# On the two-distance embedding in real Euclidean space of coherent
configuration of type (2,2;3)

Eiichi Bannai, Etsuko Bannai, Chin-Yen Lee, Ziqing Xiang, Wei-Hsuan Yu

###### Abstract.

Finding the maximum cardinality of a $2$-distance set in Euclidean space is a
classical problem in geometry. Lisoněk in 1997 constructed a maximum
$2$-distance set in $\mathbb{R}^{8}$ with $45$ points. That $2$-distance set
constructed by Lisoněk has a distinguished structure of a coherent
configuration of type $(2,2;3)$ and is embedded in two concentric spheres in
$\mathbb{R}^{8}$. In this paper we study whether there exists any other
similar embedding of a coherent configuration of type $(2,2;3)$ as a
$2$-distance set in $\mathbb{R}^{n}$, without assuming any restriction on the
size of the set. We prove that there exists no such example other than that of
Lisoněk. The key ideas of our proof are as follows: (i) study the geometry of
the embedding of the coherent configuration in Euclidean spaces and to drive
diophantine equations coming from this embedding. (ii) solve diophantine
equations with certain additional conditions of integrality of some parameters
of the combinatorial structure by using the method of auxiliary equations.

##  1\. Introduction

Let $X,Y$ be finite subsets of $\mathbb{R}^{d}$, we define
$A(X,Y)=\\{\|x-y\|:x\in X,y\in Y,\text{ and }x\neq y\\}$. A finite set
$X\subset\mathbb{R}^{d}$ is called an _$s$ -distance set_ if $|A(X,X)|=s$. By
assigning a set of linearly independent polynomials to an $s$-distance set,
Bannai, Bannai and Stanton [1] and Blokhuis [3] gave an upper bound for the
cardinality of an $s$-distance sets $X$ in $\mathbb{R}^{d}$,

$\displaystyle|X|\leq\binom{d+s}{s}.$

Szöllősi and Östergård [11] had the latest progress for the construction of
$s$-distance sets for $s\leq 6,d\leq 8$. They gave an algorithm to exhaust the
$s$-distance sets in the space of small dimension and small $s$ at least $3$.
Lisoněk [9] classified all two-distance sets for the dimension $d=4,5,6$ and
$7$. For dimension $8$, he gave an example which attains the maximum
cardinality $\binom{8+2}{2}=45$. However the classification problem is still
open. The example of Lisoněk, given below, is the only known $s$-distance set
that attain the Bannai-Bannai-Stanton and Blokhuis bound $\binom{d+s}{s}$. In
fact, these 45 points are distributed in a very symmetrical manner. It can be
divided in two parts : one part is 9 points forming a regular simplex, and the
other part is $36$ points coming from the spherical embedding of a Johnson
scheme $J(9,2)$.

######  Example 1.1.

(Lisoněk) Let $X_{1}=\\{-e_{i}+\frac{1}{3}\sum_{k=1}^{9}e_{k}:1\leq i\leq
9\\}$ and $X_{2}=\\{e_{i}+e_{j}:1\leq i<j\leq 9\\}$, where $\\{e_{i}:1\leq
i\leq 9\\}$ are the standard orthonormal basis of $\mathbb{R}^{9}$. All the
points are on the hyperplane
$H=\\{x=\sum_{i=1}^{9}x_{i}e_{i}:\sum_{i=1}^{9}x_{i}=2\\}$, and its affine
dimension is $8$. Then, $X_{1}\cup X_{2}$ is a maximum two-distance set in
$\mathbb{R}^{8}$. Notice that the distinct two distances in $X_{1}\cup X_{2}$
are $\sqrt{2}$ and $2$. The radius of $X_{1}$ and $X_{2}$ are
$\frac{2}{\sqrt{3}}$ and $\sqrt{2}$ respectively.

In fact, $X_{2}\subset\mathbb{R}^{d}$ can be interpreted as a block design
with the underlying set $X_{1}$. The union $X_{1}\cup X_{2}$ has the elegant
structure:

  1. (1)

$X_{1}$ is the regular simplex in $\mathbb{R}^{d}$;

  2. (2)

$X_{2}$ is a scaling of the spherical embedding of a strongly regular graph in
$\mathbb{R}^{d}$;

  3. (3)

distance between point and block only depends on whether the point in block or
not;

  4. (4)

$|A(X_{1},X_{1})|=1$ and $|A(X_{1},X_{2})|=|A(X_{2},X_{2})|=2$.

Condition (4) is a consequence of (1)-(3), and we list here for convenience.

A coherent configuration of type (2,2;3) is a combinatorial structure and it
consists of a point set $V$, a block set $B$ and a set of finite number of
relations $\\{R_{i}\\}$. The block design in Example 1.1 is quasi-symmetric,
and it is equivalent to the type (2,2;3) coherent configuration. We prove that
such coherent configurations can always be embedded into a Euclidean space
with the above structure.

######  Theorem 1.2.

Let $(V\cup B,\\{R_{1},\dots,R_{9}\\})$ be a coherent configuration of type
(2,2;3), where $V$ is a finite set. Then there is a map $i:V\cup
B\to\mathbb{R}^{d}$, $d=|V|-1$, such that $X_{1}=i(V)$ and $X_{2}=i(B)$
satisfying the conditions (1)-(4) in Example 1.1.

In this paper we will consider the following problem: When the embedding of
type (2,2;3) coherent configuration in Theorem 1.2 gives a two-distance set in
Euclidean space? Dropping the condition on sizes (i.e., we don’t assume
$|X_{1}\cup X_{2}|=\binom{d+2}{2}$) makes the problem very broad and
difficult. However, to our surprise, we are able to show our main theorem that
there is no other example at all.

######  Theorem 1.3.

(Main result) The example given by Lisoněk is the unique coherent
configuration of type (2,2;3) which can be embedded in Euclidean space as a
two-distance set satisfying the conditions given in Theorem 1.2.

Nozaki and Shinohara [10] study two-distance sets in $\mathbb{R}^{d}$ that
contain a regular simplex. They solve the case when Larman–Rogers–Seidel (LRS)
[8] ratio is 2. When LRS ratio is 3, they give a partial result by adding some
block design structures. So, our present paper is relevant to [10]. However,
it seems that the situation in our paper is eluded in their consideration.

The outline of our paper is as follows. In section 2, we introduce the notions
of type (2,2;3) coherent configurations and quasi-symmetric designs and we
will prove the Theorem 1.2. In section 3, we study when the embedding given by
Theorem 1.2 is a two-distance set. We derive the conditions for the embedding
of a coherent configuration forming a two-distance set into three Diophantine
equations $p_{1}(S,m,x,y)=p_{2}(S,m,x,y)=p_{2}(S,m,x,y)=0$. In section 4, we
determine the complete solutions of the system of Diophantine equations. In
section 5, we prove our main result, Theorem 1.3. Please note that we use
computers crucially. The results are rigorously proved by applying the notion
of real algebraic geometry. We can see the precedents of the basic idea of the
proof in Xiang [12], and Bannai, Bannai, Xiang, Yu and Zhu [2].

##  2\. Preliminaries

We introduce the two well-known concepts, quasi-symmetric designs and coherent
configurations of type (2,2;3). They are proved as the equivalent notions in
Higman [7]. Quasi-symmetric design is a combinatorial object whose parameters
are non-negative integers. The integral conditions between them are crucial
for classifying them.

