# On the finiteness of four-body central configurations

Xiang Yu<EMAIL_ADDRESS><EMAIL_ADDRESS>Shuqiang
Zhu2222corresponding author, ORCID: 0000-0002-6216-2326,
<EMAIL_ADDRESS>
School of Mathematics, Southwestern University of Finance and Economics,
Chengdu 611130, China

###### Abstract

The number of central configurations in the four body problem was proved to be
finite, first by Hampton and Moeckel, then by Albouy and Kaloshin, when the
masses are all positive. We prove that the four-body central configurations
are finite for any four nonzero masses.

2020AMS Subject Classification: 70F10 70F15 37Nxx .

Statements and Declarations

The authors declare that they have no conflicts of interest.

Data sharing not applicable to this article as no datasets were generated or
analysed during the current study.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Complex central configurations
  3. 3 Rules of colored diagram
    1. 3.1 New normalization. Main estimates.
  4. 4  Exclusion of 4-body diagrams
    1. 4.1 No bicolored vertex
    2. 4.2 $C=2$
    3. 4.3 $C=3$
    4. 4.4 $C=4$
    5. 4.5 $C=5$
    6. 4.6 $C=6$
  5. 5 Problematic diagrams with non-vanishing total mass
    1. 5.1 Diagram I
    2. 5.2 Diagram II
    3. 5.3 Diagram III
      1. 5.3.1 The masses for $z_{14}\prec\epsilon$
      2. 5.3.2  When $\sum_{j=1}^{3}m_{j}\neq 0$
      3. 5.3.3 When $\sum_{j=1}^{3}m_{j}=0$
    4. 5.4 Diagram IV
    5. 5.5 Diagram V
    6. 5.6 Diagram VI
  6. 6 Finiteness of Central configurations with $m\neq 0$
    1. 6.1 Finiteness with $m\neq 0$, $\prod_{1\leq k\leq 4}(m-m_{k})\neq 0$ and $I\neq 0$
    2. 6.2 Finiteness with $m\neq 0$, $\prod_{1\leq k\leq 4}(m-m_{k})\neq 0$ and $I=0$
    3. 6.3 Finiteness of Central configurations with $m\neq 0$, $\prod_{1\leq k\leq 4}(m-m_{k})=0$ and $I\neq 0$
    4. 6.4 Finiteness with $m\neq 0$, $\prod_{1\leq k\leq 4}(m-m_{k})=0$ and $I=0$
  7. 7 Finiteness of Central configurations with $m=0$
    1. 7.1 Problematic diagrams with vanishing total mass
      1. 7.1.1 Diagram I
      2. 7.1.2 Diagram II
      3. 7.1.3 Diagrams with two $zw$-edges
      4. 7.1.4 Fully edged diagrams
    2. 7.2 Finiteness of Central configurations with $m=0$
    3. 7.3 Finiteness with $m_{1}=m_{2}=-m_{3}=-m_{4}$, or $m_{1}=m_{2},m_{3}=-\alpha^{2}m_{1},m_{4}=-(2-\alpha^{2})m_{1}$.
      1. 7.3.1 The collinear case
      2. 7.3.2 The planar case
  8. 8 Appendix

##  1 Introduction

We consider the finiteness of central configurations in the four-body problem.
Our approach is to extend the method due to Albouy and Kaloshin. They first
used this method to show that the number of central configurations in the
four-body problem is finite if the masses are all positive, and that the
number of central configurations in the planar five-body problem is finite,
perhaps except if the 5-tuple of positive masses belongs to a given
codimension 2 subvariety of the mass space. We focus on the four-body case and
obtain the finiteness result without regard to the sign of the masses, which
settles the open question raised by Hampton and Moeckel in [4].

The $N$-body problem consists of describing the complete behavior of solutions
of the Newton’s equations, $m_{k}\ddot{{\mathbf{r}}}_{k}=\frac{\partial
U}{\partial\mathbf{r}_{k}},k=1,2,\cdots,N$. Started by Newton and considered
by many great mathematicians through the centuries, the problem for $N>2$
remains largely unsolved. Central configurations arose in the study of the
$N$-body problem. By definition, they are just special arrangement of the
particles. However, they play an important role in the study of the dynamics
of the $N$-body problem. For instance, they are related to the homographic
solutions, the analysis of collision orbits, and the bifurcation of integral
manifold (cf. [6, 8]).

The so-called hypothesis on finiteness of central configurations was proposed
by Chazy [3] and Wintner [8], and was listed by Smale as the sixth problem on
his list of problems for the 21-st century [7]: Is the number of relative
equilibria (planar central configurations) finite, in the $N$-body problem of
celestial mechanics, for any choice of positive real numbers
$m_{1},\cdots,m_{N}$ as the masses?

Euler and Lagrange have solved this finiteness question when $N=3$. In 2005,
Hampton and Moeckel [4] studied the case of $N=4$ , and showed that for any
4-tuple of positive masses the upper bound of numbers of central
configurations is 8472. Albouy and Kaloshin [1] answered positively the
question in 2012 for $N=5$, except perhaps if the 5-tuple of positive masses
satisfies two given polynomial conditions. We refer the reader to the
excellent review on this problem by Hampton, Moeckel [4] and Albouy, Kaloshin
[1].

The outline of the method of Albouy and Kaloshin is the following: The central
configurations are solutions of the algebraic system (2) in the real domain.
By introducing an efficient set of coordinates, Albouy and Kaloshin embedded
system (2) into the system (6). The solutions of (6) in the complex domain are
called _normalized central configurations_. If there are infinitely many
normalized central configurations, Albouy and Kaloshin could extract a
sequence of them approaching some singularities, which is called a _singular
sequence_.

These singular sequences are classified according to their limiting behaviors.
This classification was done by two-colored diagrams, which consists of
colored strokes and circles and each circle represents a mass. For $N=4$, they
found that the number of diagrams is 5, while the number increases to 16 for
$N=5$.

They noted that for each diagram, or equivalently each class of singular
sequences, there is some algebraic constraint on the masses, and another
important fact that the existence of one class of singular sequence generally
implies the existence of other classes. Hence, the existence of infinitely
many normalized central configurations would imply several algebraic
constraints on the masses. The finiteness is asserted by showing that these
constraints have no solutions. In particular, this method allowed to reduce
significant amount of computations.

We study the finiteness question of central configurations for any 4-tuple of
nonzero masses. It is worth the effort based on two reasons. Firstly, negative
masses shed light on the role of masses in the N-body problem. If negative
masses are allowed in the planar five-body problem, then Roberts [5] has
constructed a continuum of central configurations. It is natural to ask
whether such a continuum is possible for the four-body problem if negative
masses are allowed, and this is an open question raised by Hampton and Moeckel
in [4]. Secondly, the method introduced by Albouy and Kaloshin is powerful,
and the general idea could be implemented for investigating other finiteness
problem where the restriction on parameters is weaker. Our study shows its
potential in this aspect.

