# Binary forms with the same value set II. The case of ${\bf D}_{4}$

Étienne Fouvry CNRS, Laboratoire de mathématiques d’Orsay, Université
Paris–Saclay, 91405 Orsay, France<EMAIL_ADDRESS>Université Paris–Saclay  Peter Koymans Institute for Theoretical Studies, ETH
Zurich, 8092 Zurich, Switzerland<EMAIL_ADDRESS>ETH Zurich

###### Abstract

Let $F,G\in\mathbb{Z}[X,Y]$ be binary forms of degree $\geq 3$, non-zero
discriminant and with automorphism group isomorphic to $D_{4}$. If
$F(\mathbb{Z}^{2})=G(\mathbb{Z}^{2})$, we show that $F$ and $G$ are ${\rm
GL}(2,\mathbb{Z})$–equivalent.

##  1 Introduction

This paper is the continuation of [1] and deals with the following question:

######  Question 1.1.

Let $d\geq 3$ and let $F(X,Y)$ and $G(X,Y)$ two binary forms of ${\rm
Bin}(d,\mathbb{Q})$ (the set of binary forms with degree $d$, with rational
coefficients and with discriminant different from zero) such that
$F(\mathbb{Z}^{2})=G(\mathbb{Z}^{2})$. Does there exist
$\gamma:=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in{\rm GL}(2,\mathbb{Z})$ such
that

$F(aX+bY,cX+dY)=G(X,Y)?$ (1.1)

If there exists no such matrix $\gamma$, the pair $(F,G)$ is called an
extraordinary pair and the form $F$ is called extraordinary.

In order to answer this question, the article [1] shows the crucial importance
of the group of automorphisms of $F$ defined as

${\rm Aut}(F,\mathbb{Q}):=\left\\{\gamma\in{\rm
GL}(2,\mathbb{Q}):F\circ\gamma=F\right\\},$

where $F\circ\gamma$ denotes the action of $\gamma$ by linear change of
variables (see (1.1)). It is known that, for any $F\in{\rm
Bin}(d,\mathbb{Q})$, the group ${\rm Aut}(F,\mathbb{Q})$ is ${\rm
GL}(2,\mathbb{Q})$–conjugate to one element among the set $\mathfrak{K}$ of
ten subgroups of ${\rm GL}(2,\mathbb{Z})$ defined by

$\mathfrak{K}=\\{{\bf C}_{1},{\bf C}_{2},{\bf C}_{3},{\bf C}_{4},{\bf
C}_{6},{\bf D}_{1},{\bf D}_{2},{\bf D}_{3},{\bf D}_{4},{\bf D}_{6}\\},$ (1.2)

where the letters ${\bf C}_{k}$ and ${\bf D}_{\ell}$ respectively correspond
to cyclic and dihedral subgroups with cardinality $k$ and $2\ell$. For the
definition of these subgroups, we refer the reader to [1, Lemma 5.1]. In [1],
we treated the cases of ${\bf C}_{1},{\bf C}_{2},{\bf C}_{3},{\bf C}_{4},{\bf
C}_{6},{\bf D}_{1},{\bf D}_{2}$, where we prove that an affirmative answer to
Question 1.1 depends on the existence, in the group of automorphisms, of
elements of order $3$ with special type. The cases of ${\bf D}_{3}$ and ${\bf
D}_{6}$ will be treated in [2], bringing to light a similar characterization
but in a more intricate context.

In the present paper, we are concerned with the case of ${\bf D}_{4}$, which
is the dihedral subgroup of ${\rm GL}(2,\mathbb{Z})$ generated by the two
matrices

$\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}\text{ and }\begin{pmatrix}0&1\\\
-1&0\end{pmatrix}.$

This subgroup has order $8$ and every element has order $1$, $2$ or $4$. Write

$G(X,Y)=a_{d}X^{d}+a_{d-1}X^{d-1}Y+\cdots+a_{1}XY^{d-1}+a_{0}Y^{d}.$

Let $G\in{\rm Bin}(d,\mathbb{Q})$. It is not difficult to show that ${\bf
D}_{4}\subseteq{\rm Aut}(G,\mathbb{Q})$ if and only if one has the equalities

$a_{k}=0\text{ if }k\equiv 1\bmod 2\text{ and }a_{k}=a_{d-k}\text{ for }0\leq
k\leq d.$ (1.3)

In particular, $d$ is even so $d\geq 4$. Furthermore, we have ${\rm
Aut}(G,\mathbb{Q})={\bf D}_{4}$, since, in the list $\mathfrak{K}$ (see (1.2))
there is no group with cardinality strictly divisible by $8$. Finally, every
form $F$ satisfying ${\rm Aut}(F,\mathbb{Q})=\lambda^{-1}{\bf D}_{4}\lambda$,
with $\lambda\in{\rm GL}(2,\mathbb{Q})$, has the shape

$F=G\circ\lambda,$

where $G$ satisfies (1.3). This follows from the general conjugation formula
${\rm Aut}(F\circ\lambda,\mathbb{Q})=\lambda^{-1}{\rm
Aut}(F,\mathbb{Q})\lambda$.

Our central result is

######  Theorem 1.2.

Let $d\geq 4$. Then there is no extraordinary form $F$ such that

${\rm Aut}(F,\mathbb{Q})\simeq_{{\rm GL}(2,\mathbb{Q})}{\bf D}_{4}.$

This theorem was already announced in [1, Theorem A]. One simple consequence
is

######  Corollary 1.3.

Let $F\in{\rm Bin}(4,\mathbb{Q})$ such that

$F(\mathbb{Z}^{2})=\\{m:m=t^{4}+u^{4}\textup{ for some
}(t,u)\in\mathbb{Z}^{2}\\}.$

Then there exists $(a,b,c,d)\in\mathbb{Z}^{4}$, with $ad-bc=\pm 1$, such that

$F(X,Y)=(aX+cY)^{4}+(bX+dY)^{4}.$

The corollary follows upon combing Theorem 1.2 with [3, Lemma 3.3], which
asserts that ${\rm Aut}(X^{4}+Y^{4},\mathbb{Q})={\bf D}_{4}$.

### Acknowledgements

The first author thanks Michel Waldschmidt for inspiring the thema of this
paper, for sharing his ideas and for his encouragements. The second author
gratefully acknowledges the support of Dr. Max Rössler, the Walter Haefner
Foundation and the ETH Zürich Foundation.

##  2 From extraordinary forms to coverings

###  2.1 Some definitions

By a lattice, we mean an additive subgroup of $\mathbb{Z}^{2}$ with rank $2$.
This lattice is proper when it is different from $\mathbb{Z}^{2}$. If
$\Lambda$ is a lattice generated by $\vec{u},\vec{v}\in\mathbb{Z}^{2}$, the
index of $\Lambda$ is the positive integer

$[\mathbb{Z}^{2}:\Lambda]=|\mathbb{Z}^{2}/\Lambda|=|\det(\vec{u},\vec{v})|.$
(2.1)

######  Definition 2.1.

Let $\gamma\in{\rm GL}(2,\mathbb{Q})$. By definition the lattice associated to
$\gamma$ is the subset $L(\gamma)$ of $\mathbb{Z}^{2}$ defined by

$L(\gamma):=\left\\{(x,y)\in\mathbb{Z}^{2}:\gamma\begin{pmatrix}x\\\
y\end{pmatrix}\in\mathbb{Z}^{2}\right\\}.$

We will use the obvious remarks

$L(\gamma)=L(-\gamma)$ (2.2)

and

$L(\gamma)=\mathbb{Z}^{2}\iff\gamma\text{ has integer coefficients}.$ (2.3)

We recall the following property, which is proved by a direct calculation (see
[1, Lemma 6.9]).

######  Lemma 2.2.

Let $\gamma\in{\rm GL}(2,\mathbb{Q})$. Then $[\mathbb{Z}^{2}:L(\gamma)]$ is an
integer multiple of $|\det\gamma|^{-1}$.

Several times we will use the next easy lemma, where $v_{2}(n)$ is the
$2$–adic valuation of the integer $n$ with the convention $v_{2}(0)=+\infty$.

######  Lemma 2.3.

Let $\alpha,\beta,\gamma\in\mathbb{Z}$ with $\gamma\neq 0$. Suppose that
$v_{2}(\alpha)<v_{2}(\beta)$ and $v_{2}(\alpha)<v_{2}(\gamma)$. Then the
lattice defined by the equation

$\alpha x_{1}+\beta x_{2}\equiv 0\bmod\gamma$

is included in the lattice

$\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod 2\\}.$

######  Definition 2.4.

Let $F_{1}$ and $F_{2}$ be two forms in ${\rm Bin}(d,\mathbb{Q})$. By
definition, an isomorphism from $F_{1}$ to $F_{2}$ is an element $\phi\in{\rm
GL}(2,\mathbb{Q})$ such that $F_{1}\circ\phi=F_{2}$. The set of all such
isomorphisms is denoted by ${\rm Isom}(F_{1}\rightarrow F_{2},\mathbb{Q})$.
Suppose ${\rm Isom}(F_{1}\rightarrow F_{2},\mathbb{Q})$ is not empty and let
$\rho$ be one of its elements. Then we have the equalities

$\displaystyle{\rm Isom}(F_{1}\rightarrow F_{2},\mathbb{Q})$
$\displaystyle=\rho\cdot{\rm Aut}(F_{2},\mathbb{Q})={\rm
Aut}(F_{1},\mathbb{Q})\cdot\rho$ $\displaystyle{\rm Isom}(F_{2}\rightarrow
F_{1},\mathbb{Q})$ $\displaystyle=\rho^{-1}\cdot{\rm
Aut}(F_{1},\mathbb{Q})={\rm Aut}(F_{2},\mathbb{Q})\cdot\rho^{-1}.$

We extract from [1] the following key proposition

######  Proposition 2.5.

Let $d\geq 3$ and let $(F_{1},F_{2})$ be a pair of extraordinary forms. Then
there exists $\rho\in{\rm GL}(2,\mathbb{Q})$, a pair of extraordinary forms
$(G_{1},G_{2})$ and a pair $(D,\nu)$ of positive integers such that

  1. 1.

we have $F_{1}=F_{2}\circ\rho$,

  2. 2.

we have

$\Bigl{(}G_{1}\sim_{{\rm GL}(2,\mathbb{Z})}F_{1}\textup{ and }G_{2}\sim_{{\rm
GL}(2,\mathbb{Z})}F_{2}\Bigr{)}\textup{ or }\Bigl{(}G_{1}\sim_{{\rm
GL}(2,\mathbb{Z})}F_{2}\textup{ and }G_{2}\sim_{{\rm
GL}(2,\mathbb{Z})}F_{1}\Bigr{)},$

  3. 3.

we have

$D,\nu\geq 1,D\nu>1,D\mid\nu\textup{ and }1\leq\nu\leq D^{2}.$ (2.4)

The matrix

$\gamma:=\begin{pmatrix}D&0\\\ 0&D/\nu\end{pmatrix}$ (2.5)

satisfies $G_{1}=G_{2}\circ\gamma$,

  4. 4.

we have

$[\mathbb{Z}^{2}:L(\gamma)]=\min\left\\{[\mathbb{Z}^{2}:L(\tau)]:\tau\in{\rm
Isom}(G_{1}\rightarrow G_{2},\mathbb{Q})\cup{\rm Isom}(G_{2}\rightarrow
G_{1},\mathbb{Q})\right\\},$ (2.6)

  5. 5.

and finally, we have the two coverings

$\mathbb{Z}^{2}=\bigcup_{\tau\in{\rm Isom}(G_{1}\rightarrow
G_{2},\mathbb{Q})}L(\tau)=\bigcup_{\tau\in{\rm Isom}(G_{2}\rightarrow
G_{1},\mathbb{Q})}L(\tau).$

This proposition essentially gathers the following contents of [1]: Lemma 2.4,
§7.1, §7.2 (particularly the relations (7.7), (7.8), (7.9)) and Proposition
9.1.

######  Comments 2.6.

  1. 1.

This proposition is quite general, since it requires no assumption concerning
the automorphism groups of $F_{1}$ or $F_{2}$.

  2. 2.

Item 1. will not be used in the sequel of the proof of Theorem 1.2. It recalls
the starting point of the construction of $\gamma$ and it implies that both
${\rm Aut}(F_{1},\mathbb{Q})$ and ${\rm Aut}(F_{2},\mathbb{Q})$ are ${\rm
GL}(2,\mathbb{Q})$–conjugate by the formula (4.1).

