In Sect. 4, we have shown that the combined number of distinct roots of P(zf) and Q(zf) is at most 4 and assumed that P(zf) has at least 2 distinct roots. Hence we have the following two possibilities:

$$\begin{aligned} {\overline{f}}^n= & {} \frac{c(f-\alpha _1)^{\kappa _1}(f-\alpha _2)^{\kappa _2}}{(f-\beta _1)^n}, \quad \alpha _1\alpha _2\not \equiv 0, \end{aligned}$$
(6.1)
$$\begin{aligned} {\overline{f}}^n= & {} \frac{c(f-\alpha _1)^{\mu _1}(f-\alpha _2)^{\mu _2}(f-\alpha _3)^{\mu _3}}{(f-\beta _1)^n}, \quad \alpha _1\alpha _2\alpha _3\not \equiv 0, \end{aligned}$$
(6.2)

where \(\kappa _i\), \(\mu _j\) are positive integers satisfying \(\kappa _1+\kappa _2=n\), \(\mu _1+\mu _2+\mu _3=n\).

We now look at Eq. (6.1). Note that \(\alpha _1,\alpha _2\) are both meromorphic functions when \(n\ge 3\) and are algebroid functions of degree at most 2 when \(n=2\). Suppose \(\kappa _1+\kappa _2=n\ge 3\). By a similar analysis on the \(\alpha _i\)-points of f in (6.1) as on the \(\alpha _i\)-points of f in (5.1) in Sect. 5, we can obtain that (6.1) is of the form

$$\begin{aligned} {\overline{f}}^{n} = \frac{c(f-\alpha _1)^{n-1}(f-\alpha _2)}{(f-\beta _1)^n}. \end{aligned}$$
(6.3)

Moreover, we have \(N(r,1/(f-\alpha _2))\not =S(r,f)\). Set \(c_1=[c(\alpha _2-\alpha _1)]^{1/n}\) and \(c_2=1/(\alpha _2-\alpha _1)\). Then \(c_1\) is an algebroid function of degree at most n. Put

$$\begin{aligned} c_1u=\frac{{\overline{f}}(f-\beta _1)}{f-\alpha _1}, \quad v=\frac{1}{f-\alpha _1}, \end{aligned}$$
(6.4)

where u is an algebroid function with at most S(rf) many branch points. Note that all the coefficients of (6.3) are small functions of v. Then we have

$$\begin{aligned} f =\frac{1}{v}+\alpha _1, \quad {\overline{f}}=\frac{c_1u}{1+(\alpha _1-\beta _1)v} \end{aligned}$$

and from (6.3) and (6.4) that

$$\begin{aligned} u^{n}=c_2-v. \end{aligned}$$
(6.5)

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\alpha }}_2[1+(\alpha _1-\beta _1)v]}{1+(\alpha _1-\beta _1)v}={\overline{f}}-{\overline{\alpha }}_2. \end{aligned}$$
(6.6)

Since zeros of \({\overline{f}}-{\overline{\alpha }}_2\) are n-fold ramified, we see that zeros of \(F=c_1u-{\overline{\alpha }}_2[(\alpha _1-\beta _1)v+1]\) are also n-fold ramified. Let \(u_0\) and \(v_0\) be two functions such that

$$\begin{aligned} u_0^{n}=c_2-v_0, \quad c_1u_0-{\overline{\alpha }}_2[(\alpha _1-\beta _1)v_0+1]=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives

$$\begin{aligned} {\overline{\alpha }}_2^{n}[(\alpha _1-\beta _1)v_0+1]^{n} +c_1^{n}v_0-c_1^{n}c_2=0, \end{aligned}$$

i.e.,

$$\begin{aligned} {\overline{\alpha }}_2^{n}[(\alpha _1-\beta _1)^nv_0^n+n(\alpha _1 -\beta _1)^{n-1}v_0^{n-1}+\ldots +1]+c_1^{n}v_0-c_1^{n}c_2=0, \end{aligned}$$
(6.7)

from which we see that \(v_0\) is an algebroid function of degree at most n. Moreover, it is seen that \(c_2\) is not a root of (6.7). If (6.7) has only one root, then, since \(n\ge 3\), by Vieta’s formulas, we see that \(v_0=-1/(\alpha _1-\beta _1)\) and it follows from (6.7) that \(-1/(\alpha _1-\beta _1)=c_2\), which gives \(\alpha _2=\beta _1\), a contradiction. Therefore, (6.7) has at least two distinct roots, say, \(v_1\) and \(v_2\). Rewrite (6.6) as

$$\begin{aligned} -\frac{{\overline{\alpha }}_2^n[1+(\alpha _1-\beta _1)v]^n +c_1^{n}v-c_1^{n}c_2}{[1+(\alpha _1-\beta _1)v]^n} ={\overline{f}}^n-{\overline{\alpha }}_2^n. \end{aligned}$$
(6.8)

By a similar analysis as to (5.8) in Sect. 5, we obtain from (6.8) that v is completely ramified at \(v_1\)-points and \(v_2\)-points. Moreover, from (6.4), we see that v is n-fold ramified at poles and \(c_2\)-points, which is impossible by (2.4) since \(n\ge 3\). Hence \(n=2\) and we therefore have

$$\begin{aligned} {\overline{f}}^{2} = \frac{c(f-\alpha _1)(f-\alpha _2)}{(f-\beta _1)^2}. \end{aligned}$$
(6.9)

In what follows, we will still use the expressions in (6.4) to transform (6.9) into (6.5) for the case \(n=2\). Moreover, we have

$$\begin{aligned} {\overline{\alpha }}_2^{2}[(\alpha _1-\beta _1)^2v_0^2+2(\alpha _1 -\beta _1)v_0+1]+c_1^{2}v_0-c_1^{2}c_2=0, \end{aligned}$$

which has two distinct roots if

$$\begin{aligned} \begin{aligned} {[}2{\overline{\alpha }}_2^{2}&(\alpha _1-\beta _1)+c_1^{2}]^2 -4{\overline{\alpha }}_2^{2}(\alpha _1-\beta _1)^2({\overline{\alpha }}_2^{2} -c_1^{2}c_2)\not \equiv 0. \end{aligned} \end{aligned}$$

Recalling that \(c_1=[c(\alpha _2-\alpha _1)]^{1/2}\) and \(c_2=1/(\alpha _2-\alpha _1)\), we have

$$\begin{aligned} \begin{aligned} {[}4{\overline{\alpha }}_2^{2}(\alpha _1-\beta _1)(\alpha _2-\beta _1) +c(\alpha _2-\alpha _1)^{2}]c_1^{2}c_2\not \equiv 0. \end{aligned} \end{aligned}$$
(6.10)