In fact, negative masses, or even complex masses are not excluded in the study
of Albouy and Kaloshin in [1] although their main results are made for the
positive masses. However, the notion of centers of mass of clusters is
employed in their study, and so it is presumed that _the subset of bodies has
total mass nonzero_. This hypothesis must be discarded in our study since we
are considering the problem for all nonzero masses. Thus, we shall proceed
somewhat differently. In particular, we need to modify or abandon some rules
in the classification of the singular sequences, see Section 3. As a result,
more diagrams appear. The number of diagrams is now 14. Generally, the
corresponding constraints on masses are more subtle, even for the classes that
also appear in Albouy and Kaloshin’s classification. We carry out the proof
separately for the case of the nonzero total mass and the case of zero total
mass.

The results proved here are stated in the following

######  Theorem 1.1.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero, then there are
finitely many central configurations in the four-body problem.

We remark that

######  Theorem 1.2.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero. Let $\alpha$ be
the unique positive root of the algebraic equation (54). A complex continuum
of relative equilibria exists only for the two groups of masses:
$m_{1}=m_{2}=-m_{3}=-m_{4}$,
$m_{1}=m_{2},m_{3}=-\alpha^{2}m_{1},m_{4}=-(2-\alpha^{2})m_{1}$. Numerically,
$\alpha\approx 1.2407$.

The paper is structured as follows. In Section 2, we introduce some notations,
definitions, the coordinate system and the notation of singular sequences due
to Albouy and Kaloshin. In Section 3, we modify the tools to classify the
diagram corresponding to singular sequences. In Section 4, we list all
possible diagrams for the four-body problem. In Section 5, we obtain the
constraints on the masses corresponding to each diagram. In Section 6 and 7,
we outline the proofs of Theorem 1.1 and Theorem 1.2. Some heavy computations
in the proofs are postponed in the Appendix.

##  2 Preliminaries

In this section, we recall, among others notations and definitions, the
coordinate system and the notation of singular sequences due to Albouy and
Kaloshin [1].

A central configuration is a solution of the following algebraic equations

$\lambda\mathbf{r}_{k}=\sum_{1\leq j\leq N,j\neq
k}\frac{m_{j}(\mathbf{r}_{j}-\mathbf{r}_{k})}{|\mathbf{r}_{j}-\mathbf{r}_{k}|^{3}},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}k=1,2,\cdots,N.$
(1)

Since we will consider the four-body problem which includes negative masses,
the multiplier $\lambda$ can be positive, negative and even zero. We will not
study the case corresponding to $\lambda=0$. Equations (1) is invariant under
rotations and dilations in the plane. Hence, we may assume that the multiplier
equals $1$ or $-1$. Set $\mathbf{r}_{k}=(x_{k},y_{k})\in\mathbb{R}^{2}$,
$k=1,...,N$. We rewrite equations (1) as

$\displaystyle\sigma\begin{pmatrix}x_{1}\\\ y_{1}\end{pmatrix}$
$\displaystyle=m_{2}r_{12}^{-3}\begin{pmatrix}x_{21}\\\
y_{21}\end{pmatrix}+m_{3}r_{13}^{-3}\begin{pmatrix}x_{31}\\\
y_{31}\end{pmatrix}+...+m_{N}r_{1N}^{-3}\begin{pmatrix}x_{N1}\\\
y_{N1}\end{pmatrix}$ (2) $\displaystyle\sigma\begin{pmatrix}x_{2}\\\
y_{2}\end{pmatrix}$ $\displaystyle=m_{1}r_{12}^{-3}\begin{pmatrix}x_{12}\\\
y_{12}\end{pmatrix}+m_{3}r_{23}^{-3}\begin{pmatrix}x_{32}\\\
y_{32}\end{pmatrix}+...+m_{N}r_{2N}^{-3}\begin{pmatrix}x_{N2}\\\
y_{N2}\end{pmatrix}$ $\displaystyle...$
$\displaystyle\sigma\begin{pmatrix}x_{N}\\\ y_{N}\end{pmatrix}$
$\displaystyle=m_{1}r_{1N}^{-3}\begin{pmatrix}x_{1N}\\\
y_{1N}\end{pmatrix}+...+m_{N-1}r_{(N-1)N}^{-3}\begin{pmatrix}x_{(N-1)N}\\\
y_{(N-1)N}\end{pmatrix},$

where $\sigma=\pm 1,x_{kl}=x_{l}-x_{k}$, $y_{kl}=y_{l}-y_{k}$ and
$r_{kl}=(x_{kl}^{2}+y_{kl}^{2})^{1/2}>0$. Denote by $f_{k}\in\mathbb{R}^{2}$,
$k=1,...,N$, the right-hand sides of the equations. System (2) can be
abbreviated as $\sigma\mathbf{r}_{k}=f_{k},k=1,...,N.$

We first embed the system (2) above into a polynomial system in
$\mathbb{C}^{2N}\times{\mathbb{C}^{N(N-1)/2}}$:

$\displaystyle\begin{pmatrix}x_{1}\\\ y_{1}\end{pmatrix}$
$\displaystyle=m_{2}\delta_{12}^{3}\begin{pmatrix}x_{21}\\\
y_{21}\end{pmatrix}+m_{3}\delta_{13}^{3}\begin{pmatrix}x_{31}\\\
y_{31}\end{pmatrix}+...$ (3) $\displaystyle\begin{pmatrix}x_{2}\\\
y_{2}\end{pmatrix}$
$\displaystyle=m_{1}\delta_{12}^{3}\begin{pmatrix}x_{12}\\\
y_{12}\end{pmatrix}+m_{3}\delta_{23}^{3}\begin{pmatrix}x_{32}\\\
y_{32}\end{pmatrix}+...$ $\displaystyle...$
$\displaystyle\delta_{12}^{2}(x_{12}^{2}+y_{12}^{2})=1$
$\displaystyle\delta_{13}^{2}(x_{13}^{2}+y_{13}^{2})=1$ $\displaystyle...$
$\displaystyle y_{12}=0.$

######  Definition 2.1 (Normalized central configuration).

A normalized central configuration is a solution of (3). A real normalized
central configuration is a normalized central configuration such that
$(x_{k},y_{k})\in\mathbb{R}^{2}$ for any $k=1,2,\cdots,N$. A positive
normalized central configuration is a real normalized central configuration
such that $\delta_{jk}=\pm 1/\sqrt{x_{jk}^{2}+y_{jk}^{2}}$ is positive for any
$j,k,j\neq k$

######  Definition 2.2.

The following quantities are defined:

$\begin{array}[]{cc}\text{Total mass}&m=\sum_{j=1}^{N}m_{j}\\\ \text{Potential
function}&U=\sum_{1\leq j<k\leq N}\frac{m_{j}m_{k}}{r_{jk}}=\sum_{1\leq
j<k\leq N}{m_{j}m_{k}}{\delta_{jk}}\\\ \text{ Moment of mass
}&M=\sum_{j=1}^{N}m_{j}\mathbf{r}_{j}\\\ \text{ Moment of inertia
}&I=\sum_{j=1}^{N}m_{j}|\mathbf{r}_{j}|^{2}=\sum_{j=1}^{N}m_{j}(x_{j}^{2}+y_{j}^{2})^{2}\end{array}$

Then it is easy to see that

$mI=\sum_{1\leq j<k\leq N}m_{j}m_{k}r_{jk}^{2}\triangleq S.$ (4)

######  Lemma 2.1 ([1]).

Let $X$ be a closed algebraic subset of $\mathbb{C}^{\mathcal{N}}$ and
$f:\mathbb{C}^{\mathcal{N}}\mapsto\mathbb{C}$ be a polynomial. Either the
image $f(X)\in\mathbb{C}$ is a finite set, or it it the complement of a finite
set. In the second case one says that $f$ is dominating.

