# Belyi maps from zeroes of hypergeometric polynomials

Raimundas Vidunas
Institute of Applied Mathematics, Vilnius University

###### Abstract

Evaluation of low degree hypergeometric polynomials to zero defines an
algebraic hypersurface in the affine space of the free parameters and the
argument. This article investigates the algebraic surfaces ${}_{2}\mbox{\rm
F}_{\\!1}(-N,b;c;z)=0$ for $N=3$ and $N=4$. As a captivating application,
these surfaces parametrize certain families of genus 0 Belyi maps.

##  1 Introduction

Systematic cataloging of Belyi maps is an active research undertaking [MSSV19,
AS22]. Contributing to this venture, this article considers comprehensively
Belyi maps $\varphi:\mathbb{P}^{1}\to\mathbb{P}^{1}$ of the form

$\varphi(x)=(1-x)^{p}(1-\lambda x)^{q}\,G_{m}(x)^{r}$ (1.1)

or

$\varphi(x)=(1+\alpha x+\beta x^{2})^{p}\,G_{m}(x)^{r},$ (1.2)

where $G_{m}(x)$ is a polynomial of degree $m$ without multiple roots, and

$\varphi(x)=1+O(x^{m+2}).$ (1.3)

This implies $G_{m}(0)=1$. The powers $p,q,r$ are allowed to be positive or
negative integers. If the powers are positive integers, then the considered
Belyi maps are called Shabat polynomials [LZ04, §2.2]. The powers $p,q,r$ are
assumed to be different.

Assuming that the point $x=\infty$ is above $\varphi\in\\{0,\infty\\}$, the
distinct points in the three canonical fibers $\varphi\in\\{0,1,\infty\\}$
are:

  * •

two roots of $(1-x)(1-\lambda x)$ or $1+\alpha x+\beta x^{2}$, of branching
order $|p|$ or $|q|$, above $\varphi=0$ or $\varphi=\infty$ depending on the
signs of $p,q$;

  * •

the roots of $G_{m}(x)$, of branching order $|r|$, above
$\varphi\in\\{0,\infty\\}$;

  * •

the point $x=\infty$, of branching order $|p+q+mr|\neq 0$ or $|2p+mr|\neq 0$
respectively for the two forms, above $\varphi\in\\{0,\infty\\}$;

  * •

the point $x=0$, of branching order $m+2$, above $\varphi=1$;

  * •

$d-(m+2)$ non-branching points above $\varphi=1$, where $d$ is the degree of
$\varphi$.

In total we count

$2+m+1+1+(d-m-2)=d+2$ (1.4)

distinct points in the 3 fibers, which is the minimal number of points in 3
fibers for a covering $\mathbb{P}^{1}\to\mathbb{P}^{1}$ by the Riemann-Hurwitz
formula [Sil09, Theorem 5.9], and the number of points that Belyi maps of
genus 0 must have in the 3 fibers. This will be recapped in Lemma 5.1.

As it turns out, the considered families of Belyi maps are parametrized by the
zero sets of hypergeometric polynomial equations, namely

${}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-m-1,\,\frac{q}{r}\atop-m-\frac{p}{r}}\right|\,\lambda\right)=0$
(1.5)

for the Belyi maps of the form (1.1), and

${}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-\frac{m+1}{2},-\frac{m}{2}\atop-m-\frac{p}{r}}\right|\frac{4\beta}{\alpha^{2}}\right)=0$
(1.6)

for the form (1.2). These equations give the generic number of distinct Belyi
maps (up to Möbius transformations) of the considered branching patterns,
namely $n+1$ for fixed $p,r,q,m$ in the form (1.1), of $\lceil n/2\rceil$ for
fixed $p,q,m$ in (1.2). As we show in Lemmas 2.1 and 2.2, this number is
smaller exactly when one of the quotients $p/r$, $q/r$ is a smaller positive
integer. It appears that the Galois orbits of Belyi maps with the considered
branching patterns have the maximal size typically.

The considered Belyi maps constitute a borderline easy case that continues
extensively the examples in [LZ04, §2.2] and [APZ20]. The new examples and
methodology will not be radically novel for active readers of [APZ20], [LZ04].
In particular, we encounter new families of Belyi maps defined over
$\mathbb{Q}$ that are parametrized by:

  * •

rational points on an elliptic curve, as in [LZ04, §2.2.4.3]; see Examples
6.6, 7.5 and 7.6 here.

  * •

integer solutions of Pell’s equation, as in [APZ20, Ch. 10]; see Example 7.10.

  * •

rational points on a cubic surface or a cubic pencil; see Examples 6.2, 6.3.

  * •

integral points on elliptic curves or elliptic surfaces; see Example 6.7.

  * •

conic curves defined over $\mathbb{Q}$ that may have no rational points; see
Example 6.5.

The investigation of this article starts with recalling the relevant
properties of hypergeometric functions. Sections 3 and 4 considers the
algebraic surfaces defined

${}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{-N,\,b\,\atop c}\right|z\right)=0,$
(1.7)

with $N=3$ and $N=4$. We modify the lower parameter $-c+1-N$, so that after
clearing the denominators we have the symmetric identity

$(c)_{m}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-N,\,b\atop-c+1-N}\right|z\right)=(b)_{m}\,z^{N}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-N,\,c\atop-b+1-N}\right|\frac{1}{z}\,\right)$ (1.8)

of the same hypergeometric summation in the opposite directions. The
parameters $b,c$ are thereby interchangeable in combination with the Möbius
transformation $z\mapsto 1/z$. These equations with $N\in\mathbb{Z}$ define
algebraic surfaces in the affine space defined by the coordinates $b,c,z$.
Further symmetries of these hypergeometric polynomials are presented in
(2.3)–(2.6).

The technical contribution of this article is careful consideration of the
hypergeometric polynomial equations, including their degenerations, and
analysis of their algebraic structure and elliptic fibrations.

The article continues immediately with a primer on hypergeometric functions
and their properties that will used. In the next section we start with several
easier examples of Belyi maps that are largely covered in [LZ04, §2.2].
Thereby we gradually build up the basic methodology, illustrate arising
technical details, and set the context of where this article goes beyond the
established understanding.

##  2 Hypergeometric details

The Gauss hypergeometric function [AAR99, Ch. 2] is defined by the series

$\displaystyle{}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{a,\,b\,\atop
c}\right|\,z\right)=\sum_{k=0}^{\infty}\frac{(a)_{k}\,(b)_{k}}{(c)_{k}\,k!}\,z^{k}.$
(2.1)

Here $(\alpha)_{k}=\alpha\,(\alpha+1)(\alpha+2)\ldots(\alpha+k-1)$ denotes the
Pochhammer symbol, or the “raising” variant of the factorial. Using the
$\Gamma$-function [AAR99, Ch. 1] we can write

$\displaystyle(\alpha)_{k}=\,\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}.$ (2.2)

The standard analytic continuation of the ${}_{2}\mbox{\rm F}_{\\!1}$-function
is onto $\mathbb{C}\setminus[1,\infty)$. This function is undefined when $c$
equals zero or a negative integer.

###  2.1 Hypergeometric polynomials

If well-defined, the hypergeometric series (2.1) is a polynomial in $z$ when
$a$ or $b$ are non-positive integers. In addition to (1.8), we have these
symmetries of hypergeometric polynomials:

$\displaystyle(c)_{m}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-N,\,b\atop-c+1-N}\right|z\right)$
$\displaystyle=(-1)^{N}(b+c)_{m}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-N,\,b\,\atop b+c}\right|1-z\right)$ (2.3)
$\displaystyle=(b+c)_{m}\,(-z)^{N}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-N,\,c\,\atop b+c}\right|1-\frac{1}{z}\,\right)$
(2.4) $\displaystyle=(c)_{m}\,(1-z)^{N}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-N,\,e\atop-c+1-N}\right|\frac{z}{z-1}\right)$ (2.5)
$\displaystyle=(b)_{m}\,(z-1)^{N}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-N,\,e\atop-b+1-N}\right|\frac{1}{1-z}\right)\\!.$
(2.6)

where $e=-b-c+1-N$. The following expression defines a polynomial of degree
$\lfloor N/2\rfloor$ for any integer $N\geqslant 0$:

${}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{-\frac{N}{2},-\frac{N-1}{2}\atop
c}\right|z\right)$ (2.7)

If $c$ is a non-negative integer, but $a$ or $b$ is a non-negative integer
$\geqslant c$, the hypergeometric summation (2.1) can be considered as a
polynomial, of degree $-a$ or $-b$. This interpretation is often used in
theory of orthogonal polynomials [KLS10]. We adopt this interpretation of
hypergeometric functions like

${}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{-n,\,b\atop-n-m}\right|z\right)$
(2.8)

with integer $n,m\geqslant 0$. The well-known identities

$\displaystyle{}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{a,\,b\,\atop
c}\right|\,z\right)=$ $\displaystyle\,(1-z)^{-a}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{a,\,c-b\,\atop c}\right|\,\frac{z}{z-1}\right)$ (2.9)
$\displaystyle=$ $\displaystyle\,(1-z)^{c-a-b}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{c-a,\,c-b\,\atop c}\right|\,z\right)$ (2.10)

of Pfaff and Euler [AAR99, Th. 2.2.5] should not be automatically applied
then; see [Vid07, Lemma 3.1]. Yet these identities are valid when all three
numbers $a,b,c$ are non-negative integers, and $-c\geqslant\max(-a,-b)$; see
[Vid07, §9].

###  2.2 Distinctive roots of hypergeometric polynomials

Distinctiveness of roots of hypergeometric polynomials is easily proved by
using contiguous relations [AAR99, §2.5, §3.7] of ${}_{2}\mbox{\rm
F}_{\\!1}$-functions. In particular, we use the observations in the following
two lemmas.

######  Lemma 2.1.

Consider the sequence of hypergeometric polynomials

$P(k)={}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{-k,\,b\atop
1-k-c}\right|z\right),$ (2.11)

with some $b,c\in\mathbb{C}$. Suppose that for a positive integer $k$ we have
$(b)_{k}\neq 0$, $(c)_{k}\neq 0$, and $(b+c)_{k}\neq 0$. Then:

  1. (i)

The polynomials $P(k)$ and $P(k-1)$ do not have common roots.

  2. (ii)

The polynomial $P(k)$ has $k$ distinct roots.

  3. (iii)

$z=1$ is not a root of $P(k)$.

###### Proof.

The sequence $P(k)$ satisfies the recurrence relation

$\frac{k(k-1+b+c)z}{k-1+c}\,P(k-1)=(kz+k+bz+c)\,P(k)-(k+c)\,P(k+1).\ $ (2.12)

This follows from the contiguous relations [Vid03]; or by applying
Zeilberger’s algorithm [AAR99, §3.11]; or by checking the series expansion. We
apply the recurrence with decreasing $k$, and use the assumption that
$(b+c)_{k}\neq 0$. The recurrence implies that a common root of $P(k)$ and
$P(k-1)$ is either $z=0$, which conflicts with the hypergeometric values $1$
at $z=0$, or it would be a common root of $P(k-2),\ldots$, and of $P(0)=1$
contradictorily.

The second statement follows from the degree of $k$ of $P(k)$ thanks to
$(b)_{k}\neq 0$, and the fact that another contiguous relation

$\displaystyle\frac{1-z}{k}\,\frac{dP(k)}{dz}+P(k)=\frac{k-1+b+c}{k-1+c}\,P(k-1)$
(2.13)

implies that a multiple root of $P(k)$ would be also a root of $P(k-1)$,
leading to the established first statement.

The last statement follows from the Chu-Vandermonde identity [AAR99, p. 67]

${}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{-k,B\atop
C}\right|1\right)=\frac{(C-B)_{k}}{(C)_{k}},$ (2.14)

giving the value $(b+c)_{k}/(c)_{k}$ for $P(k)$ evaluated at $z=1$. ∎

######  Lemma 2.2.

Consider the sequence of hypergeometric polynomials

$P(k)={}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-\frac{k}{2},\,-\frac{k-1}{2}\atop
1-k-c}\right|z\right),$ (2.15)

with some $c\in\mathbb{C}$. Suppose that for a positive integer $k$ we have
$(c)_{k}\neq 0$ and $(2c)_{k}\neq 0$. Then:

  1. (i)

The polynomials $P(k)$ and $P(k-1)$ do not have common roots.

  2. (ii)

The polynomial $P(k)$ has $\lfloor k/2\rfloor$ distinct roots.

  3. (iii)

$z=1$ is not a root of $P(k)$.

###### Proof.

The sequence $P(k)$ satisfies the recurrence relation

$P(k)-P(k+1)=\frac{kz\,(2c+k-1)}{4(c+k)(c+k-1)}\,P(k-1).$ (2.16)

Under the assumption that $(b+c)_{k}\neq 0$, the recurrence shows that a
common root of $P(k)$ and $P(k+1)$ is either the rejectable $z=0$, or it would
be a common root of $P(k-1)$, $P(k-2),\ldots$, and of $P(0)=1$
contradictorily.

The second statement follows from the degree of $k$ of $c_{k}$, thanks to
$(a)_{k}\neq 0$, and the fact that another contiguous relation

$\displaystyle\frac{2(1-z)}{k}\frac{dP(k)}{dz}+P(k)=\frac{2c+k-1}{2(k+c-1)}\,P(k-1)$
(2.17)

implies that a multiple root of $P(k)$ would be also a root of $P(k+1)$,
leading to the established first statement.