If (6.10) holds, then v must be completely ramified at poles, \(c_2\)-points, \(v_1\)-points, and \(v_2\)-points; if not, then we have

$$\begin{aligned} 4{\overline{\alpha }}_2^{2}(\alpha _1-\beta _1)(\alpha _2-\beta _1) +c(\alpha _2-\alpha _1)^{2}=0. \end{aligned}$$
(6.11)

Rewrite (6.9) as

$$\begin{aligned} {\overline{f}}^{2}(f-\beta _1)^2 = c(g^2-c_4^2), \end{aligned}$$
(6.12)

where \(g=f-c_3\), \(c_3=(\alpha _1+\alpha _2)/2\) are both meromorphic functions and \(c_4=(\alpha _2-\alpha _1)/2=1/(2c_2)\) is an algebroid function of degree at most 2. Moreover, \(c_4^2=[(\alpha _2+\alpha _1)^2-4\alpha _1\alpha _2]/4\) is a meromorphic function. Put

$$\begin{aligned} g=\frac{c_4}{2}\left( \beta +\frac{1}{\beta }\right) . \end{aligned}$$

Then substitution yields

$$\begin{aligned} {\overline{f}}^{2}(f-\beta _1)^2 =\frac{cc_4^2}{4}\left( \beta -\frac{1}{\beta }\right) ^2, \end{aligned}$$

and, without loss of generality, we may suppose that

$$\begin{aligned} \frac{c^{1/2}c_4}{2}\left( \beta -\frac{1}{\beta }\right) =\left( {\overline{\alpha }}_1+\frac{1}{{\overline{v}}}\right) \left[ \frac{c_4}{2}\left( \beta +\frac{1}{\beta }\right) +c_3-\beta _1\right] , \end{aligned}$$

i.e.,

$$\begin{aligned}&\left[ 1-\frac{1}{c^{1/2}}\left( {\overline{\alpha }}_1+\frac{1}{{\overline{v}}}\right) \right] \beta ^2-\frac{2(c_3-\beta _1)}{c^{1/2}c_4}\left( {\overline{\alpha }}_1+\frac{1}{{\overline{v}}}\right) \beta \nonumber \\&\quad -\left[ 1+\frac{1}{c^{1/2}} \left( {\overline{\alpha }}_1+\frac{1}{{\overline{v}}}\right) \right] =0. \end{aligned}$$
(6.13)

Recall that \(f=\alpha _1+1/v\) and \(f=g+c_3\). From (6.12), we have \({\overline{f}}^2=c(g^2-c_4^2)/(f-\beta _1)^2\). Then the discriminant of the quadratic Eq. (6.13) with respect to \(\beta \),

$$\begin{aligned} \begin{aligned}&\frac{4[(c_3-\beta _1)^2-c_4^2]}{cc_4^2}\cdot \left( {\overline{\alpha }}_1+\frac{1}{{\overline{v}}}\right) ^2+4 =\frac{4[(c_3-\beta _1)^2-c_4^2](g^2-c_4^2)}{c_4^2(g+c_3-\beta _1)^2}+4\\&\quad =\frac{4[(c_3-\beta _1)g+c_4^2]^2}{c_4^2(g+c_3-\beta _1)^2}, \end{aligned} \end{aligned}$$

implies that \(\beta \) is meromorphic apart from at most S(rf) many points. Moreover, roots of the equation

$$\begin{aligned} \begin{aligned}&[4(\alpha _1-\beta _1)(\alpha _2-\beta _1){\overline{\alpha }}_1^2 +c(\alpha _2-\alpha _1)^2]{\overline{v}}^2\\&\quad +8(\alpha _1-\beta _1) (\alpha _2-\beta _1){\overline{\alpha }}_1{\overline{v}} +4(\alpha _1-\beta _1)(\alpha _2-\beta _1)=0, \end{aligned} \end{aligned}$$
(6.14)

with respect to \({\overline{v}}\) must be \(c_2\) or coincide with \(v_1,v_2\).

If \(c_2=1/(\alpha _2-\alpha _1)\) is a root of (6.14), then we have

$$\begin{aligned} \begin{aligned} 4(\alpha _1-\beta _1)(\alpha _2-\beta _1)[{\overline{\alpha }}_1^2 +2{\overline{\alpha }}_1({\overline{\alpha }}_2-{\overline{\alpha }}_1) +({\overline{\alpha }}_2-{\overline{\alpha }}_1)^2]+c(\alpha _2-\alpha _1)^2=0, \end{aligned} \end{aligned}$$

which yields Eq. (6.11). On the other hand, by exchanging the roles of \(\alpha _1\) and \(\alpha _2\), we also have

$$\begin{aligned} 4{\overline{\alpha }}_1^{2}(\alpha _1-\beta _1)(\alpha _2-\beta _1) +c(\alpha _2-\alpha _1)^{2}=0. \end{aligned}$$
(6.15)

From (6.11) and (6.15), we see that \(\alpha _2=-\alpha _1\), and it follows that \(c={\overline{\alpha }}_1^2(1-\beta _1^2/\alpha _1^2)\). Set \(\delta =\beta _1/\alpha _1\). Then \(\delta \not =\pm \,1\) is an algebroid function of degree at most 2. By doing the transformation \(f\rightarrow \alpha _1f\), (6.9) becomes

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

which is the Eq. (3.7).

If \(c_2=1/(\alpha _2-\alpha _1)\) is not a root of (6.14), then roots of (6.14) must coincide with \(v_1,v_2\), which means that

$$\begin{aligned} \begin{aligned}&\frac{\overline{{\overline{\alpha }}}_2^{2}({\overline{\alpha }}_1 -{\overline{\beta }}_1)^2}{4(\alpha _1-\beta _1)(\alpha _2-\beta _1) {\overline{\alpha }}_1^2+c(\alpha _2-\alpha _1)^2} =\frac{2\overline{{\overline{\alpha }}}_2^{2}({\overline{\alpha }}_1 -{\overline{\beta }}_1)+{\overline{c}}_1^{2}}{8(\alpha _1 -\beta _1)(\alpha _2-\beta _1){\overline{\alpha }}_1}\\&\quad =\frac{\overline{{\overline{\alpha }}}_2^{2}-{\overline{c}}_1^{2} {\overline{c}}_2}{4(\alpha _1-\beta _1)(\alpha _2-\beta _1)}. \end{aligned} \end{aligned}$$
(6.16)

Recall that \(c_1=[c(\alpha _2-\alpha _1)]^{1/2}\) and \(c_2=1/(\alpha _2-\alpha _1)\). We obtain from the second equation of (6.16) that

$$\begin{aligned} ({\overline{\alpha }}_1+{\overline{\alpha }}_2){\overline{c}}= 2\overline{{\overline{\alpha }}}_2^2{\overline{\beta }}_1. \end{aligned}$$

By exchanging the roles of \(\alpha _1\) and \(\alpha _2\), we see that \({\overline{\beta }}_1\equiv 0\), and then it follows from the first equation of (6.16) that

$$\begin{aligned} {\overline{c}}c={\overline{c}}{\overline{\alpha }}_1^2+c \overline{{\overline{\alpha }}}_1^2. \end{aligned}$$

Set \(d=c/{\overline{\alpha }}_1^2\). Then d is a meromorphic function satisfying \({\overline{d}}d={\overline{d}}+d\). By doing the transformation \(f\rightarrow \alpha _1f\), (6.9) becomes

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

which is the Eq. (3.8).