######  Lemma 2.2 ([1]).

Consider the closed algebraic subset
$\mathcal{A}\in\mathbb{C}^{2N}\times\mathbb{C}^{N(N-1)/2}$ defined by system
(3) and the polynomial functions $U$, $I$ and $S$ on it. Then
$U(\mathcal{A})$, $I(\mathcal{A})$ and $S(\mathcal{A})$ are three finite sets.

###  2.1 Complex central configurations

For notation convenience, we may use the variables $r_{jk}=1/\delta_{jk}$
instead of the $\delta_{jk}$’s. But keep in mind that the variables of system
(3) are $x_{k},y_{k}$ and $\delta_{jk}$.

Set $z_{k}=x_{k}+\mathbf{i}y_{k}$ and $w_{k}=x_{k}-\mathbf{i}y_{k}$. If
$x_{k}$’s and $y_{k}$’s are real, the $z_{k}$’s form this configuration in the
complex plane, and the $w_{k}$’s is the reflection of the configuration. We
have $x_{k}^{2}+y_{k}^{2}=z_{k}w_{k}$ and
$x_{jk}^{2}+y_{jk}^{2}=z_{jk}w_{jk}$. System (3) becomes

$\displaystyle z_{1}$
$\displaystyle=m_{2}z_{21}^{-1/2}w_{21}^{-3/2}+m_{3}z_{31}^{-1/2}w_{31}^{-3/2}+...$
(5) $\displaystyle w_{1}$
$\displaystyle=m_{2}z_{21}^{-3/2}w_{21}^{-1/2}+m_{3}z_{31}^{-3/2}w_{31}^{-1/2}+...$
$\displaystyle...$ $\displaystyle z_{12}$ $\displaystyle=w_{12}.$

_The word “distance” will mean $r_{jk}=\sqrt{z_{jk}{w_{jk}}}$. The $z_{jk}$’s
(respectively the $w_{jk}$’s) in the complex plane will be called
$z$-separation (respectively $w$-separation). _

Set $Z_{jk}=z_{jk}^{-1/2}w_{jk}^{-3/2}$ and
$W_{jk}=z_{jk}^{-3/2}w_{jk}^{-1/2}$, then system (5) becomes

$\displaystyle z_{1}$ $\displaystyle=m_{2}Z_{21}+m_{3}Z_{31}+...+m_{N}Z_{N1},$
(6) $\displaystyle w_{1}$
$\displaystyle=m_{2}W_{21}+m_{3}W_{31}+...+m_{N}W_{N1},$ $\displaystyle...$
$\displaystyle z_{N}$
$\displaystyle=m_{1}Z_{1N}+m_{2}Z_{2N}+...+m_{N-1}Z_{(N-1)N},$ $\displaystyle
w_{N}$ $\displaystyle=m_{1}W_{1N}+m_{2}W_{2N}+...+m_{N-1}W_{(N-1)N},$
$\displaystyle z_{12}$ $\displaystyle=w_{12}.$

Here

$r_{jk}=r_{kj}=1/\sqrt[4]{Z_{jk}W_{jk}}$,

$Z_{jk}=z_{jk}/r_{jk}^{3}$, $W_{jk}=w_{jk}/r_{jk}^{3}$, $Z_{jk}=-Z_{kj}$,
$W_{jk}=-W_{kj}$.

Let $\mathcal{N}=N(N+1)/2$. Any gravitational configuration

$\mathcal{Q}=(z_{1},z_{2},\cdots,z_{N},w_{1},w_{2},\cdots,w_{N},\delta_{12},\delta_{13},\cdots,\delta_{(N-1)N}),$

associates with two vectors in $\mathbb{C}^{\mathcal{N}}$

$\mathcal{Z}=(z_{1},z_{2},\cdots,z_{N},Z_{12},Z_{13},\cdots,Z_{(N-1)N})$,
$\mathcal{W}=(w_{1},w_{2},\cdots,w_{N},W_{12},W_{13},\cdots,W_{(N-1)N}).$

Let $\|\mathcal{Z}\|=\max_{j=1,2,\cdots,\mathcal{N}}|\mathcal{Z}_{j}|$ be the
modulus of the maximal component of the vector
$\mathcal{Z}\in\mathbb{C}^{\mathcal{N}}$. Similarly, set
$\|\mathcal{W}\|=\max_{k=1,2,\cdots,\mathcal{N}}|\mathcal{W}_{k}|$.

Consider a sequence $\mathcal{Q}^{(n)}$, $n=1,2,\cdots$, of solutions of (6),
which are of course normalized central configurations. Take a sub-sequence
such that the maximal component of $\mathcal{Z}^{(n)}$ is fixed, i.e., there
is a $j\in\\{1,2,\cdots,\mathcal{N}\\}$ that is independent of $n$ such that
$\|\mathcal{Z}^{(n)}\|=|\mathcal{Z}^{(n)}_{j}|$. Extract again in such a way
that the sequence $\mathcal{Z}^{(n)}/\|\mathcal{Z}^{(n)}\|$ converges. Extract
again in such a way that the maximal component of $\mathcal{W}^{(n)}$ is
fixed. Finally, extract in such a way that the sequence
$\mathcal{W}^{(n)}/\|\mathcal{W}^{(n)}\|$ converges.

If the initial sequence has the property that $\mathcal{Z}^{(n)}$ or
$\mathcal{W}^{(n)}$ is unbounded, so is the extracted sequence. Note that
$\|\mathcal{Z}^{(n)}\|$ and $\|\mathcal{W}^{(n)}\|$ are bounded away from
zero: if the first $N$ components of the vector $\mathcal{Z}^{(n)}$ or
$\mathcal{W}^{(n)}$ all approach zero, then the denominator $z_{12}=w_{12}$ of
the component $Z_{12}=W_{12}$ approaches zero, hence $\mathcal{Z}^{(n)}$ and
$\mathcal{W}^{(n)}$ are unbounded. There are two possibilities for the
extracted sub-sequences above:

  * •

$\mathcal{Z}^{(n)}$ and $\mathcal{W}^{(n)}$ are bounded,

  * •

at least one of $\mathcal{Z}^{(n)}$ and $\mathcal{W}^{(n)}$ is unbounded.

