# Number of integers represented by families of binary forms

Étienne Fouvry
Univ. Paris–Sud, CNRS, Université Paris–Saclay, CNRS,
Laboratoire de Mathématiques d’Orsay,
91405 Orsay, France
E-mail<EMAIL_ADDRESS>Michel Waldschmidt
Sorbonne Université and Université de Paris,
CNRS, IMJ–PRG,
75005 Paris, France
E-mail<EMAIL_ADDRESS>
###### Abstract

We extend our previous results on the number of integers which are values of
some cyclotomic form of degree larger than a given value (see [FW]), to more
general families of binary forms with integer coefficients. Our main
ingredient is an asymptotic upper bound for the cardinality of the set of
values which are common to two non–isomorphic binary forms of degree greater
than $3$. We apply our results to some typical examples of families of binary
forms.

Nous dédions ce travail à la mémoire d’André Schinzel

avec notre profond respect et notre affectueuse admiration

††2020 _Mathematics Subject Classification_ : Primary 11E76; Secondary 11D45
11D85.†† _Key words and phrases_ : Binary forms, Representation of integers by
binary forms, Families of Diophantine equations.

##  1 Introduction

Let $d\geq 3$ be an integer. We denote by ${\mathrm{Bin}}(d,{\mathbb{Z}})$ the
set of binary forms $F=F(X,Y)$ with integer coefficients, of degree $d$ and
with discriminant different from zero. For

(1.1) $\gamma=\begin{pmatrix}a_{1}&a_{2}\\\
a_{3}&a_{4}\end{pmatrix}\in{\mathrm{GL}}(2,{\mathbb{Q}}),$

and $F\in{\mathrm{Bin}}(d,{\mathbb{Z}})$, $F\circ\gamma$ is the binary form
with rational coefficients, defined by

$\left(F\circ\gamma\right)(X_{1},X_{2})=F(a_{1}X_{1}+a_{2}X_{2},a_{3}X_{1}+a_{4}X_{2}).$

Two elements $F_{1}$ and $F_{2}$ in ${\mathrm{Bin}}(d,{\mathbb{Z}})$ are said
to be isomorphic is there is a $\gamma\in{\mathrm{GL}}(2,{\mathbb{Q}})$ such
that

$F_{1}\circ\gamma=F_{2}.$

To estimate the number of values simultaneously taken by the binary forms
$F_{1}$ and $F_{2}$, we introduce the counting function, for $N$ an integer
$\geq 1$,

(1.2) $\displaystyle{\mathcal{N}}(F_{1},F_{2};N)$
$\displaystyle:=\sharp\left(F_{1}({\mathbb{Z}}^{2})\cap
F_{2}({\mathbb{Z}}^{2})\cap[-N,+N]\right)$
$\displaystyle=\sharp\bigl{\\{}m:|m|\leq N,\text{ there exists
}(x_{1},x_{2},x_{3},x_{4})\in{\mathbb{Z}}^{4}$ $\displaystyle\hskip
56.9055pt\text{ such that }m=F_{1}(x_{1},x_{2})=F_{2}(x_{3},x_{4})\bigr{\\}}.$

Our first result gives an upper bound for this function when the two forms are
not isomorphic.

######  Theorem 1.1.

For every $d\geq 3$, there is a constant $\vartheta_{d}<2/d$ such that, for
every $\varepsilon>0$, for every pair $(F_{1},F_{2})$ of no–isomorphic forms
of ${\mathrm{Bin}}(d,{\mathbb{Z}})$, for $N\to\infty$, one has the bound

(1.3)
${\mathcal{N}}\left(F_{1},F_{2};N\right)=O_{F_{1},F_{2},\varepsilon}\left(N^{\vartheta_{d}+\varepsilon}\right).$

This theorem calls for the following comments:

######  Remark 1.2.

The point in this theorem is that the constant $\vartheta_{d}$, defined in
(2.1), satisfies the inequality $\vartheta_{d}<2/d$ (see the inequalities
(2.3) below). In fact, it is known that for any
$F\in{\mathrm{Bin}}(d,{\mathbb{Z}})$, there exists $C_{F}>0$, such that, for
$N$ tending to infinity, one has the equality

${\mathcal{N}}(F,F;N)=\left(C_{F}+o_{F}(1)\right)N^{2/d},$

(see Theorem A in §1.1, due to Stewart and Xiao [SX, Theorem 1.1]).

######  Remark 1.3.

The explicit value of $\vartheta_{d}$ given in (2.1) leads to the inequality
$\vartheta_{d}>1/d$ for all $d\geq 3$ (see (2.3)). It also shows that
$\vartheta_{d}$ is asymptotic to $1/d$ as $d\to\infty$. This value is
asymptotically optimal as shown by the two forms

$F_{1}(X,Y)=X^{d}+Y^{d}\text{ and }F_{2}(X,Y)=X^{d}+2Y^{d}.$

These two forms are not isomorphic. From the equalities
$F_{1}(n,0)=F_{2}(n,0)=n^{d}$, we deduce the lower bound

${\mathcal{N}}(F_{1},F_{2};N)\geq N^{1/d}\ (N\geq 1).$

######  Remark 1.4.

According to [FW, Corollaire 3.3], if the two forms $F_{1}$, $F_{2}$ are
positive definite and at least one of them is not divisible by a linear form
with rational coefficients, then the exponent $\vartheta_{d}$ in the
conclusion of Theorem 1.1 can be replaced by $\eta_{d}$ with
$\eta_{d}<\vartheta_{d}$ (see the definition of $\eta_{d}$ and $\vartheta_{d}$
in §2.1).

######  Remark 1.5.

We will show in §2.4 that the exponent $\vartheta_{d}$ in the conclusion of
Theorem 1.1 can be replaced by the coefficient $\eta_{d}$ quoted in the
previous remark when the binary form $F_{1}(X,Y)F_{2}(X,Y)$ has no real root.

######  Remark 1.6.

Theorem 1.1 is no more valid for $d=2$. This is well known: see for instance
[FLW, Prop. 6.1, eq. (6.3)], where, choosing $F_{1}(X,Y)=X^{2}+Y^{2}$ and
$F_{2}(X,Y)=X^{2}+XY+Y^{2}$, one has, for $B$ tending to infinity, the
asymptotic formula

${\mathcal{N}}(F_{1},F_{2};B)=\left(\beta_{0}+o(1)\right)B\big{(}\log
B)^{-3/4},$

for some constant $\beta_{0}>0$.

######  Remark 1.7.

Theorem 1.1 is immediately generalized to binary forms with rational
coefficients, it suffices to multiply by a common denominator.

######  Remark 1.8.

The following proposition shows that if $F_{1}$ and $F_{2}$ are isomorphic,
the equality (1.3) never holds.

######  Proposition 1.9.

Let $d\geq 3$ and let $F_{1}$ and $F_{2}$ be two isomorphic binary forms in
${\mathrm{Bin}}(d,{\mathbb{Z}})$. Then there is a positive constant
$C_{F_{1},F_{2}}$, such that, for $N$ tending to infinity, we have the
inequality

${\mathcal{N}}(F_{1},F_{2};N)\geq\left(C_{F_{1},F_{2}}-o_{F_{1},F_{2}}(1)\right)N^{2/d}.$

###### Proof.

Let $\gamma$ as in (1.1) such that $F_{1}=F_{2}\circ\gamma$. Let $D\geq 1$ be
an integer such that $(Da_{1},Da_{2},Da_{3},Da_{4})$ belongs to
${\mathbb{Z}}^{4}$. By homogeneity, we deduce that the two forms

$G_{1}(X_{1},X_{2}):=F_{1}(DX_{1},DX_{2})$

and

$G_{2}(X_{1},X_{2}):=F_{2}\left(Da_{1}X_{1}+Da_{2}X_{2},Da_{3}X_{1}+Da_{4}X_{2}\right)$

are equal. So we have the equality of their images

$G_{1}({\mathbb{Z}}^{2})=G_{2}({\mathbb{Z}}^{2}).$

We also have the obvious inclusions

$G_{1}({\mathbb{Z}}^{2})\subset F_{1}({\mathbb{Z}}^{2})\text{ and
}G_{2}({\mathbb{Z}}^{2})\subset F_{2}({\mathbb{Z}}^{2}),$

which lead to the inclusion

(1.4) $G_{1}({\mathbb{Z}}^{2})\subset F_{1}({\mathbb{Z}}^{2})\cap
F_{2}({\mathbb{Z}}^{2}).$

A new application of the result of Stewart and Xiao (see Theorem A below)
gives, for some constant $C_{G_{1}}>0$, the equality

(1.5)
${\mathcal{N}}(G_{1},G_{1};N)=\left(C_{G_{1}}+o_{G_{1}}(1)\right)N^{2/d},$

as $N$ tends to infinity. Gathering (1.4) and (1.5) we obtain the inequality
claimed in Proposition 1.9. ∎

Theorem 1.1 is an important tool for our generalisation of our previous study
in [FW], where we produced an asymptotic formula for the number of values $m$,
with $|m|\leq B$ taken by some cyclotomic form $\Phi_{n}$ of degree
$\varphi(n)$ greater than a fixed $d\geq 3$. Recall that $\varphi$ is the
Euler function and that to the $n$–th cyclotomic polynomial $\phi_{n}(X)$, of
degree $\varphi(n)$, is attached the cyclotomic form
$\Phi_{n}(X,Y):=Y^{\varphi(n)}\cdot\phi_{n}(X/Y)$.

Our purpose is to study the following general problem: Let $\mathcal{F}$ be an
infinite subset of $\displaystyle{\underset{d\geq
3}{\bigcup}{\mathrm{Bin}}(d,{\mathbb{Z}})}$, satisfying natural properties.
Let $A$ be a fixed non negative integer. As $B$ tends to infinity, estimate
the counting function

(1.6) $\displaystyle{\mathcal{R}}_{\geq
d}\left(\mathcal{F},B,A\right):=\sharp\,\bigl{\\{}m:0\leq|m|\leq B,\,\text{
there is }\,F\in\mathcal{F}\text{ with }\deg F\geq d$ $\displaystyle\text{ and
}(x,y)\in{\mathbb{Z}}^{2}\text{ with }\max\\{|x|,|y|\\}\geq A,\text{ such that
}F(x,y)=m\bigr{\\}}.$

The introduction of the parameter $A$ may be seen as artificial. It is enacted
to prevent from the following phenomenon encountered for instance in the case
of the family of cyclotomic forms $\Phi_{n}$, where, for every prime $p$, we
have

(1.7) $\Phi_{p}(1,1)=p$

(recall that $\Phi_{p}(X,Y)=(X^{p}-Y^{p})\big{/}(X-Y)$). We wish to avoid
counting these values, since the set of primes, by its cardinality, completely
hides the set of other values $\Phi_{n}(x,y)$ when $\max\\{|x|,|y|\\}\geq 2$
and $\varphi(n)\geq d$.

Let $\mathcal{F}$ be a set of binary forms. We denote by $\mathcal{F}_{d}$ the
subset of forms of $\mathcal{F}$ of degree $d$. We will study the set of
values taken by forms belonging to some $(A,A_{1},d_{0},d_{1},\kappa)$–regular
families $\mathcal{F}$, that we define as follows.

######  Definition 1.10.

Let $A$, $A_{1}$, $d_{0}$, $d_{1}$ be integers and let $\kappa$ be a real
number such that

(1.8) $A\geq 1,\,A_{1}\geq 1,\,d_{1}\geq d_{0}\geq 0,\,0<\kappa<A.$

Let $\mathcal{F}$ be a set of binary forms. We say that $\mathcal{F}$ is
$(A,A_{1},d_{0},d_{1},\kappa)$–regular if it satisfies the following
conditions:

  1. (i)

The set $\mathcal{F}$ is infinite,

  2. (ii)

We have the inclusion

$\mathcal{F}\subset\bigcup_{d\geq 3}{\mathrm{Bin}}(d,{\mathbb{Z}}),$

  3. (iii)

For all $d\geq 3$, one has the inequality $\sharp\mathcal{F}_{d}\leq
d^{A_{1}}$,

  4. (iv)

Two forms of $\mathcal{F}$ are isomorphic if and only if they are equal,

  5. (v)

For any $d\geq\max\\{d_{1},d_{0}+1\\}$, the following holds

$\left.\begin{array}[]{ll}F\in\mathcal{F}_{d},\\\
(x,y)\in{\mathbb{Z}}^{2}\text{ and }F(x,y)\neq 0,\\\ \max\\{|x|,|y|\\}\geq
A,\end{array}\right\\}\Rightarrow\max\\{|x|,|y|\\}\leq\kappa\left|F(x,y)\right|^{\frac{1}{d-d_{0}}}.$

The upper bound in the right hand side of (v) is trivial for
$\max\\{|x|,|y|\\}\leq\kappa$, this is why we request $A>\kappa$.

The family of cyclotomic forms

$\boldsymbol{\Phi}:=\\{\Phi_{n}:\varphi(n)\geq 4,\;n\not\equiv 2\pmod{4}\\}$

satisfies the assumptions (i), (ii), (iii), (iv), but is not
$(1,A_{1},d_{0},d_{1},\kappa)$–regular for any value of $A_{1}$, $d_{0}$,
$d_{1}$ and $\kappa$, since (1.7) shows that (v) is not satisfied. However
$\boldsymbol{\Phi}$ is $(2,2,0,4,2/\sqrt{3})$–regular, this is a consequence
of [FW, Théorème 4.10] and of the classical inequality $n/(\log\log
n)<\varphi(n)<n$.

###  1.1 Some facts on a single form

Before stating our main result concerning $\mathcal{R}_{\geq
d}(\mathcal{F},B,A)$ defined in (1.6), we recall some fundamental objects
attached to a binary form $F\in{\mathrm{Bin}}(d,{\mathbb{Z}})$ when $d\geq 3$:

  * •

The fundamental domain of $F$ is

$\mathcal{D}(F):=\left\\{(x,y)\in{\mathbb{R}}^{2}:|F(x,y)|\leq 1\right\\},$

  * •

The area of the fundamental domain of $F$ is the real number

(1.9) $A_{F}:=\iint_{\mathcal{D}(F)}{\mathrm{d}}x\,{\mathrm{d}}y.$

We always have $0<A_{F}<\infty$.

  * •

The group of automorphisms  of $F$ is

$\displaystyle{\mathrm{Aut}}$ $\displaystyle(F;{\mathbb{Q}}):=$
$\displaystyle\left\\{\begin{pmatrix}a_{1}&a_{2}\\\
a_{3}&a_{4}\end{pmatrix}\in{\mathrm{GL}}(2,{\mathbb{Q}}):F(X,Y)=F(a_{1}X+a_{2}Y,a_{3}X+a_{4}Y)\right\\}.$

This is a finite subgroup of ${\mathrm{GL}}(2,{\mathbb{Q}})$. We now recall
the important result of Stewart and Xiao, that we already mentioned above [SX,
Theorems 1.1 and 1.2]:

######  Theorem A.

_For every $d\geq 3$, there is a constant $\kappa_{d}<2/d$ such that, for all
$F\in{\mathrm{Bin}}(d,{\mathbb{Z}})$ and for all $\varepsilon>0$, the
following equality_

$\mathcal{N}(F,F;B)=A_{F}\cdot W_{F}\cdot
B^{2/d}+O_{F,\varepsilon}\left(B^{\kappa_{d}+\varepsilon}\right),$

_holds uniformly for $B\to\infty$, where
$W_{F}=W\left({\mathrm{Aut}}(F;{\mathbb{Q}})\right)$ depends only on the group
${\mathrm{Aut}}(F;{\mathbb{Q}})$._

For $G$ a finite subgroup of ${\mathrm{GL}}(2,{\mathbb{Q}})$ which is the
group of automorphisms of an element of ${\mathrm{Bin}}(d,{\mathbb{Z}})$, the
constant $W(G)$ is a rational number which is defined in [SX, Theorem 1.2].
This definition is subtle since it depends on the denominators of the entries
of the matrices belonging to $G$. However for the families $\mathcal{F}$ that
we will meet in this paper, we will only need the equalities

(1.10)
$W\left(\\{\mathrm{Id}\\}\right)=1,\,W\left(\\{\mathrm{Id},\,-{\mathrm{Id}}\\}\right)=1/2\text{
and }W\left(\left\\{\begin{pmatrix}\pm 1&0\\\ 0&\pm
1\end{pmatrix}\right\\}\right)=1/4.$

Finally, the exponent $\kappa_{d}$ in Theorem A is defined by

(1.11) $\kappa_{d}=\begin{cases}\displaystyle{\frac{12}{19}}&\text{ if
}d=3,\\\ \displaystyle{\frac{3}{(d-2)\sqrt{d}+3}}&\text{ if }4\leq d\leq 8,\\\
\displaystyle{\frac{1}{d-1}}&\text{ if }d\geq 9.\end{cases}$

Actually the value of this exponent is improved when $F(X,Y)$ does not have a
linear factor over ${\mathbb{R}}[X,Y]$, see [SX, formula (1.11)].

###  1.2 An asymptotic formula for ${\mathcal{R}}_{\geq d}(\mathcal{F},B,A)$

Our central result is the following. The exponent $\vartheta_{d}$ is defined
in (2.1).

######  Theorem 1.11.

Let $(A,A_{1},d_{0},d_{1},\kappa)$ satisfying the conditions (1.8). Let
$\mathcal{F}$ be a $(A,A_{1},d_{0},d_{1},\kappa)$–regular family of binary
forms. Then for every $d\geq\max\\{3,d_{1}\\}$ and every positive
$\varepsilon$, one has the equality

$\mathcal{R}_{\geq
d}(\mathcal{F},B,A)=\left(\sum_{F\in\mathcal{F}_{d}}A_{F}W_{F}\right)\cdot
B^{2/d}+O_{\mathcal{F},A,d,\varepsilon}\bigl{(}B^{\vartheta_{d}+\varepsilon}\bigr{)}+O_{\mathcal{F},A,d}\bigl{(}B^{2/d^{\dagger}}\bigr{)},$

uniformly for $B\to\infty$. The integer $d^{\dagger}$ is defined by

$d^{\dagger}:=\inf\\{d^{\prime}:d^{\prime}>d\text{ such that
}\mathcal{F}_{d^{\prime}}\neq\emptyset\\}.$

Recall that $\mathcal{F}_{d}$ is not empty for infinitely many values of $d$
since the set $\mathcal{F}$ is infinite.

The assumption (v) in the definition 1.10 of a regular family cannot be
omitted, even in the case of totally imaginary forms (homogeneous versions of
polynomials without real roots), as show by the sequence of positive definite
forms $(X-Y)^{2}(X-2Y)^{2}\cdots(X-dY)^{2}+dY^{2d}$, the value of which at the
points $(x,y)=(n,1)$, $1\leq n\leq d$, is $d$.

The following is a direct application of (1.10):

######  Corollary 1.12.

Suppose that $\mathcal{F}$ satisfies the hypothesis of Theorem 1.11 and that,
for every $d\geq 3$, $\mathcal{F}_{d}$ satisfies one of the three following
conditions:

  * C1:

for all $F\in\mathcal{F}_{d}$, we have
${\mathrm{Aut}}(F,{\mathbb{Q}})=\\{{\mathrm{Id}}\\}$,

  * C2:

for all $F\in\mathcal{F}_{d}$, we have
${\mathrm{Aut}}(F,{\mathbb{Q}})=\\{\pm{\mathrm{Id}}\\}$ (cyclic group of order
$2$),

  * C3:

for all $F\in\mathcal{F}_{d}$, we have
${\mathrm{Aut}}(F,{\mathbb{Q}})=\left\\{\begin{pmatrix}\pm 1&0\\\ 0&\pm
1\end{pmatrix}\right\\}$ (Klein group of order $4$).

Then we have the equality

(1.12) $\mathcal{R}_{\geq
d}(\mathcal{F},B,A)=C_{d}\cdot\left(\sum_{F\in\mathcal{F}_{d}}A_{F}\right)\cdot
B^{2/d}+O_{\mathcal{F},A,d,\varepsilon}\bigl{(}B^{\vartheta_{d}+\varepsilon}\bigr{)}+O_{\mathcal{F},A,d}\bigl{(}B^{2/d^{\dagger}}\bigr{)},$

where the coefficient $C_{d}$ has respectively the values $1$, $1/2$ or $1/4$
according to the condition C1, C2, C3 respectively satisfied by
$\mathcal{F}_{d}$.

###  1.3 Some applications

We now give a list of regular families $\mathcal{F}$ in order to illustrate
our results.

The first example of course is given by the sequence of cyclotomic binary
forms [FLW]. We do not repeat it.

Our second example is given by a family of binomials $ax^{d}+by^{d}$ where $d$
is even while $a$, $b$ have the same sign: these restrictions allow us to
check easily the assumption (v) in the definition 1.10 of a regular family.
Since the proof is easy, we give it right away.

The three other examples below will require more work; for them we restrict
ourselves to families $\mathcal{F}$ satisfying the conditions of Corollary
1.12 in order to apply (1.12).

There are a lot of variations on these constructions.