We now look at Eq. (6.2). By using a similar analysis on the \(\alpha _i\)-points of f as on the \(\alpha _i\)-points of f in (5.2) in Sect. 5, we obtain the following three possibilities:

$$\begin{aligned} {\overline{f}}^3= & {} \frac{c(f-\alpha _1)(f-\alpha _2)(f-\alpha _3)}{(f-\beta _1)^3}, \quad \alpha _1\alpha _2\alpha _3\not \equiv 0, \end{aligned}$$
(6.17)
$$\begin{aligned} {\overline{f}}^4= & {} \frac{c(f-\alpha _1)^{2}(f-\alpha _2)(f-\alpha _3)}{(f-\beta _1)^4}, \quad \alpha _1\alpha _2\alpha _3\not \equiv 0, \end{aligned}$$
(6.18)
$$\begin{aligned} {\overline{f}}^6= & {} \frac{c(f-\alpha _1)^{3}(f-\alpha _2)^{2}(f-\alpha _3)}{(f-\beta _1)^6}, \quad \alpha _1\alpha _2\alpha _3\not \equiv 0. \end{aligned}$$
(6.19)

Note that \(\alpha _1\) in (6.18) and \(\alpha _1,\alpha _2,\alpha _3\) in (6.19) and \(\beta _1\) in these three equations are all meromorphic functions.

Let’s first consider Eq. (6.17). Put

$$\begin{aligned} \frac{{\overline{f}}(f-\beta _1)}{f-\alpha _1}=c_1u, \quad \frac{1}{f-\alpha _1}=b_2v+b_1, \end{aligned}$$
(6.20)

where u and v are two algebroid functions with at most S(rf) many branch points and the coefficients \(c_1,b_1,b_2\) take the same form as in (5.20) which are algebroid functions of degrees at most 9, 3, 3, respectively. Note that all the coefficients of (6.17) are small functions of v. Then we have

$$\begin{aligned} f=\alpha _1+\frac{1}{b_2v+b_1}, \quad {\overline{f}}=\frac{c_1u}{1+(\alpha _1-\beta _1)(b_2v+b_1)} \end{aligned}$$

and from (6.17) and (6.20) that

$$\begin{aligned} v^2=4u^3+1. \end{aligned}$$
(6.21)

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\alpha }}_2[1+(\alpha _1-\beta _1) (b_2v+b_1)]}{1+(\alpha _1-\beta _1)(b_2v+b_1)}={\overline{f}} -{\overline{\alpha }}_2. \end{aligned}$$
(6.22)

Since zeros of \({\overline{f}}-{\overline{\alpha }}_2\) are threefold ramified, we see that zeros of \(F=c_1u-{\overline{\alpha }}_2[1+(\alpha _1-\beta _1)(b_2v+b_1)]\) are also threefold ramified. Let \(u_0\) and \(v_0\) be two functions such that

$$\begin{aligned} v_0^2=4u_0^3+1, \quad c_1u_0-{\overline{\alpha }}_2[1+(\alpha _1-\beta _1)(b_2v_0+b_1)]=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives

$$\begin{aligned} 4{\overline{\alpha }}_2^3[1+(\alpha _1-\beta _1)(b_2v_0 +b_1)]^3-c_1^3(v_0^2-1)=0, \end{aligned}$$
(6.23)

from which we see that \(v_0\) is an algebroid function of degree at most 27. Moreover, it is seen that \(\pm \,1\) are not roots of (6.23), for otherwise we have \(1+(\alpha _1-\beta _1)(b_2v_0+b_1)=0\) and it follows that \(\pm b_2+b_1=-1/(\alpha _1-\beta _1)\), which gives \(\alpha _2=\beta _1\) or \(\alpha _3=\beta _1\), a contradiction. Rewrite (6.22) as

$$\begin{aligned} -\frac{4{\overline{\alpha }}_2^3[1+(\alpha _1-\beta _1) (b_2v+b_1)]^3-c_1^3(v^2-1)}{4[1+(\alpha _1-\beta _1) (b_2v+b_1)]^3}={\overline{f}}^3-{\overline{\alpha }}_2^3. \end{aligned}$$
(6.24)

From Eq. (6.21), we see that v is threefold ramified over \(\pm \,1,\infty \). This implies that (6.23) can admit only one root, for otherwise by a similar analysis as to (5.8) in Sect. 5 we obtain from (6.24) that (6.23) has at least two roots, say, \(v_1\) and \(v_2\), distinct from \(\pm \,1,\infty \) such that v is completely ramified at \(v_1\)-points and \(v_2\)-points, which is impossible by Theorem 1. Expanding (6.23) gives

$$\begin{aligned} \begin{aligned}&4{\overline{\alpha }}_2^3(\alpha _1-\beta _1)^3b_2^3v_0^3+12 {\overline{\alpha }}_2^3(\alpha _1-\beta _1)^2b_2^2[b_1(\alpha _1 -\beta _1)+1]v_0^2-c_1^3v_0^2\\&\quad +12{\overline{\alpha }}_2^3(\alpha _1-\beta _1) b_2[b_1(\alpha _1-\beta _1)+1]^2v_0+4{\overline{\alpha }}_2^3 [b_1(\alpha _1-\beta _1)+1]^3+c_1^3=0, \end{aligned} \end{aligned}$$

and so by Vieta’s formulas, we get

$$\begin{aligned} \frac{12{\overline{\alpha }}_2^3(\alpha _1-\beta _1)^2b_2^2 [b_1(\alpha _1-\beta _1)+1]-c_1^3}{4{\overline{\alpha }}_2^3 (\alpha _1-\beta _1)^3b_2^3}=-3v_0 \end{aligned}$$
(6.25)

and

$$\begin{aligned} \frac{3[b_1(\alpha _1-\beta _1)+1]^2}{(\alpha _1-\beta _1)^2b_2^2}=3v_0^2. \end{aligned}$$
(6.26)