######  Definition 2.3 (Singular sequence).

Consider a sequence of normalized central configurations. A sub-sequence
extracted by the above process, in the unbounded case, is called a _singular
sequence_.

##  3 Rules of colored diagram

In this section, we state the rules of the colored diagram, which will be the
tool for the classification of the singular sequences. Most rules are the same
as that in [1]. We remark that it is presumed that the sum, and partial sum of
the masses are nonzero in [1] and that here this hypothesis is discarded. As a
result, there is no center of mass. Hence, Rule 1c, 1d, and 1e of [1] are
modified into our Rule III, IV, and V, and Rule 2a of [1] has been dropped.

######  Definition 3.1 (Notation of asymptotic estimates).

  * $a\sim b$

means $a/b\rightarrow 1$;

  * $a\prec b$

means $a/b\rightarrow 0$;

  * $a\preceq b$

means $a/b$ is bounded;

  * $a\approx b$

means $a\preceq b$ and $a\succeq b$.

######  Definition 3.2 (Strokes and circles.).

Given a singular sequence, the indices of the bodies will be written in a
figure and we use two colors for edges and vertices.

We use the first color, called the $z$-color, to mark the maximal order
components of

$\mathcal{Z}=(z_{1},z_{2},\cdots,z_{N},Z_{12},Z_{13},\cdots,Z_{(N-1)N})$.

They are the components of the converging vector sequence
$\mathcal{Z}^{(n)}/\|\mathcal{Z}^{(n)}\|$ that do not go to zero. If the term
$z^{(n)}_{j}$ is of maximal order among all the components of
$\mathcal{Z}^{(n)}$, we draw a circle around the name of vertex j. If the term
$Z^{(n)}_{kl}$ is of maximal order among all the components of
$\mathcal{Z}^{(n)}$, we draw a stroke between the names k and l.

If there is a maximal order term in an equation, there should be another one.
Hence, we have the first Rule.

Rule I

At each end of any $z$-stroke, there is another $z$-stroke or/and a $z$-circle
drawn around the name of the body. A $z$-circle is not isolated; a $z$-stroke
must emanate from it. There exist at least one $z$-stroke in the $z$-diagram.

######  Definition 3.3 ($z$-close).

Given a singular sequence, we say that bodies k and l are close in
$z$-coordinate, or $z$-close, or that $z_{k}$ and $z_{l}$ are close, if
$z^{(n)}_{kl}\prec\|\mathcal{Z}^{(n)}\|$.

The following statement is obvious.

Rule II

If bodies k and l are close in $z$-coordinate, they are both $z$-circled or
both not $z$-circled.

######  Definition 3.4 (Isolated component).

An isolated component of the $z$-diagram is a subset of vertices such that
there is no $z$-stroke between a vertex of this subset and a vertex of its
complement.

Rule III

The moment of mass of a set of bodies forming an isolated component of the
$z$-diagram is $z$-close to the origin.

Rule IV

Consider the $z$-diagram or an isolated component of it. If there is a
$z$-circle, there is another one. The $z$-circled bodies can not all be
$z$-close together except that the sum of the masses of these bodies is zero.

######  Definition 3.5 (Maximal $z$-stroke).

Consider a $z$-stroke between the names k and l. It is called a maximal
$z$-stroke if $z_{k}$ and $z_{l}$ are not close.

Rule V

Consider a maximal $z$-stroke. At least one of the two ends is $z$-circled.

We also draw $w$-strokes and $w$-circles on the same diagram. To distinguish
from the $z$-strokes and $z$-circles, we use another color. The previous rules
and definitions apply to $w$-strokes and $w$-circles. The superposition of the
$z$-diagram and the $w$-diagram will be called simply the diagram. The
definitions about the $z$-diagram will be adapted for the $w$-digram. For
example, a subset of bodies is an isolated component of the diagram if and
only if it forms an isolated component of the $z$-diagram and an isolated
component of the $w$-diagram.

######  Definition 3.6 (Edges and strokes).

If there is either a $z$-stroke, or a $w$-stroke, or both between vertex k and
vertex l, we say that there is an edge between them. There are three types of
edges, $z$-edges, $w$-edges and $zw$-edges, and two types of strokes,
represented by two different colors.

Figure 1: A $z$-stroke, a $z$-stroke plus a $w$-stroke, a $w$-stroke, forming
respectively a z-edge, a $zw$-edge, a $w$-edge.

###  3.1 New normalization. Main estimates.

Note that a central configuration will not change by multiplying the $z$
coordinates by $a\in\mathbb{C}\backslash\\{0\\}$ and the $w$ coordinates by
$a^{-1}$. The diagram is also invariant under such an operation, because the
$z$-coordinates and the $w$-coordinates are considered separately.

The normalization $z_{12}=w_{12}$ was used in the previous sections. In the
following we normalize instead with $\|\mathcal{Z}\|=\|\mathcal{W}\|$. For a
central configuration normalized with the condition $z_{12}=w_{12}$, we
multiply the $z$-coordinates by a proper $a>0$, the $w$-coordinates by
$a^{-1}$, so that the maximal component of $\mathcal{Z}$ and the maximal
component of $\mathcal{W}$ have the same modulus, i.e.,
$\|\mathcal{Z}\|=\|\mathcal{W}\|$.

A singular sequence was defined by the condition either
$\|\mathcal{Z}^{(n)}\|$ or $\|\mathcal{W}^{(n)}\|$ tends to $\infty$. Recall
that both $\|\mathcal{Z}^{(n)}\|$ and $\|\mathcal{W}^{(n)}\|$ were bounded
away from zero. With the new normalization, a singular sequence is simply
characterized by
$\|\mathcal{Z}^{(n)}\|=\|\mathcal{W}^{(n)}\|\rightarrow\infty$. From now on we
only discuss singular sequences.

Set $\|\mathcal{Z}^{(n)}\|=\|\mathcal{W}^{(n)}\|=1/\epsilon^{2}$, then
$\epsilon\rightarrow 0$.

######  Proposition 3.1 ([1] Estimate 1).

For any $(k,l)$, $1\leq k<l\leq N$, we have $\epsilon^{2}\preceq
z_{kl}\preceq\epsilon^{-2}$, $\epsilon^{2}\preceq w_{kl}\preceq\epsilon^{-2}$
and $\epsilon\preceq r_{kl}\preceq\epsilon^{-2}$. There is a $zw$-edge between
k and l if and only if $r_{kl}\approx\epsilon$. There is a maximal $z$-edge
between k and l if and only if $w_{kl}\approx\epsilon^{2}$.