####  1.3.1 Binomial forms

For each even integer $d\geq 4$, let $\mathcal{E}_{d}$ be a finite subset of
${\mathbb{Z}}_{>0}\times{\mathbb{Z}}_{>0}$. Assume $\mathcal{E}_{d}$ is not
empty for infinitely many $d$ and has at most $d^{A_{1}}$ elements for some
$A_{1}>0$ and all $d$. Let $\mathcal{B}_{d}$ denote the family of binary forms
$aX^{d}+bY^{d}$ with $(a,b)\in\mathcal{E}_{d}$ and let
$\mathcal{B}=\cup_{d\geq 4}\mathcal{B}_{d}$. We assume that for
$(a,b)\not=(a^{\prime},b^{\prime})$ in $\mathcal{E}_{d}$, one at least of
$a/a^{\prime}$, $b/b^{\prime}$ is not a $d$-th power of a rational number, and
also one at least of $a/b^{\prime}$, $b/a^{\prime}$ is not a $d$-th power of a
rational number.

######  Theorem 1.13.

The family $\mathcal{B}$ is $(2,A_{1},0,4,1)$–regular.

Further, for every $d\geq 4$ and for every $\varepsilon>0$ we have the
equality

(1.13) $\mathcal{R}_{\geq
d}(\mathcal{B},B,2)=\left(\sum_{F\in\mathcal{B}_{d}}A_{F}W_{F}\right)B^{1/d}+O_{\mathcal{B},d,\varepsilon}\left(B^{\max\\{\vartheta_{d}+\varepsilon,2/d^{{\dagger}}\\}}\right),$

uniformly for $B\to\infty$. The integer $d^{\dagger}$ is defined by

$d^{\dagger}:=\inf\\{d^{\prime}:d^{\prime}>d\text{ such that
}\mathcal{B}_{d^{\prime}}\neq\emptyset\\}.$

We will check the hypothesis (iv) of the definition 1.10 of a regular family
by means of the following auxiliary result.

######  Lemma 1.14.

Let $d\geq 4$ be even and let $a,b,a^{\prime},b^{\prime}$ be positive
integers. Then the two binary forms $aX^{d}+bY^{d}$ and
$a^{\prime}X^{d}+b^{\prime}Y^{d}$ are isomorphic if and only if either
$a/a^{\prime}$, $b/b^{\prime}$ are both $d$-th power of rational numbers, or
$a/b^{\prime}$, $b/a^{\prime}$ are both $d$-th power of rational numbers.

###### Proof.

If $a/a^{\prime}=u^{d}$ and $b/b^{\prime}=v^{d}$, then the two forms
$aX^{d}+bY^{d}=a^{\prime}(uX)^{d}+b^{\prime}(vY)^{d}$ and
$a^{\prime}X^{d}+b^{\prime}Y^{d}$ are isomorphic. Also if $a/b^{\prime}=u^{d}$
and $b/a^{\prime}=v^{d}$, then the two forms
$aX^{d}+bY^{d}=a^{\prime}(vX)^{d}+b^{\prime}(uY)^{d}$ and
$a^{\prime}X^{d}+b^{\prime}Y^{d}$ are isomorphic. It remains to prove the
converse.

Assume that the two binary forms $aX^{d}+bY^{d}$ and
$a^{\prime}X^{d}+b^{\prime}Y^{d}$ are isomorphic. Let
$\gamma=\begin{pmatrix}a_{1}&a_{2}\\\
a_{3}&a_{4}\end{pmatrix}\in\mathrm{GL}(2,{\mathbb{Q}})$ satisfy

$a(a_{1}X+a_{2}Y)^{d}+b(a_{3}X+a_{4}Y)^{d}=a^{\prime}X^{d}+b^{\prime}Y^{d}.$

We have

$aa_{1}^{d}+ba_{3}^{d}=a^{\prime},\quad aa_{2}^{d}+ba_{4}^{d}=b^{\prime}$

and, for $i=1,\dots,d-1$,

$aa_{1}^{i}a_{2}^{d-i}+ba_{3}^{i}a_{4}^{d-i}=0.$

$\bullet$ Assume $a_{2}=0$. From

$a(a_{1}X)^{d}+b(a_{3}X+a_{4}Y)^{d}=a^{\prime}X^{d}+b^{\prime}Y^{d}$

we deduce $ba_{4}^{d}=b^{\prime}$, $a_{4}\not=0$, hence $a_{3}=0$, and
therefore $aa_{1}^{d}=a^{\prime}$.
$\bullet$ Assume $a_{1}=0$. From

$a(a_{2}Y)^{d}+b(a_{3}X+a_{4}Y)^{d}=a^{\prime}X^{d}+b^{\prime}Y^{d}$

we deduce $ba_{3}^{d}=a^{\prime}$, $a_{3}\not=0$, hence $a_{4}=0$, and
therefore $aa_{2}^{d}=b^{\prime}$.
$\bullet$ Finally let us check that the case $a_{1}a_{2}\not=0$ is not
possible. Write

$aa_{1}a_{2}^{d-1}+ba_{3}a_{4}^{d-1}=0,\quad
aa_{1}^{2}a_{2}^{d-2}+ba_{3}^{2}a_{4}^{d-2}=0.$

We deduce $a_{3}a_{4}\not=0$,

$\frac{a_{1}}{a_{3}}=-\frac{b}{a}\left(\frac{a_{4}}{a_{2}}\right)^{d-1},\quad\left(\frac{a_{1}}{a_{3}}\right)^{2}=-\frac{b}{a}\left(\frac{a_{4}}{a_{2}}\right)^{d-2},$

hence

$\left(\frac{a_{4}}{a_{2}}\right)^{d}=-\frac{a}{b}$

which is impossible for $a$, $b$ positive and $d$ even.

∎

###### Proof of Theorem 1.13.

The conditions (i), (ii) and (iii) in the definition 1.10 of a regular family
are satisfied by hypothesis.

For $(a,b)\not=(a^{\prime},b^{\prime})$ in $\mathcal{E}_{d}$, the two binary
forms $aX^{d}+bY^{d}$ and $a^{\prime}X^{d}+b^{\prime}Y^{d}$ are not
isomorphic, as shown by Lemma 1.14. Finally, for $(a,b)\in\mathcal{E}_{d}$ and
$(x,y)\in{\mathbb{Z}}^{2}$, we have

$ax^{d}+by^{d}\geq\max\\{|x|,|y|\\}^{d}.$

This completes the proof of condition (v) in the definition 1.10 of a regular
family.

The second assertion of Theorem 1.13 then follows from Theorem 1.11. ∎

Our assumptions do not allow any upper bound for $\mathcal{R}_{\geq
d}(\mathcal{B},B,1)$ better than $B$: the set of all $a$, $b$ and $a+b$ for
$(a,b)$ in $\cup_{d^{\prime}\geq d}E_{d^{\prime}}$ may contain all positive
integers.

Explicit values for $W_{F}$ and $A_{F}$ for $F\in\mathcal{F}_{d}$ are given in
Corollary 1.3 of [SX]. The values of $W_{F}$ and $A_{F}$ are computed without
the two assumptions $a,b$ of the same sign and $d$ even, but none of these two
hypotheses can be omitted for our theorem, as shown by the two sequences
$X^{d}-(d^{d}-d)Y^{d}$ ($d$ even) and $X^{d}+(d^{d}-d)Y^{d}$ ($d$ odd).

####  1.3.2 Products of positive quadratic forms

Let $(\mu_{n})_{n\geq 1}$ be an increasing sequence of positive squarefree
integers; assume that there exists $\lambda>0$ such that

(1.14) $\mu_{n}\leq\lambda\,n\text{ for all }n\geq 1.$

If we choose $\mu_{n}=q_{n}$ where $(q_{n})_{n\geq 1}$ is the full sequence

$1,\,2,\,3,\,5,\,6,\,7,\,10,\,11,\,13,\,14,\,15,...$

of positive squarefree integers, written in ascending order, then, as it is
well known (see [HW, Theorem 333] and https://oeis.org/A005117), we have

$\sharp\\{q_{n}\leq x\\}=\sum_{n\leq
x}\mu(n)^{2}=\frac{6}{\pi^{2}}x+O(\sqrt{x}),$

which implies that

$q_{n}\sim\frac{\pi^{2}}{6}n\;(n\rightarrow\infty).$

Since $\mu_{230}\geq q_{230}=381$, we have $\lambda\geq\frac{381}{230}\cdotp$
As a matter of fact we have

(1.15) $\sup_{n\geq 1}\frac{q_{n}}{n}=\frac{381}{230}\cdotp$

Hence, in the special case $\mu_{n}=q_{n}$ ($n\geq 1$), $\lambda=381/230$ is
an admissible value.

For $d\geq 2$ and $1\leq\nu\leq d+1$, we denote by $Q^{+}_{d,\nu}$ the binary
form of degree $2d$ defined by the formula

(1.16) $Q^{+}_{d,\nu}(X,Y):=\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
d+1}{n\neq\nu}}\left(X^{2}+{\mu}_{n}Y^{2}\right).$

The associated family is

$\mathcal{Q}^{+}:=\left\\{Q^{+}_{d,\nu}:d\geq 2,\,1\leq\nu\leq d+1\right\\}$

with $\mathcal{Q}^{+}_{d}=\emptyset$ for $d$ odd and
$\mathcal{Q}^{+}_{2d}=\left\\{Q^{+}_{d,\nu}:1\leq\nu\leq d+1\right\\}$ for
$d\geq 2$. With $\lambda$ defined in (1.14), we have

######  Theorem 1.15.

The family $\mathcal{Q}^{+}$ is $(2,1,0,4,1)$–regular.

Furthermore, for every $d~{}\geq 2$, $\mathcal{Q}_{2d}^{+}$ satisfies the
condition C3 of Corollary 1.12.

Finally, for every $d\geq 2$ and for every $\varepsilon>0$ we have the
equality

(1.17) $\mathcal{R}_{\geq
2d}(\mathcal{Q}^{+},B,0)=\frac{1}{4}\left(\sum_{F\in\mathcal{Q}_{2d}^{+}}A_{F}\right)B^{1/d}+O_{\lambda,d,\varepsilon}\left(B^{\max\\{\vartheta_{2d}+\varepsilon,1/(d+1)\\}}\right),$

uniformly for $B\to\infty$, and the inequalities

(1.18)
$\frac{\pi}{\sqrt{\lambda}}\cdot\sqrt{d}<\left(\sum_{F\in\mathcal{Q}_{2d}^{+}}A_{F}\right)<\pi\sqrt{\mathrm{e}}(\sqrt{d}+1).$

See (2.5) for a simplification of the exponent in the error term of (1.17).

######  Remark 1.16.

(Thanks to Jean-Baptiste Fouvry). Consider the two quartic forms

$Q_{2,3}^{+}(X,Y)=(X^{2}+Y^{2})(X^{2}+2Y^{2})\quad\text{and}\quad
Q_{2,1}^{+}(U,V)=(U^{2}+2V^{2})(U^{2}+3V^{2}).$

One checks

$Q_{2,3}^{+}(X,Y)-Q_{2,1}^{+}(U,Y)=(-U^{2}+X^{2}-Y^{2})(U^{2}+X^{2}+4Y^{2}).$

The Pythagorean triples $(y,u,x)$, namely the solutions of the equation
$y^{2}+u^{2}=x^{2}$, produce solutions $(m,x,y,u)$ to the equations

$m=Q_{2,3}^{+}(x,y)=Q_{2,1}^{+}(u,y).$

It follows that the exponent $\vartheta_{4}=0.448$ in Theorem 1.1 cannot be
replaced with an exponent $<0.25$.

####  1.3.3 Products of indefinite quadratic forms

With the above notations, including the definition of $\lambda$ in (1.14), we
assume $\mu_{1}\geq 2$ and we consider, for $d\geq 2$ and $1\leq\nu\leq d+1$,
the binary form of degree $2d$

$Q^{-}_{d,\nu}(X,Y):=\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
d+1}{n\neq\nu}}\left(X^{2}-{\mu}_{n}Y^{2}\right).$

The associated family is

$\mathcal{Q}^{-}:=\left\\{Q^{-}_{d,\nu}:d\geq 2,\,1\leq\nu\leq d+1\right\\}$

with $\mathcal{Q}^{-}_{d}=\emptyset$ for $d$ odd and
$\mathcal{Q}^{-}_{2d}=\left\\{Q^{-}_{d,\nu}:1\leq\nu\leq d+1\right\\}$ for
$d\geq 2$.

From (1.15) one deduces

$\sup_{n\geq 1}\frac{q_{n+1}}{n}=2,$

hence $\lambda\geq 2$. In the special case $\mu_{n}=q_{n+1}$ ($n\geq 1$), an
admissible value for $\lambda$ is $\lambda=2$.

######  Theorem 1.17.

For $A>2\mathrm{e}\lambda$, the family $\mathcal{Q}^{-}$ is
$(A,1,2,2,2\mathrm{e}\lambda)$–regular and satisfies the condition C3 of
Corollary 1.12. Furthermore, for $d\geq 2$, we have

$\mathcal{R}_{\geq
2d}(\mathcal{Q}^{-},B,0)=\frac{1}{4}\left(\sum_{F\in\mathcal{Q}_{2d}^{-}}A_{F}\right)B^{1/d}+O_{\lambda,A,d,\varepsilon}\left(B^{\max\\{\vartheta_{2d}+\varepsilon,1/(d+1)\\}}\right),$

uniformly for $B\to\infty$. Further, we have

(1.19)
$\frac{\pi}{\sqrt{\lambda}}\cdot\sqrt{d}\leq\sum_{F\in\mathcal{Q}_{2d}^{-}}A_{F}\leq
22\lambda\sqrt{d}$

where the lower bound is valid for all $d\geq 2$ and the upper bound for $d$
sufficiently large.

####  1.3.4 Products of linear factors

We reserve the letter $p$ to prime numbers and we consider for $5\leq d\leq
p$, the binary form $L_{d,p}\in{\mathrm{Bin}}(d,{\mathbb{Z}})$ defined by

$L_{d,p}(X,Y):=\left(X-pY\right)\cdot\prod_{0\leq n\leq
d-2}\left(X-nY\right).$

The associated family is

$\mathcal{L}:=\left\\{L_{d,p}:d\geq 5,d\leq p<2d\right\\}.$

We have

######  Theorem 1.18.

The family $\mathcal{L}$ is $(10,1,1,5,9)$–regular.

Furthermore, for $d\geq 5$, $\mathcal{L}_{d}$ respectively satisfies the
condition C1 of Corollary 1.12 for $d$ odd or the condition C2 for $d$ even.

Finally, for every $d\geq 5$ and for every $\varepsilon>0$, one has the
equality

(1.20) $\mathcal{R}_{\geq d}(\mathcal{L},B,0)=\frac{1}{(2,d)}\left(\sum_{d\leq
p<2d}A_{L_{d,p}}\right)B^{2/d}+O_{d,\varepsilon}\left(B^{\max\\{\vartheta_{d},2/(d+1)\\}}\right),$

uniformly for $B\to\infty$, and the inequalities

$\frac{\mathrm{e}^{2}-o(1)}{\log d}\leq\sum_{d\leq
p<2d}A_{L_{d,p}}\leq\frac{5\,\mathrm{e}^{2}+2\,\mathrm{e}+o(1)}{\log d}$

uniformly for $d\to\infty$.

The numerical values are $\mathrm{e}^{2}=7.389\dots$ and
$5\mathrm{e}^{2}+2\mathrm{e}=42.381\dots$

See (2.6) for a simplification of the exponent in the error term of (1.20).

######  Remark 1.19.

We now give some hints on the construction of the family $\mathcal{L}$. More
generally consider the binary form of degree $d$

$L_{{\bf n},d}(X,Y):=\prod_{1\leq i\leq d}\left(X-n_{i}Y\right),$

where ${\bf n}:=\\{n_{1}<n_{2}<\cdots<n_{d}\\}$ is a set of $d$ integers. Fix
$d\geq 5$, then for almost all $\bf n$ (in the meaning of Zariski topology),
the group of automorphisms of $L_{{\bf n},d}$ is trivial, which means equal to
$\\{{\mathrm{Id}}\\}$ or to $\\{\pm{\mathrm{Id}}\\}$, according to the parity
of $d$. Similarly, for fixed $d\geq 5$, for almost all $({\bf m},{\bf n})$ the
binary forms $L_{{\bf m},d}$ and $L_{{\bf n},d}$ are not isomorphic. For
statements of that type, see [FK], for instance. The strategy of choosing
$n_{1}=0$ and $n_{d}=p$, where $p$ is a large prime ensures that the group of
automorphisms is trivial and that the binary forms that we meet are not
isomorphic. These statements are proved by appealing to the classical
properties of the cross ratio (see §6.1 and §6.2).

Finally, we choose for $n_{1}$,…,$n_{d-1}$ the $d-1$ first integers. This
enables us to estimate the area $A_{L_{d,p}}$ (see §6.6) via Stirling’s
formula:

(1.21) $N^{N}\mathrm{e}^{-N}\sqrt{2\pi N}<N!<N^{N}\mathrm{e}^{-N}\sqrt{2\pi
N}\mathrm{e}^{1/(12N)},$

which is valid for all $N\geq 1$. In particular, as $N\rightarrow\infty$, we
have

$\log N-1<\frac{1}{N}\log(N!)<\log N-1+o(1).$

It would be interesting to further investigate the explicit construction of
other regular families of forms, which are products of ${\mathbb{Z}}$–linear
forms.

######  Remark 1.20.

A natural way to generalize the construction of the families $\mathcal{B}$,
$\mathcal{Q}^{-}$ and $\mathcal{Q}^{+}$ is to consider sets of forms which are
products of binomials of the shape

${\mathrm{B}}_{a,n}(X,Y)=X^{a}+nY^{a}.$

The key point is to choose the integers $n$ and the exponents $a\geq 2$ in
such a way that we are able to control the homographies in
${\mathrm{PGL}}(2,{\mathbb{Q}})$ which exchange the set of zeroes of the
products of ${\mathrm{B}}_{a,n}$.

##  2 Proof of Theorem 1.1

###  2.1 Beginning of the proof

The starting point is [FW, Théorème 3.1]. To state this result we use the
following notations:

$\bullet$ If $F_{1}$ and $F_{2}$ belong to ${\mathrm{Bin}}(d,{\mathbb{Z}})$
and if $B\geq 1$, we put

$\displaystyle{\mathcal{M}}(F_{1},$ $\displaystyle F_{2};B)=$
$\displaystyle\sharp\left\\{(x_{1},x_{2},x_{3},x_{4})\in{\mathbb{Z}}^{4}:\max|x_{i}|\leq
B,F_{1}(x_{1},x_{2})=F_{2}(x_{3},x_{4})\right\\},$

and

$\displaystyle{\mathcal{M}}^{*}(F_{1},$ $\displaystyle F_{2};B)=$
$\displaystyle\sharp\left\\{(x_{1},x_{2},x_{3},x_{4})\in{\mathbb{Z}}^{4}:\max|x_{i}|\leq
B,F_{1}(x_{1},x_{2})=F_{2}(x_{3},x_{4})\neq 0\right\\},$

$\bullet$ for $d\geq 3$, we introduce

$\eta_{d}=\begin{cases}\frac{2}{9}+\frac{73}{108\sqrt{3}}&\text{ for }d=3,\\\
\frac{1}{2d}+\frac{9}{4d\sqrt{d}}&\text{ for }4\leq d\leq 20,\\\
\frac{1}{d}&\text{ for }d\geq 21,\end{cases}$ (2.1)
$\vartheta_{d}=\frac{d\eta_{d}}{d\eta_{d}+d-2}$

and

(2.2) $\eta^{\prime}_{d,F_{1},F_{2}}=\begin{cases}\eta_{d}&\text{if the binary
form $F_{1}(X,Y)F_{2}(X,Y)$}\\\ &\text{has no zero in
${\mathbb{P}}^{1}({\mathbb{R}})$,}\\\
\vartheta_{d}&\text{otherwise.}\end{cases}$

Here are the first approximate values for $\eta_{d}$, $\vartheta_{d}$ and
$\kappa_{d}$ (recall (1.11)):

$\begin{matrix}d&\eta_{d}&\vartheta_{d}&\kappa_{d}\\\ 3&0.612&0.647&0.631\\\
4&0.406&0.448&0.428\hfill\\\ 5&0.301&0.334&0.309\hfill\\\
6&0.236&0.261&0.234\\\ 7&0.192&0.211&0.184\\\ 8&0.161&0.177&0.150\\\
\end{matrix}$

For $d\geq 3$ and for $F_{1}$ and $F_{2}$ belonging to
${\mathrm{Bin}}(d,{\mathbb{Z}})$, one has the inequalities

(2.3) $1/d\leq\eta_{d}\leq\eta^{\prime}_{d,F_{1},F_{2}}\leq\vartheta_{d}<2/d$

and, in particular, for $d\geq 21$, we have : $\eta_{d}=1/d$ and
$\vartheta_{d}=1/(d-1)$.