From (6.26), we have \(v_0=\pm [b_1(\alpha _1-\beta _1)+1]/[(\alpha _1-\beta _1)b_2]\). If \(v_0=-[b_1(\alpha _1-\beta _1)+1]/[(\alpha _1-\beta _1)b_2]\), then substitution into (6.23) yields \(v_0^2=1\), a contradiction. Hence \(v_0=[b_1(\alpha _1-\beta _1)+1]/[(\alpha _1-\beta _1)b_2]\). Combining this equation with (6.25), we get

$$\begin{aligned} c_1^3=24{\overline{\alpha }}_2^3(\alpha _1-\beta _1)^2b_2^2 [b_1(\alpha _1-\beta _1)+1]. \end{aligned}$$

Similarly, the above equation also holds when replacing \({\overline{\alpha }}_2\) in (6.22) with \({\overline{\alpha }}_1\) and \({\overline{\alpha }}_3\), which implies that \(\alpha _1^3=\alpha _2^3=\alpha _3^3\). Thus, \(\alpha _2=\eta \alpha _1\), \(\alpha _3=\eta ^2\alpha _1\), where \(\eta \) is the cubic root of 1 satisfying \(\eta ^2+\eta +1=0\). Moreover, we have

$$\begin{aligned} c_1^3=4cb_2^2(\alpha _2-\alpha _1)(\alpha _3-\alpha _1)= 24{\overline{\alpha }}_1^3(\alpha _1-\beta _1)^2b_2^2 [b_1(\alpha _1-\beta _1)+1]. \end{aligned}$$

Set \(\gamma _1=\beta _1/\alpha _1\), \(\gamma _1^3\not \equiv 1\). Using equation \(\eta ^2+\eta +1=0\), we get from the above equation that

$$\begin{aligned} c={\overline{\alpha }}_1^3(1-\gamma _1)^2(1+\gamma _1). \end{aligned}$$
(6.27)

Setting \(\gamma _2=\beta _1/\alpha _2=\beta _1/(\eta \alpha _1)\), \(\gamma _3=\beta _1/\alpha _3=\beta _1/(\eta ^2\alpha _1)\), we have \(\gamma _2\not \equiv 1\), \(\gamma _3\not \equiv 1\). Exchanging the role of \(\alpha _1\) to \(\alpha _2\) and \(\alpha _3\), respectively, we also have

$$\begin{aligned} c={\overline{\alpha }}_3^3(1-\eta \gamma _1)^2(1+\eta \gamma _1) ={\overline{\alpha }}_2^3(1-\eta ^2\gamma _1)^2(1+\eta ^2\gamma _1), \end{aligned}$$

which together with the equation \(\eta ^2+\eta +1=0\) yields

$$\begin{aligned} (\eta ^2-\eta )\gamma _1(\gamma _1-1)=0. \end{aligned}$$

Therefore, we have \(\gamma _1\equiv 0\) and so \(\beta _1\equiv 0\) and then from (6.27) that \(c={\overline{\alpha }}_1^3\). By doing the transformation \(f\rightarrow \alpha _1f\), (6.17) becomes

$$\begin{aligned} {\overline{f}}^3=1-f^{-3}, \end{aligned}$$

which is the Eq. (3.14).

We now consider Eq. (6.18). Put

$$\begin{aligned} \frac{{\overline{f}}(f-\beta _1)}{f-\alpha _1}=c_1u, \quad \frac{1}{f-\alpha _1}=b_2v+b_1, \end{aligned}$$
(6.28)

where u and v are two algebroid functions with at most S(rf) many branch points and the coefficients \(c_1,b_1,b_2\) take the same form as in (5.27) which are algebroid functions of degrees at most 8, 2, 2, respectively. Note that all the coefficients of (6.18) are small functions of v. Then we have

$$\begin{aligned} f=\frac{1}{b_2v+b_1}+\alpha _1, \quad {\overline{f}}=\frac{c_1u}{1+(\alpha _1-\beta _1)(b_2v+b_1)} \end{aligned}$$

and from (6.18) and (6.28) that

$$\begin{aligned} v^2=1-u^4. \end{aligned}$$
(6.29)

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\alpha }}_2[1+(\alpha _1-\beta _1)(b_2v+b_1)]}{1+(\alpha _1-\beta _1)(b_2v+b_1)}={\overline{f}}-{\overline{\alpha }}_2. \end{aligned}$$
(6.30)

Since zeros of \({\overline{f}}-{\overline{\alpha }}_2\) are fourfold ramified, we see that zeros of \(F=c_1u-{\overline{\alpha }}_2[1+(\alpha _1-\beta _1)(b_2v+b_1)]\) are also fourfold ramified. Let \(u_0\) and \(v_0\) be functions such that

$$\begin{aligned} v_0^2=1-u_0^4, \quad c_1u_0-{\overline{\alpha }}_2[1+(\alpha _1-\beta _1)(b_2v_0+b_1)]=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives

$$\begin{aligned} {\overline{\alpha }}_2^4[1+(\alpha _1-\beta _1)(b_2v_0+b_1)] ^4-c_1^4(1-v_0^2)=0, \end{aligned}$$
(6.31)

from which we see that \(v_0\) is an algebroid function of degree at most 16. Moreover, it is seen that \(\pm \,1\) are not roots of (6.31), for otherwise we have \(1+(\alpha _1-\beta _1)(b_2v_0+b_1)=0\) and it follows that \(\pm b_2+b_1=-1/(\alpha _1-\beta _1)\), which gives \(\alpha _2=\beta _1\) or \(\alpha _3=\beta _1\), a contradiction. Rewrite (6.30) as

$$\begin{aligned} \frac{c_1^4(1-v^2)-{\overline{\alpha }}_2^4[1+(\alpha _1-\beta _1) (b_2v+b_1)]^4}{[1+(\alpha _1-\beta _1)(b_2v+b_1)]^4}= {\overline{f}}^4-{\overline{\alpha }}_2^4. \end{aligned}$$
(6.32)

From Eq. (6.29), we see that v is completely ramified over \(\pm \,1,\infty \). This implies that (6.31) can admit only one root, for otherwise by a similar analysis as to (5.8) in Sect. 5 we obtain from (6.32) that (6.31) has at least two roots, say, \(v_1\) and \(v_2\), distinct from \(\pm \,1,\infty \) such that v is completely ramified at \(v_1\)-points and \(v_2\)-points, which is impossible by Theorem 1. Expanding (6.31) gives

$$\begin{aligned} \begin{aligned}&{\overline{\alpha }}_2^4\{(\alpha _1-\beta _1)^4b_2^4v_0^4+4 (\alpha _1-\beta _1)^3b_2^3[1+(\alpha _1-\beta _1)b_1]v_0^3\\&\quad +\cdots +[1+(\alpha _1-\beta _1)b_1]^4\}-c_1^4(1-v_0^2)=0, \end{aligned} \end{aligned}$$

and so by Vieta’s formulas, we obtain \(4[1+(\alpha _1-\beta _1)b_1]/(\alpha _1-\beta _1)b_2=-4v_0\), which gives \(1+(\alpha _1-\beta _1)(b_2v_0+b_1)=0\) and it follows from (6.31) that \(v_0^2=1\), a contradiction. Therefore, (6.18) cannot admit any meromorphic solutions.