######  Proposition 3.2 ([1] Estimate 2).

We assume that there is a $z$-stroke between k and l. Then

$\epsilon\preceq r_{kl}\preceq 1,\ \epsilon\preceq
z_{kl}\preceq\epsilon^{-2},\ \epsilon\succeq w_{kl}\succeq\epsilon^{2}.\ $

Under the same hypothesis the “equality case” are characterized as follows:

$\displaystyle{\rm Left:\ }$ $\displaystyle
r_{kl}\approx\epsilon\Leftrightarrow z_{kl}\approx\epsilon\Leftrightarrow
w_{kl}\approx\epsilon\Leftrightarrow zw{\rm-edge\ between\ }k\ {\rm and}\ l,$
$\displaystyle{\rm Right:\ }$ $\displaystyle r_{kl}\approx 1\Leftrightarrow
z_{kl}\approx\epsilon^{-2}\Leftrightarrow
w_{kl}\approx\epsilon^{2}\Leftrightarrow{\rm maximal\ }z{\rm-edge\ between\
}k\ {\rm and}\ l.$

######  Remark 3.1.

By the estimates above, the strokes in a $zw$-edge are not maximal. A maximal
$z$-stroke is exactly a maximal $z$-edge.

Rule VI

_Two consecutive $zw$-edges._ If two consecutive $zw$-edges are present, there
is a third $zw$-edge closing the triangle.

_Clusters_. At the limit when following a singular sequence, the $z_{k}$’s
form clusters. For example, if bodies 1, 2 and 3 are such that $z_{12}\prec
z_{23}$, we say that 1 clusters with 2 in $z$-coordinate, relatively to the
bodies 1,2,3. If there is a fourth body such that $z_{24}\prec z_{12}\prec
z_{23}$, we say the fourth body form a sub-cluster, e.g., together with body
2.

We can write a _clustering scheme_ in each coordinate. For instance, the
situation considered above is simply $z:42.1...3$, where three dots means the
largest separation within the group, one dot means the intermediate
separation, and no dot means the smallest separation.

In the following rule clustering relation inside a sub-system of three bodies
is considered. Note that these three bodies may form, e.g., in $z$-coordinate,
a cluster relatively to the whole configuration.

Rule VII

_Skew clustering._ Suppose that there is an edge from vertex 1 to vertex 2, an
edge from vertex 2 to vertex 3, and that there is no edge from vertex 1 to
vertex 3. Then the clustering schemes are $z:1.2...3,w:1...2.3$, or
$z:1...2.3,w:1.2...3$. We say there is “skew clustering”.

######  Corollary 3.1 ([1]).

Two consecutive $z$-edges cannot be maximal if they are not part of a triangle
of edges.

Rule VIII

_Cycles._ Consider a cycle of edges, the list of $z$-separations corresponding
to the edges, and the maximal oder of the $z$-separations within this list.
Two or more of the $z$-separations are of this order. The corresponding edges
have the same type. If there are only two, the corresponding separations are
not only of the same order, but equivalent.

Rule IX

_Triangles._ Consider a triangle of edges in the diagram. Then the edges are
of the same type (all $z$-edges or all $w$-edges or all $zw$-edges), all the
$z$-separations are of the same order, all the $w$-separations are of the same
order.

######  Corollary 3.2 ([1]).

Consider three vertices. There are 6, 3, 2, 1, or 0 strokes joining them. If
there are three forming a triangle, they are of the same color.

Rule X

_Fully edged sub-diagram._ Suppose that in the diagram there is a triangle of
edges, plus a fourth vertex attached to the triangle by at least two edges,
plus a fifth vertex attached to the four previous vertices by at least two
edges, and so on up to a $p$-th vertex, $p\geq 3$. Then there is indeed an
edge, of the same type, between any pair of the $p$ vertices, all the
$z$-separations are of the same order, all the $w$-separations are of the same
order.

Rule XI

If four edges form a quadrilateral, then the opposite edges are of the same
type.

Rule XII

_Bounded potential._ Consider a singular sequence. Let $k_{0}$ and $l_{0}$ be
such a pair of bodies that $r_{k_{0}l_{0}}\preceq r_{kl}$ for any $k,l,1\leq
k<l\leq n$. If $r_{k_{0}l_{0}}\longrightarrow 0$, then there is another pair
of bodies $(k_{1},l_{1})$ such that $r_{k_{0}l_{0}}\approx r_{k_{1}l_{1}}$.

######  Corollary 3.3.

If a $zw$-edge is present in the diagram, there is another one.

##  4  Exclusion of 4-body diagrams

A _bicolored vertex_ of the diagram is a vertex which connects at least a
$z$-stroke with at least a $w$-stroke. The number of edges from a bicolored
vertex is at least 1 and at most $3$. The number of strokes from a bicolored
vertex is at least 2 and at most $6$. Given a diagram, we define $C$ to be the
maximal number of strokes from a bicolored vertex. This number will be used to
classify all possible diagrams.

Recall that the $z$-diagram indicates the maximal order terms and it is
nonempty. If there is a circle, there is an edge of the same color adjacent to
the circle. So there is at least a $z$-stroke, and at least a $w$-stroke.

###  4.1 No bicolored vertex

If there does not exist a bicolored vertex, then there are at most two strokes
and they are “parallel”. Thus the only possible diagram is the one in Figure
2.

Figure 2: No bicolored vertex

###  4.2 $C=2$

There are two cases: a $zw$-edge exists or not.

If it is present, it should be isolated. On the other hand, there should be
another $zw$-edge by Corollary 3.3. Then the only possible diagram is the one
in Figure 3. By Rule IV, we have $m_{1}+m_{2}=0$ since body 1 and 2 are both
$z$-circled and $z$-close. Similarly, we get $m_{3}+m_{4}=0$, then
$\sum_{i=1}^{4}m_{i}=0$.

Figure 3: $C=2$, $zw$-edge appears

If it is not present, there are adjacent $z$-edges and $w$-edges. From any
such adjacency there is no other edge. By trying to continue it, we see that
the only diagram is the one in Figure 4.

Figure 4: $C=2$, no $zw$-edge

###  4.3 $C=3$

Consider a bicolored vertex with three strokes. There are two cases: a
$zw$-edge exists or not.

If it is not present, it is Y-shaped. Suppose that vertex 1 connects with
vertex 2 and vertex 3 by $z$-edges, and connects with vertex 4 by a $w$-edge.
By Rule I, vertex 1 is $w$-circled, then vertex 2 and vertex 3 are also
$w$-circled by Estimate 2. By Rule I again, there is a $w$-stroke emanating
from vertex 2 and vertex 3, which leads to a triangle with edges of different
types. We exclude this case by Rule IX.

If it is present, let vertex 2 be the bicolored vertex with three strokes.
Suppose it connects with vertex 1 by a $zw$-edge, with vertex 3 by a $z$-edge.
By circling method, we circle the three vertices by $w$-color. Then there is
$w$-stroke from vertex 3, which can not connect to 1 by Rule IX. Then there
are two cases: an edge between vertex 1 and vertex 4 or not. If it is not
present, then we continue it to the first one of Figure 5. If it is present,
then they form a quadrilateral. By Rule XI, we see the types of edges. There
is no diagonal by Rule IX. Then circling method gives the next three diagrams
in Figure 5. Consider the isolated components of the $w$-diagram. We see
$m_{1}+m_{2}=0$ and $m_{3}+m_{4}=0$ by Rule IV, for all the four diagrams in
Figure 5.