The last statement follows from applying the Chu-Vandermonde identity (2.14).
The evaluation at $z=1$ is

$\left.\left({\textstyle
c+\frac{1}{2}}\,\right)_{\lfloor\frac{k}{2}\rfloor}\,\right/\left(c+\left\lceil{\textstyle\frac{k}{2}}\right\rceil\right)_{\lfloor\frac{k}{2}\rfloor}$
(2.18)

for both even and odd $k$. ∎

##  3 Cubic hypergeometric polynomials

After clearing the denominators in the equation $\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-3,\,b\atop\\!-c-2}\right|z\right)=0\;$ we obtain

$b(b+1)(b+2)z^{3}+3bc(b+1)z^{2}+3bc(c+1)z+c(c+1)(c+2)=0,$ (3.1)

or rather more compactly,

$(bz+c)^{3}+3(bz+c)(bz^{2}+c)+2(bz^{3}+c)=0.$ (3.2)

Let ${\mathcal{S}}_{3}$ denote the algebraic surface defined by (3.1), of
degree 6.

Fixing $z$ leads to a cubic equation in $b,c$. For $z\in\\{0,1\\}$, the
equation factorizes into 3 linear factors. For $z\in\\{-1,2,\frac{1}{2}\\}$,
it factorizes into a quadratic and a linear parts. For other fixed $z$, we get
an irreducible cubic curve of genus $0$. Here is a parametrization of those
cubic curves by $e=bz+c$:

$b=\frac{e\,(e+1)\,(e+2)}{z\,(1-z)\,(3e+2z+2)},\qquad
c=\frac{e\,(e+z)\,(e+2z)}{(z-1)\,(3e+2z+2)}.$ (3.3)

This also gives a birational parametrisation of ${\mathcal{S}}_{3}$ by $z$ and
$e$. If $3e+2z+2=0$ in the denominators, then the polynomial in (3.1) can be
factored as

$\textstyle-\frac{4}{27}(z+1)(z-2)(2z-1).$ (3.4)

Both numerators in (3.3) equal 0 as well at the exceptional 3 points
$(e,z)\in\\{(0,-1),(-2,2),(-1,\frac{1}{2})\\}$. The 3 points are blown up to
these lines on ${\mathcal{S}}_{3}$:

$\textstyle b=c,z=-1;\quad 2b+c+2=0,z=2;\quad b+2c+2=0,z=\frac{1}{2}.$ (3.5)

We will be interested in the points on ${\mathcal{S}}_{3}$ with both $b,c$
positive, while $z\not\in\\{0,1\\}$ and $b\neq c$. Dependence of the signs of
$b,c$ on the parameters $e,z$ by the parametrization (3.3) is depicted in
Figure 1(i). There are two regions marked by + where $b$ and $c$ are positive;
they are delineated by

$z<0\qquad\mbox{and}\qquad e\,(3e+2z+2)<0.$ (3.6)

The exceptional lines (3.4) do not apply to this interest.

$e$$z$$t$$y$$=$$\pm$$=$$+$$+$$=$$\mp$$=$$\pm$$=$$\pm$$\mp$$\mp$$\pm$$=$$\mp$$\mp$$=$$\pm$$\mp$$\mp$$\pm$$=$$\pm$$\pm$$\mp$$=$$\pm$$\mp$$=$$\mp$$=$$\mp$$\pm$$=$$=$$\pm$$\mp$$\pm$$=$$\mp$$=$$\mp$$=$$\mp$$\pm$$=$$\mp$$=$$\pm$(i)(ii)
Figure 1: The signs of $b$ and $c$ depending: (i) on $e,z$ in formula (3.3);
(ii) on $t,y$ in formula (3.14). The upper sign in $\pm,\mp,=$ is for $b$,
while the lower one is for $c$. There sign + indicates both $b,c$ positive.

###  3.1 Complete factorization for some $b,c$

Suppose that the parameters $b,c$ have the form (3.3) for some
$z=s\in\mathbb{Q}$ and $e\in\mathbb{Q}$. Considering (3.1) as a cubic
polynomial in $z$, it then has the root $z=s$. For generic values of $e,s$,
the other two roots solve the quadratic equation

$\displaystyle\\!\frac{e^{2}\\!-\\!2es\\!-\\!2s^{2}\\!+\\!3e\\!+\\!2}{e+2s-2}\,z^{2}-(2e+s+1)sz+\frac{e^{2}\\!+\\!3es\\!+\\!2s^{2}\\!-\\!2e\\!-\\!2}{e-2s+2}\,s^{2}$
$\displaystyle=0.$ (3.7)

We are interested when those other roots are in $\mathbb{Q}$ as well. For
that, the discriminant

$\frac{3s^{2}(s-1)^{2}(e+2s+2)(3e+2s+2)}{(e+2s-2)\,(e-2s+2)}.$ (3.8)

must be a full square. We apply the birational transformation

$e=2\,\frac{\xi-1}{\eta-\xi},\qquad s=\frac{\eta+1}{\eta-\xi},$ (3.9)

and the discriminant becomes

$\frac{3\eta(1-\xi-\eta)(\xi+1)^{2}(\eta+1)^{2}}{\xi\,(\eta-\xi)^{4}}.$ (3.10)

The values $\xi,\eta$ should therefore determine a $\mathbb{Q}$-rational point
on the cubic surface

$3\varrho^{2}=\xi\eta(1-\xi-\eta).$ (3.11)

This cubic surface has three singularities
$(\xi,\eta,\varrho)\in\\{(0,0,0),(0,1,0),(0,0,1)\\}$. By intersecting this
surface with the pencil of lines $\xi=t\varrho$, $\eta=y\varrho$ through the
singularity $(0,0,0)$ we obtain this birational parametrization by
$(t,y)\in\mathbb{P}^{2}$:

$\xi=\frac{ty-3}{y(y+t)},\qquad\eta=\frac{ty-3}{t(y+t)},\qquad\varrho=\frac{ty-3}{ty(y+t)}.$
(3.12)

This translates to

$e=\frac{2t\,(y^{2}+3)}{(t-y)(ty-3)},\qquad
s=\frac{y\,(t^{2}+2ty-3)}{(y-t)(ty-3)},$ (3.13)

and ultimately to

$b=-\frac{(t^{2}+3)(y^{2}+3)\,(ty+3)}{3(t^{2}+2ty-3)(y^{2}+2ty-3)},\quad
c=\frac{(y^{2}+3)(t^{2}y-6t-3y)}{3(y-t)(y^{2}+2ty-3)}.$ (3.14)

This parametrizes the cases when the cubic polynomial (3.1) factors
completely. The special case $y=-t$ gives the following neat family:

$\textstyle b=\frac{1}{3}\,t^{2}-1,\qquad c=-\frac{1}{6}\,t^{2}-\frac{1}{2}.$
(3.15)

The algebraic map (3.13) is generically 2-to-1 as the central symmetry
$(t,y)\mapsto(-t,-y)$ keeps $e,s$ invariant and permutes the roots of (3.7).
Up to this symmetry, the inverse map to (3.13) is

$t=\sqrt{\frac{3(e+2s-2)(e-2s+2)}{(e+2s+2)(3e+2s+2)}},\quad
y=\sqrt{\frac{3(e+2s+2)(e-2s+2)}{(e+2s-2)(3e+2s+2)}}.$ (3.16)

There is no point $(t,y)\in\mathbb{R}^{2}$ where both $b$ and $c$ are positive
by the parametrization (3.14); see Figure 1(ii).

The following map moves $z=s$ to another root of (3.1):

$(t,y)\mapsto\left(\frac{t-3}{t+1},\,\frac{y-3}{y+1}\right).$ (3.17)

By composition with the central symmetry (on either or both sides) one
realizes all 6 permutations of the roots of (3.1) by fractional-linear
transformations of the parameters $t,y$.

###  3.2 Elliptic fibration by $b$ or $c$

For fixed $b\not\in\\{0,-1,-2\\}$, equation (3.1) defines a cubic curve of
genus 1. Its $j$-invariant equals

$.-\frac{27\,b^{2}\,(9b^{2}+32b+32)^{3}}{64\,(b+1)^{3}\,(b+2)^{2}}.$ (3.18)

We can consider ${\mathcal{S}}_{3}$ as an elliptic surface [Sil94, III.11]
over $\mathbb{Q}(b)$. Here is an isomorphic Weierstrass form for the elliptic
surface, and a characterization of rational curves on it.

######  Proposition 3.1.

For $b\not\in\\{0,-1,-2\\}$, the cubic relation $(\ref{eq:hpg3})$ between $c$
and $z$ defines a curve of genus $1$, isomorphic to the elliptic curve

${\mathcal{E}}_{3}:\quad v^{2}=u^{3}+b^{2}\,(3u-16b-16)^{2}$ (3.19)

by the isomorphism

$\displaystyle c=-(b+2)\frac{v+3bu-16b(b+1)}{2v},\quad
z=\frac{v+(3b+4)u-16b(b+1)}{2v}.$ (3.20)

The Mordell-Weil group for $(\ref{eq:gell})$ of rational points over
$\mathbb{Q}(b)$ is isomorphic to $(\mathbb{Z}/3\mathbb{Z})\times\mathbb{Z}$.
It is generated the torsion point $\big{(}0,16b(b+1)\big{)}$ and the point
$\big{(}8b,8b(b+2)\big{)}$.

###### Proof.

It is straightforward to check the isomorphism (3.20). The inverse map is
given

$\displaystyle u=\frac{16b(b+1)(bz+2z+c)}{bc+(b+2)(3bz+2c+2)},\quad
v=\frac{32b(b+1)(b+2)}{bc+(b+2)(3bz+2c+2)}.$ (3.21)

As a side note, the elliptic involution $v\mapsto-v$ corresponds to the
hypergeometric symmetry (2.3) under this isomorphism.

The canonical Weierstrass form $v^{2}=u^{3}+Au+B$ can be obtained by applying
the shift $u\mapsto u-3b^{2}$ in (3.19). The discriminant [Sil94, III.1] of
the elliptic surface equals $\Delta=2^{18}3^{3}b^{4}(b+1)^{3}(b+2)^{2}$. By
considering $A,B,\Delta$ and the $j$-invariant in (3.18), we can use [SS19,
Table 5.1] and conclude that the singular fibers $b\in\\{0,-1,-2,\infty\\}$
have the Kodaira types IV, I3, I2, I3, respectively. By [SS19, Fig. 5.1–5.2]
they have, respectively, $m_{v}=3,3,2,3$ irreducible components in the
Kodaira-Neron model. By Shioda-Tate formula [SS19, Corollary 6.7], the
Mordell-Weil rank equals $10-2-7=1$ as the Neron-Severi rank $\rho$ is $10$
for rational elliptic surfaces [SS19, §7.2]. By [SS19, Theorem 8.33], the
Mordell-Weil group is generated by polynomial points $(u,v)$ with $\deg
u\leqslant 2$, $\deg v\leqslant 3$. Direct computation with undetermined
coefficients gives 8 possibilities for such $u$:

$\displaystyle\textstyle\\!u\in\\{0,8b,-16b,4(b\\!+\\!1),16(b\\!+\\!1),-8b(b\\!+\\!1),16b(b\\!+\\!1),\frac{16}{9}(1\\!-\\!b)(1\\!+\\!2b)\\}.$
(3.22)

The points with $u=0$ are flex points with the tangents $\pm v=3bu-16b(b+1)$,
hence they are 3-torsion points. An investigation of the group relations
between all candidate points shows that we can take a point with $u\in\\{8b$,
$16(b\\!+\\!1)$, $-8b(b\\!+\\!1)\\}$ for a free generator. ∎

Taking $b\in\mathbb{Q}\setminus\\{0,-1,-2\\}$ in (3.19) gives an elliptic
curve over $\mathbb{Q}$, with the same Mordell-Weil group typically [Sil94,
III.11]. But the Mordell-Weil group group over $\mathbb{Q}$ could be larger
than the projected set of points from the elliptic surface. For example,
$b=-7$ gives the Mordell-Weil group
$(\mathbb{Z}/3\mathbb{Z})\times\mathbb{Z}^{2}$, as we will consider in Example
6.6.

It is instructive to consider several simplest sections $(u,v)$ on
${\mathcal{E}}_{3}$, parametrized by $b$, and map them to the original surface
${\mathcal{S}}_{3}$ defined by (3.1). Here are a few obtained non-degenerate
$c$-values of low degree in $b$:

$\frac{b+2}{4b-1},-\frac{4b(b+2)}{5b+1},-\frac{27b(b+1)}{16b^{2}+19b+1},-\frac{(2b+1)^{3}}{7b^{2}+b+1},-\frac{2(b-1)^{3}}{2b^{2}+17b-1}.$
(3.23)

After substituting these values into (3.1), there is a linear factor in $z$.
That parametrizes fully the corresponding section in $(c,z)$ on
${\mathcal{S}}_{3}$.

If we have a root described by (3.20) of the cubic hypergeometric polynomial
with fixed $c$ (and possibly $b$), the other two roots are equal to

$z_{2,3}=\frac{1}{2v}\left(v+(3b-2)u-16b(b+1)\pm
u\,\sqrt{\frac{3u}{b+1}-12}\right).$ (3.24)

To seek rationality of these roots, we may replace $u=(b+1)(w^{2}+12)/3$ and
obtain a family of hyperelliptic curves $(w,v)$ of genus 2. One may go through
a limited list of Mordell-Weil points on ${\mathcal{S}}_{3}$ (be it
specialized, with $b=-5/2$ for example) and check that apparently complete
factorization of the cubic polynomial happens generically only in degenerate
cases such as $c\in\\{-1,-b-1\\}$. We need $b$ to be expressible as in (3.14).
The family (3.15) corresponds to the infinite point on the elliptic surface.

For fixed $c\not\in\\{0,-1,-2\\}$ equation (3.1) has degree 6, but it defines
an isomorphic elliptic surface by the symmetry (1.8) of hypergeometric
polynomials. One can take $b\leftrightarrow c$, $z\mapsto 1/z$ in Proposition
3.1.