Furthermore, by comparison with $\kappa_{d}$ defined in (1.11), we check that

(2.4) $\begin{cases}\kappa_{d}<\vartheta_{d}&\text{ if }3\leq d\leq 20,\\\
\kappa_{d}=\vartheta_{d}&\text{ if }d\geq 21.\end{cases}$

Finally, by a direct computation we have the inequalities

(2.5) $\begin{cases}\vartheta_{2d}>1/(d+1)&\text{ if }d=2,\,3,\\\
\vartheta_{2d}<1/(d+1)&\text{ if }d\geq 4,\end{cases}$

and

(2.6) $\begin{cases}\vartheta_{d}>2/(d+1)&\text{ if }d=4,\,5,\\\
\vartheta_{d}<2/(d+1)&\text{ if }d\geq 6.\end{cases}$

We now recall (see [FW, Théorème 3.1])

######  Proposition 2.1.

Let $d\geq 3$ and let $F_{1}$ and $F_{2}$ be two non–isomorphic forms of
${\mathrm{Bin}}(d,{\mathbb{Z}})$, such that at least one of them is not
divisible by a linear form with rational coefficients. Then for all
$\varepsilon>0$ and all $B\geq 1$ one has

${\mathcal{M}}(F_{1},F_{2};B)=O_{F_{1},F_{2},\varepsilon}\left(B^{d\eta_{d}+\varepsilon}\right).$

As it is shown by [FW, Remarque 3.2], the above bound may not hold if one of
the binary forms is divisible by a linear form over ${\mathbb{Q}}$. One
eliminates this hypothesis by studying the counting function
${\mathcal{M}^{*}}$ rather than ${\mathcal{M}}$. In other words one has the
following variant for Proposition 2.1

######  Proposition 2.2.

Let $d\geq 3$ and let $F_{1}$ and $F_{2}$ be two non–isomorphic forms of
${\mathrm{Bin}}(d,{\mathbb{Z}})$. Then for every $\varepsilon>0$ and for all
$B\geq 1$ one has the bound

${\mathcal{M}}^{*}(F_{1},F_{2};B)=O_{F_{1},F_{2},\varepsilon}\left(B^{d\eta_{d}+\varepsilon}\right).$

###### Proof.

We refer to the original proof of [FW, Théorème 3.1]. The hypothesis that at
least one of the two forms $F_{1}$ and $F_{2}$ has no ${\mathbb{Q}}$–linear
factor is only used in [FW, eq. (22)] (which is equation (3.8) in the Arxiv
version). This case has no longer to be considered when one studies
${\mathcal{M}}^{*}$ instead of ${\mathcal{M}}$. ∎

###  2.2 Lemmas in diophantine approximation

Firstly we prove the following

######  Lemma 2.3.

Let $f\in{\mathbb{Z}}[t]$ be a polynomial of degree $d\geq 1$ and with
discriminant different from zero. Let $\xi_{1},\dots,\xi_{d}$ be the complex
roots of $f$. Then there are real constants $c_{1}>0$ and $c_{2}$ such that

  1. (i)

For every $t\in{\mathbb{C}}$, one has the inequality $\min_{1\leq j\leq
d}|t-\xi_{j}|\leq c_{2}|f(t)|,$

  2. (ii)

For every $t\in{\mathbb{R}}$, the condition $|f(t)|<c_{1}$ implies the
existence of a real root $\xi_{i}$ such that $|t-\xi_{i}|\leq c_{2}|f(t)|$.

###### Proof.

This statement is trivial when $d=1$. We now suppose $d\geq 2$.

We suppose that $a_{0}$ (the leading coefficient of $f$) is $\geq 1$ and we
factor $f$ into

$f(t)=a_{0}\prod_{j=1}^{d}(t-\xi_{j}).$

Let $\delta:=\min_{1\leq i<j\leq d}|\xi_{i}-\xi_{j}|$. Since the discriminant
of $f$ is different from zero, we have $\delta>0$. Let $i$ be an index such
that $|t-\xi_{i}|=\min_{1\leq j\leq d}|t-\xi_{j}|$. The triangular inequality
gives, for $j\neq i$, the lower bound

$|t-\xi_{j}|\geq\frac{|t-\xi_{j}|+|t-\xi_{i}|}{2}\geq\frac{1}{2}\,|\xi_{j}-\xi_{i}|\geq\frac{\delta}{2}\cdotp$

We write the sequence of inequalities

$|f(t)|\geq\prod_{1\leq j\leq
d}|t-\xi_{j}|\geq|t-\xi_{i}|\left(\frac{\delta}{2}\right)^{d-1},$

which leads to the point (i) with $c_{2}=(2/\delta)^{d-1}.$

For the item (ii), we now suppose that $t$ is real. We decompose the proof
into three cases.

$\bullet$ If all the $\xi_{j}$ are real, there is nothing to prove as a
consequence of (i). We choose $c_{1}=1$ for instance. $\bullet$ If no
$\xi_{j}$ is real, we set

$c_{1}:=\inf_{x\in{\mathbb{R}}}|f(x)|$

which is $>0$. $\bullet$ If $f$ has at least one real root and at least one
non real root, we put

$c_{1}=\frac{1}{c_{2}}\min\bigl{\\{}\,|{\mathrm{Im}}(\xi_{i})|:1\leq i\leq
d,\,\xi_{i}\not\in{\mathbb{R}}\,\bigr{\\}}.$

Applying the item (i) of Lemma 2.3, we notice that for $t\in{\mathbb{R}}$ the
inequality $|f(t)|<c_{1}$ implies the existence of a root $\xi_{j}$ such that

$|t-\xi_{j}|<c_{1}c_{2}=\min\bigl{\\{}\,|{\mathrm{Im}}(\xi_{i}-t)|:1\leq i\leq
d,\,\xi_{i}\not\in{\mathbb{R}}\,\bigr{\\}}.$

If $\xi_{j}$ were not real, we would deduce the inequality
$|t-\xi_{j}|<|{\mathrm{Im}}(t-\xi_{j})|$, which is impossible. Hence $\xi_{j}$
is real. ∎

The following lemma provides an upper bound for the tail of the series
defining the Riemann $\zeta$–function.

######  Lemma 2.4.

For all real $\delta>1$ and all positive integer $B$, one has the inequality

$\sum_{n\geq B}\frac{1}{n^{\delta}}\leq\zeta(\delta)B^{1-\delta}.$

###### Proof.

By dividing the interval of summation in intervals with length $B$ and by
using the inequality $Bq+r\geq Bq$, we write

$\sum_{n\geq B}\frac{1}{n^{\delta}}=\sum_{q\geq
1}\sum_{r=0}^{B-1}\frac{1}{(Bq+r)^{\delta}}\leq B^{1-\delta}\sum_{q\geq
1}\frac{1}{q^{\delta}}=\zeta(\delta)\,B^{1-\delta}.$

∎

The next lemma was inspired by [Ho, p. 34–36].

######  Lemma 2.5.

Let $\xi$, $\kappa$, $s$, $Q_{1}$ and $Q_{2}$ be real numbers such that $s>2$,
$\kappa>0$, $Q_{2}>Q_{1}\geq 1$. Then the number of rational numbers
$\frac{p}{q}$ such that

$\Bigl{|}\xi-\frac{p}{q}\Bigr{|}\leq\frac{\kappa}{q^{s}}\text{ and }Q_{1}\leq
q\leq Q_{2},$

is bounded by

$\frac{2^{s+1}\kappa}{(2^{s-2}-1)Q_{1}^{s-2}}+\left\lceil{\frac{\log\frac{Q_{2}}{Q_{1}}}{\log
2}}\right\rceil.$

###### Proof.

Firstly we consider the case when $Q_{2}\leq 2Q_{1}$ and we prove the result
with the coefficient $\frac{2^{s+1}}{2^{s-2}-1}$ replaced by $8$. Two distinct
rational numbers $\frac{p}{q}$, $\frac{p^{\prime}}{q^{\prime}}$ such that
$Q_{1}\leq q,q^{\prime}\leq Q_{2}$ satisfy the inequalities

$\left|\frac{p}{q}-\frac{p^{\prime}}{q^{\prime}}\right|\geq\frac{1}{qq^{\prime}}\geq\frac{1}{Q_{2}^{2}}\geq\frac{1}{4Q_{1}^{2}}\cdotp$

If they also satisfy

$\Bigl{|}\xi-\frac{p}{q}\Bigr{|}\leq\frac{\kappa}{q^{s}}\text{ and
}\Bigl{|}\xi-\frac{p^{\prime}}{q^{\prime}}\Bigr{|}\leq\frac{\kappa}{{q^{\prime}}^{s}},$

then they belong to the interval

$\left[\xi-\frac{\kappa}{Q_{1}^{s}},\xi+\frac{\kappa}{Q_{1}^{s}}\right],$

the length of which is $2\kappa/Q_{1}^{s}$. So the number of such
$\frac{p}{q}$ is less than

$4Q_{1}^{2}\frac{2\kappa}{Q_{1}^{s}}+1=\frac{8\kappa}{Q_{1}^{s-2}}+1.$

In the case where $Q_{2}>2Q_{1}$, we cover the interval $[Q_{1},Q_{2}]$ by
$\ell$ intervals $[2^{h}Q_{1},2^{h+1}Q_{1}]$, $0\leq h\leq\ell-1$ with
$2^{\ell-1}Q_{1}<Q_{2}\leq 2^{\ell}Q_{1}$; thus $\ell$ satisfies the
inequalities

$\frac{\log\frac{Q_{2}}{Q_{1}}}{\log
2}\leq\ell<1+\frac{\log\frac{Q_{2}}{Q_{1}}}{\log 2}\cdotp$

As we have seen, in the interval $[2^{h}Q_{1},2^{h+1}Q_{1}]$, the number of
rational numbers $\frac{p}{q}$ satisfying our assumption is bounded by

$\frac{8\kappa}{2^{h(s-2)}Q_{1}^{s-2}}+1.$

The total number of fractions $\frac{p}{q}$ satisfying our assumption is less
than

$\sum_{h=0}^{\ell-1}\left(\frac{8\kappa}{2^{h(s-2)}Q_{1}^{s-2}}+1\right)=\frac{8\kappa}{Q_{1}^{s-2}}\sum_{h=0}^{\ell-1}\frac{1}{2^{h(s-2)}}+\ell<\frac{8\kappa}{Q_{1}^{s-2}}\cdot\frac{2^{s-2}}{2^{s-2}-1}+\left\lceil{\frac{\log\frac{Q_{2}}{Q_{1}}}{\log
2}}\right\rceil.$

∎

###  2.3 On the set of the values taken by a binary form when one of the
variables is large

As a consequence of the three lemmas proved in §2.2 we will deduce

######  Proposition 2.6.

Let $d\geq 3$ and let $F\in{\mathrm{Bin}}(d,{\mathbb{Z}})$. Then there are two
constants $c_{3}$ and $c_{4}$, effectively computable and depending on $F$
only, such that, for all $\Delta>c_{3}$ and all $A>0$ one has the following
inequality

$\sharp\left\\{(x,y)\in{\mathbb{Z}}^{2}:0<|F(x,y)|\leq A,|y|\geq
A^{1/d}\Delta\right\\}\leq c_{4}\left(A^{2/d}\Delta^{2-d}+A^{1/(d-1)}\right).$

The proof of this proposition will use the following effective refinement of
Liouville’s inequality, due to N. I. Fel’dman [F]:

######  Lemma 2.7.

Let $\xi$ be an algebraic number of degree $d\geq 3$. There are two
effectively computable positive constants $c_{5}=c_{5}(\xi)$ and
$c_{6}=c_{6}(\xi)$ such that, for every fraction $p/q\in{\mathbb{Q}}$ with
$q\geq 1$, one has the inequality

$\left|\xi-\frac{p}{q}\right|\geq\frac{c_{5}}{q^{d-c_{6}}}\cdotp$

A completely explicit version of this inequality can be found in [GP, (13) p.
248].

We deduce from this lemma the following one.

######  Lemma 2.8.

Let $P(X)\in{\mathbb{Z}}[X]$ be a polynomial, of degree $d\geq 3$. There are
two effectively computable positive constants
$c^{\prime}_{5}=c^{\prime}_{5}(P)$ and $c^{\prime}_{6}=c^{\prime}_{6}(P)$ such
that, for every root $\xi$ of $P$, for every rational number $p/q$ such that
$q\geq 1$ and $p/q\neq\xi$, the following inequality holds

(2.7)
$\left|\xi-\frac{p}{q}\right|\geq\frac{c^{\prime}_{5}}{q^{d-c^{\prime}_{6}}}\cdotp$

We stress that there is no assumption on whether the polynomial $P$ is
irreducible or not, nor on whether the root $\xi$ is real or not.

###### Proof.

Let $\delta$ be the degree of $\xi$. We split the argument according to the
value of $\delta$ and to the nature of $\xi$.

  * •

If $\xi$ is not real, the inequality (2.7) is trivial since we have
$|\xi-p/q|\geq|{\mathrm{Im}}\,\xi|$, for every rational number $p/q$. We now
suppose that $\xi$ is a real number.

  * •

If $\delta=1$. We put $\xi=a/b$ with $a$ and $b$ integers and $b\geq 1$. We
have $|a/b-p/q|=|aq-bp|/bq\geq 1/bq$, since $\xi$ is different from $p/q$. We
obtain (2.7), with the choices $c^{\prime}_{5}=1/b$ and $c^{\prime}_{6}=1$
since $d\geq 3$.

  * •

If $\delta=2$. The real number $\xi$ is quadratic. Liouville’s inequality for
quadratic real numbers is optimal: there exists $\alpha=\alpha(\xi)>0$ such
that

$\left|\xi-\frac{p}{q}\right|\geq\frac{\alpha}{q^{2}}\cdotp$

By the hypothesis $d\geq 3,$ we deduce (2.7) with the choice
$c^{\prime}_{5}=\alpha$ and $c^{\prime}_{6}=1/2$.

  * •

If $\delta\geq 3$. We apply Lemma 2.7 in the form

$\left|\xi-\frac{p}{q}\right|\geq\frac{c_{5}}{q^{\delta-c_{6}}}\cdotp$

Since $\delta\leq d$, we obtain (2.7) with the choice $c^{\prime}_{5}=c_{5}$
and $c^{\prime}_{6}=c_{6}$.

We choose for $c^{\prime}_{5}=c^{\prime}_{5}(P)$ and for
$c^{\prime}_{6}=c^{\prime}_{6}(P)$ the least values $c^{\prime}_{5}$ and
$c^{\prime}_{6}$ corresponding to the various $\xi$ that we met above to
complete the proof of Lemma 2.8. ∎

Proof of Proposition 2.6. Let $f(t)=F(t,1)$, thus we have
$F(x,y)=y^{d}f(x/y)$. Let $d^{\prime}$ be the degree of $f$. Since the
discriminant of $F$ is different from zero, we have

$d^{\prime}=d\text{ or }d-1.$

If $f$ has no real root, then, for sufficiently large $\Delta$ (more
precisely, for $\Delta>\left(\inf_{t\in{\mathbb{R}}}|f(t)|\right)^{-1/d}$),
the set

$\left\\{(x,y)\in{\mathbb{Z}}^{2}:0<|F(x,y)|\leq A,|y|\geq
A^{1/d}\Delta\right\\}$

is empty.

Let $r\geq 1$ be the number of real roots of $f$, that we denote by
$\xi_{1}$,…, $\xi_{r}$. By hypothesis these roots are simple. Let
$(x,y)\in{\mathbb{Z}}^{2}$ with $y\neq 0$. The condition $0<|F(x,y)|\leq A$
implies

$0<\left|f\left(\frac{x}{y}\right)\right|\leq\frac{A}{|y|^{d}}\cdotp$

We suppose $|y|\geq A^{1/d}\Delta$ and $\Delta>c_{1}^{-1/d}$, and we apply
Lemma 2.3 (ii). We deduce the existence of some $i\in\\{1,\dots,r\\}$ such
that

(2.8) $0<\left|\frac{x}{y}-\xi_{i}\right|\leq\frac{c_{2}A}{|y|^{d}},$

which is equivalent to

(2.9) $0<|x-y\xi_{i}|\leq\frac{c_{2}A}{|y|^{d-1}}\cdotp$

When the integer $y$ is fixed, the number of integers $x$ satisfying the
inequality (2.9) is equal to

$\frac{2c_{2}A}{|y|^{d-1}}+O(1).$

We fix $Y_{0}=A^{1/(d-1)}$ and we sum over $i=1,\dots,r$. We apply Lemma 2.4
with $B=A^{1/d}\Delta$ and $\delta=d-1$, to deduce that the number of $(x,y)$
with $0<|F(x,y)|\leq A$ and $A^{1/d}\Delta\leq|y|\leq Y_{0}$ is bounded by

(2.10) $O\bigl{(}A^{2/d}\Delta^{2-d}\bigr{)}+O(Y_{0}).$

To complete the proof, we use Lemma 2.8 which implies the lower bound

(2.11)
$\left|\xi_{i}-\frac{p}{q}\right|\geq\frac{c^{\prime}_{5}}{q^{d^{\prime}-c^{\prime}_{6}}}\geq\frac{c^{\prime}_{5}}{q^{d-c^{\prime}_{6}}}\cdotp$

Combining (2.8) with (2.11), we deduce the upper bound $|y|\leq Y_{1}$ with
$Y_{1}=(\frac{c_{2}}{c^{\prime}_{5}}A)^{1/{c^{\prime}_{6}}}$. It remains to
compute the number of solutions of (2.8) satisfying $Y_{0}<|y|\leq Y_{1}$. We
use Lemma 2.5, with $s=d$, $\kappa=c_{2}A$, $Q_{1}=Y_{0}$, $Q_{2}=Y_{1}$: this
number is bounded by

$O\left(A/Y_{0}^{d-2}\right)+O\left(\log
Y_{1}\right)=O\left(A^{1/(d-1)}\right).$

By adding (2.10) we obtain the upper bound announced in Proposition 2.6.
$\square$

###  2.4 End of the proof of Theorem 1.1

We split the end of the proof into two different cases: $\bullet$ Assume the
binary form $F_{1}(X,Y)F_{2}(X,Y)$ has no zero in
${\mathbb{P}}^{1}({\mathbb{R}})$. This hypothesis holds true if and only if
the polynomial $F_{1}(t,1)F_{1}(1,t)F_{2}(t,1)F_{2}(1,t)$ has no real root. By
homogeneity, there is a constant $c_{7}>0$ such that for all
$(x_{1},x_{2},x_{3},x_{4})\in{\mathbb{R}}^{4}$, one has the inequalities

$|F_{1}(x_{1},x_{2})|\geq c_{7}\max\\{|x_{1}|^{d},|x_{2}|^{d}\\}\text{ and
}|F_{2}(x_{3},x_{4})|\geq c_{7}\max\\{|x_{3}|^{d},|x_{4}|^{d}\\}.$

This leads to the existence of a constant $c_{8}$ such that the inequalities

$|F_{1}(x_{1},x_{2})|\leq N\text{ and }|F_{2}(x_{3},x_{4})|\leq N$

imply $\max(|x_{1}|,|x_{2}|,|x_{3}|,|x_{4}|)\leq B$ with $B:=(c_{8}N)^{1/d}$.
We apply Proposition 2.2 under the form

${\mathcal{N}}(F_{1},F_{2};N)\leq
1+{\mathcal{M}}^{*}(F_{1},F_{2};B)=O_{F_{1},F_{2}}\left(B^{d\eta_{d}+\varepsilon}\right)=O_{F_{1},F_{2}}\left(N^{\eta_{d}+\varepsilon}\right).$

By the inequality (2.3), the proof of Theorem 1.1 is complete in that case,
including the refinement quoted in Remark 1.5. $\bullet$ Assume the binary
form $F_{1}(X,Y)F_{2}(X,Y)$ has at least one zero in
${\mathbb{P}}^{1}({\mathbb{R}})$. This is equivalent to the assumption that
the polynomial

$F_{1}(t,1)F_{1}(1,t)F_{2}(t,1)F_{2}(1,t)$

has at least one real root. The constant $\eta^{\prime}_{d,F_{1},F_{2}}$ is
now defined by the second formula of (2.2), that is
$\eta^{\prime}_{d,F_{1},F_{2}}=\vartheta_{d}$. Let

$\tau:=\frac{\frac{2}{d}-\eta_{d}}{d\eta_{d}+d-2},$

so we have the equalities

$\frac{2}{d}-(d-2)\tau=\eta_{d}(1+d\tau)=\eta^{\prime}_{d,F_{1},F_{2}}.$

Let $\Delta:=N^{\tau}$. To bound from above ${\mathcal{N}}(F_{1},F_{2};N)$,
which is the number of $m\in{\mathbb{Z}}$, $|m|\leq N,$ such that there is at
least one $(x_{1},x_{2},x_{3},x_{4})\in{\mathbb{Z}}^{4}$ satisfying the
equality

(2.12) $F_{1}(x_{1},x_{2})=F_{2}(x_{3},x_{4})=m,$

we first consider those $m$ such that at least one of
$(x_{1},x_{2},x_{3},x_{4})$ associated to $m$ by (2.12) satisfies the
inequality

$\max\\{|x_{1}|,|x_{2}|,|x_{3}|,|x_{4}|\\}<N^{1/d}\Delta.$

Proposition 2.2 with $B=N^{\frac{1}{d}+\tau}$ shows that the number of these
$m$ is bounded by

(2.13)
$O_{F_{1},F_{2},\varepsilon}\left(B^{d\eta_{d}+\varepsilon}\right)=O_{F_{1},F_{2},\varepsilon}\left(N^{\eta_{d}(1+d\tau)+\varepsilon}\right)=O_{F_{1},F_{2},\varepsilon}\left(N^{\eta^{\prime}_{d,F_{1},F_{2}}+\varepsilon}\right).$

Next, we estimate the number of those $m$ such that all the $4$–tuples
$(x_{1},x_{2},x_{3},x_{4})$ associated to $m$ by (2.12) satisfy the inequality

$\max\\{|x_{1}|,|x_{2}|,|x_{3}|,|x_{4}|\\}\geq N^{1/d}\Delta.$

For simplicity, we study the case where $|x_{1}|\geq N^{1/d}\Delta$, since the
other cases are similar. We only consider the values taken by the binary form
$F_{1}$ and we apply Proposition 2.6. With the choices $F=F_{1}$ and $A=N$,
using $\vartheta_{d}\geq 1/(d-1)$, we deduce that the number of corresponding
$m$ is bounded by

$O_{F_{1},F_{2}}\left(N^{2/d}\Delta^{2-d}+N^{1/(d-1)}\right)=O_{F_{1},F_{2}}\left(N^{\eta^{\prime}_{d,F_{1},F_{2}}}\right).$

By (2.13), this completes the proof of Theorem 1.1.