Figure 5: $C=3$

###  4.4 $C=4$

Consider a bicolored vertex (let us say, 1 ) with four strokes.

In the first case, vertex 1 has $zw$-edges connected with vertex 2 and vertex
3 separately. A third $zw$-edge closes the triangles by Rule VI. As $C=4$,
there is no other stroke, thus vertex 4 is neither $z$-circled nor
$w$-circled. Hence, the possible diagrams are those in Figure 6.

Figure 6: $C=4$, three $zw$-edges

In the second case, vertex 1 has one adjacent $zw$-edge connected with vertex
2, a $z$-edges connected with vertex 3 and a $w$-edges connected with vertex
4. Any other edge in this diagram would close a triangle, which would
contradicts Rule IX. Hence, 1 is $z$ and $w$-circled, and so is vertex 3 by
Rule I and Rule II. Then there is a $w$-edge emanating from 3 by Rule I,
contradiction.

In the third case, vertex 1 has one adjacent $zw$-edge connected with vertex
2, two $z$-edges connected with vertex 3 and 4. If there are more strokes, it
should be a $z$-stroke between vertex 3 and 4 by Rule IX. However, 2 is $z$
and $w$-circled, and so is vertex 1 by Rule II. Then both of vertex 3 and 4
are $w$-circled by Rule II. Then there should be a $w$-stroke emanating from
3, contradiction.

###  4.5 $C=5$

Consider a bicolored vertex (let us say, 1 ) with five strokes.

Suppose vertex 1 has one adjacent $w$-edge connected with vertex 4, and two
adjacent $zw$-edges connected with vertex 2 and vertex 3. Then there is a
fully $zw$-edged triangle between vertexes 1,2,3 by Rule VI. Rule VI also
implies that there are no more edges. By Rule I and IV, all vertices are
$w$-circled. There is no $z$-circle, otherwise all vertices are $z$-circled by
Rule IV and II. Then there would be a $z$-edge emanating from vertex 4, a
contradiction.

Figure 7: $C=5$

###  4.6 $C=6$

Consider a bicolored vertex with six strokes. Then this is a fully $zw$-edged
diagram by Rule VI. According to if there are $z$-circle or $w$-circle at
vertices, the possible diagrams are those in Figure 8.

Figure 8: $C=6$

##  5 Problematic diagrams with non-vanishing total mass

_Notations._ From now on, $x^{\frac{1}{n}}$ (or $\sqrt[n]{x}$) is understood
as one appropriate value of the $n$-th root of $x$. We denote the set $f=0$ by
$\mathcal{V}_{f}$, and $f=0,g=0$ by $\mathcal{V}_{f}\bigcap\mathcal{V}_{g}$ or
simply $\mathcal{V}_{f}\mathcal{V}_{g}$; the set $f=0$ or $g=0$ (i.e., $fg=0$)
by $\mathcal{V}_{f}\bigcup\mathcal{V}_{g}$. Recall that $m$ is the total mass
and it is nonzero.

It is easy to see that all diagrams in Figure 3 and Figure 5 and the last two
diagrams in Figure 8 must have $m=0$. Now, We show that the second diagram in
Figure 6 is also impossible.

First we have

$m_{1}+m_{2}+m_{3}=0.$ (7)

The equations $w_{k}=\sum_{j\neq k}m_{j}W_{jk},k=1,2,3,$ imply

$\ \frac{W_{12}}{m_{3}}\sim\frac{W_{23}}{m_{1}}\sim\frac{W_{31}}{m_{2}}\sim
a\epsilon^{-2},\ \ a\neq 0.$ (8)

Note that

$-m_{4}w_{4}=\sum_{j=1}^{3}m_{j}w_{j}=m_{2}w_{12}+m_{3}w_{13}=m_{1}w_{21}+m_{3}w_{23}=m_{1}w_{31}+m_{2}w_{32}\preceq\epsilon.$

We claim that $w_{4}\approx\epsilon$. Otherwise, the above equation leads to

$\frac{w_{12}}{m_{3}}\sim\frac{w_{23}}{m_{1}}\sim\frac{w_{31}}{m_{2}}\sim
b\epsilon,\ \ b\neq 0.$

Then equation (8) and the identity $W_{kl}=w_{kl}^{-1/2}z_{kl}^{-3/2}$ imply
that

$z_{12}\sim(\frac{1}{m_{3}^{3}a^{2}b})^{1/3}\epsilon,\
z_{23}\sim(\frac{1}{m_{1}^{3}a^{2}b})^{1/3}\epsilon,z_{31}\sim(\frac{1}{m_{2}^{3}a^{2}b})^{1/3}\epsilon.$

Then the identity $z_{12}+z_{23}+z_{31}=0$ implies that

$\frac{1}{m_{1}}1^{\frac{1}{3}}+\frac{1}{m_{2}}1^{\frac{1}{3}}+\frac{1}{m_{3}}1^{\frac{1}{3}}=0,$

which contradicts with equation (7).

Note that

$\sum_{j=1}^{3}m_{j}w_{j4}=mw_{4}\approx\epsilon,\
w_{14},w_{24},w_{34}\prec\epsilon^{-2},\ z_{14}\approx\epsilon^{-2},\
z_{24}-z_{14}\approx z_{34}-z_{14}\approx\epsilon.$

Hence,

$\displaystyle mI/m_{4}$ $\displaystyle\sim\sum_{j=1}^{3}m_{j}z_{j4}w_{j4}$
$\displaystyle=z_{14}\sum_{j=1}^{3}m_{j}w_{j4}+(z_{24}-z_{14})m_{2}w_{24}+(z_{34}-z_{14})m_{3}w_{34}$
$\displaystyle\approx\epsilon^{-1}.$

This is a contradiction since the momentum of inertia $I$ should be bounded.

We could not exclude the diagrams in Figure 9. Some singular sequence could
still exist and approach any of these diagrams.