  3. 3.

The pair $(G_{1},G_{2})$ is extraordinary with automorphism groups ${\rm
GL}(2,\mathbb{Z})$–conjugate with the automorphism groups of $F_{1}$ and
$F_{2}$.

  4. 4.

The initial problem is symmetrical in $(F_{1},F_{2})$. We eventually break
this symmetry in item 4. to ensure the minimality of the index in (2.6). Note
that $[\mathbb{Z}^{2}:L(\gamma)]=\nu D^{-1}$.

  5. 5.

Since $\gamma$ belongs to ${\rm Isom}(G_{2}\rightarrow G_{1},\mathbb{Q})$, we
can explicitly write the isomorphisms appearing in item 5. in terms of
$\gamma$ and ${\rm Aut}(G_{1},\mathbb{Q})$ or ${\rm Aut}(G_{2},\mathbb{Q})$
(see Definition 2.4).

  6. 6.

In fact, we will only use the first equality written in item 5. The second one
will be used in [2].

##  3 About coverings

###  3.1 General notions

Let $a$, $b$, $c$ and $d$ be four integers. To shorten notations, we write

$\mathbb{Z}\begin{pmatrix}a\\\ b\end{pmatrix}+\mathbb{Z}\begin{pmatrix}c\\\
d\end{pmatrix}=:\begin{bmatrix}a&c\\\ b&d\end{bmatrix}$

for the subgroup of $\mathbb{Z}^{2}$ generated by $(a,b)^{T}$ and $(c,d)^{T}$.
Recall that if $ad-bc\neq 0$, this subgroup is a lattice. Furthermore, if
$|ad-bc|\geq 2$, this is a proper lattice.

######  Definition 3.1.

Let $k\geq 1$ be an integer and let $(\Lambda_{i})_{1\leq i\leq k}$ be $k$
lattices. We say that

$\mathcal{C}=\\{\Lambda_{1},\ldots,\Lambda_{k}\\}$

is a covering of $\mathbb{Z}^{2}$ (or a covering) if and only if

$\bigcup_{1\leq i\leq k}\Lambda_{i}=\mathbb{Z}^{2}.$

######  Definition 3.2 (Minimal covering).

Let $k\geq 1$, let $\Lambda_{i}$ be lattices and let
$\mathcal{C}=\\{\Lambda_{1},\ldots,\Lambda_{k}\\}$ be a covering. We say that
$\mathcal{C}$ is a minimal covering of length $k$ if and only if replacing any
$\Lambda_{i}$ by some proper sublattice
$\Lambda^{\prime}_{i}\varsubsetneq\Lambda_{i}$, the set

$\\{\Lambda_{1},\ldots,\Lambda_{i-1},\Lambda^{\prime}_{i},\Lambda_{i+1},\ldots,\Lambda_{k}\\}$

is not a covering.

If $\mathcal{C}$ is a minimal covering and if $1\leq i\neq j\leq k$, we never
have $\Lambda_{i}\subseteq\Lambda_{j}$. In particular, for $k\geq 2$, every
$\Lambda_{i}$ is a proper lattice. The following lemma asserts that from any
covering one can extract a minimal covering

######  Lemma 3.3.

Let $k\geq 1$ and let

$\mathcal{C}:=\\{\Lambda_{1},\ldots,\Lambda_{k}\\}$

be a covering. Then there exists an integer $1\leq k^{\prime}\leq k$, an
injection $\phi:\\{1,\ldots,k^{\prime}\\}\rightarrow\\{1,\ldots,k\\}$, and
lattices $\Lambda^{\prime}_{j}$ ($1\leq j\leq k^{\prime}$), such that

$\mathcal{C}^{\prime}:=\\{\Lambda^{\prime}_{1},\ldots,\Lambda^{\prime}_{k^{\prime}}\\}$

is a minimal covering, and for all $1\leq j\leq k^{\prime}$ one has
$\Lambda^{\prime}_{j}\subseteq\Lambda_{\phi(j)}$.

###### Proof.

By reordering and suppressing some $\Lambda_{i}$ if necessary, we suppose that

$\bigcup_{1\leq i\leq s}\Lambda_{i}=\mathbb{Z}^{2}\quad\text{ and
}\quad\bigcup_{\begin{subarray}{c}1\leq i\leq s\\\ i\neq
j\end{subarray}}\Lambda_{i}\neq\mathbb{Z}^{2}\quad\text{ for all }1\leq j\leq
s$ (3.1)

for some $1\leq s\leq k$. The case where $s=1$ is trivial, so we suppose
$s\geq 2$. Let

$\mathcal{L}_{s}:=\\{M_{s}\text{ lattice}:M_{s}\subseteq\Lambda_{s},\
\Lambda_{1}\cup\Lambda_{2}\cup\dots\cup\Lambda_{s-1}\cup
M_{s}=\mathbb{Z}^{2}\\}.$

We then have the equality

$\mathbb{Z}^{2}=\Lambda_{1}\cup\dots\cup\Lambda_{s-1}\cup\Lambda^{\prime}_{s},$
(3.2)

with $\Lambda^{\prime}_{s}:=\bigcap_{M_{s}\in\mathcal{L}_{s}}M_{s}$. The set
$\Lambda^{\prime}_{s}$ is a subgroup of $\mathbb{Z}^{2}$. Its rank can not be
$0$ or $1$, because we would have the equality $\cup_{1\leq i\leq
s-1}\Lambda_{i}=\mathbb{Z}^{2}$, which contradicts (3.1). So
$\Lambda^{\prime}_{s}$ is a lattice and, by construction, it is the smallest
lattice included in $\Lambda_{s}$ and satisfying (3.2). We continue the same
process for the covering

$\\{\Lambda_{1},\Lambda_{2},\ldots,\Lambda_{s-1},\Lambda^{\prime}_{s}\\}$

to replace $\Lambda_{s-1}$ by a smaller lattice $\Lambda^{\prime}_{s-1}$. By
iterating this process, we prove the existence of lattices
$(\Lambda^{\prime}_{1},\dots,\Lambda^{\prime}_{s})$ with the following three
properties

  1. 1.

(inclusion) $\Lambda^{\prime}_{i}\subseteq\Lambda_{i}$ for $1\leq i\leq s$,

  2. 2.

(covering) $\bigcup_{1\leq i\leq s}\Lambda^{\prime}_{i}=\mathbb{Z}^{2}$,

  3. 3.

(partial minimality) let $1\leq k\leq s$ and let $M_{k}$ be a lattice such
that $M_{k}\subseteq\Lambda^{\prime}_{k}$ and such that

$\Lambda_{1}\cup\dots\cup\Lambda_{k-1}\cup
M_{k}\cup\Lambda^{\prime}_{k+1}\cup\dots\cup\Lambda^{\prime}_{s}=\mathbb{Z}^{2},$
(3.3)

then $M_{k}=\Lambda^{\prime}_{k}$.

We now prove that $(\Lambda^{\prime}_{1},\ldots,\Lambda^{\prime}_{s})$ is a
minimal covering associated with $\\{\Lambda_{1},\ldots,\Lambda_{s}\\}$. So we
consider lattices $M_{i}$ such that $M_{i}\subseteq\Lambda^{\prime}_{i}$ and
such that

$\mathbb{Z}^{2}=M_{1}\cup\cdots\cup M_{s}.$ (3.4)

Our aim is to show that $M_{i}=\Lambda^{\prime}_{i}$. The equality (3.4)
implies

$\mathbb{Z}^{2}=\Lambda_{1}\cup\cdots\cup\Lambda_{s-1}\cup M_{s}.$

Therefore we deduce from the partial minimality property (see (3.3)) that
$M_{s}=\Lambda^{\prime}_{s}$. We now have the equality

$\mathbb{Z}^{2}=M_{1}\cup M_{2}\cup\cdots\cup
M_{s-1}\cup\Lambda^{\prime}_{s},$

which implies

$\mathbb{Z}^{2}=\Lambda_{1}\cup\Lambda_{2}\cup\cdots\cup\Lambda_{s-2}\cup
M_{s-1}\cup\Lambda^{\prime}_{s}.$

From the partial minimality property (see (3.3)) we obtain the equality
$M_{s-1}=\Lambda^{\prime}_{s-1}$. The end of the proof is by induction. ∎

###  3.2 Minimal coverings with length at most $4$

We now list all the minimal coverings with length bounded by $4$.

######  Theorem 3.4.

The following is a complete list (up to permutation) of the minimal coverings
of $\mathbb{Z}^{2}$ of length at most four

$\displaystyle\left\\{\begin{bmatrix}1&0\\\ 0&1\end{bmatrix}\right\\}$ (3.5)
$\displaystyle\left\\{\begin{bmatrix}1&0\\\
0&2\end{bmatrix},\begin{bmatrix}2&0\\\ 0&1\end{bmatrix},\begin{bmatrix}1&0\\\
1&2\end{bmatrix}\right\\}$ (3.6) $\displaystyle\left\\{\begin{bmatrix}1&0\\\
0&2\end{bmatrix},\begin{bmatrix}4&0\\\ 0&1\end{bmatrix},\begin{bmatrix}1&0\\\
1&2\end{bmatrix},\begin{bmatrix}2&0\\\ 1&2\end{bmatrix}\right\\}$ (3.7)
$\displaystyle\left\\{\begin{bmatrix}1&0\\\
0&4\end{bmatrix},\begin{bmatrix}2&0\\\ 0&1\end{bmatrix},\begin{bmatrix}1&0\\\
1&2\end{bmatrix},\begin{bmatrix}1&0\\\ 2&4\end{bmatrix}\right\\}$ (3.8)
$\displaystyle\left\\{\begin{bmatrix}1&0\\\
0&2\end{bmatrix},\begin{bmatrix}2&0\\\ 0&1\end{bmatrix},\begin{bmatrix}1&0\\\
1&4\end{bmatrix},\begin{bmatrix}1&0\\\ 3&4\end{bmatrix}\right\\}$ (3.9)
$\displaystyle\left\\{\begin{bmatrix}1&0\\\
0&3\end{bmatrix},\begin{bmatrix}3&0\\\ 0&1\end{bmatrix},\begin{bmatrix}1&0\\\
1&3\end{bmatrix},\begin{bmatrix}1&0\\\ 2&3\end{bmatrix}\right\\}.$ (3.10)

###### Proof.

By [1, Lemma 6.6 & 6.7], there is no minimal covering with length two and only
one with length three. So we are left with finding all the minimal coverings
with length four.

$\bullet$ Checking that the sets of lattices (3.7), (3.8), (3.9) and (3.10)
are coverings. One way to prove this is to give explicit equations of these
lattices. Let us focus on (3.9) since the other cases are similar. The
equations of the four corresponding lattices are respectively

$y\equiv 0\bmod 2,\,x\equiv 0\bmod 2,\,3x+y\equiv 0\bmod 4,\,x+y\equiv 0\bmod
4.$

It remains to check that any pair $(a,b)$ of congruence classes modulo $4$
satisfies at least one of the four equations above.

$\bullet$ Construction of the minimal coverings. Let $L_{1},\dots,L_{4}$ be
such that

$L_{1}\cup L_{2}\cup L_{3}\cup L_{4}=\mathbb{Z}^{2}.$

Ruling out the trivial covering (3.5), we may assume that $L_{1}$, $L_{2}$,
$L_{3}$ and $L_{4}$ are all proper subgroups of $\mathbb{Z}^{2}$. The argument
will proceed by repeatedly looking at points outside of $L_{1}\cup L_{2}\cup
L_{3}\cup L_{4}$ and then analyzing in which possible $L_{i}$ such a point can
be. For convenience, these points will be chosen with small coordinates.

By permuting the $L_{i}$ if necessary, we may assume without loss of
generality that

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1}.$

Since $L_{1}\neq\mathbb{Z}^{2}$, we see that $\begin{pmatrix}0\\\
1\end{pmatrix}\not\in L_{1}$. By permuting the $L_{i}$ again if necessary, we
may assume without loss of generality that

$\begin{pmatrix}0\\\ 1\end{pmatrix}\in L_{2}.$

Finally, a similar argument shows that we may assume without loss of
generality that

$\begin{pmatrix}1\\\ 1\end{pmatrix}\in L_{3}.$

Summarizing, we have so far

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad\begin{pmatrix}0\\\
1\end{pmatrix}\in L_{2},\quad\begin{pmatrix}1\\\ 1\end{pmatrix}\in L_{3}.$

We will now consider the vector $(1,-1)^{T}$, and the proof naturally splits
into two cases (case 1 and case 2). Note that we must have $(1,-1)^{T}\in
L_{3}$ (case 1) or $(1,-1)^{T}\in L_{4}$ (case 2).

