Composite Rational Functions and Arithmetic Progressions

TL;DR

This paper studies composite rational functions with zeros and poles in arithmetic progression, correcting previous results and exploring their structural properties.

Contribution

It introduces new insights into the structure of composite rational functions with zeros and poles in arithmetic progression and corrects earlier published results.

Findings

Characterization of zeros and poles in arithmetic progression

Correction of previous theoretical results

Structural properties of composite rational functions

Abstract

In this paper we deal with composite rational functions having zeros and poles forming consecutive elements of an arithmetic progression. We also correct a result published earlier related to composite rational functions having a fixed number of zeros and poles.

Equations

$$f=g_1\circ g_2\circ \cdots \circ g_r=h_1\circ h_2\circ \cdots \circ h_s,$$

$$f=\frac{x^3(x+6{)}^3(x^2-6x+36{)}^3}{(x-3{)}^3(x^2+3x+9{)}^3},$$

$$f=g_1\circ g_2\circ g_3=x^3\circ \frac{x(x-12)}{x-3}\circ \frac{x(x+6)}{x-3},$$

$$f=h_1\circ h_2=\frac{x^3(x+24)}{x-3}\circ \frac{x(x^2-6x+36)}{x^2+3x+9}.$$

$$f=\frac{x^4-8x}{x^3+1}$$

$$g=\frac{x^2+4x}{x+1}\text{mbox}andh=\frac{x^2-2x}{x+1}.$$

$$f(x)=\prod _{j=1}^{t_i}({\omega }_j/{\omega }_{\infty }{)}^{k_j},g(x)=\prod _{j=1}^{t_i}(x-{\beta }_j{)}^{k_j}$$

$$h(x)=\{\begin{array}{ll}{\beta }_j+\frac{{\omega }_j}{{\omega }_{\infty }}(j=1,\dots ,t_i),& \text{mbox}if{k}_1+k_2+\dots +k_{t_i}\ne 0\\ \frac{{\beta }_{j_1}{\omega }_{j_2}-{\beta }_{j_2}{\omega }_{j_1}}{{\omega }_{j_2}-{\omega }_{j_1}}(1\le j_1<j_2\le t_i),& \text{mbox}otherwise,\end{array}$$

$${\omega }_j=\prod _{m\in S_{{\beta }_j}}(x-{\alpha }_m{)}^{l_m},j=1,\dots ,t_i$$

$${\omega }_{\infty }=\prod _{m\in S_{\infty }}(x-{\alpha }_m{)}^{l_m}.$$

$${\alpha }_i={\alpha }_0+T_{i}d$$

$$d^N=M$$

$$\sum _{r\in S_{{\beta }_i}}{l}_r=\sum _{s\in S_{{\beta }_j}}{l}_s,i,j\in \{1,\dots ,t\},i\ne j.$$

$$d^{{\sum }_{m_1\in S_{{\beta }_{j_1}}}{l}_{m_1}},d^{{\sum }_{m_2\in S_{{\beta }_{j_2}}}{l}_{m_2}},d^{{\sum }_{m_3\in S_{{\beta }_{j_3}}}{l}_{m_3}}$$

$$(x-{\alpha }_0{)}^{k_1}(x-{\alpha }_0-d{)}^{k_2}(x-{\alpha }_0-2d{)}^{k_2}(x-{\alpha }_0-3d{)}^{k_1},$$

$${\alpha }_u^{(1)}=a_{u,1}{\alpha }_1^{(2)}+a_{u,2}{\alpha }_2^{(2)}+\dots +a_{u,n}{\alpha }_n^{(2)}+a_{u,n+1},$$

$$\text{mbox}(a):{\alpha }_1-\frac{1}{4},{\alpha }_1,{\alpha }_1+\frac{1}{4}\text{mbox}difference:\frac{1}{4},$$

$$\text{mbox}(b):{\alpha }_1,{\alpha }_2,-{\alpha }_1+2{\alpha }_2\text{mbox}difference:{\alpha }_2-{\alpha }_1,$$

$$\text{mbox}(c):{\alpha }_1,\frac{{\alpha }_1+{\alpha }_2}{2},{\alpha }_2\text{mbox}difference:\frac{{\alpha }_2-{\alpha }_1}{2}.$$

$$f(x)=\prod _{i=1}^n(x-{\alpha }_i{)}^{f_i}$$

$$g(x)=\prod _{j=1}^t(x-{\beta }_j{)}^{k_j}$$

$$\prod _{i=1}^n(x-{\alpha }_i{)}^{f_i}=f(x)=g(h(x))=\prod _{j=1}^t(h(x)-{\beta }_j{)}^{k_j}.$$

$$q(x)=\prod _{m\in S_{\infty }}(x-{\alpha }_m{)}^{l_m}$$

$$h(x)={\beta }_j+\frac{1}{q(x)}\prod _{m\in S_{{\beta }_j}}(x-{\alpha }_m{)}^{l_m},$$