###  2.1. Quasi-symmetric designs

######  Definition 2.1.

[4] A _$t$ -design_ with parameters $(m,S,\Lambda)$ (or a _$t$
-$(m,S,\Lambda)$ design_) is a collection of subsets $B$ (called _blocks_) of
a set $V$ of $m$ points such that

  1. (1)

every member of $B$ contains $S$ points;

  2. (2)

any set of $t$ points is contained in exactly $\Lambda$ members of $B$.

A $2$-design is called a _quasi-symmetric design_ if the cardinality of the
intersection of two different blocks takes just two distinct values. These two
numbers are called the _intersection numbers_. We express these intersection
numbers by $\alpha$ and $\beta$, with $\beta<\alpha$. By definition each
element $v\in V$ contains in $T$ distinct blocks. For each point $v\in V$ and
each block $b\in B$, the following condition is satisfied:

$\displaystyle|\\{w\in B:|b\cap w|=\alpha\text{ and }v\in
w\\}|=\begin{cases}N,&\text{ if }v\in b,\\\ P,&\text{ if }v\not\in
b.\end{cases}$

We denote this number by $N$ when $v\in b$ and by $P$ when $v\not\in b$. The
cardinality of quasi-symmetric design is bounded above by ${m\choose 2}$ (see
Proposition 3.4 in Cameron and van Lint [4]).

######  Theorem 2.2.

[4, Proposition 3.6] For a $2$-$(m,S,\Lambda)$ design $\mathcal{D}$ with
$4\leq S\leq m-4$, any two of the following imply the third:

  1. (1)

$\mathcal{D}$ is quasi-symmetric;

  2. (2)

$\mathcal{D}$ is a $4$-design;

  3. (3)

$|B|={m\choose 2}$ .

The 4-design attaining the lower bound ${m\choose 2}$ is called a _tight_
4-design. Enomoto-Ito-Noda [5] classified the tight 4-designs.

######  Theorem 2.3.

[5] Up to complementation, the $4$-$(23,7,1)$ design is the only tight
$4$-$(m,S,\Lambda)$ design with $2<S<m-2$.

By Theorem 2.2, if $\mathcal{D}$ is quasi-symmetric with $|B|={m\choose 2}$
($4\leq S\leq m-4$), then it is the $4$-$(23,7,1)$ design by Theorem 2.3. This
fact will be used in the proof of the Main Theorem.

###  2.2. Coherent configurations

Here, we review the notion of coherent configuration given by Higman in [7].
The concept of the coherent configuration is a purely combinatorial
axiomatization of permutation groups.

######  Definition 2.4.

Let $X$ be a finite set and let $\\{R_{i}\\}_{i\in I}$ be a set of relations
on $X$ such that:

  1. (1)

$\\{R_{i}\\}_{i\in I}$ is a partition of $X\times X$;

  2. (2)

$R_{i}^{t}=R_{i^{*}}$ for some $i^{*}\in I$ where $R_{i}^{t}=\\{(y,x):(x,y)\in
R_{i}\\}$;

  3. (3)

there is a subset $\Omega\subset I$ such that $\\{(x,x):x\in
X\\}=\bigcup_{i\in\Omega}R_{i}$;

  4. (4)

given $(x,y)\in R_{k}$, $|\\{z:(x,z)\in R_{i},(z,y)\in R_{j}\\}|$ is a
constant $p_{ij}^{k}$ which depends only on $i,j,k$.

Then $(X,\\{R_{i}\\}_{i\in I})$ is said to be a _coherent configuration_.

By the third condition, $X$ is split into several parts : $X_{i}=\\{x:(x,x)\in
R_{i}\\}$ for $i\in\Omega$ and each part is called a _fiber_. Let $s_{i,j}$
denote the number of relations in $X_{i}\times X_{j}$. The _type_ of coherent
configuration is the matrix with size $|\Omega|\times|\Omega|$ and the
$(i,j)$-entry is $s_{i,j}$. When $|\Omega|=1$, it is an _association scheme_.
Each fiber carries a structure of association scheme.

######  Example 2.5.

Let $G=(V,E)$ be a regular graph that is neither complete nor empty. Then $G$
is said to be _strongly regular_ with _parameters_ $(n,k,\lambda,\mu)$ if it
is of order $n$, $k$-regular, every pair of adjacent vertices has $\lambda$
common neighbors, and every pair of distinct nonadjacent vertices has $\mu$
common neighbors. The concept of strongly regular graph and symmetric
association scheme with three relations are equivalent. The eigenvalues of $G$
are listed from large to small by $k>r>s$. The eigenvalue $s$ must be a
negative number. The reader can find details for strongly regular graphs in
[6].

######  Example 2.6.

Let $\mathcal{D}$ be a quasi-symmetric design with intersection number
$\alpha,\beta$. Define $X=V\cup B$ and

$\displaystyle R_{1}$ $\displaystyle=\\{(x,x):x\in V\\};$ $\displaystyle
R_{2}$ $\displaystyle=\\{(x,x):x\in B\\};$ $\displaystyle R_{3}$
$\displaystyle=\\{(x,y):x,y\in V,x\neq y\\};$ $\displaystyle R_{4}$
$\displaystyle=\\{(x,y)\in B\times B:|x\cap y|=\alpha\\};$ $\displaystyle
R_{5}$ $\displaystyle=\\{(x,y)\in B\times B:|x\cap y|=\beta\\};$
$\displaystyle R_{6}$ $\displaystyle=\\{(x,y)\in V\times B:x\in y\\};$
$\displaystyle R_{7}$ $\displaystyle=\\{(x,y)\in V\times B:x\not\in y\\};$
$\displaystyle R_{8}$ $\displaystyle=R_{6}^{t};\,\,\,\,\,R_{9}=R_{7}^{t}.$

Then $(X,\\{R_{1},\dots,R_{9}\\})$ is a coherent configuration of type
$\begin{pmatrix}2&2\\\ 2&3\end{pmatrix}$ (abbreviated as (2,2;3)). The fibers
are $V$ and $B$.

###  2.3. Connection between type (2,2;3) and quasi-symmetric designs

Higman showed in [7] that a coherent configuration $(X_{1}\cup
X_{2},\\{R_{1},\dots,R_{9}\\})$ of type (2,2;3) is equivalent to a
complementary pair of quasi-symmetric design. The relation $R_{6}$ restricted
in $X_{1}\times X_{2}$ carries a structure of quasi-symmetric design
$\mathcal{D}$. There is a relation $R$ in $\\{R_{4},R_{5}\\}$ restricted in
$X_{2}\times X_{2}$ equivalent to $\\{(x,y)\in B\times B:|x\cap y|=\alpha\\}$
where $\alpha$ is the largest intersection number of $\mathcal{D}$. Without
loss of generality, we assume $R=R_{4}$. The $R_{4}$ restricted in
$X_{2}\times X_{2}$ gives a structure of strongly regular graph (which is
called the _block graph_ of $\mathcal{D}$).

######  Proposition 2.7.

[7, section 9C] Let $\mathcal{D}$ be a $2$-$(m,S,\Lambda)$ quasi-symmetric
design with intersection numbers $\beta<\alpha$. We can express for the
parameters of the block graph in terms of the parameters $S,m,\alpha,\beta$ of
$\mathcal{D}$.