### Case 1

Let us assume that $(1,-1)^{T}\in L_{3}$. Then we have

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad\begin{pmatrix}0\\\
1\end{pmatrix}\in L_{2},\quad L_{3}=\begin{bmatrix}1&1\\\ 1&-1\end{bmatrix}.$

Note that equality must indeed hold for $L_{3}$, since the subgroup

$\begin{bmatrix}1&1\\\ 1&-1\end{bmatrix}$

has prime index in $\mathbb{Z}^{2}$ and $L_{3}\neq\mathbb{Z}^{2}$ by
assumption. For the rest of the proof, we will often use the above observation
implicitly. Looking at the vector $(2,1)^{T}$, we get two further cases,
namely $(2,1)^{T}\in L_{2}$ (case 1.1) or $(2,1)^{T}\in L_{4}$ (case 1.2),
since the cases $(2,1)^{T}\in L_{1}$ or $(2,1)^{T}\in L_{3}$ are impossible
(because this forces $L_{1}=\mathbb{Z}^{2}$ respectively
$L_{3}=\mathbb{Z}^{2}$).

### Case 1.1

In this case we have $(2,1)^{T}\in L_{2}$ and therefore

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad L_{2}=\begin{bmatrix}0&2\\\
1&1\end{bmatrix},\quad L_{3}=\begin{bmatrix}1&1\\\ 1&-1\end{bmatrix}.$

Considering the vector $(1,2)^{T}$ yields two subcases: $(1,2)^{T}\in L_{1}$
(case 1.1.1) or $(1,2)^{T}\in L_{4}$ (case 1.1.2).

### Case 1.1.1

We have

$L_{1}=\begin{bmatrix}1&1\\\ 0&2\end{bmatrix},\quad
L_{2}=\begin{bmatrix}0&2\\\ 1&1\end{bmatrix},\quad L_{3}=\begin{bmatrix}1&1\\\
1&-1\end{bmatrix},$

which corresponds to the covering (3.6).

### Case 1.1.2

Currently, we know that

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad L_{2}=\begin{bmatrix}0&2\\\
1&1\end{bmatrix},\quad L_{3}=\begin{bmatrix}1&1\\\
1&-1\end{bmatrix},\quad\begin{pmatrix}1\\\ 2\end{pmatrix}\in L_{4}.$

We analyze the possibilities for the vector $(1,-2)^{T}$. If we add this
vector to $L_{1}$, then the resulting covering is not minimal, as $L_{4}$ may
be removed to obtain the covering (3.6). Therefore we arrive at the covering

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad L_{2}=\begin{bmatrix}0&2\\\
1&1\end{bmatrix},\quad L_{3}=\begin{bmatrix}1&1\\\
1&-1\end{bmatrix},\quad\begin{bmatrix}1&1\\\ 2&-2\end{bmatrix}\subseteq
L_{4}.$

To finish the argument for this case, we consider the vector $(1,4)^{T}$. If
we add it to $L_{4}$, then the corresponding covering is not minimal, as
$L_{1}$ may be removed to get the covering (3.6). Thus we obtain

$L_{1}=\begin{bmatrix}1&1\\\ 0&4\end{bmatrix},\quad
L_{2}=\begin{bmatrix}0&2\\\ 1&1\end{bmatrix},\quad L_{3}=\begin{bmatrix}1&1\\\
1&-1\end{bmatrix},\quad L_{4}=\begin{bmatrix}1&1\\\ 2&-2\end{bmatrix},$

which is the covering (3.8).

### Case 1.2

At this point we have

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad\begin{pmatrix}0\\\
1\end{pmatrix}\in L_{2},\quad L_{3}=\begin{bmatrix}1&1\\\
1&-1\end{bmatrix},\quad\begin{pmatrix}2\\\ 1\end{pmatrix}\in L_{4}.$

We have that $(-2,1)^{T}\in L_{2}$ (case 1.2.1) or $(-2,1)^{T}\in L_{4}$ (case
1.2.2).

### Case 1.2.1

The following information is available to us

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad
L_{2}=\begin{bmatrix}0&-2\\\ 1&1\end{bmatrix}\quad L_{3}=\begin{bmatrix}1&1\\\
1&-1\end{bmatrix},\quad\begin{pmatrix}2\\\ 1\end{pmatrix}\in L_{4}.$

Since the vectors $(2,1)^{T},(1,-2)^{T},(1,2)^{T}$ together generate
$\mathbb{Z}^{2}$, at least one of $(1,-2)^{T}$ or $(1,2)^{T}$ must be in
$L_{1}$. Thus we get the covering (3.6) after removing $L_{4}$.

### Case 1.2.2

We obtain that

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad\begin{pmatrix}0\\\
1\end{pmatrix}\in L_{2},\quad L_{3}=\begin{bmatrix}1&1\\\
1&-1\end{bmatrix},\quad\begin{bmatrix}2&-2\\\ 1&1\end{bmatrix}\subseteq
L_{4}.$

We must have that $(1,-2)^{T}\in L_{1}$. We will now consider the vector
$(4,1)^{T}$. We either have $(4,1)^{T}\in L_{2}$, in which case we have the
covering (3.7), or $(4,1)\in L_{4}$, in which case we get the covering (3.6)
after removing $L_{2}$.

### Case 2

We will now assume that $(1,-1)^{T}\in L_{4}$. Then we are in the situation

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad\begin{pmatrix}0\\\
1\end{pmatrix}\in L_{2},\quad\begin{pmatrix}1\\\ 1\end{pmatrix}\in
L_{3},\quad\begin{pmatrix}1\\\ -1\end{pmatrix}\in L_{4}.$

Inspecting the possibilities for the point $(2,1)^{T}$, we come to the
conclusion that $(2,1)^{T}\in L_{2}$ (case 2.1) or $(2,1)^{T}\in L_{4}$ (case
2.2).

### Case 2.1

We have arrived at the following configuration

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad L_{2}=\begin{bmatrix}0&2\\\
1&1\end{bmatrix},\quad\begin{pmatrix}1\\\ 1\end{pmatrix}\in
L_{3},\quad\begin{pmatrix}1\\\ -1\end{pmatrix}\in L_{4}.$

Considering the vector $(1,2)^{T}$, we will split into the cases $(1,2)^{T}\in
L_{1}$ (case 2.1.1) or $(1,2)^{T}\in L_{4}$ (case 2.1.2).

### Case 2.1.1

Gathering the information so far, we have

$L_{1}=\begin{bmatrix}1&1\\\ 0&2\end{bmatrix},\quad
L_{2}=\begin{bmatrix}0&2\\\ 1&1\end{bmatrix},\quad\begin{pmatrix}1\\\
1\end{pmatrix}\in L_{3},\quad\begin{pmatrix}1\\\ -1\end{pmatrix}\in L_{4}.$

We inspect the different locations for the vector $(3,1)^{T}$. If
$(3,1)^{T}\in L_{3}$, we obtain the cover (3.6) after dropping the lattice
$L_{4}$. Suppose instead that $(3,1)^{T}\in L_{4}$. Then we have

$L_{1}=\begin{bmatrix}1&1\\\ 0&2\end{bmatrix},\quad
L_{2}=\begin{bmatrix}0&2\\\ 1&1\end{bmatrix},\quad\begin{pmatrix}1\\\
1\end{pmatrix}\in L_{3},\quad\begin{bmatrix}1&3\\\ -1&1\end{bmatrix}\subseteq
L_{4}.$

At last we consider the vector $(3,-1)^{T}$. If $(3,-1)^{T}\in L_{3}$, we get

$L_{1}=\begin{bmatrix}1&1\\\ 0&2\end{bmatrix},\quad
L_{2}=\begin{bmatrix}0&2\\\ 1&1\end{bmatrix},\quad L_{3}=\begin{bmatrix}1&3\\\
1&-1\end{bmatrix},\quad L_{4}=\begin{bmatrix}1&3\\\ -1&1\end{bmatrix},$

which corresponds to the covering (3.9). If instead $(3,-1)^{T}\in L_{4}$, we
get the covering (3.6) upon removing $L_{3}$.

### Case 2.1.2

At this stage we may write

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad L_{2}=\begin{bmatrix}0&2\\\
1&1\end{bmatrix},\quad\begin{pmatrix}1\\\ 1\end{pmatrix}\in
L_{3},\quad\begin{bmatrix}1&1\\\ -1&2\end{bmatrix}\in L_{4}.$

Looking at $(3,1)^{T}$, we obtain $(3,1)^{T}\in L_{3}$. Then looking at
$(3,2)^{T}$, we conclude that $(3,2)^{T}\in L_{1}$. Once we discard $L_{4}$,
we get the covering (3.6).

### Case 2.2

We know that

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad\begin{pmatrix}0\\\
1\end{pmatrix}\in L_{2},\quad\begin{pmatrix}1\\\ 1\end{pmatrix}\in L_{3},\quad
L_{4}=\begin{bmatrix}1&2\\\ -1&1\end{bmatrix}.$

Considering the vector $(3,1)^{T}$, the argument splits in two cases, namely
$(3,1)^{T}\in L_{2}$ (case 2.2.1) and $(3,1)^{T}\in L_{3}$ (case 2.2.2).

### Case 2.2.1

We have arrived at

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad L_{2}=\begin{bmatrix}0&3\\\
1&1\end{bmatrix},\quad\begin{pmatrix}1\\\ 1\end{pmatrix}\in L_{3},\quad
L_{4}=\begin{bmatrix}1&2\\\ -1&1\end{bmatrix}.$

This forces $(4,1)^{T}\in L_{3}$ and then $(1,3)^{T}\in L_{1}$, thus giving

$L_{1}=\begin{bmatrix}1&1\\\ 0&3\end{bmatrix},\quad
L_{2}=\begin{bmatrix}0&3\\\ 1&1\end{bmatrix},\quad L_{3}=\begin{bmatrix}1&4\\\
1&1\end{bmatrix},\quad L_{4}=\begin{bmatrix}1&2\\\ -1&1\end{bmatrix}.$

This is precisely the covering (3.10).

### Case 2.2.2

Finally, we have to consider the configuration

$\begin{pmatrix}1\\\ 0\end{pmatrix}\in L_{1},\quad\begin{pmatrix}0\\\
1\end{pmatrix}\in L_{2},\quad L_{3}=\begin{bmatrix}1&3\\\
1&1\end{bmatrix},\quad L_{4}=\begin{bmatrix}1&2\\\ -1&1\end{bmatrix}.$

Considering the vectors $(2,3)^{T}$ and $(3,2)^{T}$ simultaneously, this gives
two cases

$L_{1}=\begin{bmatrix}1&3\\\ 0&2\end{bmatrix},\quad
L_{2}=\begin{bmatrix}0&2\\\ 1&3\end{bmatrix},\quad L_{3}=\begin{bmatrix}1&3\\\
1&1\end{bmatrix},\quad L_{4}=\begin{bmatrix}1&2\\\ -1&1\end{bmatrix}$

and

$L_{1}=\begin{bmatrix}1&2\\\ 0&3\end{bmatrix},\quad
L_{2}=\begin{bmatrix}0&3\\\ 1&2\end{bmatrix},\quad L_{3}=\begin{bmatrix}1&3\\\
1&1\end{bmatrix},\quad L_{4}=\begin{bmatrix}1&2\\\ -1&1\end{bmatrix}.$

In the first case, the lattice $L_{4}$ is redundant, and we get the covering
(3.6) upon discarding $L_{4}$. The last case does not correspond to any non-
trivial minimal covering. Indeed, all the $L_{i}$ have prime index, but their
union does not contain the point $(1,4)^{T}$. ∎

Theoretically speaking, the method employed above leads to the complete list
of minimal coverings with length $\leq k$, where $k$ is a given integer. The
case $k\leq 6$ is vital for our work [2], which handles the case where the
automorphism group is conjugate to ${\bf D}_{3}$ or ${\bf D}_{6}$. However,
when $k=5$ and $k=6$, the number of cases is huge and the method is no longer
feasible to execute by hand. This is the reason why we have written algorithms
to produce the list of 9 minimal coverings with length equal to 5, and the
list of 49 minimal coverings with length equal to 6. To make this paper
independent of computer calculations, we have decided to prove Theorem 3.4 by
hand, but the results of Theorem 3.4 are rapidly reproduced by our algorithms
that will be published in [2].