##  4 Quartic hypergeometric polynomials

A compact polynomial form of the equation $\>{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-4,\,b\atop-c-3}\right|z\right)=0\;$ is

$\big{(}(bz+c)^{2}+3bz^{2}+3c\big{)}^{2}+2(bz^{2}+c)^{2}+8bcz(z-1)^{2}+6(bz^{4}+c)=0.$
(4.1)

Let us denote this surface by ${\mathcal{S}}_{4}$. We will associate two
elliptic surfaces ${\mathcal{E}}_{4}$ and ${\mathcal{E}}^{*}_{4}$ to it.

###  4.1 Elliptic fibration by $z$

For fixed $z$, the curve has genus 1 generically. We identify
${\mathcal{S}}_{4}$ directly as an elliptic surface. As such, it is isomorphic
to

$\displaystyle{\mathcal{E}}_{4}:\quad
v^{2}=u^{3}-20Z\,u^{2}+108Z^{2}\,u-648(Z-1)^{2}.$ (4.2)

where $Z=z^{2}-z+1$. The $j$-invariant is rather untidy in the denominator:

$-\frac{2^{8}\,19^{3}\,Z^{6}}{3^{4}\,(72Z^{6}-860Z^{5}+3907Z^{4}-9608Z^{3}+13122Z^{2}-8748Z+2187)}.$
(4.3)

An isomorphism is given by

$b=-1+\varphi_{4}(z,u,v),\qquad
c=-1+\varphi_{4}\\!\left(\frac{1}{z},\frac{u}{z^{2}},\frac{v}{z^{3}}\right),$
(4.4)

with

$\varphi_{4}(z,u,v)=\frac{2\big{(}u^{2}-4Z^{2}\,u+8Z^{3}-12Z^{2}\big{)}(1-2z)v+2W_{1}}{z(1-z)\,\big{(}u^{2}-12Zu+12Z^{2}\big{)}^{2}},$
(4.5)

where

$\displaystyle\\!\\!W_{1}=$
$\displaystyle\;(5Z-3)u^{3}+6\,(2Z^{3}-24Z^{2}+27Z-9)u^{2}$ (4.6)
$\displaystyle-4Z(20Z^{3}-159Z^{2}+171Z-5)u+72Z^{2}(Z-1)(2Z^{2}-12Z+9).\quad$

The infinite point on (4.2) is mapped to the central point $(b,c)=(-1,-1)$
among the degenerations. The parametric expression for $bz+c$ simplifies to

$bz+c=\frac{2Zv-(z+1)\big{(}u^{2}-2(8Z-3z)\,u+36(z-1)^{2}Z\big{)}}{u^{2}-12Zu+12Z^{2}}.$
(4.7)

As we consider in the next subsection, ${\mathcal{E}}_{4}$ is a rational
elliptic surface. Its Mordell-Weil group has no torsion and the maximal rank 8
for elliptic surfaces over $\mathbb{C}\mathbb{P}^{1}$, as evident [SS19, §7.3]
from the irreducible degree 12 denominator of the $j$-invariant (4.3), meaning
that the surface ${\mathcal{E}}_{4}$ has 12 singular fibers of Kodaira type
$I_{1}$. By [SS19, Theorem 7.12(i)], there are 240 candidate points with a
polynomial coordinate $u(z)\in\mathbb{C}[z]$ of degree $\leqslant 2$ to
generate the Mordell-Weil group. Computations show that 60 of them are defined
over $\mathbb{Q}(z)$. They have

$u\in\\{6,9,18,54,9z^{2}-6z+9,18z^{2}+12z+18,2z^{2}+4z+6,6z^{2}+12\\},$ (4.8)

or could be obtained by further applying the hypergeometric symmetries (1.8),
(2.3)–(2.6), generated by $u(z)\mapsto z^{2}\,u(1/z)$ and $u(z)\mapsto
u(1-z)$. Further, the two points with $u=8z^{2}-8z+8$ are defined over
$\mathbb{Q}(\sqrt{6},z)$, and the points with $u=0$ are defined over
$\mathbb{Q}(\sqrt{-2},z)$. Additionally, 64 points have
$u\in\mathbb{Q}(\sqrt{6})[z]$, 64 points have $u\in\mathbb{Q}(\sqrt{-2})[z]$,
and 48 points have $u\in\mathbb{Q}(\sqrt{6},\sqrt{-2})[z]$. The Mordell-Weil
group over $\mathbb{Q}(z)$ has rank 6. It is generated by points with
$u\in\\{6,6z^{2},6(z-1)^{2},9,9z^{2},9(z-1)^{2}\\}$. Examples of simpler
rational sections on the original surface (4.1) have these non-degenerate
values of $b$:

$\displaystyle\frac{(z+1)(z+2)}{3z\,(1-z)},\ \frac{3(z+1)(3z+2)}{z\,(1-z)},\
\frac{6(z+1)(3z-2)}{25z\,(1-z)},\
\frac{2(2z+1)(4z-3)}{z\,(1-z)(2z^{2}-3)^{2}},\;$
$\displaystyle\frac{(2-z)(z^{2}-4z+1)(z^{2}-6z+3)}{z\,(5z^{2}-2z-1)^{2}},\
\frac{2z\,(2z+1)(6z^{3}+z^{2}-2)}{(1-z)(3z^{2}-2)^{2}}.$ (4.9)

The corresponding $c\,$-coordinate is obtained from a linear factor of (4.1)
that arrises after substituting $b$. Due to the hypergeometric symmetry (1.8),
some possible values for $c$ can be obtained by substituting $z\mapsto 1/z$ in
(4.1).

###  4.2 The rational surface

The discriminant of the elliptic surface (4.2) is of degree 12 in $z$; see the
denominator of (4.3). Therefore it is a rational surface [SS19, §7.5] that can
be obtained from a pencil of cubic curves in $\mathbb{P}^{2}$ (linearly
parametrized by $z$) by blowing up the 9 intersection points of the pencil.
Rather equivalently [Sch05], it is a Del Pezzo surface in the weighted
projective space with weights$(z:1:u:v)=(1:1:2:3)$, of degree 1.

The surface $S_{4}$ is birational to the elliptic surface (4.2) in the
Weierstrass form, hence it can also be obtained from the pencil of cubic
curves in $\mathbb{P}^{2}$. To obtain a rational parametrization of $S_{4}$,
we first notice the simpler defining equation

$f\,(3f+6(z^{2}\\!+\\!z\\!+\\!1)+8ez+6e^{2}+14e)+e(e+1)(e+2)(e+3)=0$ (4.10)

in the coordinates

$e=bz+c,\qquad f=bz(z-1).$ (4.11)

The new equation defines a Del Pezzo surface in the weighted projective space
with weights$(z:e:1:f)=(1:1:1:2)$, of degree 2. It contains 4 lines in the
hyperplane $f=0$. Choosing the line $f=e=0$ to blow down, we apply a standard
step (called unprojection in [Sch05, §5]) in resolving Del Pezzo surfaces by
introducing the coordinate

$w=\frac{f+2(z^{2}+z+1)}{e}$ (4.12)

We can eliminate $f$ straightaway and obtain the following non-singular cubic
surface in $\mathbb{P}^{3}$:

$(ew-2z^{2}-2z-2)(3w+8z+6e+14)+(e+1)(e+2)(e+3)=0.$ (4.13)

Subsequently, we can blow down one of the lines on the plane $3w+8z+6e+14=0$,
or apply a classical parametrization recipe [SS19, §10.5.3] using two skew
lines on the cubic surface, say $e=3w+8z=-2$ and $w=e+1=-z-1$. Eventually, a
suitable simplified cubic pencil is defined by $U+zV=0$, where

$\displaystyle U=$ $\displaystyle\;2t^{2}y-6t^{2}+4ty-3y^{2}+3y,$ (4.14)
$\displaystyle V=$ $\displaystyle\;ty^{2}+4t^{2}-2ty+3t-6y.$ (4.15)

A parametrization of $S_{4}$ is given by $z=-U/V$ and

$\displaystyle b=$
$\displaystyle\;\frac{3(y^{2}-2t+y)(ty^{2}-8t^{2}+4ty-3y^{2}+3t+3y)}{U\,(U+V)},$
(4.16) $\displaystyle c=$
$\displaystyle\;-\frac{6(4t^{2}+2t-3y-3)(t^{2}y-t^{2}+ty-y^{2})}{V\,(U+V)}.$
(4.17)

The inverse map is given by

$\displaystyle t=$
$\displaystyle\;\frac{3(b+1)}{2}-\frac{3(c+1)(c+2)}{2bz}+\frac{3(b+c+1)(b+c+2)}{2b\,(z-1)},$
(4.18) $\displaystyle y=$
$\displaystyle\,-c+\frac{(b+2)(b+3)z}{c}+\frac{(b+c+2)(b+c+3)z}{c\,(z-1)}.$
(4.19)

Computations with two versions of $(t,y)$ in (4.14)–(4.16) show that the
hypergeometric symmetries (1.8), (2.3)–(2.6) are realized by non-linear
Cremona transformations of the parameters $t,y$. For example, (1.8) is
realized by

$(t,y)\mapsto\left(\frac{t(y^{2}-ty^{2}-3t+3y)}{2(t^{2}y-t^{2}+ty-y^{2})},\frac{(3-2t)(ty^{2}-y^{2}+3t-3y)}{ty^{2}-8t^{2}+4ty-3y^{2}+3t+3y}\right)\\!,$
(4.20)

and (2.3) is realized by

$(t,y)\mapsto\left(t,\,\frac{ty-4t^{2}-3t+3y}{ty-t-3}\right).$ (4.21)

###  4.3 The fibration by $b$

Equation (4.1) with fixed $b$ defines a quartic curve of the generic genus 3.
We can reduce the genus to 1 by factoring out the hypergeometric symmetry
(2.3). This gives an elliptic surface isomorphic to

${\mathcal{E}}^{*}_{4}:\quad
v^{2}=u\,\big{(}u^{2}-4b(5b+9)u+108\,b(b+1)^{2}(b+2)\big{)}.$ (4.22)

The symmetry invariants are parametrized as follows:

$\displaystyle z(1-z)=$
$\displaystyle-6\,\frac{v+2(2b+3)u-36\,b(b+1)(b+2)}{W_{2}},$ (4.23)
$\displaystyle c\,(b+c+3)=$
$\displaystyle-216\,\frac{b\,(b+1)(b+2)(b+3)^{2}}{W_{2}},$ (4.24)
$\displaystyle\\!cz\\!+\\!(b\\!+\\!c\\!+\\!3)(z\\!-\\!1)=$
$\displaystyle-(b+3)\times$ (4.25)
$\displaystyle\,\frac{2(4b\\!+\\!3)v+u^{2}\\!-\\!4b(b\\!+\\!3)u+108b(b\\!+\\!1)(b\\!+\\!2)(b\\!+\\!5)}{W_{2}}.$

where $W_{2}=8bv+u^{2}-4b(b+9)u+108\,b(b+1)(b+2)(b+9)$. The inverse projection
is given by

$\displaystyle u=$
$\displaystyle\;-6b(b+1)(b+2)\frac{(bz+c+3z)^{2}}{c\,(b+c+3)},$ (4.26)
$\displaystyle v=$
$\displaystyle\,12b\,(b+1)(b+2)(b+3)\frac{(bz+c+3z)(2bz+2c+3)}{c\,(b+c+3)}.$
(4.27)

Rational points on the quartic curves can be found by trying to lift from the
rational fibers on the elliptic surface. Its Mordell-Weil group can be
determined similarly as in the proof of Theorem 3.1. The discriminant of the
elliptic surface equals $2^{13}3^{6}b^{3}(b+1)^{4}(b+2)^{2}(b+3)^{3}$, and the
$j$-invariant equals

$-\frac{32\,(19b^{3}+36b^{2}-81b-162)^{3}}{729\,(b+1)^{4}(b+2)^{2}(b+3)^{3}}.$
(4.28)

Using [SS19, Table 5.1] we conclude that the singular fibers
$b\in\\{0,-1,-2,-3\\}$ have the Kodaira types III, I4, I2, I3, respectively.
They have, respectively, $m_{v}=2,4,2,3$ irreducible components. The Mordell-
Weil rank equals $10-2-7=1$. The candidates for the generators have

$\displaystyle u\in\\{$ $\displaystyle\textstyle
0,9(b+1)^{2},\frac{9}{4}(b-1)^{2},6b(b+1),54b(b+1),12b(b+2),$ (4.29)
$\displaystyle 2(b+1)(b+2),18(b+1)(b+2)\\}.$

The Mordell-Weil group is isomorphic to
$(\mathbb{Z}/2\mathbb{Z})\times\mathbb{Z}$. The group is generated by the
2-torsion point $(0,0)$ and a point having
$u\in\\{6b(b\\!+\\!1),18(b\\!+\\!1)(b\\!+\\!2)\\}$. The point

$\big{(}6b(b+1),12b(b+1)(b+3)\big{)}$ (4.30)

can be taken as a free generator.

The discriminant of the quadratic equation (4.23) for $z$ equals

$\frac{(v+4bu+12u)^{2}}{(v+4bu)^{2}+864b(b+1)(b+2)u}$ (4.31)

on ${\mathcal{E}}^{*}_{4}$. To find rational points on the genus 3 curve, we
need the denominator

$(v+4bu)^{2}+864b(b+1)(b+2)u$ (4.32)

to be a full square on ${\mathcal{E}}^{*}_{4}$. The point $(u,v)=0$ does not
always suit, because it gives

$c=-\frac{b+3}{2}\left(1\pm\sqrt{\frac{b+1}{b+9}}\right),\qquad
z=\frac{1}{2}\pm\frac{1}{2}\,\sqrt{\frac{b+1}{b+9}}.$ (4.33)

A search with generically specialized $b$ yielded only a few rational sections
corresponding to degenerate Belyi maps.