##  3 Proof of Theorem 1.11

By similarity with (1.6), we put

${\mathcal{R}}_{=d}(\mathcal{F},B,A):=\sharp\\{m:0\leq|m|\leq B,\text{ there
is }F\in\mathcal{F}_{d},(x,y)\in{\mathbb{Z}}^{2},\\\ \text{ such that
}\max(|x|,|y|)\geq A\text{ and }m=F(x,y)\\}.$

The lower bound for ${\mathcal{R}}_{\geq d}(\mathcal{F},B,A)$ is obtained as
follows

$\displaystyle{\mathcal{R}}_{\geq d}(\mathcal{F},B,A)$
$\displaystyle\geq{\mathcal{R}}_{=d}(\mathcal{F},B,A)$
$\displaystyle\geq\sum_{F\in\mathcal{F}_{d}}{\mathcal{N}}(F,F;B)-\underset{\genfrac{}{}{0.0pt}{}{F,\
F^{\prime}\in\mathcal{F}_{d}}{F\neq F^{\prime}}}{\sum\
\sum}{\mathcal{N}}(F,F^{\prime};B)-(2A+1)^{2}d^{A_{1}},$

where the counting function ${\mathcal{N}}$ is defined by (1.2). Condition
(iii) in Definition 1.10 of a regular family implies
$\sharp\mathcal{F}_{d}=O_{d}(1)$; thanks to condition (iv) and to the
inequality $\kappa_{d}\leq\vartheta_{d}$ (see (2.4)), Theorems 1.1 and A give
the inequality

(3.1) ${\mathcal{R}}_{\geq
d}(\mathcal{F},B,A)\geq\left(\sum_{F\in\mathcal{F}_{d}}A_{F}W_{F}\right)\cdot
B^{2/d}-O_{\mathcal{F},A,\varepsilon}\left(B^{\vartheta_{d}+\varepsilon}\right).$

For the upper bound, we recall that the parameters $d_{0},\,\kappa$ and
$A_{1}$ appear in Definition 1.10. We start from the inequality

(3.2) ${\mathcal{R}}_{\geq
d}(\mathcal{F},B,A)\leq\sum_{F\in\mathcal{F}_{d}}{\mathcal{N}}(F,F;B)+\sum_{n=d^{\dagger}}^{d^{\dagger}+d_{0}}\sum_{F\in\mathcal{F}_{n}}{\mathcal{N}}(F,F;B)\\\
+\sharp\,\left(\underset{n>d^{\dagger}+d_{0}}{\bigcup}\,\underset{F\in\mathcal{F}_{n}}{\bigcup}\left(F(\mathcal{Z}_{A})\cap[-B,B]\right)\right),$

with

$\mathcal{Z}_{A}={\mathbb{Z}}^{2}\setminus\bigl{(}[-A,A]\times[-A,A]\bigr{)}.$

Applying one more time Theorem A, we have the equality

(3.3)
$\sum_{F\in\mathcal{F}_{d}}{\mathcal{N}}(F,F;B)=\left(\sum_{F\in\mathcal{F}_{d}}A_{F}W_{F}\right)\cdot
B^{2/d}+O_{\mathcal{F},d,\varepsilon}\left(B^{\kappa_{d}+\varepsilon}\right),$

and the upper bound

(3.4) ${\mathcal{N}}(F,F;B)=O_{F}\left(B^{2/d^{\dagger}}\right)\text{ if }\deg
F\geq d^{\dagger}.$

Hence the second term on the right–hand side of (3.2) is bounded as follows

$\sum_{n=d^{\dagger}}^{d^{\dagger}+d_{0}}\sum_{F\in\mathcal{F}_{n}}{\mathcal{N}}(F,F;B)=O_{\mathcal{F},d}\left(B^{2/d^{\dagger}}\right).$

To deal with the third term on the right–hand side of (3.2), we interchange
the summations to write

(3.5)
$\sharp\,\left(\underset{n>d^{\dagger}+d_{0}}{\bigcup}\,\underset{F\in\mathcal{F}_{n}}{\bigcup}\left(F(\mathcal{Z}_{A})\cap[-B,B]\right)\right)\\\
\leq\sharp\,\left\\{(n,F,x,y):n>d^{\dagger}+d_{0},F\in\mathcal{F}_{n},(x,y)\in\mathcal{Z}_{A},|F(x,y)|\leq
B\right\\}.$

The condition (v) in Definition 1.10 of the
$(A,A_{1},d_{0},d_{1},\kappa)$–regularity of $\mathcal{F}$ produces a bound
for $n$, by the sequence of inequalities

(3.6)
$\kappa<A\leq\max\\{|x|,|y|\\}\leq\kappa|F(x,y)|^{\frac{1}{n-d_{0}}}\leq\kappa
B^{\frac{1}{n-d_{0}}}\leq\kappa B^{\frac{1}{d^{\dagger}+1}},$

which implies the inequality

$n\leq d_{0}+\frac{\log B}{\log(A/\kappa)}\cdotp$

Furthermore the inequalities (3.6) implies

$\max\\{|x|,|y|\\}\leq\kappa B^{1/(d^{\dagger}+1)}.$

Combining the above inequalities, we deduce that the cardinality of the
quadruples $(n,F,x,y)$ in the right–hand side of (3.5) is bounded from above
by

(3.7) $\left(d_{0}+\frac{\log B}{\log(A/\kappa)}\right)^{A_{1}}\left(1+2\kappa
B^{1/({d^{\dagger}+1})}\right)^{2}=o_{\mathcal{F}}\left(B^{2/{d^{\dagger}}}\right).$

Gathering (3.2), (3.3), (3.4) and (3.7), we finally obtain the upper bound

(3.8) ${\mathcal{R}}_{\geq
d}(\mathcal{F},B,A)\leq\left(\sum_{F\in\mathcal{F}_{d}}A_{F}W_{F}\right)\cdot
B^{2/d}+O_{\mathcal{F},A,d,\varepsilon}\left(B^{\kappa_{d}+\varepsilon}\right)+O_{\mathcal{F},d}\left(B^{2/d^{\dagger}}\right).$

Comparing (3.1) and (3.8) and recalling the inequality (2.4), we complete the
proof of Theorem 1.11.

##  4 Proof of Theorem 1.15

###  4.1 The family $\mathcal{Q}^{+}$ is $(2,1,0,4,1)$–regular

Our first purpose is to prove the following

######  Proposition 4.1.

The family $\mathcal{Q}^{+}$ is $(2,1,0,4,1)$–regular.

###### Proof.

Several times, we will use the following property satisfied by two positive
distinct squarefree numbers

(4.1) $n\neq
n^{\prime}\Rightarrow{\mathbb{Q}}\left(i\sqrt{{\mu}_{n}}\right)\neq{\mathbb{Q}}\left(i\sqrt{{\mu}_{n^{\prime}}}\right).$

We now check each of the items of Definition 1.10 of a regular family.

$\bullet$ The items (i) and (ii) are trivial.

$\bullet$ The family $\mathcal{Q}^{+}$ contains no element with odd degree
$d$. By contrast if this degree $d\geq 4$ is even, the family contains $d/2+1$
binary forms of degree $d$.

Thus the item (iii) is verified with $A_{1}=1$.

$\bullet$ For the item (iv) we proceed as follows. Suppose that there are two
isomorphic forms $F$ and $F^{\prime}$ in $\mathcal{Q}^{+}$. Necessarily they
have the same degree $2d$. So there exist $1\leq\nu<\nu^{\prime}\leq d+1$ and
a matrix $\gamma\in{\mathrm{GL}}(2,{\mathbb{Q}})$, written as in (1.1), such
that

$Q_{d,\nu}^{+}=Q_{d,\nu^{\prime}}^{+}\circ\gamma.$

Let $\tilde{\gamma}$ be the homography attached to $\gamma$. This homography

(4.2)
$\tilde{\gamma}:z\in{\mathbb{P}}^{1}({\mathbb{C}})\mapsto\frac{a_{1}z+a_{2}}{a_{3}z+a_{4}}$

induces a bijection between the set of zeroes $\mathcal{Z}(Q_{d,\nu}^{+})$ (in
${\mathbb{P}}^{1}({\mathbb{C}})$) of $Q_{d,\nu}^{+}$ and the set of zeroes
$\mathcal{Z}(Q_{d,\nu^{\prime}}^{+})$. So,
$\tilde{\gamma}(i\sqrt{{\mu}_{\nu^{\prime}}})$ is a zero of
$Q_{d,\nu^{\prime}}^{+}$, hence is one of $\pm i\sqrt{{\mu}_{n}}$ with
$n\not=\nu^{\prime}$, which contradicts (4.1).

$\bullet$ The definition (1.16) implies that $Q_{d,\nu}^{+}(x,y)=0$ if and
only if $(x,y)=(0,0)$. Furthermore, by positivity, we have the lower bound

$\left|Q_{d,\nu}^{+}(x,y)\right|\geq\left(\max\\{|x|^{2},|y|^{2}\\}\right)^{d}=\left(\max\\{|x|,|y|\\}\right)^{\deg
Q_{d,\nu}^{+}}.$

The above inequality implies

$\max\\{|x|,|y|\\}\leq\left|Q_{d,\nu}^{+}(x,y)\right|^{1/{\deg
Q_{d,\nu}^{+}}},$

which means the item (v) is satisfied for $A=2$, $d_{0}=0$, $d_{1}=4$ and
$\kappa=1$. ∎

###  4.2 Triviality of the group ${\mathrm{Aut}}(Q_{d,\nu}^{+},{\mathbb{Q}})$

We now prove

######  Proposition 4.2.

For every $d\geq 2$ and $1\leq\nu\leq d+1$, one has

${\mathrm{Aut}}(Q_{d,\nu}^{+},{\mathbb{Q}})=\left\\{\begin{pmatrix}\pm 1&0\\\
0&\pm 1\end{pmatrix}\right\\}$

(Klein group of order $4$).

###### Proof.

The four elements

$\begin{pmatrix}1&0\\\ 0&1\end{pmatrix},\quad\begin{pmatrix}-1&0\\\
0&-1\end{pmatrix},\quad\begin{pmatrix}-1&0\\\
0&1\end{pmatrix},\quad\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}$

in ${\mathrm{GL}}(2,{\mathbb{Q}})$ clearly belong to
${\mathrm{Aut}}(Q_{d,\nu}^{+},{\mathbb{Q}})$. Conversely, let
$\gamma=\begin{pmatrix}a_{1}&a_{2}\\\
a_{3}&a_{4}\end{pmatrix}\in{\mathrm{GL}}(2,{\mathbb{Q}})$ and let
$Q_{d,\nu}^{+}$ be such that

(4.3) $Q_{d,\nu}^{+}\circ\gamma=Q_{d,\nu}^{+}.$

The set of zeroes $\mathcal{Z}(Q_{d,\nu}^{+})$ is stable by the homography
$\tilde{\gamma}$ attached to $\gamma$. Appealing to (4.1), we deduce

$\tilde{\gamma}\left(i\sqrt{{\mu}_{n}}\right)=\varepsilon_{n}i\sqrt{{\mu}_{n}},(1\leq
n\leq d+1,\,n\neq\nu),$

where $\varepsilon_{n}=\pm 1$. We now prove that the value of
$\varepsilon_{n}$ is independent from $n$. Indeed, suppose that there exist
$m$ and $n$ such that $\varepsilon_{m}=1$ and $\varepsilon_{n}=-1$. Returning
to the explicit expression of $\tilde{\gamma}$ (see (4.2)), we obtain

$\begin{cases}a_{1}i\sqrt{{\mu}_{m}}+a_{2}=i\sqrt{{\mu}_{m}}\left(a_{3}i\sqrt{{\mu}_{m}}+a_{4}\right)\\\
a_{1}i\sqrt{{\mu}_{n}}+a_{2}=-i\sqrt{{\mu}_{n}}\left(a_{3}i\sqrt{{\mu}_{n}}+a_{4}\right).\\\
\end{cases}$

Since $a_{1}$, $a_{2}$, $a_{3}$ and $a_{4}$ are rational numbers, we deduce
the four equalities

$\begin{cases}a_{2}=-a_{3}\,{\mu}_{m}\\\ a_{2}=a_{3}\,{\mu}_{n}\\\
a_{1}=a_{4}\\\ a_{1}=-a_{4}.\end{cases}$

These equalities imply $(a_{1},a_{2},a_{3},a_{4})=(0,0,0,0)$ which is
forbidden. So we have $\tilde{\gamma}(z)=\varepsilon z$, for some fixed
$\varepsilon\in\\{\pm 1\\}$. This means that for some $\tau\in{\mathbb{Q}}$,
we have

$\gamma=\begin{pmatrix}\varepsilon\tau&0\\\ 0&\tau\end{pmatrix}.$

By identification in (4.3), we find that $\tau=\pm 1$. ∎

###  4.3 Estimating the number of images by $\mathcal{Q}^{+}$ of $(x,y)$ with
$\max\\{|x|,|y|\\}\geq 2$

For the family $\mathcal{Q}^{+}$, one has $(2d)^{\dagger}=2d+2$. Combining
Corollary 1.12, Propositions 4.1 and 4.2 and the equality (1.10), we proved
the following

######  Proposition 4.3.

For every $d\geq 2$, one has the equality

$\mathcal{R}_{\geq
2d}(\mathcal{Q}^{+},B,2)=\frac{1}{4}\left(\sum_{F\in\mathcal{Q}_{2d}^{+}}A_{F}\right)\cdot
B^{1/d}+O_{\lambda,d,\varepsilon}\left(B^{\vartheta_{2d}+\varepsilon}\right)+O_{\lambda,d}\left(B^{1/(d+1)}\right).$

###  4.4 Estimating the number of images by $\mathcal{Q}^{+}$ of $(x,y)$ with
$\max\\{|x|,|y|\\}<2$

The difference

(4.4) $\mathcal{R}_{\geq 2d}(\mathcal{Q}^{+},B,0)-\mathcal{R}_{\geq
2d}(\mathcal{Q}^{+},B,2)$

is bounded from above by the cardinality of the set

(4.5) $\\{m\;:\;0\leq m\leq B,\,m=Q_{d^{\prime},\nu}^{+}(\pm 1,\pm
1),\,d^{\prime}\geq d,\,1\leq\nu\leq d^{\prime}+1\\}\\\ \cup\\{m\;:\;0\leq
m\leq B,\,m=Q_{d^{\prime},\nu}^{+}(0,\pm 1),\,d^{\prime}\geq d,\,1\leq\nu\leq
d^{\prime}+1\\}\cup\\{0,1\\}.$

For every $d^{\prime}$ and $1\leq\nu\leq d^{\prime}+1$, one has the equality

$Q_{d^{\prime},\nu}^{+}(\pm 1,\pm 1)\geq\prod_{1\leq n\leq
d^{\prime}}\left(1+n^{2}\right)\geq(d^{\prime}\,!)^{2}.$

This implies that the inequality $Q_{d^{\prime},\nu}^{+}(\pm 1,\pm 1)\leq B$
can only hold if $d^{\prime}=O\left(\log B\right)$. So the cardinality of the
first set in (4.5) is bounded by $O(\log^{2}B)$. The same bound also applies
to the second set. Combining Proposition 4.3 with (4.4) we obtain

######  Proposition 4.4.

For every $d\geq 2$ and for every $\varepsilon>0$, one has the equality

$\mathcal{R}_{\geq
2d}(\mathcal{Q}^{+},B,0)=\frac{1}{4}\left(\,\sum_{F\in\mathcal{Q}_{2d}^{+}}A_{F}\right)\cdot
B^{1/d}+O_{\lambda,d,\varepsilon}\left(B^{\vartheta_{2d}+\varepsilon}\right)+O_{\lambda,d}\left(B^{1/(d+1)}\right).$

###  4.5 Some results on $A_{F}$ for $F\in\mathcal{Q}^{+}$

By the definition (1.9), the fundamental domain attached to $Q_{d,\nu}^{+}$ is

(4.6)
$\mathcal{D}(Q_{d,\nu}^{+}):=\bigl{\\{}(x,y)\in{\mathbb{R}}^{2}:\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq d+1}{n\neq\nu}}\left(x^{2}+{\mu}_{n}y^{2}\right)\leq 1\bigr{\\}}.$

Our purpose is to estimate the sum

${\mathrm{Coef}}(\mathcal{Q}^{+},2d):=\sum_{F\in\mathcal{Q}_{2d}^{+}}A_{F}$

as $d\to\infty$. We use integration techniques to express this sum of
fundamental areas as follows.

######  Lemma 4.5.

For any $d\geq 2$ and $1\leq\nu\leq d+1$, one has the equality

$A_{Q_{d,\nu}^{+}}=\int_{-\infty}^{\infty}\frac{(u^{2}+\mu_{\nu})^{1/d}}{G_{d}(u)^{1/d}}\mathrm{d}u,$

where

$G_{d}(u):=\prod_{n=1}^{d+1}\left(u^{2}+\mu_{n}\right).$

Hence

${\mathrm{Coef}}(\mathcal{Q}^{+},2d)=\int_{-\infty}^{\infty}\frac{\sum_{1\leq
n\leq d+1}\left(u^{2}+\mu_{n}\right)^{1/d}}{G_{d}(u)^{1/d}}\,{\mathrm{d}}u.$

###### Proof.

By (4.6) and by the change of variables $x=uv$, $y=v$ we have the equalities

$\displaystyle A_{Q_{d,\nu}^{+}}$
$\displaystyle=\iint_{\mathcal{D}(Q_{d,\nu}^{+})}{\mathrm{d}}x\,{\mathrm{d}}y$
$\displaystyle=\iint_{v^{2d}\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
d+1}{n\neq\nu}}(u^{2}+\mu_{n})\leq 1}|v|\,{\mathrm{d}}u\,{\mathrm{d}}v$
$\displaystyle=\int_{-\infty}^{\infty}\frac{{\mathrm{d}}u}{\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq d+1}{n\neq\nu}}(u^{2}+\mu_{n})^{1/d}}\cdotp$

Compare with [B] p. 122. Summing over all the
$Q_{d,\nu}^{+}\in\mathcal{Q}_{2d}^{+}$, we obtain the second formula of Lemma
4.5. ∎

We first give a lower bound of ${\mathrm{Coef}}(\mathcal{Q}^{+},2d)$. We have

$\displaystyle{\mathrm{Coef}}(\mathcal{Q}^{+},2d)$
$\displaystyle\geq(d+1)\int_{-\infty}^{\infty}\frac{(u^{2}+\mu_{1})^{1/d}}{G_{d}(u)^{1/d}}{\mathrm{d}}u$
$\displaystyle\geq(d+1)\int_{-\infty}^{\infty}\frac{{\mathrm{d}}u}{\prod_{2\leq
n\leq d+1}(u^{2}+\mu_{n})^{1/d}}$
$\displaystyle\geq(d+1)\int_{-\infty}^{\infty}\frac{{\mathrm{d}}u}{u^{2}+\mu_{d+1}}$
(4.7) $\displaystyle\geq\pi\cdot\frac{d+1}{\sqrt{\mu_{d+1}}}\cdotp$

From our assumption $\mu_{d+1}\leq\lambda(d+1)$ we deduce from (4.7) the lower
bound

(4.8)
${\mathrm{Coef}}(\mathcal{Q}^{+},2d)>\frac{\pi}{\sqrt{\lambda}}\,\sqrt{d}.$

For the upper bound, we write

$\displaystyle{\mathrm{Coef}}(\mathcal{Q}^{+},2d)$
$\displaystyle\leq(d+1)\int_{-\infty}^{\infty}\frac{(u^{2}+\mu_{d+1})^{1/d}}{G_{d}(u)^{1/d}}{\mathrm{d}}u$
$\displaystyle\leq(d+1)\int_{-\infty}^{\infty}\frac{{\mathrm{d}}u}{\prod_{1\leq
n\leq d}(u^{2}+\mu_{n})^{1/d}}\cdotp$

Using Hölder’s inequality we deduce

$\displaystyle{\mathrm{Coef}}(\mathcal{Q}^{+},2d)$
$\displaystyle\leq(d+1)\prod_{n=1}^{d}\left(\int_{-\infty}^{\infty}\frac{{\mathrm{d}}u}{u^{2}+\mu_{n}}\right)^{1/d}$
$\displaystyle\leq\pi(d+1)\prod_{n=1}^{d}\mu_{n}^{-1/(2d)}\leq\pi\frac{d+1}{D}$

with $D:=(d\,!)^{1/(2d)}$. Using Stirling formula (1.21), we deduce

${\mathrm{Coef}}(\mathcal{Q}^{+},2d)\leq\pi\sqrt{\mathrm{e}}(\sqrt{d}+1).$

Combining with (4.8) we complete the proof of (1.18). Recalling Propositions
4.2 and 4.4, we conclude that the proof of Theorem 1.15 is now complete.