##  4 Proof of Theorem 1.2

###  4.1 Preparation of the covering

The proof of this theorem is accomplished by contradiction by exploiting
Proposition 2.5. We start from a pair of extraordinary forms $(F_{1},F_{2})$
such that ${\rm Aut}(F_{1},\mathbb{Q})\simeq_{{\rm GL}(2,\mathbb{Q})}{\bf
D}_{4}$. By Comment 2 of 2.6 we also have ${\rm
Aut}(F_{2},\mathbb{Q})\simeq_{{\rm GL}(2,\mathbb{Q})}{\bf D}_{4}$. By item 2.
of Proposition 2.5, we also have

${\rm Aut}(G_{1},\mathbb{Q}),\,{\rm Aut}(G_{2},\mathbb{Q})\simeq_{{\rm
GL}(2,\mathbb{Q})}{\bf D}_{4}$

thanks to the conjugation formula

${\rm Aut}(F\circ\lambda,\mathbb{Q})=\lambda^{-1}{\rm
Aut}(F,\mathbb{Q})\,\lambda,\text{ for all }\lambda\in{\rm
GL}(2,\mathbb{Q})\text{ and for all }F\in{\rm Bin}(d,\mathbb{Q}).$ (4.1)

Recall that

$G_{1}\circ\gamma^{-1}=G_{2},$

where $\gamma$ is defined in (2.5). We write ${\bf D}_{4}$ explicitly as

${\bf D}_{4}=\\{{\rm id},A_{1},A_{2},A_{3},-{\rm id},-A_{1},-A_{2},-A_{3}\\},$

with

$A_{1}=\begin{pmatrix}0&1\\\ -1&0\end{pmatrix},A_{2}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\text{ and }A_{3}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$

So we have the equality

${\rm Aut}(G_{2},\mathbb{Q})=T_{2}^{-1}{\bf D}_{4}T_{2},$

where $T_{2}$ is some matrix of ${\rm GL}(2,\mathbb{Q})$ that we write as

$T_{2}=\begin{pmatrix}t_{1}&t_{2}\\\ t_{3}&t_{4}\end{pmatrix},$

where the $t_{i}$ are coprime integers. An important integer is

$d_{2}:=|\det T_{2}|=|t_{1}t_{4}-t_{2}t_{3}|.$ (4.2)

Thus we split ${\rm Aut}(G_{2},\mathbb{Q})$ into two disjoint sets

${\rm Aut}(G_{2},\mathbb{Q})=\mathfrak{S}\cup(-\mathfrak{S}),$

with

$\mathfrak{S}:=\\{{\rm
id},T_{2}^{-1}A_{1}T_{2},T_{2}^{-1}A_{2}T_{2},T_{2}^{-1}A_{3}T_{2}\\}.$

Furthermore, we have ${\rm Isom}(G_{1}\rightarrow
G_{2},\mathbb{Q})=\gamma^{-1}{\rm Aut}(G_{2},\mathbb{Q})$, by Definition 2.4
and we put

$\Lambda(\sigma):=\\{{\bf x}\in\mathbb{Z}^{2}:\gamma^{-1}\sigma({\bf
x})\in\mathbb{Z}^{2}\\}=L(\gamma^{-1}\sigma),$ (4.3)

see Definition 2.1. The equality $\Lambda(-\sigma)=\Lambda(\sigma)$ follows
from (2.2). Combining these remarks with the first equality of item 5. of
Proposition 2.5, we have the covering of $\mathbb{Z}^{2}$

$\mathbb{Z}^{2}=\bigcup_{\sigma\in\mathfrak{S}}\Lambda(\sigma)$ (4.4)

by at most four lattices. We will require the explicit equations defining the
$\Lambda(\sigma)$. By a direct computation we have

######  Lemma 4.1.

For $\sigma\in\mathfrak{S}$, the lattice $\Lambda(\sigma)$ is the set of
$(x_{1},x_{2})\in\mathbb{Z}^{2}$ such that

$\Lambda({\rm id}):\begin{cases}x_{1}&\equiv 0\bmod D\\\ \nu x_{2}&\equiv
0\bmod D\end{cases}$ (4.5)

$\Lambda(T_{2}^{-1}A_{1}T_{2}):\begin{cases}(t_{1}t_{2}+t_{3}t_{4})x_{1}+(t_{2}^{2}+t_{4}^{2})x_{2}&\equiv
0\bmod d_{2}D\\\
\nu((t_{1}^{2}+t_{3}^{2})x_{1}+(t_{1}t_{2}+t_{3}t_{4})x_{2})&\equiv 0\bmod
d_{2}D\end{cases}$ (4.6)

$\Lambda(T_{2}^{-1}A_{2}T_{2}):\begin{cases}(t_{3}t_{4}-t_{1}t_{2})x_{1}+(t_{4}^{2}-t_{2}^{2})x_{2}&\equiv
0\bmod d_{2}D\\\
\nu((t_{1}^{2}-t_{3}^{2})x_{1}+(t_{1}t_{2}-t_{3}t_{4})x_{2})&\equiv 0\bmod
d_{2}D\end{cases}$ (4.7)

$\Lambda(T_{2}^{-1}A_{3}T_{2}):\begin{cases}(t_{2}t_{3}+t_{1}t_{4})x_{1}+(2t_{2}t_{4})x_{2}&\equiv
0\bmod d_{2}D\\\ \nu((2t_{1}t_{3})x_{1}+(t_{2}t_{3}+t_{1}t_{4})x_{2})&\equiv
0\bmod d_{2}D.\end{cases}$ (4.8)

By (2.4), the second equation of (4.5) is redundant and thus we have the
equality

$[\mathbb{Z}^{2}:\Lambda({\rm id})]=D.$ (4.9)

Several times we will use the following important lemma.

######  Lemma 4.2.

We adopt the hypotheses of Proposition 2.5 and the notations above. Then the
covering (4.4) satisfies

$\Lambda(\sigma)\neq\mathbb{Z}^{2}\textup{ for all }\sigma\in\mathfrak{S}.$
(4.10)

###### Proof.

For the sake of contradiction, suppose that $\Lambda(\sigma)=\mathbb{Z}^{2}$
for some $\sigma\in\mathfrak{S}$. By the definition (4.3) and the remark
(2.3), the matrix associated with $\gamma^{-1}\sigma$ (with $\sigma\in{\rm
Aut}(G_{2},\mathbb{Q})$) has integer coefficients. This means that there
exists a matrix $M_{1}$ with integer coefficients such that

$G_{1}\circ M_{1}=G_{2}.$ (4.11)

We use the minimality of the index of $L(\gamma)$ (see item 4. of Proposition
2.5) to write

$1=[\mathbb{Z}^{2}:\Lambda(\sigma)]=[\mathbb{Z}^{2}:L(\gamma^{-1}\sigma)]\geq[\mathbb{Z}^{2}:L(\gamma)]\geq
1.$

Again by the remark (2.3) we deduce that the matrix $M_{2}$ associated with
$\gamma$ (see (2.5)), has integer coefficients. So we have the equality

$G_{2}\circ M_{2}=G_{1}.$

Combining with (4.11), we obtain

$G_{1}\circ(M_{1}\cdot M_{2})=G_{2}\circ M_{2}=G_{1},$

which implies that $|\det(M_{1}\cdot M_{2})|=1$, hence $|\det M_{1}|=|\det
M_{2}|=1$. So the matrices $M_{1}$ and $M_{2}$ belong to ${\rm
GL}(2,\mathbb{Z})$. Therefore the equality (4.11) shows that $G_{1}$ and
$G_{2}$ are ${\rm GL}(2,\mathbb{Z})$–equivalent, which is contrary to the
hypothesis that $(G_{1},G_{2})$ is an extraordinary pair (see Comment 3 of
2.6). ∎

We now initiate the proof of Theorem 1.2 by successively restricting the
values of $D$.

###  4.2  The integer $D$ satisfies $2\leq D\leq 4$

We want to prove the inequality

$2\leq D\leq 4.$ (4.12)

We separate our discussion into two subcases based on the value of the union
of the lattices $\Lambda(\sigma)$ with $\sigma\neq{\rm id}$.

####  4.2.1  If the last three lattices do not cover $\mathbb{Z}^{2}$

We suppose that

$\mathbb{Z}^{2}\neq\bigcup_{\sigma\in\mathfrak{S}-\\{{\rm
id}\\}}\Lambda(\sigma).$ (4.13)

This condition means that $\Lambda({\rm id})$ is essential to obtain the
covering (4.4). By Lemma 4.2 we know that none of the lattices appearing in
this covering are trivial. By the existence of a minimal covering (see Lemma
3.3), there exist four subgroups $\Lambda^{\prime}_{i}\ (0\leq i\leq 3)$ such
that

$\begin{cases}\Lambda^{\prime}_{i}\text{ is either }\\{0\\}\text{ or a proper
lattice,}\\\ \Lambda^{\prime}_{0}\subseteq\Lambda({\rm id})\text{ and
}\Lambda^{\prime}_{i}\subseteq\Lambda(T_{2}^{-1}A_{i}T_{2})\ (1\leq i\leq
3),\\\ \cup_{0\leq i\leq 3}\Lambda^{\prime}_{i}=\mathbb{Z}^{2},\\\
\mathcal{C}:=\\{\Lambda^{\prime}_{i}:\Lambda^{\prime}_{i}\neq\\{0\\}\\}\text{
is a minimal covering of }\mathbb{Z}^{2}.\end{cases}$

In particular, by (4.4) and by (4.13), we deduce that
$\Lambda^{\prime}_{0}\neq\\{0\\}.$ Thus the covering $\mathcal{C}$ contains
three or four lattices and $\mathcal{C}$ appears as one of the minimal
coverings $\eqref{ec2},\dots,\eqref{ec6}$ of Theorem 3.4. Thus the lattice
$\Lambda^{\prime}_{0}$ is necessarily one of the lattices of these five
minimal coverings. By (2.1), all these lattices have an index equal to $2$,
$3$ or $4$. In particular, we have
$[\mathbb{Z}^{2}:\Lambda^{\prime}_{0}]\in\\{2,3,4\\}$. Finally,
$[\mathbb{Z}^{2}:\Lambda({\rm id})]$ is a divisor (different from $1$) of
$[\mathbb{Z}^{2}:\Lambda^{\prime}_{0}]$. By (4.9), we deduce the inequality
$2\leq D\leq 4$, and (4.12) is proved.

####  4.2.2  If the last three lattices cover $\mathbb{Z}^{2}$

We now suppose that

$\mathbb{Z}^{2}=\bigcup_{\sigma\in\mathfrak{S}-\\{{\rm
id}\\}}\Lambda(\sigma).$ (4.14)

By (4.10), the equality (4.14) exhibits a non-trivial covering of
$\mathbb{Z}^{2}$ by three lattices. By Theorem 3.4 and by Lemma 3.3 we have
the equality

$\\{\Lambda(T_{2}^{-1}A_{1}T_{2}),\Lambda(T_{2}^{-1}A_{2}T_{2}),\Lambda(T_{2}^{-1}A_{3}T_{2})\\}=\\{\Lambda_{0},\Lambda_{1},\Lambda_{2}\\},$
(4.15)

where the covering (3.6) is written as
$\\{\Lambda_{0},\Lambda_{1},\Lambda_{2}\\}$ respectively.