Finally, let’s consider Eq. (6.19). Put

$$\begin{aligned} \frac{{\overline{f}}(f-\beta _1)}{f-\alpha _1}=c_1u, \quad \frac{1}{f-\alpha _1}=b_2v-b_1, \end{aligned}$$
(6.33)

where u is an algebroid function with at most S(rf) many branch points and the coefficients \(c_1,b_1,b_2\) take the same form as in (5.33) and \(c_1\) is an algebroid function of degree at most 6. Note that all the coefficients of (6.19) are small functions of v. Then we have

$$\begin{aligned} f=\frac{1}{b_2v-b_1}+\alpha _1, \quad {\overline{f}}=\frac{c_1u}{1+(\alpha _1-\beta _1)(b_2v-b_1)} \end{aligned}$$

and from (6.19) and (6.33) that

$$\begin{aligned} u^6=(v-1)^2(v+2)=v^3-3v+2. \end{aligned}$$
(6.34)

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\alpha }}_3[1+(\alpha _1-\beta _1)(b_2v-b_1)]}{1+(\alpha _1-\beta _1)(b_2v-b_1)}={\overline{f}}-{\overline{\alpha }}_3. \end{aligned}$$
(6.35)

Since zeros of \({\overline{f}}-{\overline{\alpha }}_3\) are sixfold ramified, we see that zeros of \(F=c_1u-{\overline{\alpha }}_3[1+(\alpha _1-\beta _1)(b_2v-b_1)]\) are also sixfold ramified. Let \(u_0\) and \(v_0\) be functions such that

$$\begin{aligned} u_0^6=v_0^3-3v_0+2, \quad c_1u_0-{\overline{\alpha }}_3[1+(\alpha _1-\beta _1)(b_2v_0-b_1)]=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives

$$\begin{aligned} {\overline{\alpha }}_3^6[1+(\alpha _1-\beta _1)(b_2v_0-b_1)]^6 -c_1^6(v_0^3-3v_0+2)=0, \end{aligned}$$
(6.36)

from which we see that \(v_0\) is an algebroid function of degree at most 6. Moreover, it is easy to see that 1 and \(-2\) are not roots of (6.36), for otherwise we have \(1+(\alpha _1-\beta _1)(b_2-b_1)=0\), which gives \(\alpha _2=\beta _1\), a contradiction; or \(1+(\alpha _1-\beta _1)(-2b_2-b_1)=0\), which gives \(\alpha _3=\beta _1\), a contradiction. Rewrite (6.35) as

$$\begin{aligned} \frac{c_1^6(v^3-3v+2)-{\overline{\alpha }}_3^6[1+(\alpha _1 -\beta _1)(b_2v-b_1)]^6}{[1+(\alpha _1-\beta _1)(b_2v-b_1)]^6} ={\overline{f}}^6-{\overline{\alpha }}_3^6. \end{aligned}$$
(6.37)

From Eq. (6.34), we see that v is completely ramified over \(1,-2,\infty \). This implies that (6.36) can admit only one root, for otherwise by a similar analysis as to (5.8) in Sect. 5 we obtain from (6.37) that (6.36) has at least two roots, say, \(v_1\) and \(v_2\), distinct from \(1,-2,\infty \) such that v is completely ramified at \(v_1\)-points and \(v_2\)-points, which is impossible by Theorem 1. Expanding (6.36) gives

$$\begin{aligned} \begin{aligned}&{\overline{\alpha }}_3^6\{(\alpha _1-\beta _1)^6b_2^6v_0^6+6 (\alpha _1-\beta _1)^5b_2^5v_0^5[1-(\alpha _1-\beta _1)b_1]\\&\quad +\cdots +[1-(\alpha _1-\beta _1)b_1]^6\}-c_1^6(v_0^3-3v_0+2)=0, \end{aligned} \end{aligned}$$

and so by Vieta’s formulas, we obtain \(-6[1-(\alpha _1-\beta _1)b_1]/(\alpha _1-\beta _1)b_2=-6v_0\), which gives \(1+(\alpha _1-\beta _1)(b_2v_0-b_1)=0\) and it follows from (6.36) that \(v_0^3-3v_0+2=0\), a contradiction. Therefore, (6.19) cannot admit any meromorphic solutions.

6.2 Q(zf) has Two Distinct Roots

Now P(zf) can have one or two roots. We discuss these two cases separately as follows:

Case 1:P(zf) has only one root. Then (3.2) assumes the following form:

$$\begin{aligned} {\overline{f}}^{n}=\frac{c(f-\alpha _1)^{n}}{(f-\beta _1)^{\nu _1} (f-\beta _2)^{\nu _2}}, \end{aligned}$$
(6.38)

where \(\nu _1,\nu _2\) are positive integers satisfying \(\nu _1+\nu _2=n\). By doing a bilinear transformation \(f\rightarrow 1/f\), if \(\alpha _1\equiv 0\), we obtain (5.1); if \(\alpha _1\not \equiv 0\) and \(\beta _1,\beta _2\not \equiv 0\), we obtain (6.1). So in what follows we only consider the case when \(\alpha _1\not \equiv 0\) and \(\beta _1\equiv 0\). Note that \((\nu _1,\nu _2)=1\). If \(n=2\), then obviously \(\nu _1=\nu _2=1\); if \(n\ge 3\), then by using a similar analysis on the \(\beta _j\)-points of f in (6.38) as to \(\alpha _i\)-points of f in (5.1) in Sect. 5, we obtain that \(\nu _1=1\) or \(\nu _2=1\). Thus (6.38) is one of the following forms:

$$\begin{aligned} {\overline{f}}^{n}= & {} \frac{c(f-\alpha _1)^{n}}{f(f-\beta _2)^{n-1}}, \end{aligned}$$
(6.39)
$$\begin{aligned} {\overline{f}}^{n}= & {} \frac{c(f-\alpha _1)^{n}}{f^{n-1}(f-\beta _2)}. \end{aligned}$$
(6.40)