##  5 Belyi maps

Recall that Belyi maps are algebraic coverings $\varphi:C\to\mathbb{P}^{1}$
that branch in the 3 fibers $\varphi\in\\{0,1,\infty\\}$. We will consider
Belyi of genus 0 only. The following characteristic property of Belyi maps of
genus 0 follows from the Riemann-Hurwitz formula

######  Lemma 5.1.

A Belyi map of genus $0$ and degree $d$ has exactly $d+2$ distinct points in
the $3$ fibers $\\{0,1,\infty\\}$.

As summed up in (1.4), the considered maps (1.1) and (1.2) satisfy the
condition of this lemma. The degree of (1.1) equals

$d=\max(|p|,|q|,|p+q|,m|r|,|p+mr|,|q+mr|,|p+q+mr|),$

while the degree of (1.2) equals $\max(2|p|,m|r|,|2p+mr|)$.

Without loss of generality, we may assume $\max(|p|,|q|\leqslant|p+q+mr|$.
Anything else?

###  5.1 Easy maps

The simplest Belyi maps are the power functions $\varphi(x)=x^{p}$. They have
just 2 points in the two fibers $\varphi=0$ and $\varphi=\infty$. Our
considered rational maps can be viewed as a close neighborhood of this
exemplar in the landscape of Belyi maps.

######  Example 5.2.

The Belyi maps with exactly 3 points in two fibers are easy to find [LZ04].
They necessarily have two points with some branchong orders $k,q$ in one fiber
(say $\varphi=0$) and one point (say $x=\infty$) of branching order $k+q$ in
other fiber (say $\varphi=\infty$). There must be then $k+q-1$ distinct points
in third fiber $\varphi=1$ by Lemma 5.1, hence there we have exactly one
branching point, of order 2. The most compact expression for the Belyi map is
obtained after choosing this branching point to be $x=0$:

$\displaystyle\varphi(x)=$ $\displaystyle\;(1-qx)^{p}(1+px)^{q}$ (5.1)
$\displaystyle=$ $\displaystyle\;1+O(x^{2}).$ (5.2)

This form matches the case $m=0$, $\lambda=p/q$ of (1.1), where we take $p,q$
to be positive, and rescale $x\mapsto x/q$.

######  Example 5.3.

Consider Belyi maps of the more general form

$\displaystyle\varphi(x)=$ $\displaystyle\;(1-qx)^{p}\,F_{m}(x)^{q}$ (5.3)
$\displaystyle=$ $\displaystyle\;1+O(x^{m+1}),$ (5.4)

where $F_{m}(x)$ is a polynomial or degree $m$. If $p,q$ are positive
integers, then these maps have one point of order $p$ and $m$ points of order
$q$ above $\varphi=0$, a single point of order $p+mq$ above $\varphi=\infty$,
and a branching point of order $m+1$ above $\varphi=1$. The other points above
$\varphi=1$ are non-branching by Lemma 5.1. By the equality of (5.3) and (5.4)
we have

$F_{m}(x)=(1-qx)^{-p/q}\quad\mbox{mod}\ x^{m+1}.$ (5.5)

This means that the polynomial $F_{m}(x)$ equals the truncated Taylor series
of $(1-qx)^{-p/q}$ at $x=0$. Explicitly,

$\displaystyle
F_{m}(x)=\sum_{k=0}^{m}\frac{q^{k}\big{(}p/q\big{)}_{k}}{k!}\,x^{k}.$ (5.6)

Following our interpretation of hypergeometric polynomials (2.8), we can write

$F_{m}(x)={}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-m,\,\frac{p}{q}\,\atop-m}\right|qx\right).$ (5.7)

######  Remark 5.4.

We may allow $p$ or $q$ to be negative integers in (5.3). That just means that
the corresponding point(s) are above $\varphi=\infty$ rather than above
$\varphi=0$. Formulas (5.5) still hold then. The point $x=\infty$ remains
above $\varphi=\infty$ if $p+mq>0$, and it moves to the fiber $\varphi=0$ if
$p+mq<0$.

######  Remark 5.5.

If we take $p=0$ or $p+mq=0$, then we have $m+1$ (rather than $m+2$) points in
the two fibers $\varphi\in\\{0,\infty\\}$. The branching order of $x=0$ can be
at most $m$ then, and (5.5) should be modified to $F_{m}(x)=1+O(x^{m})$. We
obtain the Belyi maps $\varphi(x)=(1-x^{m})^{q}$, after scaling $x$
additionally.

######  Example 5.6.

Now we look at the Belyi maps of the form

$\varphi(x)=(1-x)^{p}(1-\lambda x)^{q}(1-\mu x)^{r},$ (5.8)

with $\varphi(x)=1+O(x^{3})$. This is the case $m=1$ of (1.1). The series
expansion of (5.8) starts with

$1-(p+\lambda q+\mu r)\,x+\frac{1}{2}\left((p+\lambda q+\mu
r)^{2}-p-\lambda^{2}q-\mu^{2}r\right)x^{2}+\ldots.$ (5.9)

The coefficients to $x,x^{2}$ must vanish. Eliminating $\lambda$ we get a
quadratic equation for $\mu$. Its discriminat equals $-pqr(p+q+r)$, hence the
Belyi maps are expressed with $\sigma=\pm\sqrt{-pqr(p+q+r)}$. Explicitly,

$\varphi(x)=\big{(}1-x)^{p}\,\left(1+\frac{pq+\sigma}{q(q+r)}\,x\right)^{q}\,\left(1+\frac{pr-\sigma}{r(q+r)}\,x\right)^{r}.$
(5.10)

These Belyi maps with the definition field $\mathbb{Q}(\sigma)$ are obtained
in [LZ04, Example 2.2.25]. If $p,q,r$ are positive integers, the definition
field is an imaginary quadratic extension of $\mathbb{Q}$. But the Belyi maps
could be defined over $\mathbb{Q}$ if both positive and negative powers are
prescribed. For example, here is a nice family of paired Belyi maps with
$p={n+1\choose 2}$, $q={n\choose 2}$, $r=-1$:

$\displaystyle\frac{(1+x)^{p}\,(1-x)^{q}}{1+nx},\qquad\frac{\big{(}1+(n-2)x\big{)}^{p}\,\big{(}1-(n+2)x\big{)}^{q}}{1-n^{2}x}.$
(5.11)

A general rational parametrization of the triples $(p,q,r)$ with
$\rho\in\mathbb{Q}$ is obtained by parametrizing the singular cubic surface
$y^{2}+st(s+t+1)=0$, identifying $s=p/r$, $t=q/r$. Here is such a
parametrization by $u=\sigma/qr$, $v=\sigma/pr$, up to simultaneous scaling of
$p,q,r$:

$p=u(uv+1),\quad q=v(uv+1),\quad r=-u-v.$ (5.12)

Then $\sigma=-uv(u+v)(uv+1)$. After rescaling $x$ in the two forthcoming Belyi
maps, we obtain the expressions

$\displaystyle\big{(}1-(v+1)x)^{u}\,\big{(}1+(u-1)x\big{)}^{v}\,\big{(}1-(uv+1)x\big{)}^{-(u+v)/(uv+1)},$
(5.13)
$\displaystyle\big{(}1+(v-1)x)^{u}\,\big{(}1-(u+1)x\big{)}^{v}\,\big{(}1-(uv+1)x\big{)}^{-(u+v)/(uv+1)},$
(5.14)

to be raised to a common power so to make the three powers integral. The cases
with $\sigma=0$ fall outside the considered shape (5.8), and the cases
$(u+v)(u^{2}-1)(v^{2}-1)=0$ with coalescing branching points require special
attention. In the latter situation we have $(p+q)(q+r)(q+r)=0$.

######  Remark 5.7.

If $p+q=0$ in the last example, then $\sigma=\pm pr$. Taking $\sigma=pr$ gives
the trivial function $\varphi(x)=1$. Taking $\sigma=-pr$ gives

$\varphi(x)=\frac{\big{(}1-(p-r)x)^{p}\,\left(1-2px\right)^{r}}{\left(1-(p+r)x\right)^{p}}.$
(5.15)

In the cases $p+r=0$ or $q+r=0$ we also have just one Belyi map analogously.
If simultaneously $p+q=p+r=0$, both candidate functions collapse to
$\varphi(x)=1$.

##  6 Belyi maps of the form (1.1)

Belyi maps of the shape (1.1) generalize of Example 5.6 from $m=1$. We assume
that the numbers $p,q,r,p+q+mr$ are non-zero, so that there are indeed $m+3$
distinct points above $\varphi=0$ and $\varphi=\infty$. The points $x=1$,
$x=1/\lambda$, $x=\infty$ and their branching orders $p$, $q$, $-p-q-mr$ are
permutable. We also assume $p\neq q$, $p\neq q+mr$, $q\neq p+mr$, as these
cases are considered in Section 7 with the smallest field of definition.

The condition (1.3) translates into the power series relation

$G_{m}(x)=(1-x)^{-p/r}(1-\lambda x)^{-q/r}\quad\mbox{mod }x^{m+2}.$ (6.1)

The power series term with $x^{m+1}$ has to equal $0$ then. The polynomial
$G_{m}$ is determined uniquely by the power series

$(1-x)^{-p/r}(1-\lambda x)^{-q/r}=\sum_{k=0}^{\infty}h_{k}x^{k}$ (6.2)

truncated at the $(m+1)$-the term.

######  Lemma 6.1.

No Belyi map of the form $(\ref{eq:g11hm})$, $(\ref{eq:bmpws})$ exist when
$p=-\ell_{1}r$ and $q=-\ell_{2}r$ for some positive integers
$\ell_{1},\ell_{2}$ satisfying $\ell_{1}+\ell_{2}\leqslant m$.

###### Proof.

The left-hand side of (6.2) is the polynomial
$(1-x)^{\ell_{1}}(1-ax)^{\ell_{2}}$ under the stated conditions. When
$\ell_{1}+\ell_{2}<m$, the polynomial $G_{m}$ is forced to be of smaller
degree than $m$. If $\ell_{1}+\ell_{2}=m$ then $G_{m}$ has the undue roots
$x=1$, $x=1/\lambda$. ∎

The coefficients $h_{k}$ in (6.2) can be expressed as follows:

$\displaystyle h_{k}=$
$\displaystyle\,\sum_{j=0}^{k}\frac{(p/r)_{k-j}\,(q/r)_{j}\,\lambda^{j}}{(k-j)!\,j!},$
(6.3)

or

$\displaystyle h_{k}$ $\displaystyle=\frac{(p/r)_{k}}{k!}\,{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-k,\,\frac{q}{r}\atop
1-k-\frac{p}{r}}\right|\lambda\right)$ (6.4)
$\displaystyle=\frac{(q/r)_{k}\lambda^{k}}{k!}\,{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-k,\,\frac{p}{r}\atop
1-k-\frac{q}{r}}\right|\frac{1}{\lambda}\right).$ (6.5)

We must have $h_{m+1}=0$, which leads to the equation (1.5) generically.

###  6.1 Full sets of Belyi maps

Here we consider examples of the generic case (6.6) with $m+1$ Belyi maps. We
are interested in the cases $\lambda\not\in\\{0,1\\}$ and $p\neq q$. The
degenerate value $\lambda=1$ would coalesce the points of branching order $p$
and $q$, but this point is excluded by Lemma 2.1(iii). By Lemma 2.1(ii), the
hypergeometric polynomial $h_{m+1}(\lambda)$ has $m+1$ distinct roots when

$\left(\,\frac{p}{r}\,\right)_{\\!m+1}\neq
0,\quad\left(\,\frac{q}{r}\,\right)_{\\!m+1}\neq
0,\quad\left(\frac{p+q}{r}\right)_{\\!m+1}\neq 0.$ (6.6)

The condition $((p+q)/r)_{m+1}\neq 0$ is equivalent to

$\left(-\frac{p+q+mr}{r}\right)_{\\!m+1}\neq 0,$ (6.7)

underscoring the symmetry of the branching orders $p$, $q$, $-p-q-mr$.

######  Example 6.2.

In the case $m=2$ we consider the equation (3.1) specialized by $b=q/r$,
$c=p/r$, $z=\lambda$:

$q(q+r)(q+2r)\lambda^{3}+3pq(q+r)\lambda^{2}+3pq(p+r)\lambda+p(p+r)(p+2r)=0.$
(6.8)

The discriminant

$-108p^{2}q^{2}r^{3}\,(p+r)\,(q+r)(p+q+r)(p+q+2r)^{2}$

is proportional to $(p/r)_{3}(q/r)_{3}((p+q)/r)_{3}$, and it indeed vanishes
only when we have less than 3 distinct roots, and those roots
$\lambda\not\in\\{0,1\\}$ Lemma 2.1(ii).