$\displaystyle\Lambda=$
$\displaystyle\frac{S(S-1)(S-\alpha)(S-\beta)}{S^{4}-2S^{3}-((\alpha+\beta-1)(m-1)-1)S^{2}+\alpha\beta
m(m-1)};$ $\displaystyle T=$
$\displaystyle\frac{(m-1)\Lambda}{S-1};\,\,\,\,\,N=\frac{\alpha(m-S)(S(S-1)-\beta(m-1))\Lambda}{S(\alpha-\beta)(S-\alpha)(S-1)};$
$\displaystyle P=$
$\displaystyle\frac{(S(S-1)-\beta(m-1))\Lambda}{(\alpha-\beta)(S-1)};\,\,\,\,\,r=\frac{1}{\alpha-\beta}\left(\frac{(m-S)\Lambda}{S-1}-(S-\beta)\right);$
$\displaystyle k=$
$\displaystyle\frac{(m-S)(S(S-1)-\beta(m-1))\Lambda}{(\alpha-\beta)(S-\alpha)(S-1)};\,\,\,\,\,n=\frac{mT}{S};\,\,\,\,\,s=\frac{\beta-S}{\alpha-\beta}.$

######  Remark 2.8.

The parameters $S,m,\alpha,\beta$ are integers with

$\displaystyle 0\leq\beta<\alpha<S<m.$

By the definition of $2$-design, $\Lambda\in\mathbb{Z}_{\geq 1}$. By the
definition of $N,P$, they belong to $\mathbb{Z}_{\geq 0}$.

###  2.4. Representation of coherent configurations

The _adjacency matrices_ of a coherent configuration are the $|X|\times|X|$
matrices $A_{i}$ whose $(x,y)$-entry is $1$ if $(x,y)\in R_{i}$ and $0$
otherwise. The _adjacency algebra_ is
$\mathcal{A}=\text{span}_{\mathbb{C}}\\{A_{i}:i\in I\\}$. Let
$\triangle_{1},\dots,\triangle_{p}$ be the non-isomorphic irreducible
representations of degree, $e_{1},\dots,e_{p}$ correspondingly. There is a
basis $\\{\epsilon_{ij}^{s}\\}$ of $\mathcal{A}$ defined by
$\triangle_{t}(\epsilon_{ij}^{s})=\delta_{st}E_{ij}^{s}$, where $E_{ij}^{s}$
is the $e_{s}\times e_{s}$ matrix with $(i,j)$-entry 1 and all other entries
0. They satisfy the equation
$\epsilon_{ij}^{s}\epsilon_{kl}^{t}=\delta_{st}\delta_{jk}\epsilon_{il}^{s}$.

Higman showed in [7] that for a type (2,2;3) coherent configuration, the
adjacency algebra $\mathcal{A}$ has three non-isomorphic irreducible
representations. For $x=c_{1}A_{1}+\cdots+c_{9}A_{9}\in\mathcal{A}$, the three
representations
$\triangle_{1},\triangle_{2}:\mathcal{A}\to\text{Mat}_{2}(\mathbb{C}),\triangle_{3}:\mathcal{A}\to\mathbb{C}$
are defined by

$\displaystyle\triangle_{1}(x)$
$\displaystyle=\begin{pmatrix}c_{1}+(m-1)c_{3}&\alpha_{1}c_{6}+\alpha_{2}c_{7}\\\
\alpha_{1}c_{8}+\alpha_{2}c_{9}&c_{2}+kc_{4}+(n-k-1)c_{5}\end{pmatrix},$
$\displaystyle\triangle_{2}(x)$
$\displaystyle=\begin{pmatrix}c_{1}-c_{3}&\beta_{1}c_{6}+\beta_{2}c_{7}\\\
\beta_{1}c_{8}+\beta_{2}c_{9}&c_{2}+rc_{4}-(r+1)c_{5}\end{pmatrix},$
$\displaystyle\triangle_{3}(x)$ $\displaystyle=(c_{2}+sc_{4}-(s+1)c_{5}),$

where
$\alpha_{1}=\sqrt{ST},\alpha_{2}=P^{-1}(k-N)\sqrt{ST},\beta_{1}=-\beta_{2}=\sqrt{T-\Lambda}$.

Let $\epsilon_{11}^{2}=c_{1}A_{1}+\cdots+c_{9}A_{9}$. Since
$\triangle_{s}(\epsilon_{11}^{2})=\delta_{s,2}E^{2}_{11}$, we have
$(c_{2}+sc_{4}-(s+1)c_{5})=0$,

$\displaystyle\begin{pmatrix}c_{1}+(m-1)c_{3}&\alpha_{1}c_{6}+\alpha_{2}c_{7}\\\
\alpha_{1}c_{8}+\alpha_{2}c_{9}&c_{2}+kc_{4}+(n-k-1)c_{5}\end{pmatrix}=\begin{pmatrix}0&0\\\
0&0\end{pmatrix},$
$\displaystyle\begin{pmatrix}c_{1}-c_{3}&\beta_{1}c_{6}+\beta_{2}c_{7}\\\
\beta_{1}c_{8}+\beta_{2}c_{9}&c_{2}+rc_{4}-(r+1)c_{5}\end{pmatrix}=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}.$

Now, we have $9$ variables and $9$ conditions, so we can solve
$c_{1},\dots,c_{9}$. Others $\epsilon_{ij}^{s}$ also have $9$ variables and
$9$ conditions, so we can explicitly determine them. We list the solution of
$\epsilon_{ij}^{2}$ below,

$\displaystyle\epsilon_{11}^{2}=\frac{(m-1)A_{1}-A_{3}}{m},\,\,\,\,\,\epsilon_{12}^{2}=\frac{\alpha_{2}A_{6}-\alpha_{1}A_{7}}{\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}},\,\,\,\,\,\epsilon_{21}^{2}=\frac{\alpha_{2}A_{8}-\alpha_{1}A_{9}}{\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}},$
(2.1)
$\displaystyle\epsilon_{22}^{2}=\frac{-((n-k-1)s+ks+k)}{n(r-s)}A_{2}+\frac{n-k+s}{n(r-s)}A_{4}+\frac{s-k}{n(r-s)}A_{5},$

where $n,k,r,s$ are the parameters of the strongly regular graph given by
$A_{4}|_{X_{2}\times X_{2}}$.

###  2.5. Proof of Theorem 1.2

###### Proof of Theorem 1.2:.

Let
$E:=(\epsilon_{11}^{2}+\epsilon_{12}^{2}+\epsilon_{21}^{2}+\epsilon_{22}^{2})/2$
where $\epsilon_{ij}^{2}$ are given in (2.4). Then $E^{2}=E$ and $E^{t}=E$.
The trace of $E$ is $m-1$. Namely, $E$ is a symmetric idempotent of rank
$m-1$. Let $\\{e_{x}:x\in V\cup B\\}$ be the standard basis of
$\mathbb{R}^{V\cup B}$. Observe that $\langle
Ee_{x},Ee_{y}\rangle=e_{x}^{t}E^{t}Ee_{y}=e_{x}^{t}Ee_{y}$ is the
$(x,y)$-entry of $E$. In other words,

(2.2) $\displaystyle\langle
Ee_{x},Ee_{y}\rangle=\begin{cases}(m-1)/(2m),&x,y\in V,x=y,\\\ -1/(2m),&x,y\in
V,x\neq y,\\\ \alpha_{2}/(2\alpha_{1}\beta_{2}-2\alpha_{2}\beta_{1}),&x\in
V,y\in B,x\in y,\\\
-\alpha_{1}/(2\alpha_{1}\beta_{2}-2\alpha_{2}\beta_{1}),&x\in V,y\in
B,x\not\in y,\\\ -((n-k-1)s+ks+k)/(2n(r-s)),&x,y\in B,x=y,\\\
(n-k+s)/(2n(r-s)),&x,y\in B,x\cap y=a,\\\ (s-k)/(2n(r-s)),&x,y\in B,x\cap
y=b.\end{cases}$