$${\beta }_i+\frac{1}{q(x)}\prod _{r\in S_{{\beta }_i}}(x-{\alpha }_r{)}^{l_r}={\beta }_j+\frac{1}{q(x)}\prod _{s\in S_{{\beta }_j}}(x-{\alpha }_s{)}^{l_s}.$$

$$(p(x)-{\beta }_{j}q(x){)}^{k_j}=\prod _{m\in S_{{\beta }_j}}(x-{\alpha }_m{)}^{f_m}.$$

$$v_{j_1,j_2,j_3}+v_{j_3,j_1,j_2}+v_{j_2,j_3,j_1}=0,$$

$$v_{j_1,j_2,j_3}=({\beta }_{j_1}-{\beta }_{j_2})\prod _{m\in S_{{\beta }_{j_3}}}(x-{\alpha }_m{)}^{l_m}.$$

$${\beta }_i-{\beta }_j$$

$$C_1(i,j,r_1,s_1)=d^{{\sum }_{r\in S_{{\beta }_i}}{l}_r-{\sum }_{s\in S_{{\beta }_j}}{l}_s},$$

$$\sum _{r\in S_{{\beta }_i}}{l}_r=\sum _{s\in S_{{\beta }_j}}{l}_s,i,j\in \{1,\dots ,t\},i\ne j.$$

$$h(x)={\beta }_i+\frac{1}{q(x)}={\beta }_j+\frac{1}{q(x)}.$$

Full text

Research supported in part by the OTKA grants NK104208 and K100339

1. Introduction

We consider a problem related to decompositions of polynomials and rational functions. In this subject a classical result obtained by Ritt [13] says that if there is a polynomial $f\in\mathbb{C}[X]$ satisfying certain tameness properties and

[TABLE]

then $r=s$ and $\{\deg g_{1},\ldots,\deg g_{r}\}=\{\deg h_{1},\ldots,\deg h_{r}\}.$ Ritt’s fundamental result has been investigated, extended and applied in various wide-ranging contexts (see e.g. [4, 6, 7, 9, 10, 11, 14, 15]). The above mentioned result is not valid for rational functions. Gutierrez and Sevilla [9] provided the following example

[TABLE]

To determine decompositions of a given rational function there were developed algorithms (see e.g. [1, 2, 3]). In [2], Ayad and Fleischmann implemented a MAGMA [5] code to find decompositions, they provided the following example

[TABLE]

and they obtained that $f(x)=g(h(x)),$ where

[TABLE]

Fuchs and Pethő [8] proved the following theorem.

Theorem A**.**

Let $k$ be an algebraically closed field of characteristic zero. Let $n$ be a positive integer. Then there exists a positive integer $J$ and, for every $i\in\{1,\ldots,J\}$, an affine algebraic variety $V_{i}$ defined over ${\mathbb{Q}}$ and with $V_{i}\subset\mathbb{A}^{n+t_{i}}$ for some $2\leq t_{i}\leq n$, such that:

(i) If $f,g,h\in k(x)$ with $f(x)=g(h(x))$ and with $\deg g,\deg h\geq 2,g$ not of the shape $(\lambda(x))^{m},m\in{\mathbb{N}},\lambda\in PGL_{2}(k),$ and $f$ has at most $n$ zeros and poles altogether, then there exists for some $i\in\{1,\ldots,J\}$ a point $P=(\alpha_{1},\ldots,\alpha_{n},\beta_{1},\ldots,\beta_{t_{i}})\in V_{i}(k)$, a vector $(k_{1},\ldots,k_{t_{i}})\in{\mathbb{Z}}^{t_{i}}$ with $k_{1}+k_{2}+\ldots+k_{t_{i}}=0$ depending only 111in [8] it is written as ”or not depending”, this typo is corrected here. on $V_{i}$ , a partition of $\{1,\ldots,n\}$ in $t_{i}+1$ disjoint sets $S_{\infty},S_{\beta_{1}},\ldots,S_{\beta_{t_{i}}}$ with $S_{\infty}=\emptyset$ if $k_{1}+k_{2}+\ldots+k_{t_{i}}=0$, and a vector $(l_{1},\ldots,l_{n})\in\{0,1,\ldots,n-1\}^{n}$, also both depending only on $V_{i}$, such that

[TABLE]

and

[TABLE]

where

[TABLE]

and

[TABLE]

Moreover, we have $deg~{}h\leq(n-1)/\max\{t_{i}-2,1\}\leq n-1.$

(ii) Conversely for given data $P\in V_{i}(k),(k_{1},\ldots,k_{t_{i}}),$ $S_{\infty},S_{\beta_{1}},\ldots,S_{\beta_{t_{i}}},$ $(l_{1},\ldots,l_{n})$ as described in (i) one defines by the same equations rational functions $f,g,h$ with $f$ having at most $n$ zeros and poles altogether for which $f(x)=g(h(x))$ holds.