##  5 Proof of Theorem 1.17

Recall that for $d\geq 2$ and $1\leq\nu\leq d+1$, $Q^{-}_{d,\nu}$ denotes the
following binary form of degree $2d$

$Q^{-}_{d,\nu}(X,Y)=\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
d+1}{n\neq\nu}}\left(X^{2}-{\mu}_{n}Y^{2}\right)$

and $\mathcal{Q}^{-}$ denotes the family

$\mathcal{Q}^{-}=\left\\{Q^{-}_{d,\nu}:d\geq 2,\,1\leq\nu\leq d+1\right\\}.$

###  5.1 The family $\mathcal{Q}^{-}$ is
$(A,1,2,2,2\mathrm{e}\lambda)$–regular

Our goal in this subsection is to prove the following

######  Proposition 5.1.

For $A>2\mathrm{e}\lambda$, the family $\mathcal{Q}^{-}$ is
$(A,1,2,2,2\mathrm{e}\lambda)$–regular.

The proofs of items (i), (ii), (iii) and (iv) are the same as for Proposition
4.1: one only replaces (4.1) with the remark that for two positive distinct
squarefree numbers

$n\neq
n^{\prime}\Rightarrow{\mathbb{Q}}\left(\sqrt{{\mu}_{n}}\right)\neq{\mathbb{Q}}\left(\sqrt{{\mu}_{n^{\prime}}}\right).$

It remains to check the condition (v) in Definition 1.10 of a regular family.
We start with an auxiliary lemma.

######  Lemma 5.2.

For $m$ and $d$ integers satisfying $1\leq m<d$, we have

$\left(\frac{d}{m}-1\right)^{d-m}\geq\mathrm{e}^{-\mathrm{e}^{-1}m};$

further, for $n$ an integer in the range $1\leq n\leq d$, we have

$\frac{n!(d-n)!}{n^{d}}\geq\mathrm{e}^{-(1+\mathrm{e}^{-1})d}.$

The numerical value for $\mathrm{e}^{1+\mathrm{e}^{-1}}$ is
$3.927\dots<\frac{79}{20}\cdotp$

###### Proof of Lemma 5.2.

Set $t=d-m$, $f_{m}(t)=\left(\frac{t}{m}\right)^{t}$, $g_{m}(t)=\log
f_{m}(t)=t\log t-t\log m$. The derivative $g_{m}^{\prime}(t)=1+\log t-\log m$
of $g_{m}$ vanishes at $t=m/\mathrm{e}$, the minimum of $f_{m}(t)$ on the
interval $0<t\leq d-1$ is reached at $t=m/\mathrm{e}$, giving the value
$(t/m)^{t}=\mathrm{e}^{-t}=\mathrm{e}^{-m/\mathrm{e}}$.

The last part of Lemma 5.2 follows from the first one thanks to Stirling’s
formula (1.21):

$\frac{n!(d-n)!}{n^{d}}\geq\frac{n^{n}}{\mathrm{e}^{n}}\cdot\frac{(d-n)^{d-n}}{\mathrm{e}^{d-n}}\cdot\frac{1}{n^{d}}=\frac{(d-n)^{d-n}}{n^{d-n}\mathrm{e}^{d}}\geq\mathrm{e}^{-d}\mathrm{e}^{-\mathrm{e}^{-1}n}\geq\mathrm{e}^{-(1+\mathrm{e}^{-1})d}.$

∎

The last inequality of Lemma 5.2 implies

(5.1)
$\bigl{(}n!(d-n)!\bigr{)}^{1/d}\geq\mathrm{e}^{-1-\mathrm{e}^{-1}}\max\\{n,d-n\\}\geq\frac{d}{2\mathrm{e}^{1+\mathrm{e}^{-1}}}\geq\frac{d}{2\mathrm{e}^{2}}\cdotp$

###### End of the proof of Proposition 5.1.

Let $d\geq 2$, $1\leq\nu\leq d+1$,
$(x,y)\in{\mathbb{Z}}^{2}\setminus\\{(0,0)\\}$. Set $Q=Q^{-}_{d,\nu}(x,y)$.
Our goal it to prove

(5.2) $|Q|>(2\mathrm{e}\lambda)^{-2d+2}\max\\{|x|,|y|\\}^{2d-2}.$

We consider three cases depending on the sign of the factors
$x^{2}-\mu_{n}y^{2}$.
$\bullet$ If $x^{2}<\mu_{1}y^{2}$, all factors are negative. For $2\leq n\leq
d+1$ we have

$|x^{2}-{\mu}_{n}y^{2}|={\mu}_{n}y^{2}-x^{2}>({\mu}_{n}-\mu_{1})y^{2}.$

When $\nu\geq 2$, we use the lower bound $\mu_{1}y^{2}-x^{2}\geq 1$ and obtain

$|Q|>(\mu_{2}-\mu_{1})\cdots(\mu_{d+1}-\mu_{1})(\mu_{\nu}-\mu_{1})^{-1}y^{2d-2}\geq(d-1)!y^{2d-2}.$

For $\nu=1$ the stronger lower bound $|Q|>(d-1)!y^{2d}$ holds. Hence for
$1\leq\nu\leq d+1$ we have

$|Q|>\frac{(d-1)!}{\mu_{1}^{d-1}}\,x^{2d-2}\geq\frac{(d-1)!}{\lambda^{d-1}}\,x^{2d-2}.$

The desired estimate (5.2) follows.
$\bullet$ If $x^{2}>\mu_{d+1}y^{2}$, all factors are positive and
$\max\\{|x|,|y|\\}=|x|$. For $m=1,\dots,d$ we have

$x^{2}-\mu_{m}y^{2}>(\mu_{d+1}-\mu_{m})\frac{x^{2}}{\mu_{d+1}},$

while for $m=d+1$ we have $x^{2}-\mu_{d+1}y^{2}\geq 1$. Hence for
$1\leq\nu\leq d$ we have

$|Q|=Q>(\mu_{d+1}-\mu_{1})(\mu_{d+1}-\mu_{2})\cdots(\mu_{d+1}-\mu_{d})(\mu_{d+1}-\mu_{\nu})^{-1}\frac{x^{2d-2}}{\mu_{d+1}^{d-1}}\geq\frac{d!x^{2d-2}}{\mu_{d+1}^{d-1}}\cdotp$

The lower bound $|Q|>\frac{d!x^{2d-2}}{\mu_{d+1}^{d-1}}$ is also true when
$\nu=d+1$ since in this case we have

$|Q|=Q>(\mu_{d+1}-\mu_{1})(\mu_{d+1}-\mu_{2})\cdots(\mu_{d+1}-\mu_{d})\frac{x^{2d}}{\mu_{d+1}^{d}}\geq\frac{d!x^{2d}}{\mu_{d+1}^{d}}$

and $x^{2}>\mu_{d+1}y^{2}\geq\mu_{d+1}$. Since $d!>d^{d}\mathrm{e}^{-d}$ (see
(1.21)) and $\mu_{d+1}\leq\lambda(d+1)$, we have

$\frac{d!}{\mu_{d+1}^{d-1}}>\frac{d}{\mathrm{e}\left(1+\frac{1}{d}\right)^{d-1}(\lambda\mathrm{e})^{d-1}}>\frac{1}{(2\mathrm{e}\lambda)^{2d-2}}\cdotp$

This implies (5.2).
$\bullet$ Finally, assume that there is an $n$ in the interval $1\leq n\leq d$
such that

$x^{2}-\mu_{n+1}y^{2}<0<x^{2}-\mu_{n}y^{2}.$

Hence $y\not=0$ and $\max\\{|x|,|y|\\}=|x|$. We have

(5.3) $\displaystyle|Q|=$
$\displaystyle(x^{2}-\mu_{1}y^{2})(x^{2}-\mu_{2}y^{2})\cdots(x^{2}-\mu_{n}y^{2})(\mu_{n+1}y^{2}-x^{2})\cdots(\mu_{d+1}y^{2}-x^{2})|x^{2}-\mu_{\nu}y^{2}|^{-1}$

with

(5.4)
$\displaystyle(x^{2}-\mu_{1}y^{2})(x^{2}-\mu_{2}y^{2})\cdots(x^{2}-\mu_{n-1}y^{2})$
$\displaystyle>(\mu_{n}-\mu_{1})(\mu_{n}-\mu_{2})\cdots(\mu_{n}-\mu_{n-1})y^{2n-2}$
$\displaystyle\geq(n-1)!y^{2n-2}$

and

(5.5) $\displaystyle(\mu_{n+2}y^{2}-x^{2})\cdots(\mu_{d+1}y^{2}-x^{2})$
$\displaystyle>(\mu_{n+2}-\mu_{n+1})\cdots(\mu_{d+1}-\mu_{n+1})y^{2d-2n}$
$\displaystyle\geq(d-n)!y^{2d-2n}.$

For $1\leq\nu\leq n-1$, we use the lower bound

(5.6)
$(x^{2}-\mu_{1}y^{2})(x^{2}-\mu_{2}y^{2})\cdots(x^{2}-\mu_{n-1}y^{2})(x^{2}-\mu_{\nu}y^{2})^{-1}>(n-2)!y^{2n-4},$

while for $n+2\leq\nu\leq d+1$, we use the lower bound

(5.7)
$(\mu_{n+2}y^{2}-x^{2})\cdots(\mu_{d+1}y^{2}-x^{2})(\mu_{\nu}y^{2}-x^{2})^{-1}>(d-n-1)!y^{2d-2n-2}.$

It remains to estimate the product
$(x^{2}-\mu_{n}y^{2})(\mu_{n+1}y^{2}-x^{2})$ of the two terms of the middle in
(5.3). We consider two cases.

$\diamond$ Assume $|y|\geq 2$. If $\nu\in\\{n,n+1\\}$, we use the trivial
lower bound

(5.8)
$(x^{2}-\mu_{n}y^{2})(\mu_{n+1}y^{2}-x^{2})|x^{2}-\mu_{\nu}y^{2}|^{-1}\geq 1,$

while if $\nu\leq n-1$ or $\nu\geq n+2$ we use the lower bound

(5.9) $\displaystyle(x^{2}-\mu_{n}y^{2})(\mu_{n+1}y^{2}-x^{2})$
$\displaystyle\geq(x^{2}-\mu_{n}y^{2})+(\mu_{n+1}y^{2}-x^{2})-1$
$\displaystyle=(\mu_{n+1}-\mu_{n})y^{2}-1\geq y^{2}-1\geq\frac{3}{4}y^{2}.$

$\circ$ For $\nu\in\\{n,n+1\\}$, we deduce from (5.3), (5.4), (5.5), (5.8),

$|Q|\geq(n-1)!(d-n)!y^{2d-2}.$

$\circ$ For $1\leq\nu\leq n-1$, we deduce from (5.3), (5.5), (5.6), (5.9),

$|Q|\geq\frac{3}{4}(n-2)!(d-n)!y^{2d-2}.$

$\circ$ For $n+2\leq\nu\leq d+1$ we deduce from (5.3), (5.4), (5.7), (5.9),

$|Q|\geq\frac{3}{4}(n-1)!(d-n-1)!y^{2d-2}.$

In the three cases, namely for $1\leq\nu\leq d+1$, we have, thanks to Lemma
5.2,

$|Q|\geq\frac{3n!(d-n)!}{4n(d-1)}y^{2d-2}\geq\frac{3n^{d-1}}{4(d-1)\mathrm{e}^{(1+\mathrm{e}^{-1})d}}y^{2d-2}.$

From $x^{2}<\mu_{n+1}y^{2}\leq\lambda(n+1)y^{2}\leq 2\lambda ny^{2}$ we deduce

$|Q|>\frac{3}{4(d-1)\mathrm{e}^{(1+\mathrm{e}^{-1})d}(2\lambda)^{d-1}}x^{2d-2}.$

Finally, since $\lambda\geq 2$, we have

(5.10)
$\frac{3}{4(d-1)\mathrm{e}^{(1+\mathrm{e}^{-1})d}}>\frac{1}{(2\mathrm{e}^{2}\lambda)^{d-1}}$

for $d\geq 2$, and (5.2) follows,

$\diamond$ If $y^{2}=1$, hence $\mu_{n}<x^{2}<\mu_{n+1}$, using the trivial
lower bound

$(x^{2}-\mu_{n})(\mu_{n+1}-x^{2})\geq 1,$

and a combination of the above lower bounds (5.3), (5.4), (5.5), (5.6), (5.7)
yields

$|Q|\geq\begin{cases}(n-1)!(d-n)!&\hbox{if $\nu\in\\{n,n+1\\}$},\\\
(n-2)!(d-n)!&\hbox{if $1\leq\nu\leq n-1$},\\\ (n-1)!(d-n-1)!&\hbox{if
$n+2\leq\nu\leq d+1$}.\end{cases}$

For $1\leq\nu\leq d+1$, we obtain, thanks to Lemma 5.2,

$|Q|\geq\frac{n!(d-n)!}{n(d-1)}\geq\frac{n^{d-1}}{(d-1)\mathrm{e}^{(1+\mathrm{e}^{-1})d}}\cdotp$

If $x^{2}\leq 2\lambda$, using $n\geq 1$, we deduce

$|Q|\geq\frac{n^{d-1}}{(d-1)\mathrm{e}^{(1+\mathrm{e}^{-1})d}}\left(\frac{x^{2}}{2\lambda}\right)^{d-1},$

while if $x^{2}\geq 2\lambda$ we have, by (1.14), the inequalities
$n>\frac{x^{2}}{\lambda}-1\geq\frac{x^{2}}{2\lambda}$, hence again

$|Q|\geq\frac{x^{2d-2}}{(d-1)\mathrm{e}^{(1+\mathrm{e}^{-1})d}(2\lambda)^{d-1}}\cdotp$

From (5.10) we deduce the estimate (5.2) also when $|y|=1$.

This completes the proof of Proposition 5.1. ∎

###  5.2 Triviality of the group ${\mathrm{Aut}}(Q_{d,\nu}^{-},{\mathbb{Q}})$

The following result is the analog of Proposition 4.2. The proof is the same,
since $\mu_{1}\geq 2$ and the roots of $Q_{d,\nu}^{-}$ are all irrational
numbers.

######  Proposition 5.3.

For every $d\geq 2$ and $1\leq\nu\leq d+1$, one has

${\mathrm{Aut}}(Q_{d,\nu}^{-},{\mathbb{Q}})=\left\\{\begin{pmatrix}\pm 1&0\\\
0&\pm 1\end{pmatrix}\right\\}$

(Klein group of order $4$).

###  5.3 Estimating the number of images by $\mathcal{Q}^{-}$ of $(x,y)$ with
$\max\\{|x|,|y|\\}~{}\geq A$

From Corollary 1.12, the equality (1.10) and Propositions 5.1 and 5.3, we
deduce:

######  Proposition 5.4.

For every $A>2\mathrm{e}\lambda$, for every $d\geq 2$ and for every
$\varepsilon>0$, one has the equality

$\mathcal{R}_{\geq
2d}(\mathcal{Q}^{-},B,A)=\frac{1}{4}\left(\,\sum_{F\in\mathcal{Q}_{2d}^{-}}A_{F}\right)\cdot
B^{1/d}+O_{\lambda,A,d,\varepsilon}\left(B^{\vartheta_{2d}+\varepsilon}\right)+O_{\lambda,A,d}\left(B^{1/(d+1)}\right).$

###  5.4 Estimating the number of images by $\mathcal{Q}^{-}$ of $(x,y)$ with
$\max\\{|x|,|y|\\}~{}<A$

The difference

$\mathcal{R}_{\geq 2d}(\mathcal{Q}^{-},B,0)-\mathcal{R}_{\geq
2d}(\mathcal{Q}^{-},B,A)$

is at most the cardinality of the set

$\left\\{m\;:\;0\neq|m|\leq B,\,m=Q_{d^{\prime},\nu}^{-}(x,y),\,d^{\prime}\geq
d,\,1\leq\nu\leq d^{\prime}+1,\,\max\\{|x|,|y|\\}\leq A\right\\}.$

Given $d^{\prime}$, the number of such $m$ in this set is bounded by
$(d^{\prime}+1)(2A+1)^{2}$. Hence we only need to bound from above the value
of $d^{\prime}$ when $|m|\geq 2$.

We first consider the integers of the form $Q_{d^{\prime},\nu}^{-}(x,0)$.
Since $Q_{d^{\prime},\nu}^{-}(\pm 1,0)=1$, we may assume $|x|\geq 2$. From

$Q_{d^{\prime},\nu}^{-}(x,0)=x^{2d^{\prime}}\leq B$

we deduce that $d^{\prime}$ is bounded by $O(\log B)$.

Next let $m=Q_{d^{\prime},\nu}^{-}(x,y)$ with $d^{\prime}\geq d$,
$1\leq\nu\leq d^{\prime}+1$, $\max\\{|x|,|y|\\}\leq A$, $|y|\geq 1$ and
$0<|m|\leq B$. Without loss of generality we may assume $d^{\prime}>2A^{2}$.
We split the product

$\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
d^{\prime}+2}{n\neq\nu}}\left|x^{2}-{\mu}_{n}y^{2}\right|$

the value of which is $|m|$, as $P_{1}P_{2}$ where

$P_{1}=\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
2A^{2}}{n\neq\nu}}\left|x^{2}-{\mu}_{n}y^{2}\right|,\qquad
P_{2}=\prod_{\genfrac{}{}{0.0pt}{}{2A^{2}<n\leq
d^{\prime}+2}{n\neq\nu}}\left|x^{2}-{\mu}_{n}y^{2}\right|.$

The product $P_{1}$ is $\geq 1$. For $2A^{2}<n\leq d^{\prime}+2$, since
$\mu_{n}>n$, $|x|\leq A$ and $|y|\geq 1$, we have $\mu_{n}y^{2}-x^{2}\geq
A^{2}$, hence

$(A^{2})^{d^{\prime}-2A^{2}}\leq P_{2}\leq P_{1}P_{2}=|m|\leq B,$

which yields

$d^{\prime}\leq 2A^{2}+\frac{\log B}{2\log A}=O_{A}(\log B).$

Hence

$\mathcal{R}_{\geq 2d}(\mathcal{Q}^{-},B,0)-\mathcal{R}_{\geq
2d}(\mathcal{Q}^{-},B,A)=O_{A}((\log B)^{2}).$

Thanks to Proposition 5.4, this completes the proof of the estimate for
$\mathcal{R}_{\geq 2d}(\mathcal{Q}^{-},B,0)$ in Theorem 1.17.

###  5.5 Some results on $A_{F}$ for $F\in\mathcal{Q}^{-}$

By the definition (1.9), the fundamental domain attached to $Q_{d,\nu}^{-}$ is

(5.11)
$\mathcal{D}(Q_{d,\nu}^{-}):=\bigl{\\{}(x,y)\in{\mathbb{R}}^{2}:\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq d+1}{n\neq\nu}}\left|x^{2}-{\mu}_{n}y^{2}\right|\leq 1\bigr{\\}}.$

Our purpose is to estimate the sum

${\mathrm{Coef}}(\mathcal{Q}^{-},2d):=\sum_{F\in\mathcal{Q}_{2d}^{-}}A_{F}$

as $d\to\infty$ by proving (1.19).