Define $i:V\cup B\to\mathbb{R}^{V\cup B}$ by $i(x)=Ee_{x}$. Note that $i(V)$
is on one sphere and $i(B)$ is on another sphere. So, $|A(i(V),i(B))|$ is the
number of different inner product $\langle i(v),i(b)\rangle$ for $v\in V$ and
$b\in B$. From equation (2.2) we know that it is $2$. And the distance between
$i(v)$ and $i(b)$ only depends on $v$ in $b$ or not. Similarly, $i(V)$ is an
one-distance set on a sphere which is a regular simplex in $\mathbb{R}^{m-1}$.
Finally, the $F=\epsilon_{22}^{2}$ is an idempotent of rank $m-1$ on the
algebra $\text{span}_{\mathbb{C}}\\{A_{2},A_{4},A_{5}\\}$. The inner product
between points in $\mathbb{R}^{B}$ is

$\displaystyle\langle
i(x),i(y)\rangle=\frac{1}{2}e_{x}^{t}2Ee_{y}=\frac{1}{2}e_{x}^{t}Fe_{y}=\frac{1}{2}e_{x}^{t}F^{t}Fe_{y}=\frac{1}{2}\langle
Fe_{x},Fe_{y}\rangle.$

So, the $i(B)$ is the scaling of the spherical embedding of a strongly regular
graph. Now, $i$ is an embedding satisfying the conditions in the statement of
theorem. ∎

######  Remark 2.9.

In general, this embedding has at most 5 distinct distances. If we fix the
regular simplex and rescale the sphere that containing $i(B)$, it is still an
embedding satisfying conditions of Theorem 1.2. We will need smartly choosing
the radius $R_{2}$ to make the embedding as a two-distance set. How to
determine the $R_{2}$ will be discussed in the next section.

######  Example 2.10.

For the Lisoněk’s example, the parameters of $\mathcal{D}$ are
$(S,m,\alpha,\beta,\Lambda,T,N,P)=(2,9,1,0,1,8,7,2)$ and the parameters of the
block graph are $(n,k,\lambda,\mu)=(36,14,7,4)$. Now,

(2.3) $\displaystyle\langle Ee_{x},Ee_{y}\rangle=\begin{cases}4/9,&x,y\in
V,x=y,\\\ -1/18,&x,y\in V,x\neq y,\\\ -\sqrt{7}/18,&x\in V,y\in B,x\in y,\\\
\sqrt{7}/63,&x\in V,y\in B,x\not\in y,\\\ 1/9,&x,y\in B,x=y,\\\ 5/126,&x,y\in
B,x\cap y=a,\\\ -2/63,&x,y\in B,x\cap y=b.\end{cases}$

Hence, the radius $R_{1}$ of the sphere containing $EV$ is $2/3$ and the
radius $R_{2}$ of the sphere containing $EB$ is $1/3$. $A(EV,EV)=\\{1\\}$,
$A(EV,EB)=\left\\{\frac{\sqrt{\sqrt{7}+5}}{3},\sqrt{\frac{5}{9}-\frac{2\sqrt{7}}{63}}\right\\}$,
$A(EB,EB)=\\{\sqrt{1/7},\sqrt{2/7}\\}$. After rescaling $R_{1}$ with
$\sqrt{2}$ and $R_{2}$ with $\sqrt{14}$, $E(V\cup B)$ becomes a two-distance
set where $A(E(V\cup B),E(V\cup B))=\\{\sqrt{2},2\\}$.

##  3\. Embedding as a two-distance set

Let $\mathcal{D}=(V,B)$ be a quasi-symmetric design and $i$ be an embedding
satisfying conditions in Theorem 1.2. In this section, we assume that
$|A(i(X),i(X))|=|\\{\sqrt{2},\sqrt{\gamma}\\}|=2$ where $X=V\cup B$. The
following theorem is the main theorem of this section.

######  Theorem 3.1.

Let $(V\cup B,\\{R_{1},\dots,R_{9}\\})$ be a coherent configuration of type
(2,2;3). Let $\mathcal{D}=(V,B)$ be the corresponding $2$-$(m,S,\Lambda)$
quasi-symmetric design with two intersection numbers $x$ and $y$. Suppose the
embedding is a two-distance set and satisfies the conditions of Theorem 1.2.
Then $p_{1}(S,m,x,y)=p_{2}(S,m,x,y)=p_{3}(S,m,x,y)=0$, where

$\displaystyle p_{1}(S,m,x,y)=$ $\displaystyle
S^{4}-2S^{2}xm+x^{2}m^{2}-2S^{3}+2S^{2}x-2Sxm+2x^{2}m+S^{2}-2Sx+x^{2};$
$\displaystyle p_{2}(S,m,x,y)=$ $\displaystyle
S^{4}+2S^{2}xm-4S^{2}ym+x^{2}m^{2}-4xym^{2}+4y^{2}m^{2}-2S^{3}+2S^{2}x$
$\displaystyle+8S^{2}m-6Sxm-2x^{2}m-4Sym+4xym-4Sm^{2}+4xm^{2}+S^{2}-2Sx+x^{2};$
$\displaystyle p_{3}(S,m,x,y)=$ $\displaystyle
S^{2}x^{2}-Smx^{2}-2S^{2}xy+2Smxy+S^{2}y^{2}-Smy^{2}+S^{2}m$
$\displaystyle+2S^{2}x-2Smx-2Sx^{2}+mx^{2}-2S^{2}y+2Sxy+S^{2}-2Sx+x^{2}.$

###  3.1. Calculation of A(i(X),i(X))

Let $A(i(V),i(V))=\\{d_{1}\\}$, $A(i(B),i(B))=\\{d_{2},d_{3}\\}$ and
$A(i(V),i(B))=\\{d_{4},d_{5}\\}$ with $d_{2}<d_{3}$ and $d_{4}<d_{5}$.

Let $i(V)=\\{e_{1},\dots,e_{m}\\}$ be the standard orthonormal basis. It forms
a regular simplex of $\mathbb{R}^{m-1}$ located on the affine hyperplane

$H=\left\\{x\in\mathbb{R}^{m}:\sum_{i=1}^{m}x_{i}=1\right\\}\subset\mathbb{R}^{m}.$

For convenience, we define $f(x)=(x-1)/x$ for all $x\neq 0$. We will determine
the elements in $A(i(X),i(X))$ expressed by the parameters $S,m,\alpha$, and
$\beta$.

######  Theorem 3.2.

The points set $i(V)$ and $i(B)$ are located on the spheres with radius
$R_{1}$ and $R_{2}$ respectively. Then

$\displaystyle R_{1}$ $\displaystyle=\sqrt{f(m)},$ $\displaystyle R_{2}$
$\displaystyle=\begin{cases}\sqrt{2-f(m-S)}+\sqrt{f(m)-f(m-S)},&\gamma>2,\\\
\sqrt{2-f(m-S)}-\sqrt{f(m)-f(m-S)},&\gamma<2,\end{cases}$ $\displaystyle
d_{1}$
$\displaystyle=\sqrt{2},\,\,\,\,\,d_{2}=R_{2}\sqrt{\frac{2(S-\alpha)m}{(m-S)S}},\,\,\,\,\,d_{3}=R_{2}\sqrt{\frac{2(S-\beta)m}{(m-S)S}},$
$\displaystyle d_{4}$
$\displaystyle=\begin{cases}\sqrt{R_{2}^{2}-2R_{2}\sqrt{f(m)-f(m-S)}+f(m)},&\text{
if }\gamma>2,\\\ \sqrt{R_{2}^{2}-2R_{2}\sqrt{f(m)-f(S)}+f(m)},&\text{ if
}\gamma<2\end{cases}$ $\displaystyle d_{5}$
$\displaystyle=\begin{cases}\sqrt{R_{2}^{2}+2R_{2}\sqrt{f(m)-f(S)}+f(m)},&\text{
if }\gamma>2,\\\ \sqrt{R_{2}^{2}+2R_{2}\sqrt{f(m)-f(m-S)}+f(m)},&\text{ if
}\gamma<2.\end{cases}$

The Theorem 3.2 is the consequence of Lemma 3.3 and Lemma 3.5.