(iii) The integer $J$ and equations defining the varieties $V_{i}$ are effectively computable only in terms of $n.$

Pethő and Tengely [12] provided some computational experiments that they obtained by using a MAGMA [5] implementation of the algorithm of Fuchs and Pethő [8].

If the zeros and poles of a composite rational function form an arithmetic progression, then we have the following result.

Theorem 1**.**

Let $f,g,h$ be rational functions as in Theorem A. Assume that the zeros and poles of $f$ form an arithmetic progression, that is

[TABLE]

for some $\alpha_{0},d\in k$ and $T_{i}\in\{0,1,\ldots,n-1\}.$ If $k_{1}+k_{2}+\ldots+k_{t}\neq 0,$ then either the difference $d$ satisfies an equation of the form

[TABLE]

for some $N\in\mathbb{Z},M\in\mathbb{Q}$ or $(l_{1},\ldots,l_{n})\in\{0,1,\ldots,n-1\}^{n}$ satisfies a system of linear equations

[TABLE]

If $k_{1}+k_{2}+\ldots+k_{t}=0$ and $1\leq j_{1}<j_{2}<j_{3}\leq t,$ then

[TABLE]

satisfy a system of linear equations and $\beta_{j_{1}},\beta_{j_{2}},\beta_{j_{3}}$ satisfy a system of linear equations.

We will apply the above theorem to determine composite rational functions having 4 zeros and poles. We prove the following statement.

Proposition 1**.**

Let $k$ be an algebraically closed field of characteristic zero. If $f,g,h\in k(x)$ with $f(x)=g(h(x))$ and with $\deg g,\deg h\geq 2,g$ not of the shape $(\lambda(x))^{m},m\in{\mathbb{N}},\lambda\in PGL_{2}(k),$ and $f$ has 4 zeros and poles altogether forming an arithmetic progression, then $f$ is equivalent to the following rational function

[TABLE]

for some $\alpha_{0},d\in k$ and $k_{1},k_{2}\in\mathbb{Z},k_{1}+k_{2}\neq 0.$

In this paper we correct results obtained in [12], where the computations related to the case $k_{1}+k_{2}+\ldots+k_{t}\neq 0,S_{\infty}=\emptyset$ are missing. The following theorem is the corrected version of Theorem 1 from [12], where part (c) was missing. We define equivalence of rational functions. Two rational functions $f_{1}(x)=\prod_{u=1}^{n}(x-\alpha_{u}^{(1)})^{f_{u}^{(1)}}$ and $f_{2}(x)=\prod_{u=1}^{n}(x-\alpha_{u}^{(2)})^{f_{u}^{(2)}}$ are equivalent if there exist $a_{u,v}\in\mathbb{Q},u\in\{1,2,\ldots,n\},v\in\{1,2,\ldots,n+1\}$ such that

[TABLE]

for all $u\in\{1,2,\ldots,n\}.$

Theorem 2**.**

Let $k$ be an algebraically closed field of characteristic zero. If $f,g,h\in k(x)$ with $f(x)=g(h(x))$ and with $\deg g,\deg h\geq 2,g$ not of the shape $(\lambda(x))^{m},m\in{\mathbb{N}},\lambda\in PGL_{2}(k),$ and $f$ has 3 zeros and poles altogether, then $f$ is equivalent to one of the following rational functions

• (a)

$\frac{(x-\alpha_{1})^{k_{1}}(x+1/4-\alpha_{1})^{2k_{2}}}{(x-1/4-\alpha_{1})^{2k_{1}+2k_{2}}}$* for some $\alpha_{1}\in k$ and $k_{1},k_{2}\in\mathbb{Z},k_{1}+k_{2}\neq 0,$*

• (b)

$\frac{(x-\alpha_{1})^{2k_{1}}(x+\alpha_{1}-2\alpha_{2})^{2k_{2}}}{(x-\alpha_{2})^{2k_{1}+2k_{2}}}$* for some $\alpha_{1},\alpha_{2}\in k$ and $k_{1},k_{2}\in\mathbb{Z},k_{1}+k_{2}\neq 0,$*

• (c)

$\left(x-\frac{\alpha_{1}+\alpha_{2}}{2}\right)^{2k_{1}}(x-\alpha_{1})^{k_{2}}(x-\alpha_{2})^{k_{2}}$* for some $\alpha_{1},\alpha_{2}\in k$ and $k_{1},k_{2}\in\mathbb{Z},k_{1}+k_{2}\neq 0.$*

Remark**.**

The MAGMA procedure CompRatFunc.m can be downloaded from http://shrek.unideb.hu/$\sim$tengely/CompRatFunc.m. All systems in cases of $n\in\{3,4,5\}$ can be downloaded from

http://shrek.unideb.hu/$\sim$tengely/CFunc345.tar.gz.