Repeating the proof of Lemma 4.5, we obtain:

######  Lemma 5.5.

For any $d\geq 2$ and $1\leq\nu\leq d+1$, one has the equality

$A_{Q_{d,\nu}^{-}}=\int_{-\infty}^{\infty}\frac{|u^{2}-\mu_{\nu}|^{1/d}}{\prod_{1\leq
n\leq d+1}|u^{2}-\mu_{n}|^{1/d}}{\mathrm{d}}u.$

Hence

${\mathrm{Coef}}(\mathcal{Q}^{-},2d)=\int_{-\infty}^{\infty}\frac{\sum_{1\leq
n\leq
d+1}|u^{2}-\mu_{n}|^{1/d}}{\prod_{n=1}^{d+1}|u^{2}-\mu_{n}|^{1/d}}{\mathrm{d}}u.$

Since $|u^{2}-\mu_{n}|\leq u^{2}+\mu_{n}$, the lower bound of
${\mathrm{Coef}}(\mathcal{Q}^{-},2d)$ is a consequence of the lower bound of
${\mathrm{Coef}}(\mathcal{Q}^{+},2d)$. More precisely, we have, by Lemma 5.5,

$A_{Q_{d,\nu}^{-}}=\int_{-\infty}^{\infty}\frac{{\mathrm{d}}u}{\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq
d+1}{n\neq\nu}}|u^{2}-\mu_{n}|^{1/d}}\geq\int_{-\infty}^{\infty}\frac{{\mathrm{d}}u}{\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq d+1}{n\neq\nu}}(u^{2}+\mu_{n})^{1/d}},$

hence

$\displaystyle{\mathrm{Coef}}(\mathcal{Q}^{-},2d)$
$\displaystyle\geq(d+1)\int_{-\infty}^{\infty}\frac{{\mathrm{d}}u}{\prod_{2\leq
n\leq d+1}(u^{2}+\mu_{n})^{1/d}}$
$\displaystyle\geq(d+1)\int_{-\infty}^{\infty}\frac{{\mathrm{d}}u}{u^{2}+\mu_{d+1}}=\pi\cdot\frac{d+1}{\sqrt{\mu_{d+1}}}\cdotp$

This proves the lower bound

(5.12)
${\mathrm{Coef}}(\mathcal{Q}^{-},2d)>\frac{\pi}{\sqrt{\lambda}}\,\sqrt{d}.$

For the upper bound, we use once more Lemma 5.5. By the change of variable
$u^{2}=v$ we have

$A_{Q_{d,\nu}^{-}}=2\int_{0}^{\infty}\frac{{\mathrm{d}}u}{\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq
d+1}{n\neq\nu}}|u^{2}-\mu_{n}|^{1/d}}=\int_{0}^{\infty}\frac{{\mathrm{d}}v}{\sqrt{v}\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq d+1}{n\neq\nu}}|v-\mu_{n}|^{1/d}}\cdotp$

We split the integral as the sum of $d+3$ terms

$A_{Q_{d,\nu}^{-}}=\sum_{j=0}^{d+2}A_{j}$

with

$A_{0}=\int_{0}^{\mu_{1}}\frac{{\mathrm{d}}v}{\sqrt{v}\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq d+1}{n\neq\nu}}(\mu_{n}-v)^{1/d}},$
$A_{j}=\int_{\mu_{j}}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{\sqrt{v}\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq j}{n\neq\nu}}(v-\mu_{n})^{1/d}\prod_{\genfrac{}{}{0.0pt}{}{j+1\leq n\leq
d+1}{n\neq\nu}}(\mu_{n}-v)^{1/d}}\qquad(1\leq j\leq d+1)$

and

$A_{d+2}=\int_{\mu_{d+2}}^{\infty}\frac{{\mathrm{d}}v}{\sqrt{v}\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq d+1}{n\neq\nu}}(v-\mu_{n})^{1/d}}\cdotp$

$\bullet$ Upper bound for $A_{0}$.
For $\nu=1$, we use the lower bound

$\prod_{2\leq n\leq d+1}(\mu_{n}-\mu_{1})\geq d!\geq\frac{d^{d}}{e^{d}}$

which follows from Stirling’s estimate (1.21) and one deduces

$A_{0}\leq\frac{1}{\prod_{2\leq n\leq
d+1}(\mu_{n}-\mu_{1})^{1/d}}\int_{0}^{\mu_{1}}\frac{{\mathrm{d}}v}{\sqrt{v}}\leq\frac{2\mathrm{e}\sqrt{\mu_{1}}}{d}\leq\frac{2\mathrm{e}\sqrt{\lambda}}{d}\cdotp$

Similarly, for $2\leq\nu\leq d+1$ we have

$A_{0}\leq\frac{1}{\prod_{\genfrac{}{}{0.0pt}{}{2\leq n\leq
d+1}{n\neq\nu}}(\mu_{n}-\mu_{1})^{1/d}}\int_{0}^{\mu_{1}}\frac{{\mathrm{d}}v}{\sqrt{v}(\mu_{1}-v)^{1/d}}$

and

$\prod_{\genfrac{}{}{0.0pt}{}{2\leq n\leq
d+1}{n\neq\nu}}(\mu_{n}-\mu_{1})\geq(d-1)!=\frac{d!}{d},$

hence

$\prod_{\genfrac{}{}{0.0pt}{}{2\leq n\leq
d+1}{n\neq\nu}}(\mu_{n}-\mu_{1})^{1/d}\geq\frac{d}{\mathrm{e}\sqrt{2}}.$

From the upper bounds (recall $\lambda\geq 2$ and $2\leq\mu_{1}\leq\lambda$)

$\displaystyle\int_{0}^{\mu_{1}}\frac{{\mathrm{d}}v}{\sqrt{v}(\mu_{1}-v)^{1/d}}$
$\displaystyle\leq\int_{0}^{\mu_{1}}\frac{{\mathrm{d}}v}{\sqrt{v}}+\int_{0}^{\mu_{1}}\frac{{\mathrm{d}}v}{(\mu_{1}-v)^{1/d}}$
$\displaystyle=2\sqrt{\mu_{1}}+\frac{d}{d-1}\mu_{1}^{1-(1/d)}<(2+\sqrt{2})\lambda,$

we deduce

$A_{0}<\frac{5\mathrm{e}\lambda}{d}\cdotp$

$\bullet$ Upper bound for $A_{j}$, $1\leq j\leq d+1$.
$\diamond$ If $\nu\not\in\\{j,j+1\\}$, we have

$\displaystyle A_{j}$
$\displaystyle\leq\frac{1}{\sqrt{\mu}_{j}\prod_{\genfrac{}{}{0.0pt}{}{1\leq
n\leq
j-1}{n\neq\nu}}(\mu_{j}-\mu_{n})^{1/d}\prod_{\genfrac{}{}{0.0pt}{}{j+2\leq
n\leq d+1}{n\neq\nu}}(\mu_{n}-\mu_{j+1})^{1/d}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\int_{\mu_{j}}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{(v-\mu_{j})^{1/d}(\mu_{j+1}-v)^{1/d}}\cdotp$

We use (5.1): for $1\leq j\leq d$ we have

$\displaystyle\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
j-1}{n\neq\nu}}(\mu_{j}-\mu_{n})\prod_{\genfrac{}{}{0.0pt}{}{j+2\leq n\leq
d+1}{n\neq\nu}}(\mu_{n}-\mu_{j+1})$
$\displaystyle\geq\begin{cases}\frac{(j-1)!(d-j)!}{j-\nu}&\hbox{for
$1\leq\nu\leq j-1$}\\\ \frac{(j-1)!(d-j)!}{\nu-j-1}&\hbox{for $j+1\leq\nu\leq
d+1$}\end{cases}$
$\displaystyle\geq\frac{j!(d-j)!}{d^{2}}\geq\frac{1}{d^{2}}\left(\frac{d}{2\mathrm{e}^{1+\mathrm{e}^{-1}}}\right)^{d},$

while for $j=d+1$ this lower bound becomes

$\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
d}{n\neq\nu}}(\mu_{d+1}-\mu_{n})\geq\frac{d!}{d+1-\nu}\geq\frac{1}{d^{2}}\left(\frac{d}{2\mathrm{e}^{1+\mathrm{e}^{-1}}}\right)^{d}.$

Next we use the following estimate:

$\displaystyle\int_{\mu_{j}}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{(v-\mu_{j})^{1/d}(\mu_{j+1}-v)^{1/d}}\leq$
$\displaystyle\qquad\frac{2^{1/d}}{(\mu_{j+1}-\mu_{j})^{1/d}}\left(\int_{\mu_{j}}^{(\mu_{j}+\mu_{j+1})/2}\frac{{\mathrm{d}}v}{(v-\mu_{j})^{1/d}}+\int_{(\mu_{j}+\mu_{j+1})/2}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{(\mu_{j+1}-v)^{1/d}}\right).$

We have

$\int_{\mu_{j}}^{(\mu_{j}+\mu_{j+1})/2}\frac{{\mathrm{d}}v}{(v-\mu_{j})^{1/d}}=\int_{(\mu_{j}+\mu_{j+1})/2}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{(\mu_{j+1}-v)^{1/d}}=\frac{d}{d-1}\left(\frac{\mu_{j+1}-\mu_{j}}{2}\right)^{1-(1/d)}.$

Hence

$\int_{\mu_{j}}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{(v-\mu_{j})^{1/d}(\mu_{j+1}-v)^{1/d}}\leq\frac{d}{d-1}2^{2/d}(\mu_{j+1}-\mu_{j})^{1-(2/d)}.$

We deduce that for $\nu\not\in\\{j,j+1\\}$, we have

$A_{j}\leq(4d^{2})^{1/d}2\mathrm{e}^{1+\mathrm{e}^{-1}}\frac{(\mu_{j+1}-\mu_{j})^{1-(2/d)}}{(d-1)\sqrt{\mu_{j}}}\cdotp$

$\diamond$ If $\nu=j$, we have

$A_{j}\leq\frac{1}{\sqrt{\mu}_{j}\prod_{1\leq n\leq
j-1}(\mu_{j}-\mu_{n})^{1/d}\prod_{j+2\leq n\leq
d+1}(\mu_{n}-\mu_{j+1})^{1/d}}\int_{\mu_{j}}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{(\mu_{j+1}-v)^{1/d}}$

and we use the formula

$\int_{\mu_{j}}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{(\mu_{j+1}-v)^{1/d}}=\frac{d}{d-1}(\mu_{j+1}-\mu_{j})^{1-(1/d)}.$

$\diamond$ If $\nu=j+1$, we have

$A_{j}\leq\frac{1}{\sqrt{\mu}_{j}\prod_{1\leq n\leq
j-1}(\mu_{j}-\mu_{n})^{1/d}\prod_{j+2\leq n\leq
d+1}(\mu_{n}-\mu_{j+1})^{1/d}}\int_{\mu_{j}}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{(v-\mu_{j})^{1/d}}$

and we use the formula

$\int_{\mu_{j}}^{\mu_{j+1}}\frac{{\mathrm{d}}v}{(v-\mu_{j})^{1/d}}=\frac{d}{d-1}(\mu_{j+1}-\mu_{j})^{1-(1/d)}.$

We deduce that for $1\leq j\leq d+1$ and $1\leq\nu\leq d+1$, we have

(5.13)
$A_{j}\leq\bigl{(}2\mathrm{e}^{1+\mathrm{e}^{-1}}+o(1)\bigr{)}\frac{\mu_{j+1}-\mu_{j}}{d\sqrt{\mu_{j}}}\cdotp$

For $j\geq 1$, we have

$\mu_{j}\geq j+1\geq\frac{1}{\lambda}\,\mu_{j+1},$

and we deduce the inequality

$\sum_{j=1}^{d+1}\frac{\mu_{j+1}-\mu_{j}}{\sqrt{\mu_{j}}}\leq\sqrt{\lambda}\,\sum_{j=1}^{d+1}\frac{\mu_{j+1}-\mu_{j}}{\sqrt{\mu_{j+1}}}\cdotp$

Using the inequality

$\sum_{j=1}^{d+1}\frac{\mu_{j+1}-\mu_{j}}{\sqrt{\mu_{j+1}}}\leq\sum_{j=1}^{d+1}\int_{\mu_{j}}^{\mu_{j+1}}\frac{{\mathrm{d}}t}{\sqrt{t}}=\int_{\mu_{1}}^{\mu_{d+2}}\frac{{\mathrm{d}}t}{\sqrt{t}}\leq
2\sqrt{\mu_{d+2}}\leq 2\sqrt{\lambda(d+2)},$

we deduce from (5.13), that

$\displaystyle\sum_{j=1}^{d+1}A_{j}\leq$
$\displaystyle\left(\bigl{(}2\mathrm{e}^{1+\mathrm{e}^{-1}}+o_{\lambda}(1)\bigr{)}/d\right)\cdot\sqrt{\lambda}\cdot\bigl{(}2\sqrt{\lambda(d+2)}\bigr{)}$
$\displaystyle\leq\bigl{(}4\mathrm{e}^{1+\mathrm{e}^{-1}}+o_{\lambda}(1)\bigr{)}\frac{\lambda}{\sqrt{d}}\cdotp$

$\bullet$ Upper bound for $A_{d+2}$.
For $v\geq\mu_{d+2}$ and $1\leq n\leq d+1$, we have

$v-\mu_{n}\geq v\left(1-\frac{\mu_{n}}{\mu_{d+2}}\right),$

hence

$A_{d+2}\leq\frac{1}{\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
d+1}{n\neq\nu}}\left(1-\frac{\mu_{n}}{\mu_{d+2}}\right)^{1/d}}\int_{\mu_{d+2}}^{\infty}\frac{{\mathrm{d}}v}{v^{3/2}}$

with

$\int_{\mu_{d+2}}^{\infty}\frac{{\mathrm{d}}v}{v^{3/2}}=\frac{2}{\sqrt{\mu_{d+2}}}\leq\frac{2}{\sqrt{d+2}}$

and (using Stirling’s estimate (1.21) once more)

$\prod_{\genfrac{}{}{0.0pt}{}{1\leq n\leq
d+1}{n\neq\nu}}\left(1-\frac{\mu_{n}}{\mu_{d+2}}\right)^{1/d}\geq\frac{d!^{1/d}}{\mu_{d+2}}\geq\frac{d!^{1/d}}{\lambda(d+2)}\geq\frac{1}{\lambda\mathrm{e}}\left(1+\frac{2}{d}\right)^{-1}.$

We deduce

$A_{d+2}\leq\left(2\mathrm{e}+o(1)\right)\frac{\lambda}{\sqrt{d}}\cdotp$

Putting these estimates together, we obtain

$A_{Q_{d,\nu}^{-}}=\sum_{j=0}^{d+2}A_{j}\leq\left(4\mathrm{e}^{1+\mathrm{e}^{-1}}+2\mathrm{e}+o_{\lambda}(1)\right)\frac{\lambda}{\sqrt{d}}\cdotp$

Summing over all the $Q_{d,\nu}^{-}\in\mathcal{Q}_{2d}^{-}$ we conclude

${\mathrm{Coef}}(\mathcal{Q}^{-},2d)\leq\left(4\mathrm{e}^{1+\mathrm{e}^{-1}}+2\mathrm{e}+o_{\lambda}(1)\right)\lambda\sqrt{d}.$

Combining with (5.12) and with the upper bound
$4\mathrm{e}^{1+\mathrm{e}^{-1}}+2\mathrm{e}<22$, we complete the proof of
(1.19). The proof of Theorem 1.17 is now complete.

##  6 Proof of Theorem 1.18

We now use the notations of § 1.3.4. Our first purpose is to check that the
family $\mathcal{L}$ satisfies the assertions of Definition 1.10 of a regular
family. The items (i), (ii) are obvious. The item (iii) is trivially satisfied
with $A_{1}=1$. The items (iv) and (v) are more subtle.

###  6.1 Isomorphisms between two elements in $\mathcal{L}$

We will prove the following more general statement which implies that the item
(iv) is fulfilled by the family $\mathcal{L}$.

######  Proposition 6.1.

Let $d\geq 4$ be an integer, $\\{a_{i}:1\leq i\leq d-1\\}$ and $\\{b_{j}:1\leq
j\leq d-2\\}$ two sets of integers and $p$ a prime number such that

(6.1) $0<a_{1}<\cdots<a_{d-1}<p,$

and

(6.2) $0<b_{1}<\cdots<b_{d-2}<p.$

Then the binary forms

(6.3) $X\prod_{i=1}^{d-1}\left(X-a_{i}Y\right)\text{ and
}(X-pY)X\prod_{j=1}^{d-2}\left(X-b_{j}Y\right)$

are not isomorphic.

###### Proof.

The proof is based on classical properties of the cross–ratio of four points
on ${\mathbb{P}}^{1}({\mathbb{C}})={\mathbb{C}}\cup\\{\infty\\}$. Recall that
if $(x_{1},x_{2},x_{3},x_{4})$ is a quadruple of four distinct complex
numbers, the associated cross–ratio  is the complex number
$[x_{1},x_{2},x_{3},x_{4}]$ defined by

$[x_{1},x_{2},x_{3},x_{4}]:=\frac{x_{3}-x_{1}}{x_{3}-x_{2}}\,\Bigg{/}\,\frac{x_{4}-x_{1}}{x_{4}-x_{2}}\cdotp$

This definition is naturally extended to ${\mathbb{P}}^{1}({\mathbb{C}})$ when
exactly one of the elements $x_{1}$, $x_{2}$, $x_{3}$ and $x_{4}$ is equal to
$\infty$. The cross–ratio is invariant by any homography of
${\mathbb{P}}^{1}({\mathbb{C}})$. In other words, for any homography
$\mathfrak{h}$, for any quadruple $(x_{1},x_{2},x_{3},x_{4})$ of distinct
points of ${\mathbb{P}}^{1}({\mathbb{C}})$, one has the equality

(6.4)
$[x_{1},x_{2},x_{3},x_{4}]=[\mathfrak{h}(x_{1}),\mathfrak{h}(x_{2}),\mathfrak{h}(x_{3}),\mathfrak{h}(x_{4})].$

Let $a$ be a nonzero integer. The canonical decomposition of $|a|$ into prime
factors

$|a|=\prod_{p}p^{v_{p}(a)}$

defines, for each prime number $p$, the $p$–adic valuation
$v_{p}(a)\in{\mathbb{Z}}$ of $a$. Let $t=a/b\neq 0$ be a rational number,
written in its irreducible form. The $p$–adic valuation  of $t$, is the non
negative integer

$v_{p}(t):=\begin{cases}v_{p}(a)&\text{ if }p\nmid b,\\\ -v_{p}(b)&\text{ if
}p\nmid a.\end{cases}$

We now begin the proof of Proposition 6.1. This proof is by contradiction. Let
$F_{1}(X,Y)$ and $F_{2}(X,Y)$ respectively be the two binary forms introduced
in (6.3). Suppose that there is $\gamma\in{\mathrm{GL}}(2,{\mathbb{Q}})$,
written as in (1.1), such that

$F_{1}=F_{2}\circ\gamma.$

Then the homography $\mathfrak{h}$ associated with $\gamma$ has the shape

$z\mapsto\mathfrak{h}(z)=\frac{az+b}{cz+d}\cdotp$

This homography induces a bijective map between the sets of zeroes of the
polynomials $f_{1}(X):=F_{1}(X,1)$ and $f_{2}(X):=F_{2}(X,1).$ These sets of
zeroes are $\mathcal{Z}(f_{1}):=\\{0,\,a_{1},\dots,a_{d-1}\\}$ and
$\mathcal{Z}(f_{2})=\\{0,\,b_{1},\dots,b_{d-2},\,p\\}$ considered as subsets
of ${\mathbb{P}}^{1}({\mathbb{C}})$. Consider, for $j=1,\,2$, the subsets of
${\mathbb{Q}}\setminus\\{0\\}$ defined by

(6.5)
$\mathrm{Bir}(f_{j}):=\left\\{[x_{1},x_{2},x_{3},x_{4}]:x_{i}\in\mathcal{Z}(f_{j}),x_{i}\text{
distinct}\right\\}.$

The equality (6.4) implies the equality of the two sets

$\mathrm{Bir}(f_{1})=\mathrm{Bir}(f_{2}),$

and also of the two sets

$\\{v_{p}(y):y\in\mathrm{Bir}(f_{1})\\}=\\{v_{p}(z):z\in\mathrm{Bir}(f_{2})\\}.$

As a consequence of the inequalities (6.1), we have
$\\{v_{p}(y):y\in\mathrm{Bir}(f_{1})\\}=\\{0\\}$. However we also have
$1\in\\{v_{p}(z):z\in\mathrm{Bir}(f_{2})\\}$ by considering the cross ratio
$[0,b_{1},p,b_{2}]$ and the inequalities (6.2). So we reach a contradiction:
the element $\gamma$ does not exist and the binary forms $F_{1}$ and $F_{2}$
are not isomorphic. ∎

###  6.2 Triviality of the group ${\mathrm{Aut}}(L_{d,p},{\mathbb{Q}})$

In order to determine the value of the coefficient $W$ appearing in
Proposition A, we prove the following.

######  Proposition 6.2.

Let $d\geq 5$ be an integer. For every prime $p\geq d$, the automorphism group
of the binary form $L_{d,p}$ is $\\{{\mathrm{Id}}\\}$ if $d$ is odd, and
$\\{\pm{\mathrm{Id}}\\}$ if $d$ is even. In particular, the set
$\mathcal{L}_{d}$ fulfils the conditions C1 or C2 of Corollary 1.12, according
to the parity of $d$.

####  6.2.1 Two preliminary results

The proof of the following lemma is based on the analytic properties of the
homography on each of its intervals of definition.