Belyi maps defined over $\mathbb{Q}$ can be found using the parametrization
(3.3), identifying $q=br$, $p=cr$, $\lambda=z$. Polynomial Belyi maps defined
over $\mathbb{Q}$ are obtained from the + regions (3.6) in Figure 1(i). For
example, $e=-z=\frac{1}{5}$ and $e=-z=\frac{1}{6}$ produce these Shabat
polynomials, respectively:

$\displaystyle(1-5x)^{2}(1+x)^{20}(1-2x+9x^{2})^{5},$ (6.9)
$\displaystyle(1-6x)(1+x)^{20}(1-2x+6x^{2})^{7}.$ (6.10)

Their companions with the same $p,q,r$ are defined, respectively, over
$\mathbb{Q}(\sqrt{-35})$ and $\mathbb{Q}(\sqrt{-2})$. There are no triples of
Shabat polynomials defined over $\mathbb{Q}$, as there is no point
$(t,y)\in\mathbb{R}^{2}$ giving positive values for $b=q/r$ and $c=p/r$ by
Figure 1(ii). The family (3.15) with $t\in\mathbb{Q}$ provides most of the
triples of Belyi maps defined over $\mathbb{Q}$ with small absolute values of
$p,q,r$. For example, $t=2$ gives the case $(p,q,r)=(2,-7,6)$ with these 3
Belyi maps defined over $\mathbb{Q}$:

$\displaystyle\frac{(1-x)^{2}(1-2x-\frac{1}{6}x^{2})^{6}}{(1-2x)^{7}},\quad\frac{(1-x)^{2}(1+5x+\frac{10}{3}x^{2})^{6}}{(1+4x)^{7}},$
$\displaystyle\frac{(1-5x)^{2}(1-3x-\frac{2}{3}x^{2})^{6}}{(1-4x)^{7}}.$
(6.11)

Here are a few projective ratios $(p:q:r)$ giving 3 Belyi maps defined over
$\mathbb{Q}$ outside the family (3.15):

$\displaystyle\textstyle(1:-\frac{7}{8}:\frac{1}{19}),\,(2:-\frac{1}{5}:\frac{1}{37}),\,(1:-\frac{26}{31}:\frac{1}{37}),\,(1:-\frac{11}{41}:\frac{1}{43}),\,(1:-\frac{14}{17}:\frac{1}{61}),$
$\displaystyle\textstyle(7:-\frac{5}{2}:\frac{4}{13}),\;(7:-\frac{65}{19}:3),\;(19:-\frac{35}{8}:1),\;(-19:-\frac{37}{7}:\frac{1}{2}).$
(6.12)

Families of Belyi maps could be obtained from sections on the elliptic
siurface ${\mathcal{E}}_{3}$, particularly from the $c$-values in (3.22).
Taking $c=(b+2)/(4b-1)$ with the reparametrization $t=4b-1$ and independent
rescaling of $x$ and of the powers $p,q,r$ gives

$(1-2tx)^{t+9}\,\big{(}1+6x\big{)}^{t(t+1)}\left(1-(t-3)x-(t+3)^{2}x^{2}\right)^{4t}.$
(6.13)

Similarly, taking $c=-4b(b+2)/(5b+1)$ and $t=(5b+1)/(3b)$ gives

$(1-tx)^{12-8t}\,\big{(}1-2(t-1)x\big{)}^{t}\left(1-2x-\frac{2(t-2)^{2}}{3t-5}x^{2}\right)^{\\!t(3t-5)}.$
(6.14)

Or one may employ the whole 2-parameter uniformizations (3.3) or (3.15).

######  Example 6.3.

Similarly, for $m=3$ we obtain the surface ${\mathcal{S}}_{4}$ in (4.1) with
$b=q/r$, $c=p/r$, $z=\lambda$. We can use the full parametrization
(4.16)–(4.17), or a section of the elliptic surface ${\mathcal{E}}_{4}$ in
(4.2). For example, the first $b$-value from (4.1) gives the family

$\displaystyle(1-x)^{2t(t-2)(t+2)}\,\big{(}1-tx\big{)}^{(t+1)(t+2)}$
$\displaystyle\qquad\textstyle\times\left(1-(t+2)x-\frac{(t+1)(t+2)}{3}x^{2}-\frac{(t+1)(t-2)(t+2)}{9}x^{3}\right)^{3t(1-t)},$
(6.15)

with $t=z$. Here are some values of $b,c$ with one of them a positive integer,
found by an extensive search through the parametrization (4.16):

$\displaystyle\textstyle(5,-\frac{7}{2}),(5,-\frac{7}{11}),(5,-\frac{14}{13}),(13,-\frac{13}{3}),(13,-\frac{26}{5}),(14,-\frac{7}{2}),(15,-\frac{45}{43}),(19,-\frac{28}{3}),\quad$
$\displaystyle\textstyle(19,-\frac{19}{7}),(20,-\frac{92}{11}),(22,-\frac{40}{7}),(23,-\frac{25}{2}),(24,-\frac{117}{47}),(45,-\frac{256}{13}),(45,-\frac{207}{19}),$
$\displaystyle\textstyle(45,-\frac{47}{51}),(51,-\frac{442}{71}),(54,-\frac{135}{7}),(55,-\frac{35}{2}),(55,-\frac{99}{4}),(56,-\frac{29}{5}),(56,-\frac{133}{5}),$
$\displaystyle\textstyle(68,-\frac{92}{7}),(68,-\frac{355}{33}),(68,-\frac{85}{131}),(76,-\frac{143}{5}),(77,-\frac{79}{2}),(84,-\frac{192}{5}),(91,-\frac{188}{41}).$

The negative integer values of $b,c$ are considered in Example 6.7. Here are
some other found values of $b,c$, excluding the values
$\\{-1/2,-3/2,-5/2,-7/2,-9/2\\}$ with reference to Examples 7.7 and 7.8:

$\displaystyle\textstyle(\frac{1}{2},-\frac{8}{3}),(\frac{1}{2},-\frac{7}{5}),(\frac{7}{2},-\frac{26}{5}),(\frac{17}{2},-\frac{3}{10}),(\frac{4}{3},-\frac{27}{7}),(\frac{7}{3},-\frac{13}{9}),(\frac{7}{3},-\frac{26}{5}),(\frac{14}{3},-\frac{34}{9}),\quad$
$\displaystyle\textstyle(\frac{20}{3},-\frac{23}{9}),(\frac{29}{3},-\frac{63}{5}),(\frac{74}{3},-\frac{37}{2}),(\frac{11}{4},-\frac{40}{7}),(\frac{15}{4},-\frac{45}{7}),(\frac{15}{4},-\frac{46}{7}),(\frac{4}{5},-\frac{19}{7}),$
$\displaystyle\textstyle(\frac{7}{6},-\frac{35}{9}),(\frac{8}{5},-\frac{13}{4}),(\frac{1}{6},-\frac{9}{22}),(\frac{5}{6},-\frac{2}{9}),(\frac{1}{7},-\frac{1}{4}),(\frac{1}{7},-\frac{11}{5}),(\frac{3}{7},-\frac{17}{5}),(\frac{6}{7},-\frac{18}{5}),$
$\displaystyle\textstyle(-\frac{2}{3},-\frac{7}{5}),(-\frac{5}{3},-\frac{2}{9}),(-\frac{8}{3},-\frac{3}{8}),(-\frac{1}{4},-\frac{7}{5}),(-\frac{5}{4},-\frac{8}{11}),(-\frac{3}{5},-\frac{4}{7}),(-\frac{7}{6},-\frac{9}{10}).$

###  6.2 Cases with fewer Belyi maps

The number of Belyi maps (1.1) with $\lambda\not\in\\{0,1\\}$ is smaller than
$m+1$ when at least one of the conditions (6.6) is not satisfied. Indeed, if
$q=-\ell r$ for a positive integer $\ell\leqslant m$, the polynomial
$h_{m+1}(\lambda)$ has the degree $\ell$ rather than $m+1$ as the terms with
$j>\ell$ in (6.3) vanish then. If $p=-\ell r$ for a positive integer
$\ell\leqslant m$, then the first $m+1-\ell$ terms in (6.3) vanish, and

$h_{m+1}=\frac{(-1)^{\ell}\,(q/r)_{m+1-\ell}}{(m+1-\ell)!}\,\lambda^{m+1-\ell}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-\ell,\,\frac{q}{r}+m+1-\ell\atop
m+2-\ell}\right|\lambda\right).$ (6.16)

The factor $\lambda^{m+1-\ell}$ reduces the number of relevant roots to
$\ell$. If $p+q=-\ell r$ for a positive integer $\ell\leqslant m$, but
$(p/r)_{m}\neq 0$ and $(q/r)_{m}\neq 0$, then we apply Euler’s transformation
(2.10) to the hypergeometric polynomial $h_{m+1}(\lambda)$ and obtain

$\displaystyle
h_{m+1}=\frac{(p/r)_{m+1}}{(m+1)!}\,(1-\lambda)^{\ell+1}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-m+\ell,\,1+\frac{p}{r}\atop-m-\frac{p}{r}}\right|\lambda\right).$
(6.17)

The factor $(1-\lambda)^{\ell+1}$ reduces the number of relevant roots to
$m-\ell$. If $p/r\in\mathbb{Z}$ and $q/r\in\mathbb{Z}$, then these
possibilities may combine, giving lower degree of $h_{m+1}(\lambda)$ and
multiple undue roots $\lambda=0$, $\lambda=1$. Figure 2 depicts the regions
and line segments or rays on the integer lattice for the values of $p/r$ and
$q/r$ where the number of Belyi maps (1.1) is fixed $\leqslant m+1$.

$p/r$$q/r$$1$$2$$3$$m\\!-\\!1$$m$$m\\!+\\!1$$m\\!-\\!1$$m$$1$$2$$3$$m\\!-\\!1$$m$$m\\!+\\!1$$1$$2$$3$$m\\!+\\!1$$m\\!+\\!1$$0$$0$
Figure 2: The number of Belyi maps (1.1) for integer values of $p/r$ and
$q/r$. The dark region in the middle, and the lines emanating from it,
represent the cases with no Belyi maps. The light grey regions represent $m+1$
Belyi maps; the dashed lines — unique Belyi maps; the dense dotted lines —
pairs of Belyi maps; the sparser dotted lines — triples of maps; the two kinds
of dashed-dotted lines: $(m-1)$ or $m$ maps.

Let us specify $b=q/r$ and $c=p/r$, for shorthand.

######  Example 6.4.

Let us consider the cases that have single Belyi maps. They are represented by
some lattice points on the lines $p=-r$, $q=-r$, $p+q=(1-m)r$ or $p=2r$,
$q=2r$, $p+q=-(m+2)r$ in Figure 2. It is enough to consider two cases, that
is, one from both displayed triples of lines, due to the hypergeometric
symmetries (1.8), (2.3)–(2.6). When $q=-r$ and $(c-1)_{m}\neq 0$, the
hypergeometric polynomial $h_{m+1}$ is linear in $\lambda$. It gives
$\lambda=(m+c)/(m+1)$. Following (6.1), consider the power series

$(1-x)^{-c}\left(1-\frac{m+c}{m+1}\,x\right)=1+\frac{1}{m+1}\sum_{k=1}^{\infty}\frac{(c-1)_{k}}{k!}(m+1-k)x^{k}.$
(6.18)

The term with $x^{m+1}$ indeed vanishes, and the earlier terms define
$Q_{m}(x)$. When $p=2r$ and $(b)_{m+1}=0$, then (6.17) reads

${}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-m-1,\,b\atop-m-2}\right|\lambda\right)=(1-\lambda)^{-b-1}\,{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-1,-b-m-2\atop-m-2}\right|\lambda\right).$ (6.19)

The hypergeometric polynomial on the right-hand side is linear, and gives
$\lambda=(m+2)/(b+m+2)$. The implication for the power series of

$(1-x)^{-2}\left(1-\frac{m+2}{b+m+2}\,x\right)^{-b}$ (6.20)

is remarkable: the $k$-th term is divisible by $(b+1)_{k}$ when $m=k-1$. Here
is a revealing form of the starting terms:

$\displaystyle
1+\frac{m(b\\!+\\!2)+4(b\\!+\\!1)}{b+m+2}+\frac{(m\\!-\\!1)(b\\!+\\!3)\big{(}m(b\\!+\\!2)\\!+\\!9b\\!+\\!10\big{)}+27(b\\!+\\!1)(b\\!+\\!2)}{2\,(b+m+2)^{2}}$
$\displaystyle\quad+\frac{(m\\!-\\!2)(b\\!+\\!4)\big{(}m^{2}(b\\!+\\!2)(b\\!+\\!3)+\ldots\big{)}+256(b\\!+\\!1)(b\\!+\\!2)(b\\!+\\!3)}{6\,(b+m+2)^{3}}+\ldots.$
(6.21)

######  Example 6.5.

Here we consider the cases with two Belyi maps. They are represented by some
lattice points on the lines $p=-2r$, $q=-2r$, $p+q=(2-m)r$ or $p=3r$, $q=3r$,
$p+q=-(m+3)r$ in Figure 2. It is enough to consider two cases. When $q=-2r$
and $(c-2)_{m}\neq 0$, the hypergeometric polynomial $h_{m+1}$ is quadratic in
$\lambda$:

$m(m+1)\lambda^{2}-2(m+1)(m+c-1)\lambda+(m+c)(m+c-1).$ (6.22)

The two solutions are

$\lambda=\frac{m+c-1}{r}\pm\frac{\sqrt{(m+1)(c-1)(m+c-1)}}{m(m+1)}$ (6.23)

We have two Belyi maps are defined over $\mathbb{Q}$ when

$c=\frac{m+1+(m-1)u^{2}}{m+1-u^{2}}.$ (6.24)

for some $u\in\mathbb{Q}$. Then the $\lambda$-values are $(m+1\pm
u)/(m+1-u^{2})$. The power series then looks like this:

$\displaystyle(1-x)^{-c}\left(1-\frac{m+1+u}{m+1-u^{2}}\,x\right)=1+\left(c-1-\frac{m+(u+1)^{2}}{m+1-u^{2}}\right)x\quad$
$\displaystyle+\frac{1}{m}\sum_{k=2}^{\infty}\frac{(c-1)_{k-1}}{k!}\\!\left(c-1-\frac{m+k(u+1)^{2}}{m+1-u^{2}}\right)\\!(m+1-k)x^{k}.$
(6.25)

Note that if $c=2$ then $u=\pm 1$, and we have just one Belyi map with $u=1$,
a special case of (6.20).

Like in the previous example, in the case $p=3r$ and $(q/r)_{m+1}=0$ we use
(6.17):

${}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-m-1,\,b\atop-m-3}\right|\lambda\right)=(1-\lambda)^{-b-2}\,{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-2,-b-m-3\atop-m-3}\right|\lambda\right).$ (6.26)

We obtain a quadratic equation analogous to (6.22). It has solutions
$\lambda\in\mathbb{Q}$ when

$b=-\frac{(m+3)u^{2}}{m+2+u^{2}}$ (6.27)

for some $u\in\mathbb{Q}$. Then the $\lambda$-values are $(m+2+u^{2})/(m+2\pm
u)$.