Consider Eq. (6.39) first. By doing a bilinear transformation \(f\rightarrow 1/f\), we obtain

$$\begin{aligned} {\overline{f}}^{n}=\frac{d_1(f-\gamma _1)^{n-1}}{(f-\lambda _1)^{n}}, \end{aligned}$$
(6.41)

where \(d_1=-\beta _2^{n-1}/(c\alpha _1^n)\), \(\gamma _1=1/\beta _2\), and \(\lambda _1=1/\alpha _1\) are nonzero small meromorphic functions of f. Moreover, it is seen that f is completely ramified at poles, \(\gamma _1\)-points and also \(\lambda _1\)-points. Set \(c_1=d_1^{1/n}\). Then \(c_1\) is an algebroid function of degree at most n. Put

$$\begin{aligned} c_1u=\frac{{\overline{f}}(f-\lambda _1)}{f-\gamma _1}, \quad v=\frac{1}{f-\gamma _1}, \end{aligned}$$
(6.42)

where u is an algebroid function with at most S(rf) many branch points. By the Valiron–Mohon’ko identity [15, 26] (see also [13]), we have \(T(r,v)=T(r,f)+S(r,f)\) and thus all the coefficients of (6.39) are small functions of v. Then we have

$$\begin{aligned} f=\frac{1}{v}+\gamma _1, \quad {\overline{f}}=\frac{c_1u}{1+(\gamma _1-\lambda _1)v} \end{aligned}$$

and from (6.41) and (6.42) that

$$\begin{aligned} u^{n}=v. \end{aligned}$$

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\gamma }}_1[1+(\gamma _1-\lambda _1)v]}{1+(\gamma _1-\lambda _1)v}={\overline{f}}-{\overline{\gamma }}_1. \end{aligned}$$
(6.43)

Since \({\overline{f}}\) is completely ramified at \({\overline{\gamma }}_1\)-points, we see that \(F=c_1u-{\overline{\gamma }}_1[(\gamma _1-\lambda _1)v+1]\) is completely ramified at zeros. Let \(u_0\) and \(v_0\) be functions such that

$$\begin{aligned} u_0^{n}=v_0, \quad c_1u_0-{\overline{\gamma }}_1[(\gamma _1-\lambda _1)v_0+1]=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives

$$\begin{aligned} {\overline{\gamma }}_1^{n}[(\gamma _1-\lambda _1)v_0+1]^{n}-c_1^{n}v_0=0, \end{aligned}$$

i.e.,

$$\begin{aligned} {\overline{\gamma }}_1^{n}[(\gamma _1-\lambda _1)^nv_0^n +n(\gamma _1-\lambda _1)^{n-1}v_0^{n-1}+\ldots +1]-c_1^{n}v_0=0, \end{aligned}$$
(6.44)

from which we see that \(v_0\) is an algebroid function of degree at most n. Moreover, it is seen that 0 and \(1/(\lambda _1-\gamma _1)\) are not roots of (6.44). Rewrite (6.43) as

$$\begin{aligned} \frac{c_1^nv-{\overline{\gamma }}_1^n[1+(\gamma _1-\lambda _1)v]^n}{[1+(\gamma _1-\lambda _1)v]^n}={\overline{f}}^n-{\overline{\gamma }}_1^n. \end{aligned}$$
(6.45)

We see from (6.42) that v is completely ramified at zeros, poles, and \(1/(\lambda _1-\gamma _1)\)-points. This implies that (6.44) can admit only one root, for otherwise by using a similar analysis as to (5.8) in Sect. 5 we obtain from (6.45) that (6.44) has at least two roots, say, \(v_1\) and \(v_2\), distinct from \(0,\infty ,1/(\lambda _1-\gamma _1)\) such that v is completely ramified at \(v_1\)-points and \(v_2\)-points, which is impossible by Theorem 1. However, when \(n\ge 3\), by Vieta’s formulas we see that the only root is \(v_0=-1/(\gamma _1-\lambda _1)\) and it follows from (6.44) that \(c_1^{n}/(\gamma _1-\lambda _1)=0\), a contradiction. Hence \(n=2\) and we therefore have

$$\begin{aligned} {\overline{f}}^{2}=\frac{c(f-\gamma _1)}{(f-\lambda _1)^{2}}. \end{aligned}$$
(6.46)

Moreover, we conclude from the above reasoning that the quadratic equation

$$\begin{aligned} {\overline{\gamma }}_1^{2}[(\gamma _1-\lambda _1)v_0+1]^{2}-c_1^{2}v_0=0 \end{aligned}$$
(6.47)

can admit only one root. It follows that the discriminant of Eq. (6.47) with respect to \(v_0\) satisfies

$$\begin{aligned} {[}2{\overline{\gamma }}_1^{2}(\gamma _1-\lambda _1)-c_1^{2}]^2- 4{\overline{\gamma }}_1^{4}(\gamma _1-\lambda _1)^2=0, \end{aligned}$$

i.e.,

$$\begin{aligned} d_1=4{\overline{\gamma }}_1^{2}(\gamma _1-\lambda _1). \end{aligned}$$
(6.48)

On the other hand, since \({\overline{f}}\) is completely ramified at \({\overline{\lambda }}_1\)-points, if we consider equation

$$\begin{aligned} \frac{c_1u-{\overline{\lambda }}_1[1+(\gamma _1-\lambda _1)v]}{1+(\gamma _1-\lambda _1)v}={\overline{f}}-{\overline{\lambda }}_1, \end{aligned}$$

then by similar arguments as above we also have

$$\begin{aligned} d_1=4{\overline{\lambda }}_1^{2}(\gamma _1-\lambda _1). \end{aligned}$$
(6.49)

From (6.48) and (6.49), we see that \(\lambda _1=-\gamma _1\) and it follows that \(d_1=-8{\overline{\lambda }}_1^{2}\lambda _1\). By doing the transformation \(f\rightarrow \gamma _1f\), then (6.46) becomes

$$\begin{aligned} {\overline{f}}^{2}=-\frac{8(f+1)}{(f-1)^{2}} =1-\left( \frac{f+3}{f-1}\right) ^{2}, \end{aligned}$$

which is the Eq. (3.9).

Consider now Eq. (6.40). By doing a bilinear transformation \(f\rightarrow 1/f\), (6.40) becomes

$$\begin{aligned} {\overline{f}}^{n}=\frac{d_2(f-\gamma _2)}{(f-\lambda _2)^{n}}, \end{aligned}$$
(6.50)

where \(d_2=(-1)^{1-n}\beta _1/(c\alpha _1^n)\), \(\gamma _2=1/\beta _1\), and \(\lambda _2=1/\alpha _1\) are nonzero small meromorphic functions of f. Moreover, f is completely ramified at poles, \(\gamma _2\)-points and also \(\lambda _2\)-points. By a similar reasoning as above, we can also obtain \(n=2\) and this will lead (6.50) into (3.9) again. We omit all those details.