######  Lemma 3.3.

The radius $R_{1}$ and the distances $d_{1},d_{2},$ and $d_{3}$ between the
points satisfy the formula in Theorem 3.2.

###### Proof.

We assume that the vertices of the regular simplex are
$e_{1},e_{2},\dots,e_{m}$ and the center of the simplex is
$(\frac{1}{m},\frac{1}{m},\cdots,\frac{1}{m})$. With easy calculation, we can
obtain $R_{1}=\sqrt{\frac{m-1}{m}}=\sqrt{f(m)}$ and $d_{1}=\sqrt{2}$. We
require the $i(B)$ as the spherical embedding of a strongly regular graph. If
it is on the sphere with radius $R_{2}$, then the elements in $A(i(B),i(B))$
are

$\displaystyle
R_{2}\sqrt{2-2\frac{r}{k}},\,\,\,\,\,R_{2}\sqrt{2-2\frac{-1-r}{n-k-1}}$

due to the formula (2.2) (the block graph is neither complete nor empty,
$n-k-1>0$ and $k>0$). If we substitute $r,k$ by $S,m,\alpha,\beta$, assuming
$\beta<\alpha$ and $d_{2}<d_{3}$, we have
$d_{2}=R_{2}\sqrt{\frac{2(S-\alpha)m}{(m-S)S}}$ and
$d_{3}=R_{2}\sqrt{\frac{2(S-\beta)m}{(m-S)S}}$. ∎

The centroid of the regular simplex $i(V)$ is $o=\frac{1}{m}(1,\dots,1)$. Take
$v,w\in V$ and $b\in B$ with $v\in b$ and $w\not\in b$. Let

$p=\frac{1}{S}\sum_{\begin{subarray}{c}t\in V\\\ t\in
b\end{subarray}}i(t),\,\,\,\,\,q=\frac{1}{m-S}\sum_{\begin{subarray}{c}t\in
V\\\ t\not\in b\end{subarray}}i(t)$

be the centroids of points in $b$ and points not in $b$. Direct computation
shows that $\langle i(v)-p,p-o\rangle=0=\langle o-q,q-i(w)\rangle$.
Geometrically, we may regard that the centroid of $v$ will lie in the
perpendicular bisector plane. We have
$\|i(v)-p\|_{2}=\|(1,0,\cdots,0)-(\frac{1}{S},\frac{1}{S},\dots,\frac{1}{S},0,\dots,0)\|_{2}=\sqrt{\frac{S-1}{S}}=\sqrt{f(S)}$.
Similarly, we can obtain $\|i(w)-q\|_{2}=\sqrt{f(m-S)}$. To visualize our
calculation, we give the figure 1 in the following.

Figure 1. length of $\|i(v)-p\|_{2},\|p-o\|_{2},\|o-q\|_{2},\|q-i(w)\|_{2}$

We divide $i$ into $4$ cases. If $i$ belongs to case $t$, then use $\iota_{t}$
denote $i$. First, the embedding of vertices located in a smaller sphere or
the other way. Let $\iota_{1},\iota_{2}$ send vertices to the smaller sphere
and $\iota_{3},\iota_{4}$ send vertices to the larger sphere. Second,
$\iota_{1},\iota_{4}$ send elements not in block located closer than elements
in block, and $\iota_{2},\iota_{3}$ sends elements in block located closer
than elements not in block. We show the $4$ categories geometrically in the
Figure 2.

Figure 2. $\iota_{1},\iota_{2},\iota_{3},\iota_{4}$ Figure 3. $i(b)=q$

######  Remark 3.4.

It is impossible that $i(b)=q$ (see the Figure 3). We will explain it in the
following. Let $D_{1}=\|i(v)-i(w)\|_{2}$, $D_{2}=\|i(w)-i(b)\|_{2}$ and
$D_{3}=\|i(v)-i(b)\|_{2}$. We already knew $D_{1}=\sqrt{2}$ and
$D_{2}=\sqrt{f(m-S)}$. Now,

$\displaystyle D_{3}=\left\|\left(1,\underbrace{0,\dots,0}_{S-1\text{
terms}},\underbrace{0,\dots,0}_{m-S\text{
terms}}\right)-\frac{1}{m-S}\left(\underbrace{0,\dots,0}_{S\text{
terms}},\underbrace{1,\dots,1}_{m-S\text{
terms}}\right)\right\|_{2}=\sqrt{\frac{m-S+1}{m-S}}>1.$

Hence, $D_{3}>D_{2}$. According to the two-distance set assumption,
$D_{3}=D_{1}=\sqrt{2}$ implies $m-S=1$ and $D_{2}=\sqrt{f(m-S)}=0$. Hence, we
send every point not in block to the same place which is a contradiction.

######  Lemma 3.5.

The radius $R_{2}$ and the distances $d_{4}$ and $d_{5}$ between the points
satisfy the formula in Theorem 3.2.

###### Proof.

According to $\gamma>2$ or $\gamma<2$, there are total $8$ cases need to be
considered. We listed them in Table 1.

| $\iota_{1}$ | $\iota_{2}$ | $\iota_{3}$ | $\iota_{4}$
---|---|---|---|---
$\gamma>2$ | (A) | (B) | (C) | (D)
$\gamma<2$ | (E) | (F) | (G) | (H)
Table 1. $8$ cases

Now, we can determine the $d_{4}$ and $d_{5}$. Take $v,w\in V$ and $b\in B$
with $v\in b$ and $w\not\in b$. According to Figure 1, for case 1 and case 4,
we have
$d_{5}^{2}=\|i(v)-i(b)\|^{2}_{2}=\|p-i(v)\|_{2}^{2}+(R_{2}+\|o-p\|_{2})^{2}=f(S)+(R^{2}_{2}+\sqrt{f(m)-f(S)})^{2}$.
Also,
$d_{4}^{2}=\|i(w)-i(b)\|^{2}_{2}=\|q-i(w)\|_{2}^{2}+(R_{2}-\|o-q\|_{2})^{2}=f(m-S)+(R^{2}_{2}+\sqrt{f(m)-f(m-S)})^{2}$.
Hence,

$\displaystyle\|i(b)-i(v)\|_{2}=\begin{cases}d_{4}=\sqrt{R_{2}^{2}-2R_{2}\sqrt{f(m)-f(m-S)}+f(m)},&\text{
if }v\not\in b;\\\ d_{5}=\sqrt{R_{2}^{2}+2R_{2}\sqrt{f(m)-f(S)}+f(m)},&\text{
if }v\in b.\end{cases}$

For case 2 and case 3,

$\displaystyle\|i(b)-i(v)\|_{2}=\begin{cases}d_{4}=\sqrt{R_{2}^{2}-2R_{2}\sqrt{f(m)-f(S)}+f(m)},&\text{
if }v\in b;\\\ d_{5}=\sqrt{R_{2}^{2}+2R_{2}\sqrt{f(m)-f(m-S)}+f(m)},&\text{ if
}v\not\in b.\end{cases}$

Denote the complement of $\mathcal{D}$ as $\overline{\mathcal{D}}$. If the
embedding of $\mathcal{D}$ is case 1, the embedding of
$\overline{\mathcal{D}}$ is case 2. Moreover,
$\iota_{1}(\mathcal{D})=\iota_{2}(\overline{\mathcal{D}})$. The same is true
for case 3 and case 4. This reduces to four cases. Without loss of generality,
we consider cases (A), (D), (F), (G).