######  Example 6.6.

Now we consider the cases with larger $m$ that have three Belyi maps. Suppose
that $q=-3r$. Then the hypergeometric polynomial (6.16) for $h_{m+1}$ defines
a cubic relation between $\lambda$ and $p/r$. This cubic relation defines a
curve of genus 1, as stated in Proposition 3.1. The elliptic curve has the
Mordell-Weil group isomorphic to $(\mathbb{Z}/3\mathbb{Z})\times\mathbb{Z}$
for general $m$. The isomorphism (3.20) has to be adjusted with

$b=-m-1,\qquad c=3-m-\frac{p}{r}.$ (6.28)

Analysis with the data base [LMFDB] found the examples with
$m\in\\{6,11,13,17,23,25\\}$ giving the Mordell-Weil group
$(\mathbb{Z}/3\mathbb{Z})\times\mathbb{Z}^{2}\\!$, though the cases
$m\\!\in\\!\\{10,14,20,21,22\\}$ are not in the data base yet. In particular,
the elliptic curve for the case $m=6$ has the label 39690.bj2 in [LMFDB]. The
equation specialized from (3.19) is

$y^{2}=x^{3}+49(3x+96)^{2}.$ (6.29)

Besides the specialized generators $(0,672)$ and $(-56,280)$ of Proposition
3.1, its Mordell-Weil group has also the generator $(-48,48)$ over
$\mathbb{Q}$. This extra generator corresponds to the hypergeometric
evaluation ${}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{-7,-3\atop
13}\right|-1\right)=0$, and the Belyi map

$\frac{1-16x+117x^{2}-512x^{3}+1463x^{4}-2736x^{5}+2907x^{6}}{(1-x)^{19}(1+x)^{3}}.$
(6.30)

It is interesting to observe that the expanded polynomial
$(1-x)^{19}(1+x)^{3}$ does not have the terms with $x^{7}$ (as we just used)
and with $x^{11}$, $x^{15}$. Therefore we can obtain two more Belyi maps by
extending the numerator of (6.30) to polynomials of degree $m=10$ or $m=14$.
This phenomenon is typical for integer points on our hypergeometric surfaces,
as the next example suggests.

######  Example 6.7.

For $q=-4r$, the parametrization (4.16)–(4.17) has to be adjusted by (6.28).
Therefore, we seek negative values of $b$ or $c,-b-c-m$ due to the
hypergeometric symmetries (1.8), (2.3)–(2.6). Here are some cases found:

$\displaystyle\textstyle(-5,-\frac{4}{7}),(-7,-\frac{7}{5}),(-9,-\frac{9}{2}),(-14,-\frac{26}{5}),(-14,-\frac{22}{19}),(-19,-\frac{3}{20}),\quad$
$\displaystyle\textstyle(-19,-\frac{17}{31}),(-22,-\frac{5}{31}),(-22,-\frac{57}{107}),(-24,-\frac{77}{4}),(-24,-\frac{33}{10}),$
$\displaystyle\textstyle(-33,-\frac{33}{7}),(-36,-\frac{20}{3}),(-43,-\frac{43}{2}),(-43,-\frac{168}{127}),(-54,-\frac{39}{85}),$
$\displaystyle\textstyle(-54,-\frac{51}{233}),(-59,-\frac{19}{85}),(-59,-\frac{63}{139}),(-65,-\frac{52}{7}),(-68,-\frac{871}{19}),$
$\displaystyle\textstyle(-71,-\frac{180}{19}),(-78,-\frac{22}{3}),(-78,-\frac{190}{7}),(-79,-\frac{247}{5}),(-80,-\frac{231}{23}).$
(6.31)

Of particular interest are the points with both $b,c$ negative integers. Found
instances are

$\displaystyle(-7,-10),(-14,-66),(-28,-65),(-14,-52),(-30,-36),(-22,-35),\qquad$
$\displaystyle(-63,-64),(-41,-247),(-78,-210),(-115,-210),(-247,-780),$
$\displaystyle(-341,-715),(-901,-1856),(-1730,-8478),(-2795,-16512).$ (6.32)

They give two Belyi maps though for different cases of $m,p/r$. For example,
the first instance gives the hypergeometric evaluations

${}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{-7,-4\atop
7}\right|-1\right)=0,\qquad{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-10,-4\atop 4}\right|-1\right)=0.$ (6.33)

They both lead to consider the polynomial

$\displaystyle(1-x)^{4}(1+x)^{13}=1+9x+32x^{2}+48x^{3}-12x^{4}-156x^{5}-208x^{6}$
(6.34)
$\displaystyle\quad+286x^{8}+286x^{9}-208x^{11}-156x^{12}-12x^{13}+32x^{15}+9x^{16}+x^{17}\qquad$

as a power series without the terms $x^{7}$ or $x^{10}$.

##  7 Belyi maps of the form (1.2)

The form (1.2) of Belyi maps is the compacted case $q=p$ of the form (1.1),
where $(1-x)^{p}(1-\lambda x)^{p}$ is replaced by the aggregate power
$(1+\alpha x+\beta x^{2})^{p}$. This grouping of points with the same
branching order is routinely used to simplify maximally the field of
definition of Belyi maps [LZ04]. Similarly, here we are not interested in the
case $r=p$. These maps correspond to the cases $(1-x^{m+2})^{p}$ of Remark
5.5, with a quadratic factor of $1-x^{m+2}$ distinguished.

Let us denote $H_{2}(x)=1+\alpha x+\beta x^{2}$. The condition (1.3)
translates into the power series relation

$G_{m}(x)=H_{2}(x)^{-p/r}\quad\mbox{mod }x^{m+2}.$ (7.1)

The power series term with $x^{m+1}$ has to equal $0$. The polynomial $G_{m}$
is determined uniquely by the power series

$H_{2}(x)^{-p/r}=\sum_{k=0}^{\infty}g_{k}x^{k}$ (7.2)

truncated at the $(m+1)$-the term. The power series of can be computed by
expanding $(-\alpha-\beta x)^{j}$ in

$\displaystyle H_{2}(x)^{-p/r}=$
$\displaystyle\,\sum_{j=0}^{\infty}\,\frac{1}{j!}\,\Big{(}\,\frac{p}{r}\,\Big{)}_{\\!j}\,(-\alpha-\beta
x)^{j}x^{j}.$ (7.3)

Explicitly,

$g_{k}=\sum_{j=0}^{\lfloor
k/2\rfloor}\frac{(-1)^{k-j}(p/r)_{k-j}\,\alpha^{k-2j}\beta^{j}}{(k-2j)!\,j!}.$
(7.4)

If $(p/r)_{k}\neq 0$ and $\alpha\neq 0$, we have the hypergeometric expression

$g_{k}=\frac{(-\alpha)^{k}}{k!}\,\Big{(}\,\frac{p}{r}\,\Big{)}_{\\!k}\;{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-\frac{k}{2},-\frac{k-1}{2}\atop
1-k-\frac{p}{r}}\right|\frac{4\beta}{\alpha^{2}}\right).$ (7.5)

Distinct Belyi maps correspond to the roots of $g_{m+1}(\alpha\\!:\\!\beta)$,
identified up to the $(1\\!:\\!2)$-weighted homogeneous action of scaling $x$.
There is a Belyi map with $\alpha=0$ when $m$ is even. It is obtained from
Example 5.3 after the substitution $x\mapsto x^{2}$, $q=r$ (with $q,m$ over
there equal to the current $r,m/2$). The following degeneratated cases are
encountered:

  * •

There are no Belyi maps when $p=-\ell r$ for a positive integer
$\ell\leqslant\lfloor m/2\rfloor$, because then $G_{m}(x)=H_{2}(x)^{j}$, and
possibly of lesser degree than $m$.

  * •

If $p=-\ell r$ for a positive integer satisfying $\lfloor
m/2\rfloor<\ell\leqslant m$, then the first $m+1-\ell$ terms for the sum (7.4)
for $g_{m+1}$ are zero, and $g_{m+1}$ has the degenerate root $\beta=0$ of
that multiplicity. After shifting the summation index by $k-\ell$,

$g_{k}=\frac{\alpha^{2\ell-k}\beta^{k-\ell}\,\ell!}{(k-\ell)!\,(2\ell-k)!}\,{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{\frac{k}{2}-\ell,\frac{k+1}{2}-\ell\atop
1+k-\ell}\right|\frac{4\beta}{\alpha^{2}\\!}\right)\\!.$ (7.6)

There are $\ell-\lceil m/2\rceil$ proper Belyi maps then, including the
$\alpha=0$ case for even $m$.

  * •

If $p=-\ell r/2$ for a positive odd integer $\ell\leqslant m$, then

$g_{m+1}=\frac{(-\alpha)^{m+1}\big{(}\\!-\\!\frac{\ell}{2}\big{)}_{\\!m+1}}{(m+1)!}\left(1\\!-\\!\frac{4\beta}{\alpha^{2}\\!}\right)^{\\!\\!\frac{\ell+1}{2}}{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{\\!\frac{\ell-m}{2},\frac{\ell+1-m}{2}\atop\frac{\ell}{2}-m}\right|\frac{4\beta}{\alpha^{2}\\!}\right)\\!.$
(7.7)

by Euler’s formula (2.10). The root $\beta=\alpha^{2}/4$ corresponds to the
degeneration of $H_{2}(x)$ to a full square. The transformed hypergeometric
sum can be identified as $g_{m-\ell}$ with the substituted $p/r=1+\ell/2$.
There are then $\lceil(m-\ell)/2\rceil$ Belyi maps, including the $\alpha=0$
case for even $m$. In particular, there are no Belyi maps when the odd
$\ell=m$, and there is only the map with $\alpha=0$ when the odd $\ell=m-1$.

Otherwise, that is when $p\neq-\ell r$ for all positive $\ell\leqslant m$ and
$p\neq-\ell r/2$ for all positive odd $\ell\leqslant m$, the ${}_{2}\mbox{\rm
F}_{\\!1}$-factor in (7.5) for $g_{m+1}$ gives $\lceil m/2\rceil$ distinct
values with $\beta\not\in\\{0,\alpha^{2}/4\\}$ by Lemma 2.2. Together with
$\alpha=0$, in total we then have $\lceil(m+1)/2\rceil$ Belyi maps.

###  7.1 Full sets of Belyi maps

Let us enumerate the following examples by the value of $m$. We are
principally interested in Belyi maps defined over $\mathbb{Q}$.

######  Example 7.2.

For $m=2$, there is a Belyi map defined with $\alpha=0$. When $p\neq-r/2$,
there is other map. We can scale $x$ to obtain

$\displaystyle H_{2}(x)=$
$\displaystyle\;1+rx+{\textstyle\frac{1}{6}\,}r(p+2r)\,x^{2},$ (7.8)
$\displaystyle G_{2}(x)=$
$\displaystyle\;1-px+{\textstyle\frac{1}{6}\,}p(2p+r)\,x^{2}.$

######  Example 7.3.

For $m=3$, the Belyi maps are defined with the radical
$R_{3}=\pm\sqrt{3(p+2r)(2p+3r)}$:

$\displaystyle H_{2}(x)=$
$\displaystyle\;1+rx+{\textstyle\frac{1}{6}\,}r\big{(}3p+6r+R_{3}\big{)}\,x^{2},$
(7.9) $\displaystyle G_{3}(x)=$
$\displaystyle\;1-px+{\textstyle\frac{1}{6}\,}p\big{(}3r+R_{3}\big{)}\,x^{2}+{\textstyle\frac{1}{6}\,}p(p+r)\big{(}2p+4r+R_{3}\big{)}\,x^{3}.$

There are no Belyi maps for $p/r\in\\{-1,-2,-\frac{3}{2}\\}$. For $p=-r/2$ or
$p=-3r$ we have single Belyi maps, respectively:

$\displaystyle\frac{(1+2x+5x^{2})^{p}}{(1+x+2x^{2}-2x^{3})^{2p}},\qquad\frac{(1+3x-5x^{3})^{r}}{(1+x-x^{2})^{3r}}.$
(7.10)

The cases when $R_{3}\in\mathbb{Q}$ are parametrized as

$\displaystyle\frac{p}{r}=\frac{9-2s^{2}}{s^{2}-6},\qquad s\in\mathbb{Q}.$
(7.11)

######  Example 7.4.

For $m=4$, the Belyi maps with $\alpha\neq 0$ are defined with the radical
$R_{4}=\pm\sqrt{5(p+3r)(2p+3r)}$:

$\displaystyle H_{2}(x)=$
$\displaystyle\;1+rx+{\textstyle\frac{1}{30}\,}r\big{(}5p+15r+R_{4}\big{)}\,x^{2},$
(7.12) $\displaystyle G_{4}(x)=$
$\displaystyle\;1-px+{\textstyle\frac{1}{30}\,}p\big{(}10p-R_{4}\big{)}\,x^{2}+{\textstyle\frac{1}{30}\,}p(p+r)\big{(}5r+R_{4}\big{)}\,x^{3}\quad$
$\displaystyle\,-{\textstyle\frac{1}{180}\,}p(p+r)(2p+3r)\big{(}2p+6r+R_{4}\big{)}\,x^{4}.$

There are no Belyi maps for $p/r\in\\{-1,-2,-3,-\frac{3}{2}\\}$. For $p=-r/2$
or $p=-4r$ we have single Belyi maps, respectively:

$\displaystyle\frac{(1+6x+21x^{2})^{p}}{(1+3x+6x^{2}-18x^{3}+36x^{4})^{2p}},\quad\frac{(1+12x+42x^{2}-189x^{4})^{r}}{(1+3x-3x^{2})^{4q}}.$
(7.13)

The cases when $R_{4}\in\mathbb{Q}$ are parametrized as

$\displaystyle\frac{p}{r}=\frac{15-3s^{2}}{2s^{2}-5},\qquad s\in\mathbb{Q}.$
(7.14)