Remark**.**

It is interesting to note that in the above formulas the zeros and poles form an arithmetic progression

[TABLE]

2. Auxiliary results

We repeat some parts of the proof of Theorem A from [8] that will be used here later on. Without loss of generality we may assume that $f$ and $g$ are monic. Let

[TABLE]

with pairwise distinct $\alpha_{i}\in k$ and $f_{i}\in{\mathbb{Z}}$ for $i=1,\ldots,n.$ Similarly, let

[TABLE]

with pairwise distinct $\beta_{j}\in k$ and $k_{j}\in{\mathbb{Z}}$ for $j=1,\ldots,t$ and $t\in{\mathbb{N}}.$ Hence we have

[TABLE]

We shall write $h(x)=p(x)/q(x)$ with $p,q\in k[x],p,q$ coprime. Fuchs and Pethő [8] showed that if $k_{1}+k_{2}+\ldots+k_{t}\neq 0,$ then there is a subset $S_{\infty}$ of the set $\{1,\ldots,n\}$ for which

[TABLE]

and there is a partition of the set $\{1,\ldots,n\}\setminus S_{\infty}$ in $t$ disjoint non empty subsets $S_{\beta_{1}},\ldots,S_{\beta_{t}}$ such that

[TABLE]

where $l_{m}\in{\mathbb{N}}$ satisfies $l_{m}k_{j}=f_{m}$ for $m\in S_{\beta_{j}},$ and this holds true for every $j=1,\ldots,t.$ We get at least two different representations of $h,$ since we assumed that $g$ is not of the special shape $(\lambda(x))^{m}.$ Therefore we get at least one equation of the form

[TABLE]

If $k_{1}+k_{2}+\ldots+k_{t}=0,$ then we have

[TABLE]

Now we have that $t\geq 3,$ otherwise $g$ is in the special form we excluded. Siegel’s identity provides the equations in this case. That is if $1\leq j_{1}<j_{2}<j_{3}\leq t,$ then we have

[TABLE]

where

[TABLE]

3. Proofs of Theorem 1 and Theorem 2

Proof of Theorem 1.

If $k_{1}+k_{2}+\ldots+k_{t}\neq 0$ and there exist $r_{1}\in S_{\beta_{i}},s_{1}\in S_{\beta_{j}}$ for some $i\neq j$ such that $l_{r_{1}}\neq 0$ and $l_{s_{1}}\neq 0,$ then it follows from (2) that

[TABLE]

for any appropriate $\alpha_{r_{1}}\in S_{\beta_{i}}$ and $\alpha_{s_{1}}\in S_{\beta_{j}}.$ Hence we obtain that

[TABLE]

where $C_{1}(i,j,r_{1},s_{1})\in\mathbb{Q}.$ If there exist $S_{\beta_{i}}$ and $S_{\beta_{j}}$ for which $\sum_{r\in S_{\beta_{i}}}l_{r}-\sum_{s\in S_{\beta_{j}}}l_{s}\neq 0,$ then the possible values of $d$ satisfy equations of the form $x^{N}=M.$ Otherwise we get that

[TABLE]

Let us consider the special case when $l_{r}=0$ for all $r\in S_{\beta_{i}}.$ If $l_{s}=0$ for all $s\in S_{\beta_{j}},$ then we get that

[TABLE]

Hence $\beta_{i}=\beta_{j}$ for some $i\neq j,$ a contradiction. Thus we may assume that there exists $s_{1}\in S_{\beta_{j}}$ for which $l_{s_{1}}\neq 0.$ In a similar way as in the above case it follows that

[TABLE]

Therefore

[TABLE]

where $C_{2}(i,j,r_{1},s_{1})\in\mathbb{Q}.$ Since $s_{1}>0$ we have that $\sum_{s\in S_{\beta_{j}}}l_{s}\neq 0,$ that is $d$ satisfies an appropriate polynomial equation.

If $k_{1}+k_{2}+\ldots+k_{t}=0,$ then there are at least 3 partitions and for any appropriate $r_{1}\in S_{\beta_{j_{1}}},r_{2}\in S_{\beta_{j_{2}}},r_{3}\in S_{\beta_{j_{3}}}$ (that is $l_{r_{i}}\neq 0,i=1,2,3$) equation (3) implies that

[TABLE]

that is a system of linear equations in $d_{1},d_{2},d_{3},$ where $d_{i}=d^{\sum_{m_{i}\in S_{\beta_{j_{i}}}}l_{m_{i}}},i\in\{1,2,3\}$ and the statement follows. In a very similar way we obtain a system of equations if $l_{r}=0$ for all $r\in S_{\beta_{j_{3}}},$ the last two equations are as before, while on the left-hand side of the first one there is an additional term $\beta_{j_{1}}-\beta_{j_{2}}.$ ∎

Proof of Theorem 2.