If (4.14) holds, then the integer $D$ is not divisible by an odd prime $p$ For
the sake of contradiction, suppose that $D$ is divisible by some $p\geq 3$. We
will show that there exists at least one $\sigma\in\mathfrak{S}-\\{{\rm
id}\\}$ such that

$p\mid[\mathbb{Z}^{2}:\Lambda(\sigma)].$ (4.16)

Such a divisibility is impossible since, on the right–hand side of (4.15), all
the lattices have an index equal to $2$. To prove (4.16), we will argue by
contradiction. So we suppose that

$p\nmid[\mathbb{Z}^{2}:\Lambda(\sigma)]\text{ for all }\sigma\neq{\rm id}.$
(4.17)

By keeping only the first equation of the systems defining the lattices modulo
$p$, we see that the lattices $\Lambda(T_{2}^{-1}A_{1}T_{2})$,
$\Lambda(T_{2}^{-1}A_{2}T_{2})$ and $\Lambda(T_{2}^{-1}A_{3}T_{2})$ (see
(4.6), (4.7) and (4.8)) are respectively included in the following lattices
$\mathcal{L}_{p}$

$\displaystyle\mathcal{L}_{p}(T_{2}^{-1}A_{1}T_{2})$
$\displaystyle:(t_{1}t_{2}+t_{3}t_{4})x_{1}+(t_{2}^{2}+t_{4}^{2})x_{2}\equiv
0\bmod p,$ (4.18) $\displaystyle\mathcal{L}_{p}(T_{2}^{-1}A_{2}T_{2})$
$\displaystyle:(t_{3}t_{4}-t_{1}t_{2})x_{1}+(t_{4}^{2}-t_{2}^{2})x_{2}\equiv
0\bmod p,$ (4.19) $\displaystyle\mathcal{L}_{p}(T_{2}^{-1}A_{3}T_{2})$
$\displaystyle:(t_{2}t_{3}+t_{1}t_{4})x_{1}+(2t_{2}t_{4})x_{2}\equiv 0\bmod
p.$ (4.20)

The index of these three lattices is $1$ or $p$. It can not be equal to $p$,
otherwise we would have (4.16) for some $\sigma$. So the index is always equal
to $1$, which means that $p$ divides all the coefficients

$(t_{1}t_{2}+t_{3}t_{4}),\ (t_{2}^{2}+t_{4}^{2}),\ (t_{3}t_{4}-t_{1}t_{2}),\
(t_{4}^{2}-t_{2}^{2}),\ (t_{2}t_{3}+t_{1}t_{4})\text{ and }(2t_{2}t_{4}).$
(4.21)

This implies that $p$ divides $t_{2}$ and $t_{4}$. Let $g:={\rm
gcd}(t_{2},t_{4},p^{\infty})\geq
p,\tilde{t}_{2}=t_{2}/g,\tilde{t}_{4}=t_{4}/g$. We observe that $g$ divides
$d_{2}$, by its definition (4.2). We return to the original system of
equations (4.6), (4.7) and (4.8), where we keep the first equation of each
system. After division by $g$, we observe that the lattices
$\Lambda(T_{2}^{-1}A_{1}T_{2})$, $\Lambda(T_{2}^{-1}A_{2}T_{2})$ and
$\Lambda(T_{2}^{-1}A_{3}T_{2})$ are respectively included in the following
lattices

$\displaystyle\tilde{\mathcal{L}_{p}}(T_{2}^{-1}A_{1}T_{2})$
$\displaystyle:(t_{1}\tilde{t}_{2}+t_{3}\tilde{t}_{4})x_{1}\equiv 0\bmod p,$
(4.22) $\displaystyle\tilde{\mathcal{L}_{p}}(T_{2}^{-1}A_{2}T_{2})$
$\displaystyle:(t_{3}\tilde{t}_{4}-t_{1}\tilde{t}_{2})x_{1}\equiv 0\bmod p,$
(4.23) $\displaystyle\tilde{\mathcal{L}_{p}}(T_{2}^{-1}A_{3}T_{2})$
$\displaystyle:(\tilde{t}_{2}t_{3}+t_{1}\tilde{t}_{4})x_{1}\equiv 0\bmod p.$
(4.24)

We observe that ${\rm gcd}(t_{1},t_{3},p)=1$ (otherwise the integers $t_{i}$
would not be coprime altogether), and that ${\rm
gcd}(\tilde{t}_{2},\tilde{t}_{4},p)=1$.

– Suppose that $p$ does not divide $(t_{1}\tilde{t}_{2}+t_{3}\tilde{t}_{4})$,
then the lattice $\tilde{\mathcal{L}_{p}}(T_{2}^{-1}A_{1}T_{2})$ has its index
equal to $p$, and since this lattice contains $\Lambda(T_{2}^{-1}A_{1}T_{2})$,
we obtain a contradiction with (4.17).

– Same type of reasoning when $p$ does not divide
$(t_{3}\tilde{t}_{4}-t_{1}\tilde{t}_{2})$.

– Now suppose that $p$ divides $(t_{1}\tilde{t}_{2}+t_{3}\tilde{t}_{4})$ and
$(t_{3}\tilde{t}_{4}-t_{1}\tilde{t}_{2})$, then $p$ divides
$t_{1}\tilde{t}_{2}$ and $t_{3}\tilde{t}_{4}$. The above coprimality
conditions imply that $p$ does not divide the coefficient of $x_{1}$ in
(4.24), contradicting (4.17).

So $D$ has no odd prime divisor when (4.14) holds.

If (4.14) holds, then the integer $D$ is not divisible by $8$ The strategy is
the same as in §4.2.2. We suppose that $D$ is divisible by $8$. We will prove
that there is a $\sigma\in\mathfrak{S}-\\{{\rm id}\\}$ such that

$4\mid[\mathbb{Z}^{2}:\Lambda(\sigma)].$ (4.25)

Such a divisibility contradicts the equality (4.15), since all the lattices
$\Lambda_{i}$ ($1\leq i\leq 3$) have index equal to $2$. To prove (4.25), we
will argue by contradiction. So we suppose that

$4\nmid[\mathbb{Z}^{2}:\Lambda(\sigma)]\text{ for all }\sigma\neq{\rm id}.$
(4.26)

The lattices $\Lambda(T_{2}^{-1}A_{1}T_{2}),$ $\Lambda(T_{2}^{-1}A_{2}T_{2})$
and $\Lambda(T_{2}^{-1}A_{3}T_{2})$ are respectively included in the following
lattices

$\displaystyle\mathcal{L}_{8}(T_{2}^{-1}A_{1}T_{2})$
$\displaystyle:(t_{1}t_{2}+t_{3}t_{4})x_{1}+(t_{2}^{2}+t_{4}^{2})x_{2}\equiv
0\bmod 8,$ $\displaystyle\mathcal{L}_{8}(T_{2}^{-1}A_{2}T_{2})$
$\displaystyle:(t_{3}t_{4}-t_{1}t_{2})x_{1}+(t_{4}^{2}-t_{2}^{2})x_{2}\equiv
0\bmod 8,$ $\displaystyle\mathcal{L}_{8}(T_{2}^{-1}A_{3}T_{2})$
$\displaystyle:(t_{2}t_{3}+t_{1}t_{4})x_{1}+(2t_{2}t_{4})x_{2}\equiv 0\bmod
8.$

These lattices have an index equal to $1$, $2$, $4$ or $8$. It can not be
divisible by $4$ (otherwise, we would contradict (4.26)). So these indexes are
equal to $1$ or $2$, which means that $4$ divides all the coefficients listed
in (4.21). This implies that $2$ divides $t_{2}$ and $t_{4}$. Let
$g=(t_{2},t_{4},2^{\infty})\geq 2$, $\tilde{t}_{2}=t_{2}/g$,
$\tilde{t}_{4}=t_{4}/g$. We return to the initial definitions of the lattices
(4.6), (4.7) and (4.8), where we only keep the first equation. Since $g$
divides $d_{2}$, we observe that the lattices $\Lambda(T_{2}^{-1}A_{1}T_{2})$,
$\Lambda(T_{2}^{-1}A_{2}T_{2})$ and $\Lambda(T_{2}^{-1}A_{3}T_{2})$ are
respectively included in the following lattices

$\displaystyle\tilde{\mathcal{L}_{8}}(T_{2}^{-1}A_{1}T_{2})$
$\displaystyle:(t_{1}\tilde{t}_{2}+t_{3}\tilde{t}_{4})x_{1}+g(\tilde{t}_{2}^{2}+\tilde{t}_{4}^{2})x_{2}\equiv
0\bmod 8,$ (4.27) $\displaystyle\tilde{\mathcal{L}_{8}}(T_{2}^{-1}A_{2}T_{2})$
$\displaystyle:(t_{3}\tilde{t}_{4}-t_{1}\tilde{t}_{2})x_{1}+g(\tilde{t}_{4}^{2}-\tilde{t}_{2}^{2})x_{2}\equiv
0\bmod 8,$ (4.28) $\displaystyle\tilde{\mathcal{L}_{8}}(T_{2}^{-1}A_{3}T_{2})$
$\displaystyle:(\tilde{t}_{2}t_{3}+t_{1}\tilde{t}_{4})x_{1}+g(2\tilde{t}_{2}\tilde{t}_{4})x_{2}\equiv
0\bmod 8.$ (4.29)

We exploit the coprimality ${\rm gcd}(t_{1},t_{3},2)=1$ (otherwise, the
integers $t_{i}$ would not be coprime altogether) and the coprimality ${\rm
gcd}(\tilde{t}_{2},\tilde{t}_{4},2)=1$ to deduce that at least one of the
three coefficients attached to $x_{1}$ in (4.27), (4.28) and (4.29) is not
divisible by $4$. This implies that the corresponding lattice
$\tilde{\mathcal{L}_{8}}(\sigma)$ has index equal to $4$ or $8$. Hence the
associated lattice $\Lambda(\sigma)$ (contained in
$\tilde{\mathcal{L}_{8}}(\sigma)$) has index divisible by $4$. This
contradicts our assumption (4.26).

We have considered all the possible cases. The proof of (4.12) is complete.

###  4.3  The integer $D$ is different from $3$

Our task is to prove that

$D\neq 3.$ (4.30)

We will now assume that $D=3$ to derive a contradiction. We already know that
$[\mathbb{Z}^{2}:\Lambda({\rm id})]=D=3$ by (4.9) and
$\Lambda(\sigma)\neq\mathbb{Z}^{2}$ for all $\sigma\in\mathfrak{S}$ by Lemma
4.2. Reasoning as before, when we obtained (4.16), there exists
$\sigma\in\mathfrak{S}-\\{{\rm id}\\}$ such that

$3\mid[\mathbb{Z}^{2}:\Lambda(\sigma)].$

Then the minimal covering contained in the covering
$\\{\Lambda(\sigma):\sigma\in\mathfrak{S}\\}$ (see (4.4)), which exists by
Lemma 3.3, can only be the covering (3.10) by Theorem 3.4. Then the covering
$\\{\Lambda(\sigma):\sigma\in\mathfrak{S}\\}$ from (4.4) must coincide with
(3.10).

Write $g={\rm gcd}(t_{2},t_{4},3^{\infty})$, $\tilde{t}_{2}=t_{2}/g$ and
$\tilde{t}_{4}=t_{4}/g$. We now split our discussion according to the classes
modulo $3$ of the numbers $g$, $t_{2}$ and $t_{4}$.

####  4.3.1  If $3\mid g$

We follow the arguments that led to equation (4.16). In particular, consider
the three lattices $\tilde{\mathcal{L}}_{p}$ defined by (4.22), (4.23) and
(4.24) with $p=3$. At least, one of the coefficients of $x_{1}$ is non-zero
modulo $3$. Thus we get the existence of some $\sigma\in\mathfrak{S}-\\{{\rm
id}\\}$, such that

$\Lambda(\sigma)\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod 3\\}.$

But $\Lambda({\rm id})=\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod 3\\}$ (see (4.5)
and (2.4)). So the covering (4.4) can never be equal to (3.10), which is the
desired contradiction.

####  4.3.2  If $3\nmid g$, $3\nmid t_{2}$, $3\nmid t_{4}$ and $t_{2}\equiv
t_{4}\bmod 3$

Under these assumptions, we have the congruences

$t_{2}^{2}+t_{4}^{2}\equiv 2\bmod 3,\ 2t_{2}t_{4}\equiv 2\bmod 3$

and

$t_{1}t_{2}+t_{3}t_{4}\equiv t_{2}t_{3}+t_{1}t_{4}\bmod 3.$

By (4.18) and (4.20), we see that the lattices
$\mathcal{L}_{3}(T_{2}^{-1}A_{1}T_{2})$ and
$\mathcal{L}_{3}(T_{2}^{-1}A_{3}T_{2})$ coincide. Both have index 3. This
implies that the two lattices $\Lambda(T_{2}^{-1}A_{1}T_{2})$ and
$\Lambda(T_{2}^{-1}A_{3}T_{2})$ sit inside the same lattice with index $3$. We
obtain the same contradiction as in §4.3.1.