Suppose $\gamma>2$. From the formula for $d_{4},d_{5}$, we find that $R_{2}$
is a root of the degree two polynomial
$x^{2}+2x\sqrt{f(m)-f(m-S)}+f(m)-d_{4}^{2}$. But, since $f(m)<2$, the only
positive root is $\sqrt{f(m)-f(m-S)}+\sqrt{2-f(m-S)}$. Since this number
greater than $R_{1}$ we conclude that case (D) is impossible.

Suppose $\gamma<2$. From the formula for $d_{4},d_{5}$, we find that $R_{2}$
is a root of the polynomial $x^{2}-2x\sqrt{f(m)-f(S)}+f(m)-d_{4}^{2}$. The
only positive root is $\sqrt{2-f(m-S)}-\sqrt{f(m)-f(m-S)}$. Since this number
is less than $R_{1}$, we conclude that case (F) is impossible. ∎

######  Remark 3.6.

The parameters of conference graphs are $(n,(n-1)/2,(n-5)/4,(n-1)/4)$. If the
block graph is a conference graph, then $d:=m-1$ is the multiplicity of $r$.
Hence, $n=2m-1$. Now, the $|X|=m+n>2d+3$ (when $m>2$). By [8, Theorem 2] and
Theorem 3.2, $s\in\mathbb{Z}$ (and hence $r\in\mathbb{Z}$).

###  3.2. Proof of Theorem 3.1

###### Proof of Theorem 3.1.

Let $A(i(X),i(X))=\\{\sqrt{2},\sqrt{\gamma}\\}$ and let $(S,m,\alpha,\beta)$
be the parameters of $\mathcal{D}$. We will prove the following statement: For
$\gamma>2$,
$p_{1}(S,m,\alpha,\beta)=p_{2}(S,m,\alpha,\beta)=p_{3}(S,m,\alpha,\beta)=0$
and for $\gamma<2$,
$p_{1}(S,m,\beta,\alpha)=p_{2}(S,m,\beta,\alpha)=p_{3}(S,m,\beta,\alpha)=0$.

If $\gamma>2$, then the embedding is of case 1. Since
$2=d_{2}^{2}=R_{2}^{2}(2(S-\alpha)m)/(mS-S^{2})$, we have
$R_{2}=\sqrt{\frac{(m-S)S}{(S-\alpha)m}}$ and
$d_{4}^{2}=2=R_{2}^{2}-2R_{2}\sqrt{f(m)-f(m-S)}+f(m)$ and equivalent to
$2S\sqrt{S-\alpha}=-(S^{2}-\alpha m+S-\alpha)$ which implies
$p_{1}(S,m,\alpha,\beta)=0$.

Next, $d_{3}^{2}=d_{5}^{2}$ if and only if
$R_{2}^{2}(2(S-\beta)m)/(mS-S^{2})=R_{2}^{2}+2R_{2}\sqrt{f(m)-f(S)}+f(m)$ and
equivalent to $2(m-S)\sqrt{S-\alpha}=-(S^{2}+\alpha m-2\beta m+S-\alpha)$
which implies $p_{2}(S,m,\alpha,\beta)=0$. By Lemma 3.3,
$R_{2}=\sqrt{2-f(m-S)}+\sqrt{f(m)-f(m-S)}$. By Theorem 3.2,
$\frac{d_{2}^{2}}{d_{3}^{2}}=\frac{-s-1}{-s}.$ So,
$\frac{d_{2}^{2}}{d_{3}^{2}}=f(-s)$ if and only if

$\frac{2}{R_{2}^{2}+2R_{2}\sqrt{f(m)-f(S)}+f(m)}=\frac{S-\alpha}{S-\beta}$

which implies that $p_{3}(S,m,\alpha,\beta)=0$.

If $\gamma<2$, then embedding is of case 3. We apply a similar argument. Now,
$R_{2}=\sqrt{\frac{(m-S)S}{(S-\beta)m}}$. Hence, $d_{3}^{2}=d_{5}^{2}$ implies
$p_{1}(S,m,\beta,\alpha)=0$ and $d_{2}^{2}=d_{4}^{2}$ implies
$p_{2}(S,m,\beta,\alpha)=0$. Finally,
$R_{2}=\sqrt{2-f(m-S)}-\sqrt{f(m)-f(m-S)}$ and
$\frac{d_{2}^{2}}{d_{3}^{2}}=f(-s)=\frac{S-\alpha}{S-\beta}$ implies
$p_{3}(S,m,\beta,\alpha)=0$.

∎

######  Remark 3.7.

If $p_{1}(S,m,0,y)=0$, then $S\in\\{0,1\\}$. Now,
$p_{2}(1,m,0,y)=4(my+m-2)m(y-1)$ implies $m\in\\{0,2/(y+1)\\}$ or $y=1$.
Finally, $p_{3}(1,\frac{2}{y+1},0,y)=(y-1)(y-3)$. Since
$0\leq\beta<\alpha<S<m$, we can’t find any feasible solution.

##  4\. Integer solutions of the polynomials system

We want to find integer solutions of $p_{i}(S_{0},m_{0},x_{0},y_{0})=0$ for
$i=1,2,3$, in conjunction of that all the parameters
$S_{0},m_{0},y_{0},\Lambda_{0},T_{0},n_{0},k_{0},r_{0},s_{0},l_{0},N_{0},P_{0}$
are integers, where

$\displaystyle p_{1}(S,m,x,y)=$ $\displaystyle
S^{4}-2S^{2}xm+x^{2}m^{2}-2S^{3}+2S^{2}x-2Sxm+2x^{2}m+S^{2}-2Sx+x^{2};$
$\displaystyle p_{2}(S,m,x,y)=$ $\displaystyle
S^{4}+2S^{2}xm-4S^{2}ym+x^{2}m^{2}-4xym^{2}+4y^{2}m^{2}-2S^{3}+2S^{2}x$
$\displaystyle+8S^{2}m-6Sxm-2x^{2}m-4Sym+4xym-4Sm^{2}+4xm^{2}+S^{2}-2Sx+x^{2};$
$\displaystyle p_{3}(S,m,x,y)=$ $\displaystyle
S^{2}x^{2}-Smx^{2}-2S^{2}xy+2Smxy+S^{2}y^{2}-Smy^{2}+S^{2}m$
$\displaystyle+2S^{2}x-2Smx-2Sx^{2}+mx^{2}-2S^{2}y+2Sxy+S^{2}-2Sx+x^{2}.$

It would not be an easy task, but we can completely determine the solutions in
the following steps. First, by the condition $p_{1}=p_{2}=p_{3}=0$, we could
express all parameters $\Lambda,T,n,k,r,s,N,P$ given in Proposition 2.7 by the
variables $x$ and $z$. Then, we will define a magic auxiliary function
$g(x,z)$ such that the function $g(x,z)$ could be bounded values on the
unbounded domain of the $zx$ plane. Since $g$ is an integer and bounded, then
there are only finitely many cases to be discussed and analysis. Therefore, we
can completely determine the solution set.