In [12] all cases are given with $k_{1}+k_{2}+\ldots+k_{t}=0$ and also with $k_{1}+k_{2}+\ldots+k_{t}\neq 0,S_{\infty}\neq\emptyset.$ Therefore it remains to deal with those cases with $k_{1}+k_{2}+\ldots+k_{t}\neq 0,S_{\infty}=\emptyset.$ First let $t=2.$ There are 18 systems of equations. Among these systems there are two types. The first one has only a single equation, e.g. when $S_{\beta_{1}}=\{1,2\},S_{\beta_{2}}=\{3\},(l_{1},l_{2},l_{3})=(1,0,1),$ this equation is as follows

[TABLE]

Hence

[TABLE]

is a linear function. A system from the second type is given by $S_{\beta_{1}}=\{1,2\},S_{\beta_{2}}=\{3\},(l_{1},l_{2},l_{3})=(1,1,2)$ and equations as follows

[TABLE]

That is we obtain that

[TABLE]

It is a decomposition of type (c) in the theorem. Let $t=3.$ There are 6 systems of equations, all of the same type, e.g. $S_{\beta_{1}}=\{1\},S_{\beta_{2}}=\{2\},S_{\beta_{3}}=\{3\},(l_{1},l_{2},l_{3})=(1,1,1)$ and

[TABLE]

Hence the degree of $h$ is 1, that yields a trivial decomposition. ∎

4. Proof of Proposition 1

Proof of Proposition 1.

In this section we apply Theorem 1 to determine composite rational functions having zeros and poles as consecutive elements of certain arithmetic progressions. We need to handle the following cases

[TABLE]

In the proof we use the notation of Theorem 1, that is we write

[TABLE]

where $\alpha_{0},d\in k$ and $\{T_{1},T_{2},T_{3},T_{4}\}=\{0,1,2,3\}.$

$(I):t=2,\{|S_{\beta_{1}}|,|S_{\beta_{2}}|\}=\{1,3\}.$ We may assume that $S_{\beta_{1}}=\{1\},S_{\beta_{2}}=\{2,3,4\}.$ We obtain that

[TABLE]

Substituting $x=\alpha_{2},\alpha_{3},\alpha_{4}$ yields (assuming $l_{2}l_{3}l_{4}\neq 0$)

[TABLE]

Since the zeros and poles form an arithmetic progression one gets that either $d=0$ or $l_{1}=0.$ In the former case the zeros and poles are not distinct, a contradiction. In the latter case the degree of $h$ is less than 2, a contradiction as well. If two out of $l_{2},l_{3},l_{4}$ are equal to zero, then it follows that $l_{1}=1,$ hence the degree of $h$ is 1, a contradiction. If exactly one out of $l_{2},l_{3},l_{4}$ is zero, then $l_{1}=2$ and the corresponding $f$ has only 3 zeros and poles. As an example we consider the case $l_{4}=0.$ We obtain that

[TABLE]

It follows that $f(x)=\left(x-\frac{\alpha_{2}+\alpha_{3}}{2}\right)^{2}f_{1}(x),$ where $\deg f_{1}=2.$

$(I):t=2,\{|S_{\beta_{1}}|,|S_{\beta_{2}}|\}=\{2\}.$ Here we may assume that $S_{\beta_{1}}=\{1,2\},S_{\beta_{2}}=\{3,4\}.$ We get that

[TABLE]

It follows that (assuming that $0\notin\{l_{1},l_{2},l_{3},l_{4}\}$)

[TABLE]

and

[TABLE]

Using the fact that the zeros and poles form an arithmetic progression it turns out that one has to deal with 80 cases.

• •

There are 8 cases with $(l_{1},l_{2},l_{3},l_{4})=(1,1,1,1).$ We obtain equivalent solutions, so we only consider one of these. Let $\alpha_{1}=\alpha_{0},\alpha_{2}=\alpha_{0}+3d.$ It follows that $\beta_{2}=\beta_{1}-2d^{2}.$ That is we have

[TABLE]

• •

There are 16 equivalent cases with $(l_{1},l_{2},l_{3},l_{4})\in\{(1,1,2,2),(2,2,1,1)\}.$ One obtains that $d^{2}=\pm\frac{1}{2}$ and $\beta_{2}=\beta_{1}\pm 1.$ One example from this family is given by

[TABLE]

where $f_{2}(x)$ is a quadratic polynomial such that $f$ has more than 4 zeros and poles. We remark that if we use the equations related to $\beta_{2}$ we have

[TABLE]

that is we obtain a ”solution” covered by the family given by the case $(l_{1},l_{2},l_{3},l_{4})=(1,1,1,1).$

• •

There are 8 equivalent cases with $(l_{1},l_{2},l_{3},l_{4})=(2,2,2,2).$ All of these cases can be eliminated in the same way. From the equation

[TABLE]

it follows that

[TABLE]

where $\{T_{1},T_{2},T_{3},T_{4}\}=\{0,1,2,3\}.$ The left-hand side is $d^{0}=1$ and the right-hand side is -1, a contradiction.