####  4.3.3  If $3\nmid g$, $3\nmid t_{2}$, $3\nmid t_{4}$ and
$t_{2}\not\equiv t_{4}\bmod 3$

In this situation, we obtain

$t_{2}^{2}+t_{4}^{2}\equiv 2\bmod 3,2t_{2}t_{4}\equiv 1\bmod 3$

and

$t_{1}t_{2}+t_{3}t_{4}\equiv(2t_{2}t_{4})(t_{1}t_{2}+t_{3}t_{4})\equiv
2(t_{1}t_{4}+t_{2}t_{3})\bmod 3.$

Up to a factor $2$ the equations (4.18) and (4.20) coincide. Once again, this
implies that two of the lattices appearing in the covering (4.4) are
sublattices of the same lattice with index $3$. We obtain the same
contradiction as above.

####  4.3.4  If $3\nmid g$ and if either $3\mid t_{2}$ or $3\mid t_{4}$

Consider the lattices $\mathcal{L}_{3}(T_{2}^{-1}A_{1}T_{2})$ and
$\mathcal{L}_{3}(T_{2}^{-1}A_{2}T_{2})$ defined by (4.18) and (4.19). Their
equations reduce to $\pm(ax_{1}+x_{2})\equiv 0\bmod 3$ (for some integer $a$).
So these two lattices coincide, and $\Lambda(T_{2}^{-1}A_{1}T_{2})$ and
$\Lambda(T_{2}^{-1}A_{2}T_{2})$ are both included in the same lattice with
index $3$. Therefore we obtain a contradiction with the covering (4.4) in this
case as well.

The proof of (4.30) is complete. Combining (4.12) and (4.30), we have reduced
our theorem to the following situation.

###  4.4  Study when $D\in\\{2,4\\}$

Our first task is to circumscribe the possible values of $\nu$. Recall the
conditions (2.4). They lead to the following possibilities for $(D,\nu)$:
$(2,2)$, $(2,4)$, $(4,4)$, $(4,8)$, $(4,12)$ and $(4,16)$. We will restrict
this list to

$(D,\nu)\in\\{(2,2),(2,4),(4,4),(4,8)\\}.$ (4.31)

To prove (4.31), we start from the covering (4.4) with the condition (4.10).
Two possibilities occur

  * •

If for some $\sigma^{\dagger}\in\mathfrak{S}$, one has
$\cup_{\sigma\in\mathfrak{S}-\\{\sigma^{\dagger}\\}}\Lambda(\sigma)=\mathbb{Z}^{2}$.
This is a covering by three lattices. By Theorem 3.4, these three lattices
have index $2$. So we have

$[\mathbb{Z}^{2}:\Lambda(\sigma_{0})]=2\text{ for some }\sigma_{0}\neq{\rm
id}.$ (4.32)

  * •

If such a $\sigma^{\dagger}$ does not exist. The covering (4.4) corresponds to
one of the coverings (3.7), (3.8), (3.9) or (3.10) of Theorem 3.4. Since
$\Lambda({\rm id})$ has index equal to $2$ or $4$, we can eliminate the
covering (3.10) (because the lattices comprising this covering all have an
index equal to $3$). Now, in the coverings (3.7), (3.8) and (3.9), there are
exactly two lattices with index $2$. So (4.32) is also true in that case.

We now exploit the relation (2.6), with the choice $\tau=\sigma_{0}$, and the
relation $[\mathbb{Z}^{2}:L(\gamma)]=\nu/D$ by Comment 4 of 2.6 to obtain

$\nu D^{-1}\leq 2,$

which leads to the divisibility

$\nu\mid 2D,$

since $\nu D^{-1}$ is an integer. This gives the condition (4.31).

We can further restrict the set of possible values for the pair $(D,\nu)$ as
follows. By (4.3), we have $\Lambda(\sigma_{0})=L(\gamma^{-1}\sigma_{0})$,
where $\sigma_{0}$ satisfies (4.32). But
$|\det(\gamma^{-1}\sigma_{0})|=\frac{\nu}{D^{2}}$. By Lemma 2.2, we deduce
that $2$ is a multiple of $D^{2}/\nu$. This gives the inequality $D^{2}\leq
2\nu$. So we know that the set defined in (4.31) is restricted to

$(D,\nu)\in\\{(2,2),(2,4),(4,8)\\}.$ (4.33)

###  4.5  Possible values for $d_{2}$

We now investigate the possible values for $d_{2}$.

####  4.5.1  The integer $d_{2}$ has no odd prime divisor

We will prove that

$d_{2}\in\\{1,2,4,8,16,\dots\\}.$ (4.34)

Suppose that $d_{2}$ is divisible by some odd prime $p$. We recall that
$\nu\in\\{2,4,8\\}$ (see (4.33)) and we return to the definitions (4.6), (4.7)
and (4.8) to deduce that the lattices $\Lambda(T_{2}^{-1}A_{1}T_{2})$,
$\Lambda(T_{2}^{-1}A_{2}T_{2})$ and $\Lambda(T_{2}^{-1}A_{3}T_{2})$ are
respectively included in the following lattices

$\displaystyle\mathcal{M}_{1,p}$
$\displaystyle:\begin{cases}(t_{1}t_{2}+t_{3}t_{4})x_{1}+(t_{2}^{2}+t_{4}^{2})x_{2}\equiv
0\bmod p,\\\ (t_{1}^{2}+t_{3}^{2})x_{1}+(t_{1}t_{2}+t_{3}t_{4})x_{2}\equiv
0\bmod p,\end{cases}$ $\displaystyle\mathcal{M}_{2,p}$
$\displaystyle:\begin{cases}(t_{3}t_{4}-t_{1}t_{2})x_{1}+(t_{4}^{2}-t_{2}^{2})x_{2}\equiv
0\bmod p,\\\ (t_{1}^{2}-t_{3}^{2})x_{1}+(t_{1}t_{2}-t_{3}t_{4})x_{2}\equiv
0\bmod p,\end{cases}$ $\displaystyle\mathcal{M}_{3,p}$
$\displaystyle:\begin{cases}(t_{2}t_{3}+t_{1}t_{4})x_{1}+(2t_{2}t_{4})x_{2}\equiv
0\bmod p,\\\ (2t_{1}t_{3})x_{1}+(t_{2}t_{3}+t_{1}t_{4})x_{2}\equiv 0\bmod
p.\end{cases}$

We use a combinatorial lemma dealing with the antidiagonal coefficients in the
above system of equations.

######  Lemma 4.3.

Let $p$ be an odd prime. Let $(t_{1},t_{2},t_{3},t_{4})\in\mathbb{Z}^{4}$ be
such that $p\nmid{\rm gcd}(t_{1},t_{2},t_{3},t_{4})$. Consider the following
three $2$–sets of quadratic forms

$S_{1}:=\\{t_{2}^{2}+t_{4}^{2},t_{1}^{2}+t_{3}^{2}\\},\quad
S_{2}:=\\{t_{4}^{2}-t_{2}^{2},t_{1}^{2}-t_{3}^{2}\\},\quad
S_{3}:=\\{2t_{2}t_{4},2t_{1}t_{3}\\}.$

Then for all pairs $(i,j)$ with $1\leq i<j\leq 3$, there exists $P\in
S_{i}\cup S_{j}$ such that $p\nmid P$.

###### Proof.

Omitted. ∎

Lemma 4.3 implies that there exist $i$ and $j$ such that $\mathcal{M}_{i,p}$
and $\mathcal{M}_{j,p}$ have index divisible by $p$. Hence the indices of the
corresponding lattices
$\Lambda(T_{2}^{-1}A_{i}T_{2})\subseteq\mathcal{M}_{i,p}$ and
$\Lambda(T_{2}^{-1}A_{j}T_{2})\subseteq\mathcal{M}_{j,p}$ are divisible by
$p$. This is incompatible with the covering (4.4) and Theorem 3.4, since the
index of $\Lambda({\rm id})$ is $2$ or $4$. The proof of (4.34) is complete.

####  4.5.2  The case $d_{2}=1$ is impossible

Recall that $D\in\\{2,4\\}$. By keeping only the first equation in the systems
(4.5), (4.6), (4.7) and (4.8) and by replacing $d_{2}D$ by $2$, we deduce that
the four lattices $\Lambda(\sigma)$ ($\sigma\in\mathfrak{S})$ are respectively
included in the four lattices defined by the equations

$\displaystyle\mathcal{L}_{2}({\rm id})$ $\displaystyle:x_{1}\equiv 0\bmod 2,$
$\displaystyle\mathcal{L}_{2}(T_{2}^{-1}A_{1}T_{2})$
$\displaystyle:(t_{1}t_{2}+t_{3}t_{4})x_{1}+(t_{2}^{2}+t_{4}^{2})x_{2}\equiv
0\bmod 2,$ $\displaystyle\mathcal{L}_{2}(T_{2}^{-1}A_{2}T_{2})$
$\displaystyle:(t_{3}t_{4}-t_{1}t_{2})x_{1}+(t_{4}^{2}-t_{2}^{2})x_{2}\equiv
0\bmod 2,$ $\displaystyle\mathcal{L}_{2}(T_{2}^{-1}A_{3}T_{2})$
$\displaystyle:(t_{2}t_{3}+t_{1}t_{4})x_{1}\equiv 0\bmod 2.$

The assumption $d_{2}=1$ implies $t_{1}t_{4}+t_{2}t_{3}\equiv 1\bmod 2$, from
which we deduce the equality

$\mathcal{L}_{2}({\rm id})=\mathcal{L}_{2}(T_{2}^{-1}A_{3}T_{2}).$

Considerations of parities also lead to the equality

$\mathcal{L}_{2}(T_{2}^{-1}A_{1}T_{2})=\mathcal{L}_{2}(T_{2}^{-1}A_{2}T_{2}).$

Again playing with the parities of the $t_{i}$ and using $1=d_{2}\equiv
t_{1}t_{4}+t_{2}t_{3}\bmod 2$, we see that we never have
$t_{1}t_{2}+t_{3}t_{4}\equiv t_{2}^{2}+t_{4}^{2}\equiv 0\bmod 2$, which means
that $\mathcal{L}_{2}(T_{2}^{-1}A_{1}T_{2})\neq\mathbb{Z}^{2}$. These
considerations show that the covering (4.4) leads to the non-trivial covering

$\mathbb{Z}^{2}=\mathcal{L}_{2}({\rm
id})\cup\mathcal{L}_{2}(T_{2}^{-1}A_{1}T_{2}),$

which is in contradiction with Theorem 3.4.

####  4.5.3  The case $d_{2}=2$ is impossible

The strategy is the same as in the section above, but more intricate since we
will distinguish cases based on the parity of $t_{2}$ and $t_{4}$. We suppose
that $d_{2}=2$ and we will arrive at a contradiction. Since $4\mid d_{2}D$,
the four lattices $\Lambda(\sigma)$ ($\sigma\in\mathfrak{S})$ are respectively
included in the four lattices defined by the equations

$\displaystyle\mathcal{L}_{2}({\rm id})$ $\displaystyle:x_{1}\equiv 0\bmod 2,$
$\displaystyle\mathcal{L}_{4}(T_{2}^{-1}A_{1}T_{2})$
$\displaystyle:(t_{1}t_{2}+t_{3}t_{4})x_{1}+(t_{2}^{2}+t_{4}^{2})x_{2}\equiv
0\bmod 4,$ $\displaystyle\mathcal{L}_{4}(T_{2}^{-1}A_{2}T_{2})$
$\displaystyle:(t_{3}t_{4}-t_{1}t_{2})x_{1}+(t_{4}^{2}-t_{2}^{2})x_{2}\equiv
0\bmod 4,$ $\displaystyle\mathcal{L}_{4}(T_{2}^{-1}A_{3}T_{2})$
$\displaystyle:(t_{2}t_{3}+t_{1}t_{4})x_{1}+(2t_{2}t_{4})x_{2}\equiv 0\bmod
4.$

Recall the equality $|t_{1}t_{4}-t_{2}t_{3}|=d_{2}=2$.