Case 2:P(zf) has two distinct roots. Then (3.2) assumes the following form:

$$\begin{aligned} {\overline{f}}^{n}=\frac{c(f-\alpha _1)^{\kappa _1}(f-\alpha _2)^{\kappa _2}}{(f-\beta _1)^{\nu _1}(f-\beta _2)^{\nu _2}}, \quad \alpha _1\alpha _2\not \equiv 0, \end{aligned}$$

where \(\kappa _i\), \(\nu _j\) are positive integers satisfying \(\kappa _1+\kappa _2=n\), \(\nu _1+\nu _2=n\). Set \(k_1=(n,\kappa _1)\), \(k_2=(n,\kappa _2)\), \(l_1=(n,\nu _1)\), \(l_2=(n,\nu _2)\) and let \(K_1=n/k_1\), \(K_2=n/k_2\), \(L_1=n/l_1\), \(L_2=n/l_2\). Obviously, \(K_1\ge 2\), \(K_2\ge 2\), \(L_1\ge 2\), \(L_2\ge 2\). Now f is \(K_i\)-fold ramified at \(\alpha _i\)-points and \(L_j\)-fold ramified at \(\beta _j\)-points, and by (2.4) we conclude that \(K_1=K_2=L_1=L_2=2\) and \(k_1+k_2=n\), \(l_1+l_2=n\). Since \(\kappa _1+\kappa _2=n\), \(\nu _1+\nu _2=n\), we see that \(\kappa _1=\kappa _2=\nu _1=\nu _2=n/2\). Hence \(n=2\) and we therefore have

$$\begin{aligned} {\overline{f}}^2=\frac{c(f-\alpha _1)(f-\alpha _2)}{(f-\beta _1)(f-\beta _2)}, \quad \alpha _1\alpha _2\not \equiv 0. \end{aligned}$$
(6.51)

Now \(\alpha _i,\beta _j\) are algebroid functions of degree at most 2. Moreover, if \(\beta _1\equiv 0\) or \(\beta _2\equiv 0\), then f will be fourfold ramified at \(\alpha _i\)-points, which yields a contradiction to (2.4). Therefore, \(\beta _1\not \equiv 0\) and \(\beta _2\not \equiv 0\).

Set \(c_1=c^{1/2}\). Then \(c_1\) is an algebroid function of degree at most 2. Put

$$\begin{aligned} c_1u=\frac{{\overline{f}}(f-\beta _1)}{f-\alpha _1}, \quad v=\frac{f-\beta _1}{f-\alpha _1}, \end{aligned}$$
(6.52)

where u and v are two algebroid functions with at most S(rf) many branch points. By the Valiron–Mohon’ko identity [15, 26] (see also [13]), we have \(T(r,v)=T(r,f)+S(r,f)\) and thus all the coefficients of (6.51) are small functions of v. Then we have

$$\begin{aligned} f=\frac{\alpha _1v-\beta _1}{v-1}, \quad {\overline{f}}=\frac{c_1u}{v} \end{aligned}$$

and from (6.51) and (6.52) that

$$\begin{aligned} u^2=v\frac{(\alpha _1-\alpha _2)v-(\beta _1-\alpha _2)}{(\alpha _1-\beta _2)v-(\beta _1-\beta _2)}. \end{aligned}$$
(6.53)

It also follows that

$$\begin{aligned} \frac{c_1u-{\overline{\alpha }}_1v}{v}={\overline{f}}-{\overline{\alpha }}_1. \end{aligned}$$
(6.54)

Since zeros of \({\overline{f}}-{\overline{\alpha }}_1\) are twofold ramified, we see that zeros of \(F=c_1u-{\overline{\alpha }}_1v\) are also twofold ramified. Let \(u_0\) and \(v_0\) be functions such that

$$\begin{aligned} u_0^2=v_0\frac{(\alpha _1-\alpha _2)v_0-(\beta _1-\alpha _2)}{(\alpha _1-\beta _2)v_0-(\beta _1-\beta _2)}, \quad c_1u_0-{\overline{\alpha }}_1v_0=0. \end{aligned}$$

Eliminating \(u_0\) from the above two equations gives \(v_0\equiv 0\) or

$$\begin{aligned} {\overline{\alpha }}_1^2(\alpha _1-\beta _2)v_0^2+c_1^2(\beta _1-\alpha _2) -[{\overline{\alpha }}_1^2(\beta _1-\beta _2)+c_1^2(\alpha _1-\alpha _2)]v_0=0, \end{aligned}$$
(6.55)

from which we see that \(v_0\) is an algebroid function of degree at most 16. Moreover, it is seen that none of 0, \((\alpha _2-\beta _1)/(\alpha _2-\alpha _1)\), and \((\beta _2-\beta _1)/(\beta _2-\alpha _1)\) is a root of (6.55), for otherwise we get \({\overline{\alpha }}_1\equiv 0\), a contradiction. If \(v_0\equiv 0\), then for a point \(z_0\) such that \(f(z_0+1)=\alpha _1(z_0+1)\), we have \(u(z_0)=v(z_0)=0\) and from (6.53) we see that \(c_1(z_0)u(z_0)/v(z_0)=\infty \), a contradiction since \({\overline{f}}=c_1u/v\). Therefore, \(v_0\not \equiv 0\). Rewrite (6.54) as

$$\begin{aligned} \frac{c_1^2[(\alpha _1-\alpha _2)v-(\beta _1-\alpha _2)]v -[(\alpha _1-\beta _2)v-(\beta _1-\beta _2)]{\overline{\alpha }}_1^2v^2}{[(\alpha _1-\beta _2)v-(\beta _1-\beta _2)]v^2}={\overline{f}}^2 -{\overline{\alpha }}_1^2. \end{aligned}$$
(6.56)