######  Lemma 6.3.

Let $\mathfrak{h}$ be a homography belonging to
${\mathrm{PGL}}(2,{\mathbb{R}})$, $M>0$ be a real number, $t\geq 1$ be an
integer, $x_{1},\dots,x_{t}$ be $t$ real numbers satisfying
$0<x_{1}<\cdots<x_{t}<M$, $y_{1},\dots,y_{t}$ be $t$ real numbers satisfying
$0<y_{1}<\cdots<y_{t}<M$. Assume

$\begin{cases}\mathfrak{h}\left(\left\\{x_{i}:1\leq i\leq
t\right\\}\right)=\left\\{y_{j}:1\leq j\leq t\right\\},\\\
\mathfrak{h}(0)=0\text{ and }\mathfrak{h}(M)=M.\end{cases}$

Then, for every $1\leq i\leq t$, one has the equality
$\mathfrak{h}(x_{i})=y_{i}$.

###### Proof.

We split the proof in several cases depending on the nature of the homography
$\mathfrak{h}$.

  * •

If $\mathfrak{h}(\infty)=\infty$, the restriction of $\mathfrak{h}$ to the
real affine line has the shape $\mathfrak{h}(x)=ax+b$, where $a\neq 0$ and $b$
are real numbers. The conditions $\mathfrak{h}(0)=0$ and $\mathfrak{h}(M)=M$
imply $a=1$ and $b=0$. Hence the result since $\mathfrak{h}$ is the identity.

  * •

If $\mathfrak{h}(\infty)\neq\infty$, $\mathfrak{h}$ has a unique expansion as

(6.6) $\mathfrak{h}(x)=a+\frac{b}{x-c},$

where $a,$ $b$ and $c$ are real numbers such that
$c\not\in\\{0,\,x_{1},\cdots,\,x_{t},M\\}$ and $b\neq 0$. We now consider the
respective values of $b$ and $c$.

    * –

If $b>0$ the function $x\mapsto\mathfrak{h}(x)$ is decreasing on the two
intervals $(-\infty,c)$ and $(c,+\infty)$. We consider the value of $c$.

      * *

If $c<x_{t}\ (<M)$, we have the inequality
$\mathfrak{h}(x_{t})>\mathfrak{h}(M)=M,$ since $\mathfrak{h}$ is decreasing.
This contradicts the hypothesis $\mathfrak{h}(x_{t})<M$.

      * *

If $c>x_{t}\ (>0)$, we have $0=\mathfrak{h}(0)>\mathfrak{h}(x_{t})$. This
contradicts the hypothesis $\mathfrak{h}(x_{t})>0$. We conclude that
$\mathfrak{h}$ is not of the form (6.6) with $b>0$.

    * –

If $b<0$, the function $x\mapsto\mathfrak{h}(x)$ is increasing on both
intervals $(\infty,c)$ and $(c,+\infty)$. We now consider the value of $c$.

      * *

If $c\not\in[0,M]$, the function $x\mapsto\mathfrak{h}(x)$ is increasing on
$(0,M)$, so we have $\mathfrak{h}(x_{i})=y_{i}$ for $1\leq i\leq t$.

      * *

If $0<c<M$, the hyperbola
$\\{(x,\mathfrak{h}(x))\in{\mathbb{R}}^{2}:x\in{\mathbb{R}},\,x\neq c\\}$ has
two asymptotes: one with abscissa $c$ and the other one with ordinate $a$.
Elementary considerations on this hyperbola lead to the inequalities

$\mathfrak{h}(0)>a>\mathfrak{h}(M).$

This contradicts the hypothesis $\mathfrak{h}(0)=0$ and $\mathfrak{h}(M)=M$.
In conclusion $\mathfrak{h}$ is not of the form (6.6) with $b<0$ and $0<c<M$.

∎

We will require the following variant of Lemma 6.3.

######  Lemma 6.4.

Let $\mathfrak{h}$ be a homography belonging to
${\mathrm{PGL}}(2,{\mathbb{R}})$, $M>0$ be a real number, $t\geq 1$ be an
integer, $x_{1},\dots,x_{t}$ be $t$ real numbers satisfying
$0<x_{1}<\cdots<x_{t}<M$, $y_{1},\dots,y_{t}$ be $t$ real numbers satisfying
$0<y_{1}<\cdots<y_{t}<M$. Assume

$\begin{cases}\mathfrak{h}\left(\left\\{x_{i}:1\leq i\leq
t\right\\}\right)=\left\\{y_{j}:1\leq j\leq t\right\\},\\\
\mathfrak{h}(0)=M\text{ and }\mathfrak{h}(M)=0.\end{cases}$

Then for every $1\leq i\leq t$, one has the equality
$\mathfrak{h}(x_{i})=y_{t+1-i}$.

###### Proof.

Introduce the homography $\mathfrak{g}=\mathfrak{s}\circ\mathfrak{h},$ where
$\mathfrak{s}$ is the symmetry $\mathfrak{s}(x)=M-x$. The homography
$\mathfrak{g}$ fulfils the hypotheses of Lemma 6.3 provided that we replace
the points $y_{i}$ ($1\leq i\leq t)$ by the points
$y^{\prime}_{i}:=M-y_{t+1-i}$. We deduce that for all $i$ one has the equality
$\mathfrak{g}(x_{i})=y^{\prime}_{i}$, which gives
$\mathfrak{h}(x_{i})=y_{t+1-i}$. ∎

####  6.2.2 Proof of Proposition 6.2

###### Proof.

Consider the polynomial

$f(X)=L_{d,p}(X,1)$

and its set of zeroes $\mathcal{Z}(f)=\\{0,\,1,\,2,\cdots,\,d-2,\,p\\}$. In
order to prove that the group of automorphisms of $L_{d,p}$ is trivial it
suffices to prove that the unique homography
$\mathfrak{h}\in{\mathrm{PGL}}(2,{\mathbb{Q}})$, such that

(6.7) $\mathfrak{h}\left(\mathcal{Z}(f)\right)=\mathcal{Z}(f),$

is the identity as soon as the prime $p$ satisfies $p\geq d$.

As in the proof of Proposition 6.1, we will play with the $p$–adic valuation
of the elements in $\mathrm{Bir}(f)$, defined in (6.5). We first notice that
for $x$ and $y$ two distinct integers in $\\{1,\,2,\dots,d-2\\}$, the elements

$\alpha:=[0,x,p,y],\,[p,x,0,y],[x,0,y,p]\text{ and }[x,p,y,0],$

belong to $\mathrm{Bir}(f)$ and satisfy $v_{p}(\alpha)=1$. These are the only
elements in $\mathrm{Bir}(f)$ which satisfy $v_{p}(\alpha)=1$. In particular,
if four distinct elements $x,y,z,t$ in $\mathcal{Z}(f)$ satisfy
$v_{p}([x,y,z,t])=1$, then $\\{0,p\\}\subset\\{x,y,z,t\\}$.

By (6.4), we have the following equality

$v_{p}\left([\mathfrak{h}(x),\mathfrak{h}(0),\mathfrak{h}(y),\mathfrak{h}(p)]\right)=1,$

where $x$ and $y$ are integers as above. Since $d\geq 5$, there exists an
integer $x$ in $\\{1,\,2,\dots,d-2\\}$ such that
$\mathfrak{h}(x)\not\in\\{0,p\\}$. We claim that there is another integer
$y\not=x$ in $\\{1,\,2,\dots,d-2\\}$ with the same property, namely such that
$\mathfrak{h}(y)\not\in\\{0,p\\}$. This is plain for $d\geq 6$; for $d=5$, the
only case where this would not be true is when $\\{1,2,3\\}=\\{x,y,z\\}$ with
$\\{\mathfrak{h}(y),\mathfrak{h}(z)\\}=\\{0,p\\}$, but this case is not
possible since it would not be compatible with our requirement that

$\\{0,p\\}\subset\left\\{\mathfrak{h}(x),\mathfrak{h}(0),\mathfrak{h}(y),\mathfrak{h}(p)\right\\}.$

This proves our claim that there are two distinct integers $x$ and $y$ in the
set $\\{1,\,2,\dots,d-2\\}$ such that
$\\{\mathfrak{h}(x),\mathfrak{h}(y)\\}\cap\\{0,p\\}=\emptyset$. Therefore

$\\{\mathfrak{h}(0),\mathfrak{h}(p)\\}=\\{0,p\\}.$

We consider two cases.

  1. (i)

Assume

$\mathfrak{h}(0)=0\text{ and }\mathfrak{h}(p)=p.$

Since $\mathfrak{h}$ induces by restriction a bijective map of
$\mathcal{Z}(f)$ onto itself, we may apply Lemma 6.3. We deduce that
$\mathfrak{h}(t)=t$ for $0\leq t\leq d-2$ and $\mathfrak{h}(p)=p$. Since a
homography is determined by its restriction to a set with three elements, we
deduce that $\mathfrak{h}={\mathrm{Id}}$.

  2. (ii)

If

(6.8) $\mathfrak{h}(0)=p\text{ and }\mathfrak{h}(p)=0,$

we apply Lemma 6.4 to deduce that $\mathfrak{h}(i)=d-1-i$, for $1\leq i\leq
d-2$. The unique homography $\mathfrak{h}$ satisfying this property is the
symmetry defined by $\mathfrak{h}:z\mapsto d-1-z$. But such a formula is not
compatible with the fact that $\mathfrak{h}(0)=p$. So there is no homography
$\mathfrak{h}$ satisfying (6.7) and (6.8).

We conclude that the set of $\mathfrak{h}$ satisfying (6.7) is reduced to the
identity. The proof of Proposition 6.2 is complete. ∎

###  6.3 The family $\mathcal{L}$ is regular (continued)

We now investigate the condition (v) of Definition 1.10. We will prove

######  Proposition 6.5.

For every $d\geq 5$, for every $p$ with $p\geq d-1$, and for all
$(x,y)\in{\mathbb{Z}}^{2}$ such that $L_{d,p}(x,y)\neq 0$. the following
inequality holds

(6.9) $\max\\{|x|,|y|\\}\leq 9\cdot|\,L_{d,p}(x,y)\,|^{\frac{1}{d-1}}.$

The inequality (6.9) is equivalent to the lower bound

(6.10)
$|L_{d,p}(x,y)|\geq\left(\,\frac{1}{9}\cdot\max\\{|x|,|y|\\}\,\right)^{d-1},$

under the hypotheses of Proposition 6.5. We will rather work with (6.10).

The proof of (6.10) depends on the relative sizes of $|x|$ and $|y|$. However,
if we suppose that $xy\leq 0$ and $L_{d,p}(x,y)\not=0$, it is straightforward
to obtain the lower bound

$|L_{d,p}(x,y)|\geq\left(\max\\{|x|,|y|\\}\right)^{d-1}.$

Hence we may assume that $x$ and $y$ are not zero and have the same sign.
Besides, since $|L_{d,p}(-x,-y)|=|L_{d,p}(x,y)|$, we will assume that both $x$
and $y$ are positive.

The basic equality is the following one

(6.11) $|L_{d,p}(x,y)|=x\cdot|x-y|\cdot|x-2y|\cdots|x-(d-2)y|\cdot|x-py|.$

We split the argument according to the relative sizes of $x$ and $y$.

####  6.3.1 Assume $1\leq x\leq y$

Let $x$ and $y$ be positive integers such that $L_{d,p}(x,y)\neq 0$ with
$y\geq x$. Hence $y\geq x+1$. We deduce from (6.11)

$\displaystyle|L_{d,p}(x,y)|$
$\displaystyle=x\cdot(y-x)\cdot(2y-x)\cdots((d-2)y-x)\cdot(py-x)$
$\displaystyle>x\cdot(y-x)\cdot y\cdot(2y)\cdots((d-3)y)\cdot((p-1)y)$
$\displaystyle=x\cdot(y-x)\cdot(d-3)\,!\cdot(p-1)\,y^{d-2}.$

If $y\geq 2x$ we have $x(y-x)\geq y-x\geq y/2$, while for $x<y\leq 2x$ we have
$x(y-x)\geq x\geq y/2$. Hence

$|L_{d,p}(x,y)|>\frac{1}{2}(d-3)\,!(p-1)\left(\max\\{|x|,|y|\\}\right)^{d-1}.$

So we proved

######  Proposition 6.6.

For every $d\geq 3$, for every $p\geq d-1$, for every integers $x$ and $y$
such that $L_{d,p}(x,y)\neq 0$ and $|x|\leq|y|$, one has the inequality

$|L_{d,p}(x,y)|\geq\max\\{|x|,|y|\\}^{d-1}.$

####  6.3.2 Assume $(d-2)y\leq x$

Let $x$ and $y$ be positive integers such that $L_{d,p}(x,y)\neq 0$ with
$x\geq(d-2)y$, hence $x\geq(d-2)y+1$. We deduce from (6.11)

$|L_{d,p}(x,y)|=x\cdot(x-y)\cdot(x-2y)\cdots(x-(d-2)y)\cdot|x-py|.$

$\bullet$ If $y=1$, since $x\geq d-1$, we have

$x-n=x\left(1-\frac{n}{x}\right)\geq
x\left(1-\frac{n}{d-1}\right)=x\left(\frac{d-n-1}{d-1}\right)$

for $0\leq n\leq d-2$; using the trivial lower bound $|x-p|\geq 1$ together
with Stirling’s formula (1.21), we deduce

$|L_{d,p}(x,1)|\geq
x\cdot(x-1)\cdot(x-2)\cdots(x-(d-2))\geq\frac{(d-1)!}{(d-1)^{d-1}}\,x^{d-1}\geq\frac{x^{d-1}}{\mathrm{e}^{d-1}}\cdotp$

$\bullet$ We assume now $y\geq 2$. As a consequence of the hypothesis $y\leq
x/(d-2)$, we have the inequality

$x\cdot(x-y)\cdot(x-2y)\cdots(x-(d-3)y)\geq\frac{(d-2)!}{(d-2)^{d-2}}\,x^{d-2}.$

$\diamond$ If $x>py$, then

$x-(d-2)y\geq x\left(1-\frac{d-2}{p}\right)\geq
x\left(1-\frac{d-2}{d-1}\right)=\frac{x}{d-1}$

and the trivial lower bound $x-py\geq 1$ suffices to deduce

$L_{d,p}(x,y)\geq\frac{(d-2)!}{(d-1)(d-2)^{d-2}}\,x^{d-1}.$

$\diamond$ If $py>x$, then from $x-(d-2)y\geq 1$ and $py-x\geq 1$ we deduce

$(x-(d-2)y)\cdot(py-x)\geq(x-(d-2)y)+(py-x)-1\geq y(p-d+2)-1.$

If $p=d-1$ we use the assumption $y\geq 2$ which yields

$y(p-d+2)-1=y-1\geq\frac{y}{2}>\frac{x}{2p}=\frac{x}{2(d-1)},$

while for $p\geq d$ we use the lower bounds

$y(p-d+2)-1\geq y(p-d+1)\geq
py\left(1-\frac{d-1}{p}\right)>x\left(1-\frac{d-1}{d}\right)=\frac{x}{d}\cdotp$

Therefore, for $(d-2)y\leq x$ and $y\geq 2$, we have

$|L_{d,p}(x,y)|\geq\frac{(d-2)!}{2(d-1)(d-2)^{d-2}}\,x^{d-1}\geq\frac{x^{d-1}}{2d\mathrm{e}^{d-2}}\cdotp$

We deduce

######  Proposition 6.7.

For $d\geq 3$, $p$ prime $\geq d-1$ and $(x,y)\in{\mathbb{Z}}^{2}$ such that
$|x|\geq(d-2)|y|$ and $L_{d,p}(x,y)\neq 0$ we have

$\left|L_{d,p}(x,y)\right|\geq\frac{1}{d\mathrm{e}^{d}}\max\\{|x|,|y|\\}^{d-1}.$

####  6.3.3 Assume $(n-1)y\leq x\leq ny$ for some $n$ with $2\leq n\leq d-2$

We deduce from (6.11)

$|L_{d,p}(x,y)|=x\cdot(x-y)\cdots(x-(n-1)y)\cdot(ny-x)\cdots((d-2)y-x)\cdot(py-x).$

We have

$x\cdot(x-y)\cdots(x-(n-2)y)\geq(n-1)!y^{n-1}$

and

$\displaystyle((n+1)y-x)\cdots((d-2)y-x)\cdot(py-x)$
$\displaystyle\geq(d-n-2)!(p-n)y^{d-n-1}$
$\displaystyle\geq(d-n-1)!y^{d-n-1}.$

For the product of the two terms in the middle, if $y=1$ we use the trivial
lower bound $(x-(n-1)y)(ny-x)\geq 1$ which yields

$|L_{d,p}(x,y)|\geq(n-1)!(d-n-1)!y^{d-2}\geq\frac{(n-1)!(d-n-1)!}{n^{d-2}}x^{d-2},$

while for $y\geq 2$ we use

$(x-(n-1)y)(ny-x)\geq(x-(n-1)y)+(ny-x)-1=y-1\geq\frac{y}{2},$

which yields

$|L_{d,p}(x,y)|\geq\frac{1}{2}(n-1)!(d-n-1)!y^{d-1}\geq\frac{(n-1)!(d-n-1)!}{2n^{d-1}}x^{d-1}.$

We now use Lemma 5.2:

$\frac{(n-1)!(d-n-1)!}{n^{d-1}}=\frac{n!(d-n)!}{n^{d}(d-n)}\geq\mathrm{e}^{-(1+\mathrm{e}^{-1})d}\frac{1}{d-n},$

from which we deduce

$|L_{d,p}(x,y)|\geq\mathrm{e}^{-(1+\mathrm{e}^{-1})d}\frac{1}{2(d-n)}x^{d-1}.$

This proves the following result:

######  Proposition 6.8.

For $d\geq 3$, $2\leq n\leq d-2$, $p$ prime $\geq d-1$ and $x$ and $y$ such
that $(n-1)|y|\leq|x|\leq n|y|$ and $L_{d,p}(x,y)\neq 0$, we have

$\left|L_{d,p}(x,y)\right|\geq\frac{1}{2(d-2)}\cdot\frac{\max\\{|x|,|y|\\}^{d-1}}{\mathrm{e}^{(1+\mathrm{e}^{-1})d}}\cdotp$

For $d\geq 5$, we have

$2(d-2)\cdot\mathrm{e}^{(1+\mathrm{e}^{-1})d}<9^{d-1}.$

We may now gather Propositions 6.6, 6.7 and 6.8 to deduce (6.10), which
completes the proof of Proposition 6.5.

###  6.4 Estimating the number of images by $\mathcal{L}$ of $(x,y)$ with
$\max\\{|x|,|y|\\}\geq 10$

Gathering Propositions 6.1 and 6.5, we proved that the family $\mathcal{L}$ is
$(10,1,1,5,9)$–regular. Furthermore, according to the parity of $d$, the set
$\mathcal{L}_{d}$ satisfies the conditions C1 or C2 of Corollary 1.12, by
Proposition 6.2. As a consequence of Corollary 1.12 we have the following

######  Proposition 6.9.

For any $d\geq 5$, for every $\varepsilon>0$, one has the equality

$\mathcal{R}_{\geq
d}\left(\mathcal{L},B,10\right)=\frac{1}{(2,d)}\left(\sum_{d\leq
p<2d}A_{L_{d,p}}\right)B^{2/d}+O_{d,\varepsilon}\left(B^{\vartheta_{d}+\varepsilon}\right)+O_{d}\left(B^{2/(d+1)}\right).$

###  6.5 Estimating the number of images by $\mathcal{L}$ of $(x,y)$ with
$\max\\{|x|,|y|\\}<10$

The difference

(6.12) $\mathcal{R}_{\geq d}(\mathcal{L},B,0)-\mathcal{R}_{\geq
d}(\mathcal{L},B,10)$

is bounded from above by two times the cardinality of the set

$\mathfrak{Er}_{\geq d}(B)\\\ :=\\{m\;:\;0<m=|L_{d^{\prime},p}(x,y)|\leq
B,\;d\leq d^{\prime}\leq p<2d^{\prime},\,\max\\{|x|,|y|\\}\leq 9\\}.$

There are $19^{2}$ pairs $(x,y)$ with $\max\\{|x|,|y|\\}\leq 9$. We first
count the number of $m$ in $\mathfrak{Er}_{\geq d}(B)$ of the form
$|L_{d^{\prime},p}(x,0)|$, namely with $y=0$. For $x=\pm 1$ and $y=0$ we have
$m=1$; for $2\leq|x|\leq 9$ and $y=0$, we have $2^{d^{\prime}}\leq B$, hence
there are at most $O_{d}(\log B)$ such values of $m$.

We count now the number of $m$ in $\mathfrak{Er}_{\geq d}(B)$ of the form
$|L_{d^{\prime},p}(x,y)|$ with $|y|\geq 1$. We have $|x-ny|\geq n-|x|\geq
n-9\geq 2$ for $n\geq 11$, hence

$B\geq m\geq\prod_{11\leq n\leq d^{\prime}-2}(n-9)\geq 2^{d^{\prime}-12},$

and therefore $d^{\prime}\leq O(\log B)$. It follows that the number of pairs
$(d^{\prime},p)$ as above is bounded by $O_{d}(\log^{2}B)$. So we proved

$\sharp\,\mathfrak{Er}_{\geq d}(B)=O_{d}(\log^{2}B).$

Combining this bound with (6.12) and with Proposition 6.9, we obtain the
equality (1.20) of Theorem 1.18.