• •

There are 16 equivalent cases with $(l_{1},l_{2},l_{3},l_{4})\in\{(1,1,3,3),(3,3,1,1)\}.$ As an example we handle the one with $(l_{1},l_{2},l_{3},l_{4})=(3,3,1,1)$ and

[TABLE]

Equation (8) implies that either $d=0$ or $d^{4}=\frac{1}{4}.$ If $d^{2}=\frac{1}{2},$ then we get

[TABLE]

where $f_{3}(x)$ is a quartic polynomial resulting an $f$ having more than 4 zeros and poles. If $d^{2}=-\frac{1}{2},$ then we get

[TABLE]

where $f_{4}$ is a quartic polynomial and we get a contradiction in the same way as before.

• •

There are 16 equivalent cases with $(l_{1},l_{2},l_{3},l_{4})\in\{(2,2,3,3),(3,3,2,2)\}.$ We handle the case with $(l_{1},l_{2},l_{3},l_{4})=(3,3,2,2)$ and

[TABLE]

It follows from equation (8) that $d=0$ or $d^{2}=\frac{1}{2}.$ Also we have that $\beta_{2}=\beta_{1}-1.$ In a similar way as in the above cases we obtain a composite function $f$ having 4 zeros and poles forming an arithmetic progression, but an additional quartic factor appears, a contradiction.

• •

There are 8 equivalent cases with $(l_{1},l_{2},l_{3},l_{4})=(3,3,3,3).$ Here we consider the case with

[TABLE]

It follows that $\beta_{2}=\beta_{1}-8d^{6}.$ As in the previous cases $g(h(x))$ has 4 zeros and poles coming from an arithmetic progression, but there is an additional quartic factor yielding a contradiction.

If $0\in\{l_{1},l_{2},l_{3},l_{4}\},$ then we have three possibilities. Either $\{l_{1},l_{2}\}=\{l_{3},l_{4}\}=\{0,1\}$ or $\{l_{1},l_{2}\}=\{1\},\{l_{3},l_{4}\}=\{0,2\}$ or $\{l_{1},l_{2}\}=\{0,2\},\{l_{3},l_{4}\}=\{1\}.$ In the first case the degree of $h$ is 1, a contradiction. The last two cases can be handled in the same way, therefore we only deal with the case $\{l_{1},l_{2}\}=\{1\},\{l_{3},l_{4}\}=\{0,2\}.$ Without loss of generality we may assume that $l_{3}=2,l_{4}=0.$ It follows that $\alpha_{1}=2\alpha_{3}-\alpha_{2}$ and $\beta_{2}=\beta_{1}-(\alpha_{2}-\alpha_{3})^{2}.$ Thus

[TABLE]

We conclude that $f(x)$ has only 3 zeros and poles, a contradiction.

$(I):t=3,|S_{\beta_{1}}|=|S_{\beta_{2}}|=1,|S_{\beta_{3}}|=2.$ Here we may assume that $S_{\beta_{1}}=\{1\},S_{\beta_{2}}=\{2\},S_{\beta_{3}}=\{3,4\},$ that is one has

[TABLE]

where $l_{1},l_{2}\in\{2,3\}.$ Let us consider the case $l_{3}\neq 0,l_{4}\neq 0.$ Substitute $\alpha_{3},\alpha_{4}$ into the above system of equations to get

[TABLE]

These equations imply that $\alpha_{i}=\alpha_{j}$ for some $i\neq j,$ a contradiction. Now assume that $l_{4}=0,$ hence $l_{3}=2$ or 3. We can reduce the system as follows

[TABLE]

where $l_{1},l_{2},l_{3}\in\{2,3\}.$ We get a contradiction in all these cases.

$(I):t=4,S_{\beta_{1}}=\{1\},S_{\beta_{2}}=\{2\},S_{\beta_{3}}=\{3\},S_{\beta_{4}}=\{4\}.$ We obtain the system of equations

[TABLE]

where $l_{i}\geq 2$ (since $\deg h\geq 2.$) Here we prove that this type of composite rational function cannot exist. One has that for any different $i,j$

[TABLE]

If $l_{i}=l_{j}=2,$ then we have a contradiction. Assume that $l_{i}=2.$ There exist $l_{j}=l_{k}=3.$ Hence $\alpha_{i}=\alpha_{j}-1$ and $\alpha_{i}=\alpha_{k}-1,$ a contradiction. Let us deal with the case $(l_{1},l_{2},l_{3},l_{4})=(3,3,3,3).$ Substituting $\alpha_{1}+\alpha_{2}$ into the system of equations yields $\beta_{1}=\beta_{2}+\alpha_{1}^{3}-\alpha_{2}^{3}.$ We also have that $\beta_{1}=\beta_{2}+(\alpha_{1}-\alpha_{2})^{3}.$ By combining these equations we get that

[TABLE]

In a similar way we obtain

[TABLE]

It follows that for some different $i,j$ one has $\alpha_{i}=\alpha_{j},$ a contradiction.