######  Example 7.5.

The case $m=5$ is considered in [LZ04, §2.2.4.3]. The cubic hypergeometric
relation

$\displaystyle{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-3,-\frac{5}{2}\atop-\frac{p}{r}-5\,}\right|\frac{4\beta}{\alpha^{2}\\!}\right)=0$
(7.15)

can be analyzed within the context of Section 3.2 with $b=-5/2$, $c=3+p/r$,
and $z=4\beta/\alpha^{2}$. We combine Theorem 3.1 with a simple transformation

$.(u,v)\mapsto\left(\frac{u-75}{4},\,\frac{v}{8}\right).$ (7.16)

The obtained elliptic curve is the same as in [LZ04]:

$\displaystyle{\mathcal{E}}_{5}:\quad v^{2}=u^{3}-2475u-5850.$ (7.17)

The isomorphism obtained from (3.20) is simpler than in [LZ04, pg. 105]:

$\displaystyle\frac{p}{q}=\frac{645-15u-11v}{4v},\qquad z=\frac{45-7u+v}{2v}.$
(7.18)

The data base of elliptic curves [LMFDB] identifies this curve by the label
4050.y2, and confirms that the Mordell-Weil group of $E_{1}$ is isomorphic to
$\mathbb{Z}\times(\mathbb{Z}/3\mathbb{Z})$. Any $\mathbb{Q}$-rational point on
$E_{1}$ can be expressed using the addition law on the elliptic curve as

$n\,(-5,80)+\varepsilon\,(75,480),\qquad\mbox{with
}n\in\mathbb{Z},\;\varepsilon\in\\{0,1,-1\\}.$ (7.19)

Several of the rational points correspond to the degenerate cases
$p/r\in\left\\{0,-3,-5/2\right\\}$ of no Belyi maps,
$p/r\in\left\\{-4,-3/2\right\\}$ of single maps, and
$p/r\in\left\\{-5,-1/2\right\\}$ of coupled maps. The elliptic involution
represents the hypergeometric identity (2.3), and acts as

$\left(\frac{p}{r},\,z\right)\mapsto\left(-\frac{p}{r}-\frac{11}{2},\,1-z\right).$

There are rational points with $p/r\in\\{-11/2,-11/4\\}$. Here are some other
values of $p/r$ that give Belyi maps defined over $\mathbb{Q}$:

$\displaystyle-{\textstyle\frac{28}{11}\,},-{\textstyle\frac{65}{22}\,};-{\textstyle\frac{59}{23}\,},-{\textstyle\frac{135}{46}\,};-{\textstyle\frac{65}{107}\,},-{\textstyle\frac{1047}{214}\,};^{\prime}{\textstyle\frac{33}{124}\,},-{\textstyle\frac{715}{124}\,};-{\textstyle\frac{251}{169}\,},-{\textstyle\frac{1357}{338}\,}.$
(7.20)

This list has only one positive value, representing a Shabat polynomial
defined over $\mathbb{Q}$. Rarity of positive values for $p/r$ is observed in
[LZ04]. Here are the next few:

${\textstyle\frac{2008145}{1653242}\,},{\textstyle\frac{1317156026567}{649800344821}\,},{\textstyle\frac{487834953714776556005}{104793834699948131134}\,},{\textstyle\frac{1999297558019926898176821908516}{208841813498019535906845150263}\,}.$

Following (7.18), the positivity region in Figure 3(i) is cut out by the lines
$v=0$ and $15u+11v=645$. We have 3 separate small regions on
${\mathcal{E}}_{5}$ between these lines, the one with $u>0$ especially tiny.
The rational points are distributed ergodically on ${\mathcal{E}}_{5}$, with
the density proportional to invariant holomorphic differential $du/v$. The
whole measure is the real period over the finite oval or the infinite piece.
It can be computed avoiding numeric issues near $v=0$ or $u=\infty$ by
integrating $du/v$ between points that differ by a 3-torsion point, and then
multiplying by 3. For example,

$\displaystyle\varrho_{5}=3\int_{-45}^{-5}\frac{du}{v}=3\int_{51}^{315}\frac{du}{v}\approx
0.732116211.$ (7.21)

The integrals over the 3 small regions could be stably computed by shifting
the intgeration range by a 3-torsion point. The two integrals on the finite
oval evaluate to $\approx 0.0564864103+0.0524120276$. After division by
$\varrho_{5}$ we can compute the percentage and conclude that positive values
of $p/q$ asymptotically occur about $5.72$ less frequently than the negative
ones on the finite oval. The points (7.19) with $|n|<200$ give 46 and 43
positive values in the two regions. This is within the rounding error from the
asymptotic prediction. The tiny integral on the infinite branch is $\approx
0.00407438266$. This gives the odds ratio $\approx 179$ for positive values.
The points (7.19) with $|n|<200$ give 3 positive values of $p/q$, with
$n\in\\{54,162,-188\\}$. Their denominators have 736, 6640 or 8944 decimal
digits, respectively. The same search looked for complete factorizations of
the cubic polynomial in $z$, but only predictable degenerations were found. By
(3.24) and (7.16), we need $\sqrt{102-2u}\in\mathbb{Q}$ then.

(i)(ii) Figure 3: The elliptic curves (7.17) and (7.24), with the numerator
line in (7.18) or (7.25) that together with the horizontal line $v=0$
determines the sign of $p/r$.

######  Example 7.6.

Similarly, for $m=6$ we can investigate the cubic relation

$\displaystyle{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-3,-\frac{7}{2}\atop-\frac{p}{r}-6\,}\right|\,z\right)=0$
(7.22)

between $p/r$ and $z=4\beta/\alpha^{2}$ by applying Theorem 3.1 with $b=-7/2$,
$c=4+p/r$. Additionally, we apply the simple transformation

$.(u,v)\mapsto\left(\frac{u-147}{4},\,\frac{v}{8}\right)$ (7.23)

and derive the elliptic curve

$\displaystyle{\mathcal{E}}_{6}:\quad v^{2}=u^{3}-17787u+692566,$ (7.24)

with an isomorphism

$\displaystyle\frac{p}{r}=\frac{5901-63u-13v}{4v},\qquad
z=\frac{791-13u+v}{8v}.$ (7.25)

The data base of elliptic curves [LMFDB, 13230.dp1] tells that the Mordell-
Weil group is $\mathbb{Z}\times(\mathbb{Z}/3\mathbb{Z})$ here as well. The
generators of $\mathbb{Q}$-rational points are $(u,v)=(35,336)$ and a torsion
point $(147,1120)$. The elliptic involution acts as

$\left(\frac{p}{r},\,z\right)\mapsto\left(-\frac{p}{r}-\frac{13}{2},\,1-z\right).$

Here are some values of $p/r$ that give Belyi maps defined over $\mathbb{Q}$:

$\displaystyle-{\textstyle\frac{13}{4}\,}\\!,\,1,-{\textstyle\frac{15}{2}\,},-{\textstyle\frac{13}{5}\,},-{\textstyle\frac{39}{10}\,},-{\textstyle\frac{30}{11}\,},-{\textstyle\frac{83}{22}\,},-{\textstyle\frac{52}{37}\,},-{\textstyle\frac{377}{74}\,},-{\textstyle\frac{20}{29}\,},-{\textstyle\frac{337}{58}\,},{\textstyle\frac{41}{16}\,},-{\textstyle\frac{145}{16}\,}.$
(7.26)

Positive values appear more frequently than in the $m=2$ case. Besides the
discardable $p/r=1$ representing a quadratic factor of $1-16x^{8}$ in
$\mathbb{Q}[x]$, we see ${\textstyle\frac{41}{16}\,}$ as well. Here are the
next few positive values:

${\textstyle\frac{83031}{5198}\,},{\textstyle\frac{572982199}{35574034}\,},{\textstyle\frac{66779978696204}{16641846371989}\,},{\textstyle\frac{77057440650930450189}{148795415472621031586}\,},{\textstyle\frac{559984659717697802475460}{36618976557904027122191}\,}.$

Following (7.25), the positivity region in Figure 3(i) is cut out by the lines
$v=0$ and $63u+13v=5901$. Again, we have 3 separate small regions on
${\mathcal{E}}_{6}$ between these lines. The whole real period can be computed
to be

$\displaystyle\varrho_{6}=3\int_{-133}^{\,35}\frac{du}{v}=3\int_{107}^{707}\frac{du}{v}\approx
0.541858251.$ (7.27)

The two integrals on the finite oval evaluate to $\approx
0.0507070923+0.0430448636$. After division by $\varrho_{6}$ we can compute the
odds ratio for positive $p/r$ to be $\approx 4.78$ for a rational point with
$u<50$. The small integral on the infinite branch is $\approx 0.00766222865$,
giving the odds ratio $\approx 69.7$. For a complete factorization of the
cubic polynomial we need $\sqrt{30(107-u)}\in\mathbb{Q}$, but only a few
predictable degenerations were found.

######  Example 7.7.

The case $m=7$ gives a genus 3 polynomial relation

$\displaystyle{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-4,-\frac{7}{2}\atop-\frac{p}{r}-7\,}\right|\,z\right)=0$
(7.28)

between $p/r$ and $z=4\beta/\alpha^{2}$. The Faltings theorem [Bom90] implies
that the genus 3 curve has only finitely many $\mathbb{Q}$-rational points.
Section 4.3 implies that it could be projected to an elliptic curve (4.22)
with $b=-7/2$, which is

$\displaystyle{\mathcal{E}}_{7}:\quad
v^{2}=u\,\big{(}u^{2}-119u+{\textstyle\frac{14175}{4}\,}\big{)}.$ (7.29)

The data base of elliptic curves [LMFDB, 94080.el2] tells that the Mordell-
Weil group of this curve has rank 2, and is isomorphic to
$\mathbb{Z}^{2}\times(\mathbb{Z}/2\mathbb{Z})$. Beside the 2-torsion point
$(u,v)=(0,0)$ and the generator
$({\textstyle\frac{105}{2}\,},-{\textstyle\frac{105}{2}\,})$ specialized from
(4.30), another free generator is $(60,15)$. To obtain rational points on the
genus 3 curve, we need rational points $(u,v)$ on ${\mathcal{E}}_{7}$ giving a
full square in (4.32). That means $(v-14u)^{2}-11340u\,$ must be a full
square. An extensive search through $2\cdot 61\cdot 51>6000$ points on
${\mathcal{E}}_{7}$ gives the following 12 values of $p/r=c-4$, paired by
(4.24):

$\displaystyle
1,-{\textstyle\frac{17}{2}\,};10,-{\textstyle\frac{35}{2}\,};-{\textstyle\frac{7}{3}\,},-{\textstyle\frac{31}{6}\,};-{\textstyle\frac{14}{5}\,},-{\textstyle\frac{47}{10}\,};-{\textstyle\frac{79}{11}\,},-{\textstyle\frac{7}{22}\,};-{\textstyle\frac{34}{11}\,},-{\textstyle\frac{97}{22}\,}.$
(7.30)

The value $1$ should be discarded along with the degenerate
$p/r\in\\{-5,-6,-7$, $-\frac{1}{2},-\frac{3}{2},-\frac{5}{2}\\}$. The positive
value 10 gives the Shabat polynomial

$(1+2x+4x^{2})^{10\,}(1-20x+180x^{2}-880x^{3}+1760x^{4}+6336x^{5}-59840x^{6}+183040x^{7}).\,$
(7.31)

######  Example 7.8.

Similarly, the case $m=8$ gives a genus 3 polynomial relation

$\displaystyle{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-4,-\frac{9}{2}\atop-\frac{p}{r}-8\,}\right|\,z\right)=0$
(7.32)

between $p/r$ and $z=4\beta/\alpha^{2}$. Section 4.3 implies that this curve
could be projected to an elliptic curve (4.22) with $b=-9/2$, which is

$\displaystyle{\mathcal{E}}_{8}:\quad
v^{2}=u\,\big{(}u^{2}-243u+{\textstyle\frac{59535}{4}\,}\big{)}.$ (7.33)

The data base of elliptic curves [LMFDB, 40320.bf2] tells that the Mordell-
Weil group is isomorphic to $\mathbb{Z}^{2}\times(\mathbb{Z}/2\mathbb{Z})$.
Beside the generator
$({\textstyle\frac{189}{2}\,},-{\textstyle\frac{567}{2}\,})$ specialized from
(4.30), another free generator is $(\frac{945}{4},\frac{14175}{8})$. We need
rational points $(u,v)$ on ${\mathcal{E}}_{8}$ giving a full square in (4.32).
That means $(v-18u)^{2}-34020u\,$ must be a full square. An extensive search
through $2\cdot 61\cdot 41>5000$ points on ${\mathcal{E}}_{8}$ gives these 13
pertinent values of $p/r=c-5$ from (4.24):

$\displaystyle-{\textstyle\frac{17}{2}\,};-14,{\textstyle\frac{11}{2}\,};-{\textstyle\frac{8}{3}\,},-{\textstyle\frac{35}{6}\,};-{\textstyle\frac{16}{5}\,},-{\textstyle\frac{53}{10}\,};-{\textstyle\frac{17}{5}\,},-{\textstyle\frac{51}{10}\,};-{\textstyle\frac{17}{7}\,},-{\textstyle\frac{85}{14}\,};-{\textstyle\frac{371}{151}\,},-{\textstyle\frac{1825}{302}\,}.$
(7.34)

The positive value $p/r=\frac{11}{2}$ gives the Shabat polynomial

$(1+2x+5x^{2})^{11}\,(1-11x+44x^{2}-715x^{4}+2717x^{5}-572x^{6}-29172x^{7}+97240x^{8})^{2}.$
(7.35)

###  7.2 Cases with fewer Belyi maps

######  Example 7.9.

Let us examine the cases when the hypergeometric polynomials in (7.6) or (7.7)
are linear in $z=4\beta/\alpha^{2}$. For odd $m>2$ we obtain a single Belyi
map when, respectively,