###  6.6 Some results on $A_{F}$ for $F\in\mathcal{L}$

The area of the fundamental domain associated to $L_{d,p}$ is, by the
definition (1.9), equal to

$A_{L_{d,p}}=\iint_{\mathcal{D}(L_{d,p})}{\mathrm{d}}x\,{\mathrm{d}}y,$

with

$\mathcal{D}(L_{d,p}):=\bigl{\\{}(x,y)\in{\mathbb{R}}^{2}:\left|x(x-y)(x-2y)\cdots(x-(d-2)y)(x-py)\right|\leq
1\bigr{\\}}.$

By the change of variables $u=x$ and $v=y/x$, we obtain

$A_{L_{d,p}}=\iint_{\mathcal{D}^{*}(L_{d,p})}|u|\,{\mathrm{d}}u\,{\mathrm{d}}v,$

with

$\mathcal{D}^{*}(L_{d,p}):=\bigl{\\{}(u,v)\in{\mathbb{R}}^{2}:|u|^{d}\cdot\left|(1-v)(1-2v)\cdots(1-(d-2)v)(1-pv)\right|\leq
1\bigr{\\}}.$

Some elementary calculations transform $A_{L_{d,p}}$ into a single integral.

######  Lemma 6.10.

For $d\geq 5$ and $p\geq d-1$ the following equalities hold

$A_{L_{d,p}}=\int_{-\infty}^{\infty}\frac{{\mathrm{d}}v}{\left(\,|1-v|\cdot|1-2v|\cdots|1-(d-2)v|\cdot|1-pv|\,\right)^{2/d}}$

and

$A_{L_{d,p}}=\int_{-\infty}^{\infty}\frac{{\mathrm{d}}t}{\left(\,|t|\cdot|t-1|\cdot|t-2|\cdots|t-(d-2)|\cdot|t-p|\,\right)^{2/d}}\cdotp$

We will only work with the second expression of $A_{L_{d,p}}$. So we introduce
the function

$\lambda_{d,p}(t):=t(t-1)\cdots(t-(d-2))(t-p),$

which is the product of $d$ linear factors in $t$. We split the interval of
integration into $d$ intervals of length $1$ around the singularities $0$,…,
$d-2$ and $p$ and three remaining intervals to write the equality:

(6.13) $A_{L_{d,p}}:=\\\
\left(\int_{-\infty}^{-1/2}+\int_{-1/2}^{1/2}+\cdots+\int_{d-(5/2)}^{d-(3/2)}+\int_{d-(3/2)}^{p-(1/2)}+\int_{p-(1/2)}^{p+(1/2)}+\int_{p+(1/2)}^{\infty}\right)\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}}\cdotp$

We will give an upper bound and a lower bound for each of these positive
integrals in order to prove

######  Proposition 6.11.

Uniformly for $d\to\infty$ and $d\leq p<2d$ one has

$\frac{\mathrm{e}^{2}-o(1)}{d}\leq
A_{L_{d,p}}\leq\frac{5\,\mathrm{e}^{2}+2\mathrm{e}+o(1)}{d}\cdotp$

The last part of Theorem 1.18 is obtained from this proposition after a
summation over $d\leq p<2d$ and an application of the Prime Number Theorem.

####  6.6.1 An auxiliary lemma

######  Lemma 6.12.

For $d\to\infty$, we have

$\left(1\cdot 3\cdot 5\cdots(2d-3)\right)^{1/d}=(2\mathrm{e}^{-1}+o(1))d.$

###### Proof.

We write

$1\cdot 3\cdot
5\cdots(2d-3)=\frac{(2d-3)!}{2^{d-2}(d-2)!}=\frac{(2d)!}{(2d-1)2^{d}d!}$

and we use Stirling’s formula (1.21) which gives

$\Bigl{(}\frac{2d}{\rm e}\Bigr{)}^{d}\frac{\sqrt{2}}{(2d-1)\cdot{\rm
e}^{1/12d}}\leq 1\cdot 3\cdot 5\cdots(2d-3)\leq\Bigl{(}\frac{2d}{\rm
e}\Bigr{)}^{d}\frac{\sqrt{2}\cdot{\rm e}^{1/24d}}{2d-1}\cdotp$

∎

####  6.6.2 Study of $\int_{-\infty}^{-1/2}$ and of $\int_{p+1/2}^{\infty}$

######  Lemma 6.13.

For $d\to\infty$ and $p\geq d$, one has

$0\leq\int_{-\infty}^{-1/2}\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}}\leq\frac{\mathrm{e}+o(1)}{d}\cdotp$

###### Proof.

Using Hölder inequality and Lemma 6.12, we obtain

$\displaystyle\int_{-\infty}^{-1/2}\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}}\leq$
$\displaystyle\left(\int_{-\infty}^{-1/2}\frac{{\mathrm{d}}t}{|t|^{2}}\right)^{1/d}\left(\int_{-\infty}^{-1/2}\frac{{\mathrm{d}}t}{|t-1|^{2}}\right)^{1/d}\cdots\left(\int_{-\infty}^{-1/2}\frac{{\mathrm{d}}t}{|t-(d-2)|^{2}}\right)^{1/d}\left(\int_{-\infty}^{-1/2}\frac{{\mathrm{d}}t}{|t-p|^{2}}\right)^{1/d}$
$\displaystyle\leq\left(\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{2}{5}\cdots\frac{2}{2d-3}\cdot\frac{2}{2p+1}\right)^{1/d}\leq\left(\frac{2^{d}}{1\cdot
3\cdot 5\cdots(2d-3)\cdot(2p+1)}\right)^{1/d}$
$\displaystyle\leq\frac{\mathrm{e}+o(1)}{d}\cdotp$

∎

Similarly, one proves

######  Lemma 6.14.

For $d\to\infty$ and $p\geq d$, one has

$0\leq\int_{p+(1/2)}^{\infty}\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}}\leq\frac{\mathrm{e}+o(1)}{d}\cdotp$

###### Proof.

For $t>p+\frac{1}{2}$, we have

$|\lambda_{d,p}(t)|=\lambda_{d,p}(t)=t(t-1)\cdots(t-(d-2))(t-p).$

Using Hölder inequality and Lemma 6.12, we obtain

$\displaystyle\int_{p+(1/2)}^{\infty}\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}}\leq$
$\displaystyle\left(\int_{p+(1/2)}^{\infty}\frac{{\mathrm{d}}t}{t^{2}}\right)^{1/d}\left(\int_{p+(1/2)}^{\infty}\frac{{\mathrm{d}}t}{(t-1)^{2}}\right)^{1/d}\cdots\left(\int_{p+(1/2)}^{\infty}\frac{{\mathrm{d}}t}{(t-(d-2))^{2}}\right)^{1/d}\left(\int_{p+(1/2)}^{\infty}\frac{{\mathrm{d}}t}{(t-p)^{2}}\right)^{1/d}$
$\displaystyle\leq\left(\frac{2}{2p+1}\cdot\frac{2}{2p-1}\cdot\frac{2}{2p-3}\cdots\frac{2}{2p-2d+5}\cdot\frac{2}{1}\right)^{1/d}$
$\displaystyle\leq\left(\frac{2^{d}}{1\cdot 3\cdot
5\cdots(2d-3)}\right)^{1/d}$
$\displaystyle\leq\frac{\mathrm{e}+o(1)}{d}\cdotp$

∎

####  6.6.3 Study of $\int_{d-3/2}^{p-1/2}$

######  Lemma 6.15.

For $d\to\infty$ and $d\leq p<2d$, one has

$0\leq\int_{d-3/2}^{p-1/2}\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}}\leq\frac{\mathrm{e}^{2}+o(1)}{d}\cdotp$

###### Proof.

For $t$ in the interval $(d-(3/2),p-(1/2))$, we have $p-t>1/2$,

$|\lambda_{d,p}(t)|=t(t-1)\cdots(t-(d-2))(p-t)$

and, for $0\leq n\leq d-2$,

$t-n>\frac{2d-2n-3}{2},$

hence

$|\lambda_{d,p}(t)|\geq\frac{(2d-3)\cdot(2d-5)\cdots 3\cdot 1}{2^{d}}$

and therefore

$|\lambda_{d,p}(t)|^{2/d}\geq(\mathrm{e}^{-2}+o(1))d^{2}$

by Lemma 6.12. Since $d\leq p<2d$, the interval of integration has length at
most $d+1$, and so we deduce

$\int_{d-3/2}^{p-1/2}\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}}\leq\frac{\mathrm{e}^{2}+o(1)}{d}\cdotp$

∎

####  6.6.4 Study of $\int_{p-1/2}^{p+1/2}$

######  Lemma 6.16.

For $d\geq 5$ and $d\leq p<2d$, one has

$0\leq\int_{p-1/2}^{p+1/2}\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}}=O\left(\frac{1}{d^{2}}\right).$

We introduce the polynomial

${\mathcal{M}}(t):=t(t-1)\cdots(t-(d-2))$

of degree $d-1$. It is easy to see that

$\min_{|t-p|\leq
1/2}|{\mathcal{M}}(t)|=|{\mathcal{M}}(p-(1/2))|\geq{\mathcal{M}}(d-(3/2))=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdots\frac{2d-5}{2}\cdot\frac{2d-3}{2},$

hence by Lemma 6.12, we have

$\min_{|t-p|\leq 1/2}|{\mathcal{M}}(t)|^{2/d}\geq(\mathrm{e}^{-2}+o(1))d^{2}.$

Since

$\int_{p-1/2}^{p+1/2}\frac{{\mathrm{d}}t}{|t-p|^{2/d}}=O(1)$

we conclude

$\displaystyle\int_{p-1/2}^{p+1/2}\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}}$
$\displaystyle\leq\left(\int_{p-1/2}^{p+1/2}\frac{{\mathrm{d}}t}{|t-p|^{2/d}}\right)\cdot\left(\frac{1}{\min_{|t-p|\leq
1/2}|{\mathcal{M}}(t)|}\right)^{2/d}$
$\displaystyle=O\left(\frac{1}{d^{2}}\right).$

####  6.6.5 Study of the remaining integrals

We are now concerned, for $\nu=0,1\dots,d-2$, with the integrals

$\mathcal{I}_{\nu}=\mathcal{I}_{d,p,\nu}=\int_{\nu-1/2}^{\nu+1/2}\frac{{\mathrm{d}}t}{|\lambda_{d,p}(t)|^{2/d}},$

for which we want to find an upper and a lower bound. We split the product
defining $\lambda_{d,p}(t)$ into four pieces

(6.14)
$\lambda_{d,p}(t)=(t-\nu)\cdot(t-p)\cdot\lambda_{\nu}^{-}(t)\cdot\lambda_{d,\nu}^{+}(t),$

with

$\lambda_{\nu}^{-}(t):=\prod_{0\leq
k<\nu}\left(t-k\right)\quad\text{and}\quad\lambda_{d,\nu}^{+}(t):=\prod_{\nu<k\leq
d-2}\left(t-k\right).$

We have

(6.15)
$\mathcal{I}_{\nu}\leq\left(\int_{\nu-1/2}^{\nu+1/2}\frac{{\mathrm{d}}t}{|t-\nu|^{2/d}}\right)\cdot\left(\min|\lambda_{\nu}^{-}(t)|\right)^{-2/d}\cdot\left(\min|\lambda_{d,\nu}^{+}(t)|\right)^{-2/d}\cdot\left(\min|t-p|\right)^{-2/d},$

and

(6.16)
$\mathcal{I}_{\nu}\geq\left(\int_{\nu-1/2}^{\nu+1/2}\frac{{\mathrm{d}}t}{|t-\nu|^{2/d}}\right)\cdot\left(\max|\lambda_{\nu}^{-}(t)|\right)^{-2/d}\cdot\left(\max|\lambda_{d,\nu}^{+}(t)|\right)^{-2/d}\cdot\left(\max|t-p|\right)^{-2/d}$

where all the maximum and minimum are taken for $\nu-1/2\leq t\leq\nu+1/2$.
Direct computations transform (6.15) and (6.16) into

$\displaystyle(1-o(1))\left(\max|\lambda_{\nu}^{-}(t)|\right)^{-2/d}\cdot$
$\displaystyle\left(\max|\lambda_{d,\nu}^{+}(t)|\right)^{-2/d}\leq\mathcal{I}_{\nu}\leq$
$\displaystyle(1+o(1))\left(\min|\lambda_{\nu}^{-}(t)|\right)^{-2/d}\cdot\left(\min|\lambda_{d,\nu}^{+}(t)|\right)^{-2/d}$

which is also

(6.17) $\displaystyle(1-o(1))|\lambda_{\nu}^{-}(\nu+1/2)|^{-2/d}\cdot$
$\displaystyle|\lambda_{d,\nu}^{+}(\nu-1/2)|^{-2/d}\leq\mathcal{I}_{\nu}\leq$
$\displaystyle(1+o(1))|\lambda_{\nu}^{-}(\nu-1/2)|^{-2/d}\cdot|\lambda_{d,\nu}^{+}(\nu+1/2)|^{-2/d},$

uniformly for $d\to\infty$ and $d\leq p<2d$.

For $1\leq\nu\leq d-2$, we have the equalities

$\lambda_{\nu}^{-}(\nu+1/2)=\frac{(2\nu+1)(2\nu-1)\cdots
3}{2^{\nu}}=\frac{(2\nu+1)\,!}{2^{2\nu}\cdot\nu\,!}=\frac{(2\nu)\,!}{2^{2\nu}\cdot\nu\,!}\cdot\frac{1}{2\nu+1},$
$\lambda_{\nu}^{-}(\nu-1/2)=\frac{(2\nu-1)(2\nu-3)\cdots
1}{2^{\nu}}=\frac{(2\nu-1)\,!}{2^{2\nu-1}\cdot(\nu\
-1)\,!}=\frac{(2\nu)\,!}{2^{2\nu}\cdot\nu\,!},$

and for $0\leq\nu\leq d-3$, we have

$|\,\lambda_{d,\nu}^{+}(\nu+1/2)\,|=\frac{(2d^{*}-1)(2d^{*}-3)\cdots 3\cdot
1}{2^{d^{*}}}=\frac{(2d^{*}-1)\,!}{2^{2d^{*}-1}\cdot(d^{*}-1)\,!}=\frac{(2d^{*})\,!}{2^{2d^{*}}\cdot
d^{*}\,!},$
$|\,\lambda_{d,\nu}^{+}(\nu-1/2)\,|=\frac{(2d^{*}+1)(2d^{*}-1)\cdots 5\cdot
3}{2^{d^{*}}}=\frac{(2d^{*}+1)\,!}{2^{2d^{*}}\cdot
d^{*}\,!}=\frac{(2d^{*})\,!}{2^{2d^{*}}\cdot d^{*}\,!}\cdot(2d^{*}+1),$

with the notation $d^{*}=d-2-\nu$. Furthermore, since we have empty products
in the decomposition (6.14) , we have

(6.18)
$\lambda_{0}^{-}(1/2)=\lambda_{0}^{-}(-1/2)=\lambda_{d,d-2}^{+}(d-3/2)=\lambda_{d,d-2}^{+}(d-5/2)=1.$

The following lemma shows that the inequalities (6.17) are sharp.

######  Lemma 6.17.

Uniformly for $0\leq\nu\leq d-2$ and $d\rightarrow\infty$ one has

$1-o(1)\leq\left(\frac{|\lambda_{\nu}^{-}(\nu-1/2)|\cdot|\lambda_{d,\nu}^{+}(\nu+1/2)|}{|\lambda_{\nu}^{-}(\nu+1/2)|\cdot|\lambda_{d,\nu}^{+}(\nu-1/2)|}\right)^{-2/d}\leq
1+o(1)$

###### Proof.

Obvious consequence of the explicit formulas given above. ∎

For $0\leq\nu\leq d-2$, let

$\Lambda=\Lambda(d,\nu):=|\lambda_{\nu}^{-}(\nu-1/2)|^{-2/d}\cdot|\lambda_{d,\nu}^{+}(\nu+1/2)|^{-2/d}$

As a consequence of the explicit formulas of $\lambda_{\nu}^{-}$ and
$\lambda_{d,\nu}^{+}$, we have the equality

$\log\Lambda=-\frac{2}{d}\Bigl{\\{}\log((2\nu)\,!)+\log((2d^{*})\,!)\\\
-\log(\nu\,!)-\log(d^{*}\,!)-2d\log 2+o(d)\Bigr{\\}},$

uniformly for $1\leq\nu\leq d-3$ and $d\to\infty$. Using Stirling formula
(1.21), we deduce

$\displaystyle-\frac{d}{2}\cdot\log\Lambda$
$\displaystyle=\nu\log\nu+d^{*}\log d^{*}-d+o(d)$
$\displaystyle=\nu\log\nu+(d-\nu)\log(d-\nu)-d+o(d),$

hence

(6.19)
$\log\Lambda=-\frac{2}{d}\Bigl{(}\nu\log\nu+(d-\nu)\log(d-\nu)\Bigl{)}+2+o(1),$

uniformly for $1\leq\nu\leq d-3$ and $d\to\infty$. By a direct study of the
function $f_{d}$ defined by

$f_{d}:t\in[1,d-1]\mapsto f_{d}(t)=t\log t+(d-t)\log(d-t),$

we deduce that, for all $1\leq t\leq d-1$, the function $f_{d}$ satisfies the
inequality

$f_{d}(d/2)=d\log(d/2)\leq f_{d}(t)\leq f_{d}(1)=f_{d}(d-1)=(d-1)\log(d-1).$

Inserting this bound into (6.19), we obtain that

(6.20) $-2\log d+2-o(1)\leq\log\Lambda(d,\nu)\leq-2\log d+2\log 2+2+o(1),$

uniformly for $1\leq\nu\leq d-3$. Actually this formula also holds for
$\Lambda(d,0)$ and $\Lambda(d,d-2)$ thanks to the formulas (6.18).

Combining (6.17), (6.20) and Lemma 6.17, we proved

######  Lemma 6.18.

Uniformly for $d\to\infty$, $0\leq\nu\leq d-2$ and $d\leq p<2d$, one has

$\frac{{\rm e}^{2}-o(1)}{d^{2}}\leq\mathcal{I}_{d,p,\nu}\leq\frac{4\,{\rm
e}^{2}+o(1)}{d^{2}}\cdotp$

####  6.6.6 End of the proof of Proposition 6.11

We split the end of the proof in two parts. $\bullet$ For the lower bound, we
use positivity to write the inequality

$A_{L_{d,p}}\geq\sum_{\nu=0}^{d-2}\mathcal{I}_{\nu}\geq(d-1)\cdot\frac{{\rm
e}^{2}-o(1)}{d^{2}}\geq\frac{{\rm e}^{2}-o(1)}{d},$

as a consequence of (6.13) and Lemma 6.18.

$\bullet$ For the upper bound, we respectively apply Lemmas 6.13, 6.14, 6.15,
6.16 and 6.18 to bound each of these terms in (6.13), and we obtain

$A_{L_{d,p}}\leq\frac{5\,\mathrm{e}^{2}+2\,\mathrm{e}+o(1)}{d}\cdotp$

The proof of Proposition 6.11 is now complete. This concludes the proof of
Theorem 1.18.

### Acknowledgements

We are thankful to the referee for his valuable comments.

## References

  * [B] M. Bean, An isoperimetric inequality for the area of plane regions defined by binary forms.  Comp. Math. 92 (2), (1994), 115–131. Zbl 0816.11026 MR1283225
  * [F] N.I. Fel’dman, An effective refinement of the exponent in Liouville’s Theorem. (Russian). Izv. Akad. Nauk SSSR Ser. Mat. 35 (1971), 973–990. Zbl 0237.10018 MR0289418
  * [FK] É. Fouvry & P. Koymans, _Binary forms with the same value set,_ (in preparation).
  * [FLW] É. Fouvry, C. Levesque & M. Waldschmidt, _Representation of integers by cyclotomic binary forms_ , Acta Arith. 184 (2018), no. 1, 67–86. Zbl 1417.11028 MR3826641
  * [FW] É. Fouvry & M. Waldschmidt, _Sur la représentation des entiers par des formes cyclotomiques de grand degré_ , Bull. Soc. Math. France 148 (2020) no. 2, 253–282. Zbl 1455.11066 MR4124501
arXiv:1909.01892 [math.NT]

  * [GP] K. Győry & Z.Z. Papp, Norm form equations and explicit lower bounds for linear forms with algebraic coefficients, in: Studies in Pure Mathematics, (To the memory of Paul Turàn), 245–257, Publ.House of the Hungarian Academy of Sciences, 1983. Zbl 0518.10020 MR0820227
  * [HW] G.H. Hardy & E.M. Wright, _An introduction to the theory of numbers_. Sixth edition. Revised by D. R. Heath-Brown and J. H. Silverman. With a foreword by Andrew Wiles. Oxford University Press, Oxford, 2008. Zbl 1159.11001 MR2445243
  * [Ho] C. Hooley, _On binary cubic forms,_ J. Reine Angew. Math. 226 (1967), 30–87. Zbl 0163.04605 MR0213299
  * [SX] C. L. Stewart & S. Y. Xiao, _On the representation of integers by binary forms_ , Math. Ann. 375 (2019), no. 1-2, 133–163. Zbl 1464.11035 MR4000237