Notice that the subscript $|_{0}$ on these variables mean that the variables
are evaluated at some particular values. For instance, $x_{0}$ is the notation
for given value of variable $x$. Let

(4.1) $z:=\frac{x(m+1)-S(S+1)}{2S}.$

Lemma 4.1 allows us to rewrite the parameters in Proposition 2.7 as rational
functions in just two variables, $x$ and $z$.

######  Lemma 4.1.

Assume that $S,m,x\neq 0$, $p_{1}(S,m,x,y)=0$ and $p_{2}(S,m,x,y)=0$. Then,

(4.2) $\displaystyle S$ $\displaystyle=x+z^{2},$ (4.3) $\displaystyle m$
$\displaystyle=(x+z^{2}+z)^{2}/x,$ (4.4) $\displaystyle y$
$\displaystyle=\text{$y_{1}$ or $y_{2}$},$

where

$\displaystyle y_{1}$ $\displaystyle=x-z,$ $\displaystyle y_{2}$
$\displaystyle=\frac{x^{3}+2x^{2}z^{2}+x^{2}z+xz^{4}+2xz^{3}+3xz^{2}+z^{5}+2z^{4}+z^{3}}{\left(x+z^{2}+z\right)^{2}}.$

Moreover, if $S_{0},m_{0},x_{0},y_{0}$ are nonzero integers such that
$p_{1}(S_{0},m_{0},x_{0},y_{0})=p_{2}(S_{0},m_{0},x_{0},y_{0})=0$, then
$z_{0}:=z|_{S=S_{0},m=m_{0},x=x_{0},y=y_{0}}$ is an integer.

###### Proof.

From (4.1), we get

(4.5) $m=\frac{S(S+1+2z)}{x}-1.$

Equation (4.2) follows from substituting (4.5) into $p_{1}/S^{2}=0$.
Substitute equation (4.2) back into (4.5) and we get (4.3). Substitute both
equation (4.2) and (4.3) into $p_{2}/m=0$, and we get a quadratic equation in
$y$, which has two solutions of the equation (4.4).

Now let $S_{0},m_{0},x_{0},y_{0}$ be integers satisfying the assumptions. We
see from (4.2) that $z_{0}^{2}=S_{0}-x_{0}$ is an integer, and from (4.2) and
(4.3) that $(S_{0}+z_{0})^{2}=m_{0}x_{0}$ is an integer. Therefore, $z_{0}$ is
an integer. ∎

There are two cases that $y=y_{1}$ and $y=y_{2}$.

For $y=y_{1}$ we discuss in Proposition 4.2 and $y=y_{2}$ in Proposition 4.3.

######  Proposition 4.2.

Treat all parameters $\Lambda,T,n,k,r,s,N,P$ given in Proposition 2.7 as
elements in $\operatorname{\mathbb{R}}(x,z)$ by Lemma 4.1 and $y=y_{1}$. Let
$x_{0}$ be a positive integer and $z_{0}$ an integer such that
$\Lambda_{0},T_{0},n_{0},k_{0},r_{0},s_{0},l_{0},N_{0},P_{0}$ are all
integers. Then, either $x_{0}=z_{0}(z_{0}+1)/2$, or $z_{0}=0$, or $z_{0}=-1$.

###### Proof.

Consider the auxiliary polynomial

$g:=3+x+19z+16z^{2}+3k+3\Lambda-m-4n-18P+21zr.$

Since this is a polynomial of parameters with integral coefficients, $g_{0}$
is an integer. Observe that

$\mathbb{R}^{2}=\bigcup_{z,k\in\mathbb{R}}(z,z(z+1)/2+k).$

We want to find $(z,x)\in\mathbb{Z}\times\mathbb{Z}_{\geq 1}$, so
$k\in\mathbb{Z}$. Use computers to prove that:

  * •

(Region1): $g\in(0,1)$ when $z\in(-\infty,-2]\cup[1,+\infty)$, $x\geq 1$ and
$x\geq z(z+1)/2+1$;

  * •

(Region2): $g\in(0,2)$ when $x\geq 1$ and $x=z(z+1)/2-1$;

  * •

(Region3): $g\in(0,1)$ when $x\geq 1$ and $x\leq z(z+1)/2-2$.

Figure 4. Regions

The regions are showed in Figure 4. By the facts above and that $g_{0}$ is an
integer, if $(x_{0},z_{0})$ does not satisfy the conclusion of this lemma,
then the only possibility is that $x_{0}=z_{0}(z_{0}+1)/2-1$ and $g_{0}=1$.
Solving this system of equations would give us
$z_{0}=\frac{1}{2}(-1\pm\sqrt{41})$, which contradicts the fact that $z_{0}$
is an integer. ∎

######  Proposition 4.3.

Under the same assumption of Proposition 4.2 except that $y=y_{2}$, we assume
in addition that $p_{3}(S_{0},m_{0},x_{0},y_{0})=0$. Then, $z_{0}=0$.

###### Proof.

Consider the auxiliary polynomial

$g:=-72\Lambda+13m+13n+99T-45x+32y-14z-13mz+39nz-13Tz+13xz-33z^{2}+13z^{3}.$

Since it is an integer coefficient polynomial in parameters, $g_{0}$ is an
integer.

Use computers to prove that:

  * •

$g\in(31,33)$ when $z\in(-\infty,-15]\cup[10,+\infty)$ and $x\in\\{1,2\\}$.

  * •

$g\in(-1,38)$ when $x\geq 3$.

Case 1. $x_{0}\in\\{1,2\\}$.

An enumeration of small pairs $(x_{0},z_{0})$ shows that there are no small
integral pair which gives $g_{0}=32$. For $x_{0}\in\\{1,2\\}$ and
$z_{0}\in\\{-14,\dots,9\\}$, only $g(1,-1)=136,g(1,0)=0,g(2,0)=0$ are
integers. But, $\Lambda(1,-1)=-1$.

Case 2. $x_{0}\geq 3$ and $g_{0}\in\\{1,\dots,37\\}$.

For each $i\in\\{1,\dots,37\\}$, the intersection of the curves $p_{3}=0$ and
$g=i$ consists of finitely many points. An explicit calculation shows that the
only integral points are $(x_{0},z_{0})=(0,1)$ and $(x_{0},z_{0})=(-9,3)$.
Both violate the assumption that $x_{0}$ is positive.

Case 3. $x_{0}\geq 3$ and $g_{0}=0$.

The intersection of the curves $p_{3}=0$ and $g=0$ consists of a curve $z=0$,
and $8$ points, all of which are not integral points. ∎

######  Theorem 4.4.

Consider the system of equations
$p_{1}(S,m,x,y)=p_{2}(S,m,x,y)=p_{3}(S,m,x,y)=0$. The only integral solutions
$(S_{0},m_{0},x_{0},y_{0})$ such that $S_{0},m_{0}\neq 0$, $x_{0}\geq 1$ and
$\Lambda_{0},$ $T_{0},$ $n_{0},$ $k_{0},$ $r_{0},$ $s_{0},$ $l_{0},$ $N_{0},$
$P_{0}$ are all integers are on the following parametrized curves.