$(II):t=2,|S_{\infty}|=2,|S_{\beta_{1}}|=|S_{\beta_{2}}|=1.$ We may assume that $S_{\infty}=\{1,2\},S_{\beta_{1}}=\{3\},S_{\beta_{2}}=\{4\}.$ The system of equations in this case is as follows

[TABLE]

If $l_{3}=l_{4}=0,$ then it follows that $\beta_{1}=\beta_{2},$ a contradiction. Let us deal with the case $l_{3}=0,l_{4}\neq 0$ (in a similar way one can handle the case $l_{3}\neq 0,l_{4}=0$). There are only three systems to consider. If $(l_{1},l_{2},l_{3},l_{4})=(0,1,0,1)$ or $(1,0,0,1),$ then $\beta_{1}-1=\beta_{2}$ and the composite function $f$ has only 2 zeros and poles, a contradiction. If $(l_{1},l_{2},l_{3},l_{4})=(1,1,0,2),$ then $\beta_{1}-1=\beta_{2}$ and $\alpha_{4}=\alpha_{2}\pm 1,\alpha_{1}=\alpha_{2}\pm 2.$ In all these cases we obtain a composite function $f$ having only 3 zeros and poles, a contradiction. Let us consider the cases with $l_{3}\neq 0,l_{4}\neq 0.$ There are 18 systems to deal with. It turns out that $d$ satisfies the equation

[TABLE]

where $\alpha_{i}=\alpha_{0}+T_{i}d$ for some $T_{i}\in\{0,1,2,3\}.$ If $(l_{1},l_{2},l_{3},l_{4})=(1,0,2,2),$ then

[TABLE]

In all these cases we obtain a composite function $f$ having only 3 zeros and poles, a contradiction. As an example we compute $f$ when $(T_{1},T_{2},T_{3},T_{4})=(1,3,0,2).$ We get that $\beta_{2}=\beta_{1}+4d$ and

[TABLE]

We exclude the tuple $(l_{1},l_{2},l_{3},l_{4})=(0,1,2,2)$ following the same lines. If $(l_{1},l_{2},l_{3},l_{4})=(1,1,1,2),$ then we also have that $d=\frac{1}{T_{1}+T_{2}-2T_{4}}$ and $d=\frac{T_{2}-T_{3}}{(T_{2}-T_{4})^{2}},$ it is easy to check that such tuple $(T_{1},T_{2},T_{3},T_{4})$ does not exist. In a very similar way if $(l_{1},l_{2},l_{3},l_{4})=(1,1,2,1)$ we obtain that

[TABLE]

and such tuple $(T_{1},T_{2},T_{3},T_{4})$ does not exist. If $(l_{1},l_{2},l_{3},l_{4})=(2,1,2,3),$ then

[TABLE]

There is no solution in $T_{i}\in\{0,1,2,3\},T_{i}\neq T_{j},i\neq j.$ We obtain a very similar system of equations in case of $(l_{1},l_{2},l_{3},l_{4})=(1,2,3,2),(1,2,2,3),(2,1,3,2).$ If $(l_{1},l_{2},l_{3},l_{4})=(1,1,3,3),$ then we get

[TABLE]

The above system has no solution in $(T_{1},T_{2},T_{3},T_{4}).$ If $(l_{1},l_{2},l_{3},l_{4})=(1,2,3,1),$ then

[TABLE]

The system has no solution. The same argument works in case of $(l_{1},l_{2},l_{3},l_{4})=(1,2,1,3),(2,1,1,3),(2,1,3,1).$ If $(l_{1},l_{2},l_{3},l_{4})=(0,2,2,1),$ then we have

[TABLE]

hence

[TABLE]

That is $f$ has only 3 zeros and poles, a contradiction. We handle in the same way the tuples $(l_{1},l_{2},l_{3},l_{4})=(2,0,2,1),(2,0,1,2),(0,2,1,2).$ If $(l_{1},l_{2},l_{3},l_{4})=(0,0,1,1),$ then $\deg h(x)=1,$ a contradiction.