Figure 9: Problematic Diagrams

###  5.1 Diagram I

Following the argument from [1], we obtain the equation

$\frac{m_{1}m_{3}}{(m_{2}m_{4})^{\frac{1}{2}}}+\frac{m_{2}m_{3}}{(-m_{1}m_{4})^{\frac{1}{2}}}+\frac{m_{1}m_{4}}{(-m_{2}m_{3})^{\frac{1}{2}}}+\frac{m_{2}m_{4}}{(m_{1}m_{3})^{\frac{1}{2}}}=0,$

or equivalently,

$m_{1}m_{3}\sqrt{m_{1}m_{3}}+m_{2}m_{3}\sqrt{-m_{2}m_{3}}+m_{1}m_{4}\sqrt{-m_{1}m_{4}}+m_{2}m_{4}\sqrt{m_{2}m_{4}}=0.$
(9)

Note that $z_{jk},w_{jk}\approx\epsilon^{-2}$, $j=1,2,k=3,4$. It follows

$r_{jk}\approx\epsilon^{-2},j=1,2,k=3,4.$ (10)

and $Z_{jk},W_{jk}\approx\epsilon^{4}$. Combined with the following two
equations

$\begin{array}[]{c}w_{12}=(m_{1}+m_{2})W_{12}+m_{3}(W_{32}-W_{31})+m_{4}(W_{42}-W_{41}),\\\
z_{34}=(m_{3}+m_{4})Z_{34}+m_{1}(Z_{14}-Z_{13})+m_{4}(Z_{24}-Z_{23}),\end{array}$

we obtain

$(m_{1}+m_{2})(m_{3}+m_{4})\neq 0,r_{12}^{3}\sim m_{1}+m_{2},r_{34}^{3}\sim
m_{3}+m_{4}.$ (11)

Let us simplify equation (9). Note that equation (9) is homogeneous, and is
invariant under the transformation

$(m_{1},m_{2},m_{3},m_{4})\mapsto(m_{2},m_{1},m_{4},m_{3}).$

Assume that the signs of the masses are the same. It suffices to consider only
the case $(+,+,+,+)$. Then equation (9) becomes

$m_{1}m_{3}\sqrt{m_{1}m_{3}}+m_{2}m_{4}\sqrt{m_{2}m_{4}}+i(m_{2}m_{3}\sqrt{m_{2}m_{3}}+m_{1}m_{4}\sqrt{m_{1}m_{4}})=0,$

which is $m_{1}^{3}m_{3}^{3}=m_{2}^{3}m_{4}^{3}$ and
$m_{2}^{3}m_{3}^{3}=m_{1}^{3}m_{4}^{3}$. Hence, it is necessary that

$m_{1}=m_{2},\ \ m_{3}=m_{4},\ m_{1}m_{3}>0.$

Assume that only one of the signs of the masses is different from the others.
It suffices to consider only $(+,+,+,-)$ and $(+,-,+,+)$. In the first
subcase, equation (9) is equivalent to

$\begin{cases}m_{1}m_{3}\sqrt{m_{1}m_{3}}+m_{1}m_{4}\sqrt{-m_{1}m_{4}}=0,\cr
m_{2}m_{3}\sqrt{-m_{2}m_{3}}+m_{2}m_{4}\sqrt{m_{2}m_{4}}=0.\end{cases}$

Then we have $m_{3}=-m_{4}$, which contradicts with equation (11). Similarly,
the second subcase is impossible neither.

Assume that only two of the signs of the masses is different from the others.
It suffices to consider only $(+,+,-,-)$, $(+,-,+,-)$ and $(+,-,-,+)$. In the
first subcase, equation (9) is equivalent to

$\begin{cases}m_{1}m_{3}\sqrt{m_{1}m_{3}}+m_{2}m_{4}\sqrt{m_{2}m_{4}}=0,\cr
m_{2}m_{3}\sqrt{-m_{2}m_{3}}+m_{1}m_{4}\sqrt{-m_{1}m_{4}}=0,\end{cases}$

which is equivalent to

$m_{1}=m_{2},\ m_{3}=m_{4},\ m_{1}m_{3}<0.$

In the rest subcases, equation (9) can not be simplified.

To summarize, the masses of central configurations corresponding to the first
diagram belong to one of the following two sets of constraints

$\begin{array}[]{ll}\mathcal{V}_{IA}[12,34]:&\hbox{$m_{1}=m_{2},\
m_{3}=m_{4};$}\\\
\mathcal{V}_{IB}[12,34]:&\hbox{$m_{1}m_{3}\sqrt{m_{1}m_{3}}+m_{2}m_{3}\sqrt{-m_{2}m_{3}}+m_{1}m_{4}\sqrt{-m_{1}m_{4}}+m_{2}m_{4}\sqrt{m_{2}m_{4}}=0,$}\\\
&\hbox{$m_{1}m_{2}<0,\ m_{3}m_{4}<0.$}\end{array}$

Furthermore, we claim that if $I=0$, then the masses belong to the set

$\mathcal{I}_{I}[12,34]:~{}~{}~{}~{}~{}~{}~{}~{}~{}\frac{m_{1}m_{2}}{\sqrt[3]{m_{1}+m_{2}}}+\frac{m_{3}m_{4}}{\sqrt[3]{m_{3}+m_{4}}}=0.$
(12)

This is a consequence of the fact that $U=I=0$.

In fact, it is easy to see that if $I=0$, then the masses belong to the set

$\mathcal{V}_{I0}[12,34]~{}~{}~{}~{}\triangleq~{}~{}~{}~{}~{}\mathcal{V}_{IB}[12,34]\mathcal{I}_{I}[12,34].$

Note that Diagram I has five other variants where the vertices are renumbered.
Two of them are presented in Figure 10. We still call them Diagram I if there
is no confusion arise. The corresponding mass sets are denoted by

$\mathcal{V}_{IA}[ij,pk],\ \mathcal{V}_{IB}[ij,pk],\
\mathcal{I}_{I}[ij,pk],\mbox{ and }\mathcal{V}_{I0}[ij,pk],$

if there is one stroke, either $z$ or $w$-stroke, between vertices i and j,
and the other stroke is between vertices p and k. For instance, for both of
the two diagrams in Figure 10, the corresponding mass sets are denoted by
$\mathcal{V}_{IA}[13,24],\mathcal{V}_{IB}[13,24],\mathcal{I}_{I}[13,24],$ or
$\mathcal{V}_{I0}[13,24]$.

###  5.2 Diagram II

By the argument from [1] and the hypothesis $m\neq 0$ , the masses of central
configurations corresponding to the second diagram belong to the following set

$\displaystyle\mathcal{V}_{II}[13,24]:~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle m_{1}m_{3}=m_{2}m_{4},$
$\displaystyle(m_{1}+m_{2})(m_{2}+m_{3})(m_{3}+m_{4})(m_{1}+m_{4})\neq 0.$

By the facts
$z_{12}=(m_{1}+m_{2})Z_{12}+m_{3}(Z_{32}-Z_{31})+m_{4}(Z_{42}-Z_{41})$ and
$Z_{12}=z_{12}r_{12}^{-3}$, it follows $r_{12}^{3}\sim(m_{1}+m_{2})$.
Similarly, we have

$r_{12}^{3}\sim(m_{1}+m_{2}),r_{23}^{3}\sim(m_{2}+m_{3}),r_{34}^{3}\sim(m_{3}+m_{4}),r_{14}^{3}\sim(m_{1}+m_{4}).$
(13)

Note that

$r_{13}\approx\epsilon^{-2},r_{24}\approx\epsilon^{-2}.$ (14)

If $I=0$, then the masses belong further to the set

$\mathcal{I}_{II}[13,24]:~{}~{}~{}~{}~{}~{}~{}~{}~{}\frac{m_{1}m_{2}}{\sqrt[3]{m_{1}+m_{2}}}+\frac{m_{2}m_{3}}{\sqrt[3]{m_{2}+m_{3}}}+\frac{m_{3}m_{4}}{\sqrt[3]{m_{3}+m_{4}}}+\frac{m_{1}m_{4}}{\sqrt[3]{m_{1}+m_{4}}}=0.$
(15)

That is, if $I=0$, then the masses belong to the set

$\mathcal{V}_{II0}[13,24]~{}~{}~{}~{}\triangleq~{}~{}~{}~{}~{}\mathcal{V}_{II}[13,24]\mathcal{I}_{II}[13,24].$

This is a consequence of the fact that $U=I=0$.