$\ell=-\frac{p}{r}\in\left\\{\frac{m+3}{2},m-2\right\\}.$ (7.36)

Indeed, setting $\ell=(m+3)/2$ in (7.6) gives $z=-m-1$. Up to scaling,
$H_{2}(x)=1+2x-(m+1)x^{2}$. According to (7.1), the terms to $x^{2k}$ in the
power series

$\displaystyle\left(1+2x-(m+1)x^{2}\right)^{\\!\frac{m+3}{2}}=\textstyle
1+(m+3)x-\frac{(m+1)(m+2)(m+3)}{3}\,x^{3}$
$\displaystyle\qquad\textstyle-\frac{(m+1)(m+2)(m+3)}{12}(m-3)x^{4}+\frac{(m+1)(m+2)(m+3)(m-1)(m+5)}{20}\,x^{5}\quad$
$\displaystyle\qquad\textstyle+\frac{(m+1)(m+2)(m+3)^{2}(m-1)}{45}(m-5)x^{6}$
(7.37)
$\displaystyle\qquad\textstyle-\frac{(m+1)(m+2)(m+3)(m-1)(m-3)(m^{2}+19m+42)}{252}\,x^{7}$
$\displaystyle\qquad\textstyle-\frac{(m+1)(m+2)(m+3)(m-1)(m-3)(m+5)(11m+26)}{3360}(m-7)\,x^{8}+\ldots.$

must be divisible by $(m-2k+1)$. Setting $\ell=m-2$ in (7.7) gives $z=m+2$.
Here as well, the terms to $x^{2k}$ in the power series

$\displaystyle\left(1+2x+(m+2)x\right)^{\frac{m-2}{2}}=\textstyle
1+(m\\!-\\!2)x+(m\\!-\\!2)(m\\!-\\!1)x^{2}+\frac{2m(m-2)(m-4)}{3}x^{3}$
$\displaystyle\qquad\textstyle+\frac{(m-2)(m-4)(5m+2)}{12}(m-3)\,x^{4}+\frac{(m-2)(m-4)(m-6)(13m^{2}-9m-10)}{60}\,x^{5}$
$\displaystyle\qquad\textstyle+\frac{(m-2)(m-4)(m-6)(19m^{2}+5m-6)}{180}(m-5)\,x^{6}+\ldots.$
(7.38)

must be hypnotically divisible by $(m-2k+1)$. For even $m>2$ we obtain
isolated Belyi maps when

$\ell=-\frac{p}{r}\in\left\\{\frac{m+4}{2},m-3\right\\}.$ (7.39)

We obtain then $z=-m/3$ or $z=m/3+1$ from (7.6) or (7.7), respectively. The
terms to $x^{2k+1}$ in the following power series must be divisible by
$(m-2k)$:

$\displaystyle\left(1+2x-\\!{\textstyle\frac{m}{3}\,}x^{2}\right)^{\\!\frac{m+4}{2}}=\textstyle
1+(m+4)x+\frac{(m+3)(m+4)}{3}\,x^{2}-\frac{m(m+2)(m+3)(m+4)}{36}\,x^{4}$
$\displaystyle\qquad\textstyle-\frac{m(m+2)(m+3)(m+4)}{180}(m-4)\,x^{5}+\frac{m(m+2)(m+3)(m+4)(m^{2}+15m-36)}{1620}\,x^{6}$
$\displaystyle\qquad\textstyle+\frac{m(m-2)(m+2)(m+3)(m+4)(m+6)}{2835}(m-6)\,x^{7}+\ldots,$
(7.40)
$\displaystyle\left(1+2x+\\!{\textstyle\frac{m+3}{3}\,}x^{2}\right)^{\\!\frac{m-3}{2}}=\textstyle
1+(m-3)x+\frac{2(m-3)^{2}}{3}x^{2}+\frac{(m-3)(m-5)}{3}(m-2)x^{3}$
$\displaystyle\qquad\textstyle+\frac{(m-3)(m-5)(5m^{2}-33m+36)}{36}\,x^{4}+\frac{(m-3)(m-5)(m-7)(m-1)}{20}(m-4)\,x^{5}$
$\displaystyle\qquad\textstyle+\frac{(m-3)^{2}(m-5)(m-7)(13m^{2}-105m+72)}{810}\,x^{6}$
$\displaystyle\qquad\textstyle+\frac{(m-3)(m-5)(m-7)(m-9)(53m^{2}-153m+72)}{11340}(m-6)\,x^{7}+\ldots.$
(7.41)

######  Example 7.10.

Let us look at the cases when the hypergeometric polynomials in (7.6) or (7.7)
are quadratic in $z=4\beta/\alpha^{2}$. For odd $m$ we then have

$\ell=-\frac{p}{r}\in\left\\{\frac{m+5}{2},m-4\right\\}.$ (7.42)

Setting $\ell=(m+5)/2$ in (7.6) gives this equation for $z$:

$3z^{2}+6(m+1)z+m^{2}-1=0.$ (7.43)

It has roots in $\mathbb{Q}$ when the discriminant $24(m+1)(m+2)$ is a square.
An equivalent condition is existence of integer solutions of the Pell equation

$(2m+3)^{2}-6d^{2}=1,$ (7.44)

as observed in [APZ20, Ch. 10]. These solutions correspond to the units in the
field $\mathbb{Q}(\sqrt{6})$, which are generated by $5+2\sqrt{6}$. We should
express them as

$\big{(}5+2\sqrt{6}\big{)}^{n}=(2m+3)\pm\sqrt{6}\,(z+m+1),$ (7.45)

leading to two integer values of $z$ for suitable $m$. For $n\in\\{2,3,4\\}$
we get the suitable values $m\in\\{23,241,2399\\}$. The case $m=23$ gives two
Belyi maps with $p/r=-14$ defined $\mathbb{Q}$, with $H_{2}(x)=1+x-x^{2}$ or
$H_{2}(x)=1+x-11x^{2}$.

Setting $\ell=m-4$ in (7.7) gives this equation for $z$:

$3z^{2}-6(m+2)z+(m+2)(m+4)=0.$ (7.46)

It has the same discriminant as (7.43), leading to the same Pell equation
(7.44). The case $m=23$ gives two Belyi maps with $p/r=-19/2$ defined
$\mathbb{Q}$, with $H_{2}=1+2x+5x^{2}$ or $H_{2}=1+2x+45x^{2}$.

For even $m$ we have

$\ell=-\frac{p}{r}\in\left\\{\frac{m+6}{2},\frac{m-5}{2}\right\\}.$ (7.47)

Setting $\ell=(m+6)/2$ in (7.6) gives this equation for $z$:

$15z^{2}-10mz+m(m-2)=0.$ (7.48)

It has roots in $\mathbb{Q}$ when the discriminant $40m(m+3)$ is a square. An
equivalent condition is existence of integer solutions of the equation

$(2m+3)^{2}-10d^{2}=9.$ (7.49)

If $m$ is divisible by 3 (and thus by 6), we have a reduction to Pell’s
equation

$\big{(}{\textstyle\frac{2}{3}m+1}\big{)}^{2}-10\hat{d}^{2}=1.$ (7.50)

The solutions correspond to units in the field $\mathbb{Q}(\sqrt{10})$, which
are generated by $3+\sqrt{10}$. We should express the solutions as

$\big{(}3+\sqrt{10}\big{)}^{n}=\textstyle\big{(}\frac{2}{3}m+1\big{)}\pm\sqrt{10}\,\big{(}z+\frac{1}{3}m\big{)},$
(7.51)

leading to two integer values of $z$ for suitable $m$. But the norm of
$3+\sqrt{10}$ in $\mathbb{Q}(\sqrt{10})$ is $-1$ rather than $1$; thus $n$
should be even. Further, $m$ is prescribed to be even (while $n=2$ gives
$m=27$, for example). For that we need $n$ to be divisible by 4. The smallest
possibility $n=4$ gives $m=1080$.

If $m$ is not divisible by $3$, equation (7.49) is solved by considering the
numbers of the norm 3 in $\mathbb{Q}(\sqrt{10})$:

$\displaystyle(1+\sqrt{10})\big{(}3+\sqrt{10}\big{)}^{n},\qquad(1-\sqrt{10})\big{(}3+\sqrt{10}\big{)}^{n}.$
(7.52)

The two options correspond to the fact that $\mathbb{Q}(\sqrt{10})$ is not a
unique factorization domain; its class number equals 2. In the first case, we
need $n=3$ mod 4 for even $m$; the smallest $n=3$ gives $m=242$. In the second
case we need $n=1$ mod 4; the smallest interesting $n=5$ gives $m=4802$.

Setting $\ell=m-5$ in (7.7) gives this equation for $z$:

$15z^{2}-10(m+3)z+(m+3)(m+5)=0.$ (7.53)

It has the same discriminant as (7.48), leading to the same equation (7.49).

######  Example 7.11.

Let us consider hypergeometric polynomials (7.6) of degree 3 or 4. Note that
the lower parameter $1+k-\ell$ is supposed to be a positive integer for
$k=m+1$. With odd $m$ and $\ell=(m+7)/2$, the polynomial in
$z=4\beta/\alpha^{2}$ is

${}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-3,\,-\frac{5}{2}\,\atop\ell-5}\right|\,z\right)=0.$
(7.54)

Comparing with (7.15), we conclude that we need integer values $p/r=-\ell$ of
Example 7.5 satisfying $p/r<-5$. That excludes the discarded
$p/r\in\\{0,-1,-2,-3$, $-4,-5\\}$ again. The number of points on
${\mathcal{E}}_{5}$ that give integer values of $p/r$ (or any other regular
function) is finite by Siegel’s theorem [Sil09, IX.3]. In this case, there
appear to be no relevant integer values.

Similarly, consideration of integer
$\ell\in\\{\frac{m+8}{2},\frac{m+9}{2},\frac{m+10}{2}\\}$ in (7.6) leads to
the hypergeometric polynomials

${}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-3,\,-\frac{7}{2}\,\atop\ell-6}\right|z\right)\\!,\quad{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-4,\,-\frac{7}{2}\,\atop\ell-7}\right|z\right)\\!,\quad{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{-4,\,-\frac{9}{2}\,\atop\ell-8}\right|z\right)\\!,$
(7.55)

comparable to (7.22), (7.28), (7.32), respectively. We need sufficiently
negative integer values of $p/r=-\ell$ from Examples 7.6 – 7.8. Only the value
$p/r=-14$ of Example 7.8 suits us. The pivotal hypergeometric evaluation is

${}_{2}\mbox{\rm F}_{\\!1}\\!\left(\left.{-4,\,-\frac{9}{2}\,\atop
6}\right|-4\right)=0.$ (7.56)

This gives $\ell=14$, $m=18$ and $H_{2}(x)=1+x-x^{2}$. The expansion of
$H_{2}(x)^{14}$ misses the term with $x^{19}$ accordingly. The polynomial
$G_{18}(x)$ is defined by the lower degree terms:

$\displaystyle
1+14x+77x^{2}+182x^{3}-910x^{5}-1365x^{6}+1430x^{7}+5005x^{8}-10010x^{10}-3640x^{11}$
$\displaystyle+14105x^{12}+6930x^{13}-15625x^{14}-6930x^{15}+14105x^{16}+3640x^{17}-10010x^{18}.$

The terms with $x^{4},x^{9}$ are missed as well. The broken symmetry around
the term with $x^{14}$ is notable. The Belyi map is $H_{2}(x)^{14}/G_{18}(x)$.

######  Example 7.12.

Hypergeometric polynomials (7.7) of degree 3 or 4 are considered similarly.
Consideration of odd $\ell\in\\{m-6,m-7,m-8,m-9\\}$ in (7.7) leads to the
hypergeometric polynomials

${}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{\\!-3,-\frac{5}{2}\atop\\!-\frac{\ell}{2}-6}\right|z\right)\\!,\quad{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{\\!-3,-\frac{7}{2}\atop\\!-\frac{\ell}{2}-7}\right|z\right)\\!,\quad{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{\\!-4,-\frac{7}{2}\atop\\!-\frac{\ell}{2}-8}\right|z\right)\\!,\quad{}_{2}\mbox{\rm
F}_{\\!1}\\!\left(\left.{\\!-4,-\frac{9}{2}\atop\\!-\frac{\ell}{2}-9}\right|z\right)\\!,$

comparable to (7.15), (7.22), (7.28), (7.32), respectively. The relation with
$p/r$ of Examples 7.5 – 7.8 is always $p/r=1+\ell/2$. We look for positive
half-integer values $p/r$. Again, only Example 7.8 provides: $p/r=11/2$. This
gives $\ell=9$, $m=18$, and also $H_{2}(x)=1+2x+5x^{2}$. The expansion of
$H_{2}(x)^{9/2}$ misses the term with $x^{19}$ indeed. We have

$\displaystyle
G_{18}(x)=1+9x+54x^{2}+210x^{3}+630x^{4}+1386x^{5}+2394x^{6}+2394x^{7}+2655x^{8}$
$\displaystyle+1195x^{9}+252x^{10}-252x^{11}+168x^{12}-180x^{14}+228x^{15}-18x^{16}-378x^{17}+560x^{18}.$

For some reason, the coefficients $2394$ and $\pm 252$ repeat twice
consequently. The Belyi map is $G_{18}(x)^{2}/H_{2}(x)^{9}$.

## References

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  * [AS22] N.M. Adrianov, G.B. Shabat, Calculating complete lists of Belyi pairs, Mathematics 10 (2022). DOI: https://doi.org/10.3390/math10020258.
  * [AAR99] G.A. Andrews, R. Askey, R.Roy. Special Functions. Cambridge University Press, 1999.
  * [Bom90] E. Bombieri. The Mordell conjecture revisited. Ann. Scuola Norm. Sup. Pisa Cl. Sci. 90 (1990): 615–640.
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