$(II):t=3,|S_{\infty}|=|S_{\beta_{1}}|=|S_{\beta_{2}}|=|S_{\beta_{3}}|=1.$ We may assume that $S_{\infty}=\{1\},S_{\beta_{1}}=\{2\},S_{\beta_{2}}=\{3\},S_{\beta_{3}}=\{4\}.$ In this case $h(x)$ can be written as follows

[TABLE]

The only possible exponent tuple $(l_{1},l_{2},l_{3},l_{4})$ is $(0,1,1,1).$ Thus $\deg h(x)=1,$ a contradiction.

$(III):t=3,|S_{\beta_{1}}|=2,|S_{\beta_{2}}|=|S_{\beta_{3}}|=1.$ We may assume that $S_{\beta_{1}}=\{1,2\},S_{\beta_{2}}=\{3\},S_{\beta_{3}}=\{4\}.$ The only exponent tuple for which $\deg h(x)>1$ is given by $(l_{1},l_{2},l_{3},l_{4})$ is $(1,1,2,2).$ We obtain the following system of equations if $d\neq 0:$

[TABLE]

where $\{T_{1},T_{2},T_{3},T_{4}\}=\{0,1,2,3\}.$ Solving the above system of equations for all possible tuples $(T_{1},T_{2},T_{3},T_{4})$ one gets that $\beta_{i}=\beta_{j}$ for some $i\neq j,$ a contradiction.

$(III):t=3,|S_{\beta_{1}}|=|S_{\beta_{2}}|=|S_{\beta_{3}}|=|S_{\beta_{4}}|=1.$ We may assume that $S_{\beta_{1}}=\{1\},S_{\beta_{2}}=\{2\},S_{\beta_{3}}=\{3\},S_{\beta_{4}}=\{4\}.$ The only possible exponent tuple is $(l_{1},l_{2},l_{3},l_{4})=(1,1,1,1).$ Thus the corresponding $h(x)$ has degree 1, a contradiction. As an example we consider the case

[TABLE]

We use equation (3) here with $(j_{1},j_{2},j_{3})=(1,2,3)$ and $(j_{1},j_{2},j_{3})=(1,2,4).$ If $d\neq 0,$ then we have

[TABLE]

Let $k_{1},k_{2},k_{3},k_{4}\in\mathbb{Z}$ such that $k_{1}+k_{2}+k_{3}+k_{4}=0.$ Theorem A implies that

[TABLE]

∎

5. Cases with $n=4$

In this section we provide some details of the computation corresponding to cases with $n=4,t\in\{2,3,4\},k_{1}+k_{2}+\ldots+k_{t}\neq 0,S_{\infty}=\emptyset.$ These are the cases which are not mentioned in Section 5 in [12].

The case $n=4,t=2$ and $S_{\infty}=\emptyset.$ There are 134 systems to deal with. We treat only a few representative examples.

If $S_{\beta_{1}}=\{1,2\},S_{\beta_{2}}=\{3,4\}$ and $(l_{1},l_{2},l_{3},l_{4})=(2,1,2,1),$ then we have

[TABLE]

The corresponding rational functions are as follows

[TABLE]

where $k_{1}+k_{2}\neq 0.$ We note that the zeros and poles of $f$ do not form an arithmetic progression for all values of the parameters as the choice $\alpha_{1}=0,\alpha_{2}=3$ shows.

If $S_{\beta_{1}}=\{1,2\},S_{\beta_{2}}=\{3,4\}$ and $(l_{1},l_{2},l_{3},l_{4})=(1,1,0,2),$ then we get the system of equations

[TABLE]

It yields a decomposable rational function $f$ having only 3 zeros and poles altogether.

If $S_{\beta_{1}}=\{1,2\},S_{\beta_{2}}=\{3,4\}$ and $(l_{1},l_{2},l_{3},l_{4})=(1,1,1,1),$ then we obtain

[TABLE]

It yields the following solution

[TABLE]

where $k_{1}+k_{2}\neq 0.$

If $S_{\beta_{1}}=\{1,2,3\},S_{\beta_{2}}=\{4\}$ and $(l_{1},l_{2},l_{3},l_{4})=(1,1,1,3),$ then we have

[TABLE]

We obtain the following rational functions

[TABLE]

where $k_{1}+k_{2}\neq 0$ and

[TABLE]

The case $n=4,t=3$ and $S_{\infty}=\emptyset.$ There are 48 systems to handle in this case. We consider one of these. Let $S_{\beta_{1}}=\{1\},S_{\beta_{2}}=\{2,3\},S_{\beta_{3}}=\{4\}$ and $(l_{1},l_{2},l_{3},l_{4})=(1,1,0,1).$ We obtain the system of equations

[TABLE]

It follows that $h$ is a linear function, which only provides trivial decomposition. In the remaining cases we have the same conclusion.

The case $n=4,t=4$ and $S_{\infty}=\emptyset.$ Here we get 24 systems to consider. In all cases we have that

[TABLE]

and $(l_{1},l_{2},l_{3},l_{4})=(1,1,1,1).$ Therefore $h$ is linear, a contradiction.