If $t_{2}$ and $t_{4}$ have different parities We then have the equalities

${\rm gcd}(t_{2}^{2}+t_{4}^{2},4)={\rm gcd}(t_{4}^{2}-t_{2}^{2},4)=1,$

which implies

$[\mathbb{Z}^{2}:\mathcal{L}_{4}(T_{2}^{-1}A_{1}T_{2})]=[\mathbb{Z}^{2}:\mathcal{L}_{4}(T_{2}^{-1}A_{2}T_{2})]=4$

by a direct study of the equations defining these lattices. Furthermore, the
coefficient of $x_{1}$ in the equation defining
$\mathcal{L}_{4}(T_{2}^{-1}A_{3}T_{2})$ satisfies one of the following two
relations

$t_{2}t_{3}+t_{1}t_{4}=\begin{cases}d_{2}+2t_{2}t_{3}\equiv 2\bmod 4\\\
d_{2}+2t_{1}t_{4}\equiv 2\bmod 4.\end{cases}$ (4.35)

This implies the equality between lattices

$\mathcal{L}_{2}({\rm id})=\mathcal{L}_{4}(T_{2}^{-1}A_{3}T_{2}).$

The covering (4.4) leads to the covering

$\mathbb{Z}^{2}=\mathcal{L}_{2}({\rm
id})\cup\mathcal{L}_{4}(T_{2}^{-1}A_{1}T_{2})\cup\mathcal{L}_{4}(T_{2}^{-1}A_{2}T_{2}),$

which is nonsense, since the index of these three lattices are $2$, $4$ and
$4$ respectively. Therefore this covering does not correspond to a minimal
covering in Theorem 3.4.

If $t_{2}$ and $t_{4}$ are both even Under this assumption, we have the
following similarities between the coefficients of the lattices
$\mathcal{L}_{4}(T_{2}^{-1}A_{1}T_{2})$ and
$\mathcal{L}_{4}(T_{2}^{-1}A_{2}T_{2})$ defined in §4.5.3:

$t_{2}^{2}+t_{4}^{2}\equiv t_{4}^{2}-t_{2}^{2}\equiv 0\bmod 4,$

and

$t_{1}t_{2}+t_{3}t_{4}\equiv t_{3}t_{4}-t_{1}t_{2}\equiv 0\bmod 2.$

We now discuss on the class $t_{1}t_{2}+t_{3}t_{4}\equiv
t_{3}t_{4}-t_{1}t_{2}\bmod 4.$

#####  4.5.3.0.1 If $t_{2}$ and $t_{4}$ are both even and if
$t_{1}t_{2}+t_{3}t_{4}\equiv t_{3}t_{4}-t_{1}t_{2}\equiv 2\bmod 4$.

We return to the definitions of the lattices to deduce the equalities between
lattices

$\mathcal{L}_{2}({\rm
id})=\mathcal{L}_{4}(T_{2}^{-1}A_{1}T_{2})=\mathcal{L}_{4}(T_{2}^{-1}A_{2}T_{2}),$

which certainly can not lead to a covering.

#####  4.5.3.0.2 If $t_{2}$ and $t_{4}$ are both even and if
$t_{1}t_{2}+t_{3}t_{4}\equiv t_{3}t_{4}-t_{1}t_{2}\equiv 0\bmod 4$.

In (4.35), we have already seen that

$t_{2}t_{3}+t_{1}t_{4}\equiv 2\bmod 4.$ (4.36)

We split our discussion according to the value of $D$ (see (4.33)).

$\diamond$ Case 1: $D=4$. In that case, we have $\nu=8$ and $d_{2}D=8$, so the
second equations of (4.6), (4.7) and (4.8) are automatically satisfied. We
observe that

$\Lambda({\rm id})=\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod 4\\}.$

By (4.36), we have the inclusion

$\Lambda(T_{2}^{-1}A_{3}T_{2})\subseteq\Lambda({\rm id}).$

The covering (4.4) is simplified to the covering

$\mathbb{Z}^{2}=\Lambda({\rm
id})\cup\Lambda(T_{2}^{-1}A_{1}T_{2})\cup\Lambda(T_{2}^{-1}A_{2}T_{2}),$

where the first lattice has index $4$ and where the last two lattices have an
index $\geq 2$. This covering with three lattices does not resonate with
Theorem 3.4.

$\diamond$ Case 2: $D=2$. We then have

$d_{2}=D=2,\quad\nu\in\\{2,4\\},\quad t_{2}\equiv t_{4}\equiv 0\bmod
2\quad\text{ and }\quad t_{1}t_{2}+t_{3}t_{4}\equiv 0\bmod 4.$

Actually, the case $\nu=4$ can never happen. Indeed, by the formula given in
Lemma 4.1, with the values $d_{2}=D=2$, $\nu=4$ and the constraints of the
congruence modulo $4$ of the $t_{i}$, we see that
$\Lambda(T_{2}^{-1}A_{1}T_{2})=\mathbb{Z}^{2}$. This is forbidden by Lemma
4.2.

So we restrict to $\nu=2$. The equations of the lattices $\Lambda(\sigma)$
($\sigma\in\mathfrak{S}$) given in Lemma 4.1 are equivalent to a single
equation

$\begin{cases}\Lambda({\rm id})&:x_{1}\equiv 0\bmod 2,\\\
\Lambda(T_{2}^{-1}A_{1}T_{2})&:(t_{1}^{2}+t_{3}^{2})x_{1}\equiv 0\bmod 2,\\\
\Lambda(T_{2}^{-1}A_{2}T_{2})&:(t_{1}^{2}-t_{3}^{2})x_{1}\equiv 0\bmod 2,\\\
\Lambda(T_{2}^{-1}A_{3}T_{2})&:x_{1}\equiv 0\bmod 2.\end{cases}$

Since $\Lambda(\sigma)\neq\mathbb{Z}^{2}$ by Lemma 4.2, this implies that the
coefficients of $x_{1}$ in the second and the third equations are odd. We
deduce that these four equations show the equality

$\Lambda(\sigma)=\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod 2\\}\text{ for all
}\sigma\in\mathfrak{S}.$

This contradicts the covering (4.4).

If $t_{2}$ and $t_{4}$ are both odd The equality $t_{1}t_{4}-t_{2}t_{3}=\pm 2$
implies that $t_{1}$ and $t_{3}$ have the same parity. We now split the
argument according to this parity.

#####  4.5.3.0.3 If $t_{2}$ and $t_{4}$ are both odd and if $t_{1}\equiv
t_{3}\equiv 0\bmod 2$.

Since $d_{2}=2$, these conditions imply that $(t_{1},t_{3})\equiv(0,2)$ or
$(2,0)$ modulo $4$. Therefore we have

$t_{1}t_{2}+t_{3}t_{4}\equiv t_{3}t_{4}-t_{1}t_{2}\equiv
t_{2}t_{3}+t_{1}t_{4}\equiv 2\bmod 4.$

We implement these congruences into the first equations of (4.6), (4.7) and
(4.8) to deduce the inclusions

$\Lambda(T_{2}^{-1}A_{1}T_{2}),\,\Lambda(T_{2}^{-1}A_{3}T_{2})\subseteq\\{(x_{1},x_{2}):x_{1}+x_{2}\equiv
0\bmod 2\\},$

and

$\Lambda(T_{2}^{-1}A_{2}T_{2})\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod
2\\}.$

Recalling the inclusion

$\Lambda({\rm id})\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod 2\\},$ (4.37)

and returning to the covering (4.4), we obtain a covering of $\mathbb{Z}^{2}$
by two lattices with index $2$. This contradicts Theorem 3.4.

#####  4.5.3.0.4 If $t_{2}$ and $t_{4}$ are both odd and if $t_{1}\equiv
t_{3}\equiv 1\bmod 2$.

Since $d_{2}=2$, the following congruences hold

$t_{1}t_{2}+t_{3}t_{4}\equiv t_{2}t_{3}+t_{1}t_{4}\equiv 0\bmod 4\quad\text{
and }\quad t_{3}t_{4}-t_{1}t_{2}\equiv 2\bmod 4.$

We insert these congruences and the condition $4\mid d_{2}D$ into the first
equations of the systems (4.6), (4.7) and (4.8) to obtain the inclusions

$\Lambda(T_{2}^{-1}A_{1}T_{2}),\Lambda(T_{2}^{-1}A_{3}T_{2})\subseteq\\{(x_{1},x_{2}):x_{2}\equiv
0\bmod 2\\},$

and

$\Lambda(T_{2}^{-1}A_{2}T_{2})\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod
2\\}.$

These inclusions and the inclusion (4.37) contradict (4.4), since we would
once more obtain a covering of $\mathbb{Z}^{2}$ by two lattices with index
$2$.

Gathering all these cases, we proved that $d_{2}\neq 2$.

####  4.5.4  The case $4\mid d_{2}$ is impossible

We will suppose that $4\mid d_{2}$ to arrive at a contradiction. This implies
that $8\mid d_{2}D$ thanks to (4.33). We divide our proof according to the
parity of $t_{2}$ and $t_{4}$.

If $t_{2}$ and $t_{4}$ have different parities In that case, we have
$\gcd(t_{2}^{2}\pm t_{4}^{2},8)=1$ and the first equations of (4.6) and (4.7)
give the divisibility

$8\mid[\mathbb{Z}^{2}:\Lambda(T_{2}^{-1}A_{i}T_{2})]\quad\text{ for
}i\in\\{1,2\\}.$

The covering (4.4) can not hold: we would obtain a covering with four proper
lattices, with at least two with index divisible by $8$. This does not exist
by Theorem 3.4.

If $t_{2}$ and $t_{4}$ have the same parity Our first step is to show that we
necessarily have

$t_{1}\equiv t_{3}\bmod 2.$ (4.38)

Suppose that this does not hold. We always have

$t_{1}^{2}+t_{3}^{2}\equiv t_{1}^{2}-t_{3}^{2}\bmod 2.$

We argue as follows:

$\bullet$ If $t_{1}\not\equiv t_{3}\bmod 2$ and $t_{2}\equiv t_{4}\equiv
0\bmod 2$. By (4.33), we know that $2\mid d_{2}D/\nu$. By the first equation
of (4.5) and the second equation of (4.6) and (4.7), we deduce the inclusions

$\Lambda(\sigma)\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod 2\\}\text{ for
}\sigma\in\mathfrak{S}-\\{T_{2}^{-1}A_{3}T_{2}\\}.$

By (4.4), we would obtain a covering by two proper lattices, and this
contradicts Theorem 3.4. $\bullet$ If $t_{1}\not\equiv t_{3}\bmod 2$ and
$t_{2}\equiv t_{4}\equiv 1\bmod 2$. We only study the first equations of
(4.6), (4.7) and (4.8). We use the congruences

$t_{1}t_{2}+t_{3}t_{4}\equiv t_{3}t_{4}-t_{1}t_{2}\equiv
t_{2}t_{3}+t_{1}t_{4}\equiv 1\bmod 2\quad\text{ and }\quad 8\mid d_{2}D$

to deduce that

$8\mid[\mathbb{Z}^{2}:\Lambda(T_{2}^{-1}A_{i}T_{2})]\quad\text{ for }1\leq
i\leq 3.$

The covering (4.4) then would be incompatible with Theorem 3.4. So (4.38) is
proved.

To summarize, we necessarily have the congruence conditions

$t_{2}\equiv t_{4}\bmod 2\text{ and }t_{1}\equiv t_{3}\bmod 2.$

Recall that the $t_{i}$ are coprime altogether, so modulo $2$, the quadruplet
$(t_{1},t_{2},t_{3},t_{4})$ belongs to the following set $\Omega$ with three
elements

$\Omega:=\\{(1,0,1,0),(0,1,0,1),(1,1,1,1)\\}.$

These possibilities will be the basis of our discussion below.

#####  4.5.4.0.1 If $(t_{1},t_{2},t_{3},t_{4})\equiv(0,1,0,1)$ or
$(1,1,1,1)\bmod 2$.

We then have the congruences $t_{2}^{2}+t_{4}^{2}\equiv 2\bmod 8$ and
$2t_{2}t_{4}\equiv\pm 2\bmod 8$. By considering the first equations in the
systems (4.6) and (4.8), we obtain the inequality

$v_{2}([\mathbb{Z}^{2}:\Lambda(T_{2}^{-1}A_{i}T_{2})])\geq 2\text{ for
}i\in\\{1,3\\}.$

Furthermore, for this to be an equality, we must have $d_{2}D=8$, which means
$d_{2}=4$ and $D=2$.