  1. (i)

$S=\frac{1}{2}z(3z+1)$, $m=\frac{9}{2}z(z+1)$, $x=\frac{1}{2}z(z+1)$,
$y=\frac{1}{2}z(z-1)$.

  2. (ii)

$S=z$, $m=z$, $x=z$, $y=z$.

  3. (iii)

$S=z+1$, $m=z$, $x=z$, $y=z+1$.

###### Proof.

Recall that

$\displaystyle S$ $\displaystyle=x+z^{2},$ $\displaystyle m$
$\displaystyle=(x+z^{2}+z)^{2}/x,$ $\displaystyle y$
$\displaystyle=\text{$y_{1}$ or $y_{2}$},$

where

$\displaystyle y_{1}$ $\displaystyle=x-z,$ $\displaystyle y_{2}$
$\displaystyle=\frac{x^{3}+2x^{2}z^{2}+x^{2}z+xz^{4}+2xz^{3}+3xz^{2}+z^{5}+2z^{4}+z^{3}}{\left(x+z^{2}+z\right)^{2}}.$

For the case $y=y_{1}$, we have $x_{0}=z_{0}(z_{0}+1)/2$, or $z_{0}=0$, or
$z_{0}=-1$ by Proposition 4.2. If $x=z(z+1)/2$, then $S=\frac{1}{2}z(3z+1)$,
$m=\frac{9}{2}z(z+1)$, $x=\frac{1}{2}z(z+1)$, $y=\frac{1}{2}z(z-1)$. If $z=0$,
then $S=m=y=x$. If $z=-1$, then $S=y=x+1$ and $m=x$.

For the case $y=y_{2}$, we have $z_{0}=0$ by Proposition 4.3. Thus, $S=m=y=x$
holds.∎

##  5\. Proof of main Theorem

###### Proof of the Theorem 1.3.

Let $\mathcal{D}$ be a $2$-$(m_{0},S_{0},\Lambda_{0})$ quasi-symmetric design
satisfying the conditions in Theorem 3.1. It follows from Theorem 3.1 that

$p_{1}(S_{0},m_{0},x_{0},y_{0})=p_{2}(S_{0},m_{0},x_{0},y_{0})=p_{3}(S_{0},m_{0},x_{0},y_{0})=0.$

By Remark 3.7, $x_{0}\geq 1$. From Theorem 4.4, we have three possible
solutions need to be discussed. Parametric (ii) and (iii) give
$S_{0}=\alpha_{0}$ which is a contradiction. It remains to consider the
Parametrized solution (i). From Parametrized solution (i):

$\displaystyle
S=\frac{1}{2}z(3z+1),\,\,\,m=\frac{9}{2}z(z+1),\,\,\,x=\frac{1}{2}z(z+1),\,\,\,y=\frac{1}{2}z(z-1)$

By Remark 3.7 and Lemma 4.1, $z_{0}$ should be an integer.

Suppose $2\leq S_{0}\leq 3$. If $S_{0}=2$, then $z_{0}=1$ or $-4/3$. Lemma 4.1
implies $z_{0}=1$ which is corresponding to the Lisoněk’s example. If
$S_{0}=3$, then $z_{0}=\pm\sqrt{73}/6-1/6$ which is impossible.

Suppose $4\leq S_{0}\leq m_{0}-4$. Since $n_{0}={m_{0}\choose 2}$, by Theorem
2.2, $\mathcal{D}$ is the $4$-$(23,7,1)$ design. Now, $S_{0}=7$ implies
$z_{0}\in\\{-7/3,2\\}$. If $z_{0}=2$, then $m_{0}=27\neq 23$.

Suppose $S_{0}=m_{0}-3$, then $z_{0}=\pm\sqrt{13}/3-2/3$ which is impossible.

Suppose $S_{0}=m_{0}-2$, then $z_{0}=\pm\sqrt{10}/3-2/3$ which is impossible.

Suppose $S_{0}=m_{0}-1$, then $z_{0}=\pm\sqrt{7}/3-2/3$ which is impossible. ∎

##  6\. Discussions

We would like to propose some further research problems.

(1) In Nozaki-Shinohara [10], they consider the two-distance sets in
$\mathbb{R}^{d}$ which contain a regular simplex and a strongly regular graph.
We believe that the case where: the regular simplex is of size $d+1$ and the
strongly regular graph comes from a natural embedding (with respect to a
primitive idempotent of rank $d$) would be the most interesting (and extremal)
case. We discussed this problem assuming the additional condition that the
two-distance set has the structure of a coherent configuration of type
(2,2;3). We wonder whether it is possible to drop this additional condition on
the existence of the coherent configuration.

(2) It would be interesting whether there exists any two-distance set in
$\mathbb{R}^{d}$ coming from the natural embedding of a coherent configuration
of type (3,2;3).

(3) It is natural, although it is not so easy, to try to generalize the
discussion on two-distance sets to three-distance sets in some way. For
example, can we classify three-distance set in $\mathbb{R}^{d}$ coming from
the natural embedding of a coherent configuration of type (2,2;4). (There are
many other possibilities.)

(4) Although it seems to be a difficult problem, it would be interesting to
study two-distance set $X$ in $\mathbb{R}^{d}$ of the maximum cardinality
${d+2\choose 2},$ whether we can find a structure of a coherent configuration,
or some combinatorial structure close to the coherent configuration?

## References

  * [1] Eiichi Bannai, Etsuko Bannai, and Dennis Stanton, _An upper bound for the cardinality of an s-distance subset in real euclidean space, ii_ , Combinatorica 3 (1983), no. 2, 147–152.
  * [2] Eiichi Bannai, Etsuko Bannai, Ziqing Xiang, Wei-Hsuan Yu, and Yan Zhu, _Classification of spherical $2$-distance $\\{4,2,1\\}$-designs by solving diophantine equations_, Taiwanese Journal of Mathematics 25 (2021), no. 1, 1–22.
  * [3] Aart Blokhuis, _Few-distance sets_ , PhD thesis, Eindhoven University of Technology (1983).
  * [4] Peter Jephson Cameron and Jacobus Hendricus van Lint, _Graph theory, coding theory and block designs_ , vol. 19, Cambridge University Press, 1975.
  * [5] Hikoe Enomoto, Noboru Ito, and Ryuzaburo Noda, _Tight 4-designs_ , Osaka Journal of Mathematics 16 (1979), no. 1, 39–43.
  * [6] Chris Godsil and Gordon F. Royle, _Algebraic graph theory_ , vol. 207, Springer Science & Business Media, 2001.
  * [7] Donald Gordon Higman, _Coherent algebras_ , Linear algebra and its applications 93 (1987), 209–239.
  * [8] David G Larman, C Ambrose Rogers, and Johan J Seidel, _On two-distance sets in euclidean space_ , Bulletin of the London Mathematical Society 9 (1977), no. 3, 261–267.
  * [9] Petr Lisoněk, _New maximal two-distance sets_ , Journal of combinatorial theory, Series A 77 (1997), no. 2, 318–338.
  * [10] Hiroshi Nozaki and Masashi Shinohara, _Maximal 2-distance sets containing the regular simplex_ , Discrete Mathematics 343 (2020), no. 11, 112071\.
  * [11] Ferenc Szöllősi and Patric R.J. Östergård, _Constructions of maximum few-distance sets in euclidean spaces_ , arXiv preprint arXiv:1804.06040 (2018).
  * [12] Ziqing Xiang, _Nonexistence of nontrivial tight 8-designs_ , Journal of Algebraic Combinatorics 47 (2018), no. 2, 301–318.