From (6.52) we see that v is completely ramified at poles, zeros, \((\alpha _2-\beta _1)/(\alpha _2-\alpha _1)\)-points, and \((\beta _2-\beta _1)/(\beta _2-\alpha _1)\)-points. This implies that (6.55) can admit only one root, for otherwise by using a similar analysis as to (5.8) in Sect. 5 we obtain from (6.56) that (6.55) has at least two roots, say, \(v_1\) and \(v_2\), distinct from \(0,\infty ,(\alpha _2-\beta _1)/(\alpha _2-\alpha _1),(\beta _2-\beta _1) /(\beta _2-\alpha _1)\) such that v is completely ramified at \(v_1\)-points and \(v_2\)-points, which is impossible by Theorem 1. Therefore, the discriminant of the quadratic Eq. (6.55) with respect to \(v_0\) satisfies

$$\begin{aligned} {[}{\overline{\alpha }}_1^2(\beta _1-\beta _2)+c_1^2(\alpha _1 -\alpha _2)]^2-4{\overline{\alpha }}_1^2c_1^2(\alpha _1-\beta _2) (\beta _1-\alpha _2)=0, \end{aligned}$$

i.e.,

$$\begin{aligned} {\overline{\alpha }}_1^4-\frac{2c[2(\beta _1-\alpha _2)(\alpha _1-\beta _2) -(\beta _1-\beta _2)(\alpha _1-\alpha _2)]}{(\beta _1-\beta _2)^2} {\overline{\alpha }}_1^2+\frac{c^{2}(\alpha _1-\alpha _2)^2}{(\beta _1-\beta _2)^2}=0. \end{aligned}$$
(6.57)

Similarly, we obtain that (6.57) still holds when replacing \({\overline{\alpha }}_1\) by any of \({\overline{\alpha }}_2\), \({\overline{\beta }}_1\), and \({\overline{\beta }}_2\), which implies that \({\overline{\alpha }}_1^2={\overline{\alpha }}_2^2\) or \({\overline{\alpha }}_1^2={\overline{\beta }}_1^2\). Thus we need to consider the following two cases: (1), \(\alpha _1=-\alpha _2\), \(\beta _1=-\beta _2\); and (2), \(\alpha _1=-\beta _1\), \(\alpha _2=-\beta _2\).

If \(\alpha _1=-\alpha _2\), \(\beta _1=-\beta _2\), then by Vieta’s formulas we obtain from (6.57) that

$$\begin{aligned} {\overline{\alpha }}_1^2{\overline{\beta }}_1^2 =\frac{c^{2}\alpha _1^2}{\beta _1^2}, \quad {\overline{\alpha }}_1^2+{\overline{\beta }}_1^2=c\frac{\beta _1^2 +\alpha _1^2}{\beta _1^2}. \end{aligned}$$

Set \(\gamma _1=\alpha _1/\beta _1\), \(\gamma _1^2\not \equiv 0,1\) and \(\gamma _2=\theta {\overline{\gamma }}_1/\gamma _1=c/{\overline{\beta }}_1^2\), \(\theta =\pm \,1\). Then \(\gamma _1^2\) is meromorphic and \(\gamma _2\) satisfies \(({\overline{\gamma }}_1^2+1)\gamma _1=\theta {\overline{\gamma }}_1 (\gamma _1^2+1)\), which gives \({\overline{\gamma }}_1=\theta \gamma _1\) or \({\overline{\gamma }}_1\gamma _1=\theta \). It follows that \(\gamma _2\equiv 1\) or \(\gamma _2={\overline{\gamma }}_1^2\). If \(\gamma _2\equiv 1\), then \(\gamma _1^2\) is a periodic function with period 1. By doing the transformation \(f\rightarrow \beta _1f\), (6.51) becomes

$$\begin{aligned} {\overline{f}}^2=\frac{f^2-\gamma _1^2}{f^2-1}, \end{aligned}$$

which is the Eq. (3.10); if \(\gamma _2={\overline{\gamma }}_1^2\), then \({\overline{\gamma }}_1^2\gamma _1^2=1\). By doing the transformation \(f\rightarrow \beta _1f\), (6.51) becomes

$$\begin{aligned} {\overline{f}}^2=\frac{{\overline{\gamma }}_1^2f^2-1}{f^2-1}, \end{aligned}$$

which is the Eq. (3.11).

If \(\alpha _1=-\beta _1\), \(\alpha _2=-\beta _2\), then by Vieta’s formulas we obtain from (6.57) that

$$\begin{aligned} {\overline{\alpha }}_1^2{\overline{\alpha }}_2^2=c^2, \quad {\overline{\alpha }}_1^2+{\overline{\alpha }}_2^2=2c \frac{-2(\alpha _1+\alpha _2)^2+(\alpha _1-\alpha _2)^2}{(\alpha _1-\alpha _2)^2}. \end{aligned}$$

Set \(\lambda =(\alpha _1/\alpha _2)^{1/2}\), \(\lambda ^2\not \equiv 0,\pm \,1\). Let \(d=\lambda +1/\lambda \). Then \(d^2=(\alpha _1+\alpha _2)^{2}/\alpha _1\alpha _2\) is a meromorphic function satisfying

$$\begin{aligned} {\overline{d}}^2-2=2\theta \frac{-2d^2+d^2-4}{d^2-4}, \end{aligned}$$

i.e.,

$$\begin{aligned} {\overline{d}}^2(d^2-4)=2(1-\theta )d^2-8(1+\theta ), \end{aligned}$$

where \(\theta =\pm \,1\). By doing the transformation \(f\rightarrow (\alpha _1\alpha _2)^{1/2}f\), (6.51) becomes

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

which is the Eq. (3.12).

6.3 Q(zf) has Three Distinct Roots

In this case, f is completely ramified at \(\beta _j\)-points. By using a similar analysis on the \(\beta _j\)-points of f as on the \(\alpha _i\)-points of f in (5.2) in Sect. 5, we obtain the following three possibilities:

$$\begin{aligned} {\overline{f}}^3= & {} \frac{c(f-\alpha _1)^3}{(f-\beta _1)(f-\beta _2)(f-\beta _3)}, \end{aligned}$$
(6.58)
$$\begin{aligned} {\overline{f}}^4= & {} \frac{c(f-\alpha _1)^4}{(f-\beta _1)^{2}(f-\beta _2)(f-\beta _3)}, \end{aligned}$$
(6.59)
$$\begin{aligned} {\overline{f}}^6= & {} \frac{c(f-\alpha _1)^6}{(f-\beta _1)^{3}(f-\beta _2)^{2}(f-\beta _3)}, \end{aligned}$$
(6.60)

where \(\alpha _1\) is meromorphic. In (6.58), f is threefold ramified at \(\beta _j\)-points, \(j=1,2,3\). This implies that \(\beta _j\not \equiv 0\), \(j=1,2,3\), for otherwise f will also be threefold ramified at \(\alpha _1\)-points, which yields a contradiction to (2.4). Similarly, in Eqs. (6.59)–(6.60), we have \(\beta _j\not \equiv 0\), \(j=1,2,3\). By doing a bilinear transformation \(f\rightarrow 1/f\), if \(\alpha _1\equiv 0\) we obtain Eqs. (5.16)–(5.18) and if \(\alpha _1\not \equiv 0\) we obtain Eqs. (6.17)–(6.19). This completes the proof of Theorem 2.