When $(D,d_{2})\neq(2,4)$ (which is equivalent to $16\mid d_{2}D$), the
lattices $\Lambda(T_{2}^{-1}A_{i}T_{2})$ for $i\in\\{1,3\\}$ have index
divisible by $8$. The covering (4.4) is therefore impossible thanks to Theorem
3.4.

When $(D,d_{2})=(2,4)$, the congruences

$\pm d_{2}=t_{1}t_{4}-t_{2}t_{3}\equiv 4\bmod 8,$

and $t_{2}\equiv t_{4}\equiv 1\bmod 2$ imply

$4\equiv t_{1}t_{4}-t_{2}t_{3}\equiv t_{2}t_{4}(t_{1}t_{4}-t_{2}t_{3})\equiv
t_{1}t_{2}-t_{3}t_{4}\bmod 8.$

Since we also have $t_{4}^{2}-t_{2}^{2}\equiv 0\bmod 8$, we conclude that

$\Lambda(T_{2}^{-1}A_{2}T_{2})\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod
2\\}$

by (4.7) and by Lemma 2.3. Since $\Lambda({\rm id})$ satisfies the same
inclusion (see (4.37)) we obtain a contradiction with (4.4), since we would
obtain a covering of $\mathbb{Z}^{2}$ by three lattices with index $2$, $4$
and $4$.

#####  4.5.4.0.2 If $(t_{1},t_{2},t_{3},t_{4})\equiv(1,0,1,0)\bmod 2$.

This case is more delicate. We handle this case by splitting the proof in
three cases.

$\diamond$ Case 1: $v_{2}(t_{2})\neq v_{2}(t_{4})\ (\geq 1)$. We have the
sequence of relations

$2\leq
v_{2}(d_{2})=\min(v_{2}(t_{2}),v_{2}(t_{4}))=v_{2}(t_{1}t_{2}+t_{3}t_{4})=v_{2}(t_{3}t_{4}-t_{1}t_{2})=v_{2}(t_{2}t_{3}+t_{1}t_{4}),$

and the inequalities (recall that $D\in\\{2,\,4\\}$)

$2\min(v_{2}(t_{2}),v_{2}(t_{4}))=v_{2}(t_{2}^{2}+t_{4}^{2})=v_{2}(t_{4}^{2}-t_{2}^{2})\geq
v_{2}(d_{2}D)\text{ and }v_{2}(2t_{2}t_{4})\geq v_{2}(d_{2}D).$

With these remarks, we consider the first equations of the systems (4.5),
(4.6), (4.7) and (4.8) to deduce the inclusions

$\Lambda(\sigma)\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod 2\\}\text{ for
all }\sigma\in\mathfrak{S}.$

These relations obviously contradict the covering (4.4).

Now we suppose that $v_{2}(t_{2})=v_{2}(t_{4})\ (\geq 1)$. Since, by
hypothesis, the integers $t_{1}$ and $t_{3}$ are both odd, we have
$v_{2}(d_{2})>v_{2}(t_{2})=v_{2}(t_{4})$. So we split our forthcoming
discussion according to the difference between $v_{2}(d_{2})$ and
$v_{2}(t_{2})=v_{2}(t_{4})$.

$\diamond$ Case 2: $v_{2}(t_{2})=v_{2}(t_{4})\geq 1$ and
$v_{2}(d_{2})=v_{2}(t_{2})+1$. Under these hypotheses, we have

$v_{2}(t_{1}t_{2}+t_{3}t_{4})>v_{2}(d_{2})\text{ and
}v_{2}(t_{2}^{2}+t_{4}^{2})>v_{2}(d_{2}).$ (4.39)

For the first equality of (4.39), we use that

$t_{1}t_{2}+t_{3}t_{4}\equiv t_{1}t_{4}+t_{2}t_{3}\equiv\pm
d_{2}+2t_{2}t_{3}\equiv 0\bmod 2d_{2}.$

For the second equation, we use the equality
$v_{2}(t_{2}^{2}+t_{4}^{2})=1+2v_{2}(t_{2})$. By (4.33), we have three
possibilities

$(d_{2},D,\nu)=(4,2,4),\quad v_{2}(d_{2}D/\nu)>1\quad\text{ or }\quad D=4.$
(4.40)

$\bullet$ The first possibility is impossible because, with the above values
of $d_{2}$, $D$ and $\nu$ and with the conditions on the $2$-adic valuations
of the $t_{i}$, the two equations defining $\Lambda(T_{2}^{-1}A_{1}T_{2})$
(see (4.6)) are congruences modulo $8$, and all the coefficients of $x_{1}$
and $x_{2}$ are $\equiv 0\bmod 8$. So we would have
$\Lambda(T_{2}^{-1}A_{1}T_{2})=\mathbb{Z}^{2}$, which is contrary to Lemma
4.2.

$\bullet$ The second possibility of (4.40) can not hold for the following
reason: consider the second equations of the systems (4.6) and (4.8). We
benefit from the congruences

$t_{1}^{2}+t_{3}^{2}\equiv 2t_{1}t_{3}\equiv 2\bmod 4$

to deduce the two inclusions

$\Lambda(T_{2}^{-1}A_{i}T_{2})\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod
2\\}\quad\text{ for }i\in\\{1,3\\}.$

Since the same inclusion holds for $\Lambda({\rm id})$ and since
$\Lambda(T_{2}^{-1}A_{2}T_{2})\neq\mathbb{Z}^{2}$, the covering (4.4) is
impossible by Theorem 3.4.

$\bullet$ We now prove that the third possibility of (4.40) can not hold.
Indeed, suppose that $D=4$ and, as usual, let
$g:=\gcd(t_{2},t_{4}),\tilde{t}_{2}:=t_{2}/g$ and $\tilde{t}_{4}:=t_{4}/g$.
Since $g\mid d_{2}$, $g$ is necessarily a power of $2$ and we have
$v_{2}(g)=v_{2}(t_{2})=v_{2}(t_{4})=v_{2}(d_{2})-1\ (\geq 1)$ and the integers
$\tilde{t}_{2}$ and $\tilde{t}_{4}$ are odd. We then have

$2\equiv
t_{1}\tilde{t}_{4}-\tilde{t}_{2}t_{3}\equiv(t_{1}t_{3})(t_{1}\tilde{t}_{4}-\tilde{t}_{2}t_{3})\equiv
t_{3}\tilde{t}_{4}-t_{1}\tilde{t}_{2}\bmod 4,$

and therefore

$v_{2}(t_{3}t_{4}-t_{1}t_{2})=v_{2}(d_{2}).$ (4.41)

Since $v_{2}(t_{2})\geq 1$, it follows that

$t_{4}^{2}-t_{2}^{2}\equiv 0\bmod d_{2}D.$ (4.42)

To prove (4.42), write $t_{2}=2^{t}a_{2}$, $t_{4}=2^{t}a_{4}$ with odd $a_{2}$
and $a_{4}$ and $t\geq 1$. Then
$t_{4}^{2}-t_{2}^{2}=2^{2t}(a_{4}^{2}-a_{2}^{2})\equiv 0\bmod 2^{2t+2}$.
Furthermore, $d_{2}D=2^{t+1}\cdot 4=2^{t+3}$. We obtain the desired congruence
(4.42), since $2t+2\geq t+3$ for $t\geq 1$.

By the first equations of (4.5) and (4.7), by (4.41) and (4.42) and by the
hypothesis $D=4$, we obtain the inclusions

$\Lambda({\rm id}),\
\Lambda(T_{2}^{-1}A_{2}T_{2})\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod
4\\}.$

These inclusions are not compatible with the covering (4.4), since we would
obtain a covering of $\mathbb{Z}^{2}$ by three lattices containing one lattice
with index $4$ (see Theorem 3.4). Hence the case $D=4$ does not happen.

$\diamond$ Case 3: $v_{2}(t_{2})=v_{2}(t_{4})\geq 1$ and
$v_{2}(d_{2})>v_{2}(t_{2})+1$. We will first prove the two equalities

$v_{2}(t_{1}t_{2}+t_{3}t_{4})=v_{2}(t_{2})+1\quad\text{ and }\quad
v_{2}(t_{2}t_{3}+t_{1}t_{4})=v_{2}(t_{2})+1.$ (4.43)

$\bullet$ To prove the first equality, we write

$v_{2}(t_{1}t_{2}+t_{3}t_{4})=v_{2}(t_{1}t_{2}t_{3}+t_{3}^{2}t_{4})=v_{2}(t_{1}(t_{1}t_{4}\pm
d_{2})+t_{3}^{2}t_{4})=v_{2}((t_{1}^{2}+t_{3}^{2})t_{4}\pm t_{1}d_{2}).$

We now observe that

$v_{2}(\pm t_{1}d_{2})=v_{2}(d_{2})$

and

$v_{2}((t_{1}^{2}+t_{3}^{2})t_{4})=v_{2}(t_{1}^{2}+t_{3}^{2})+v_{2}(t_{4})=v_{2}(t_{2})+1,$

because $t_{1}$ and $t_{3}$ are both odd. Therefore we conclude that

$v_{2}(t_{1}t_{2}+t_{3}t_{4})=v_{2}((t_{1}^{2}+t_{3}^{2})t_{4}\pm
t_{1}d_{2})=v_{2}(t_{2})+1,$

since $v_{2}(t_{2})+1<v_{2}(d_{2})$ by assumption.

$\bullet$ To prove the second equality of (4.43), we write

$v_{2}(t_{2}t_{3}+t_{1}t_{4})=v_{2}(2t_{2}t_{3}+t_{1}t_{4}-t_{2}t_{3})=v_{2}(2t_{2}t_{3}\pm
d_{2})=v_{2}(2t_{2}t_{3})=v_{2}(2t_{2})=v_{2}(t_{2})+1.$

The proof of (4.43) is complete.

We now use (4.43) to deduce from the first equations of (4.6) and (4.8) the
two inclusions

$\Lambda(T_{2}^{-1}A_{i}T_{2})\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod
2\\}\quad\text{ for }i\in\\{1,3\\}.$ (4.44)

$\bullet$ Proof of (4.44) for $i=1$. The coefficient of $x_{1}$ in the first
equation of (4.6) satisfies
$v_{2}(t_{1}t_{2}+t_{3}t_{4})=v_{2}(t_{2})+1<v_{2}(d_{2})<v_{2}(d_{2}D)$.
Furthermore, the coefficient of $x_{2}$ satisfies
$v_{2}(t_{2}^{2}+t_{4}^{2})=1+2v_{2}(t_{2})>v_{2}(t_{2})+1=v_{2}(t_{1}t_{2}+t_{3}t_{4})$.
Lemma 2.3 gives the inclusion (4.44) when $i=1$.

$\bullet$ Proof of (4.44) for $i=3$. The coefficient of $x_{1}$ in the first
equation of (4.8) satisfies
$v_{2}(t_{2}t_{3}+t_{1}t_{4})=v_{2}(t_{2})+1<v_{2}(d_{2})<v_{2}(d_{2}D)$.
Furthermore, the coefficient of $x_{2}$ satisfies
$v_{2}(2t_{2}t_{4})=2v_{2}(t_{2})+1>v_{2}(t_{2})+1=v_{2}(t_{2}t_{3}+t_{1}t_{4})$.
Lemma 2.3 gives the inclusion (4.44) when $i=3$.

The proof of (4.44) is complete. Combining the inclusions (4.44) with the
inclusion $\Lambda({\rm id})\subseteq\\{(x_{1},x_{2}):x_{1}\equiv 0\bmod 2\\}$
and with the covering (4.4), we arrive at a covering of $\mathbb{Z}^{2}$ by
two proper lattices, which does not exist by Theorem 3.4.

We investigated all the cases to assert that $4\nmid d_{2}$. Gathering the
properties of $d_{2}$ proved in §4.5.1, §4.5.2, §4.5.3 and §4.5.4, we see that
$d_{2}$ does not exist.

The proof of Theorem 1.2 is complete.

## References

  * [1] É. Fouvry and P. Koymans, _Binary forms with the same value set I._ Preprint.
  * [2] É. Fouvry and P. Koymans, _Binary forms with the same value set III. The case of ${\bf D}_{3}$ and ${\bf D}_{6}$._ Preprint.
  * [3] C. L. Stewart and S. Y. Xiao, _On the representation of integers by binary forms._ Math. Ann. 375 (2019), no. 1–2, 133–163.