Similarly, Diagram II has five other variants. Some of them are presented in
the second row of Figure 11. We still call them Diagram II if there is no
confusion arise. The corresponding mass sets are denoted by

$\mathcal{V}_{II}[ij,pk],\ \mathcal{I}_{II}[ij,pk],\mbox{ and
}\mathcal{V}_{II0}[ij,pk],$

if one diagonal is between vertices i and j, and the other is between vertices
p and k. For instance, for both of the last two diagrams in the second row of
Figure 11, the corresponding mass sets are denoted by
$\mathcal{V}_{II}[14,23],\mathcal{I}_{II}[14,23]$ or
$\mathcal{V}_{II0}[14,23]$.

###  5.3 Diagram III

Following the argument from [1], we arrive at the estimation

$\frac{Z_{12}}{m_{3}}\sim\frac{Z_{23}}{m_{1}}\sim\frac{Z_{31}}{m_{1}},\
\frac{W_{12}}{m_{3}}\sim\frac{W_{23}}{m_{1}}\sim\frac{W_{31}}{m_{1}}.$

Since $Z_{12},W_{12}$ are both of the maximal order, we assume the first set
is asymptotic to $a\epsilon^{-2}$ and the second asymptotic to
$b\epsilon^{-2}$, $ab\neq 0$. Multiplying the two, by
$Z_{kl}W_{kl}=r_{kl}^{-4}$, we obtain

$\frac{1}{m_{3}^{2}r_{12}^{4}}\sim\frac{1}{m_{1}^{2}r_{23}^{4}}\sim\frac{1}{m_{2}^{2}r_{31}^{4}}\sim
ab\epsilon^{-4}.$

Then we have

$r_{12}\sim 1^{\frac{1}{4}}\frac{\epsilon}{\sqrt{|m_{3}|}}c,\ r_{23}\sim
1^{\frac{1}{4}}\frac{\epsilon}{\sqrt{|m_{1}|}}c,\ r_{31}\sim
1^{\frac{1}{4}}\frac{\epsilon}{\sqrt{|m_{2}|}}c,\ c\neq 0.$

Then by $z_{kl}=r_{kl}^{3}Z_{kl}$, we have

$z_{12}\sim 1^{\frac{1}{4}}\frac{\epsilon}{\sqrt{|m_{3}|}}d,\ z_{23}\sim
1^{\frac{1}{4}}\frac{\epsilon}{\sqrt{|m_{1}|}}d,\ z_{31}\sim
1^{\frac{1}{4}}\frac{\epsilon}{\sqrt{|m_{2}|}}d,\ d\neq 0.$ (16)

Then the identity $z_{12}+z_{23}+z_{31}=0$ implies that the masses belong to
the set

$\mathcal{V}_{III}[123]:~{}~{}~{}~{}~{}~{}~{}~{}~{}\frac{1}{{|m_{1}|^{\frac{1}{2}}}}+\frac{1}{{|m_{2}|^{\frac{1}{2}}}}+\frac{1}{{|m_{3}|^{\frac{1}{2}}}}=0.$

It is easy to see that

$r_{12},r_{23},r_{13}\approx\epsilon,r_{14},r_{24},r_{34}\succ\epsilon.$ (17)

We discuss the distances $\\{r_{j4}\\}_{j=1}^{3}$ further in the following.
Without loss of generality, assume that

$z_{14}\preceq z_{24},z_{34},\ {\it and}\ z_{14}\preceq w_{14}.$

Then $\epsilon\prec w_{14}\sim w_{24}\sim w_{34}$. For $z_{14}$, there are
three possibilities, namely $\prec,\approx$, or $\succ\epsilon$. Note that

$\displaystyle z_{14}\prec\epsilon,\ \Rightarrow z_{24}\approx
z_{34}\approx\epsilon,\ r_{14}\prec r_{24}\approx r_{34},\ W_{14}\succ
W_{24}\approx W_{34};$ $\displaystyle z_{14}\approx\epsilon,\ \Rightarrow
z_{14}\approx z_{24}\approx z_{34}\approx\epsilon,\ r_{14}^{2}\approx
r_{24}^{2}\approx r_{34}^{2},\ W_{14}\approx W_{24}\approx W_{34};$
$\displaystyle z_{14}\succ\epsilon,\ \Rightarrow z_{14}\sim z_{24}\sim
z_{34}\succ\epsilon,\ r_{14}^{2}\sim r_{24}^{2}\sim r_{34}^{2},\
W_{14}^{2}\sim W_{24}^{2}\sim W_{34}^{2}.$

####  5.3.1 The masses for $z_{14}\prec\epsilon$

By $w_{4}=\sum_{j=1}^{3}m_{j}W_{j4}$ and $z_{4}=\sum_{j=1}^{3}m_{j}Z_{j4}$ it
follows that $w_{4}\sim m_{1}W_{14},z_{4}\sim m_{1}Z_{14}.$ On the other hand,
note that $\sum_{j=1}^{4}m_{j}w_{4}=\sum_{j=1}^{3}m_{j}w_{j4}\preceq w_{14}$,
then $w_{4}\sim m_{1}W_{14}\preceq w_{14}$, which implies that $r_{14}\succeq
1$. Therefore $z_{4}\sim m_{1}Z_{14}\preceq z_{14}$, and

$z_{4},z_{1}\prec\epsilon\approx z_{2}\approx z_{3}.$

Then it is easy to see that $z_{12}\sim z_{2}\sim-m_{3}f\epsilon,z_{13}\sim
z_{3}\sim m_{2}f\epsilon$ and $z_{23}=z_{3}-z_{2}\sim(m_{2}+m_{3})f\epsilon$,
where $f$ is some nonzero constant. Compared with equation (16), it is
necessary that

$4|m_{1}|=|m_{2}|=|m_{3}|.$ (18)

####  5.3.2  When $\sum_{j=1}^{3}m_{j}\neq 0$

The identity $\sum_{j=1}^{4}m_{j}w_{4}=\sum_{j=1}^{3}m_{j}w_{j4}$ implies that

$w_{4}\sim\frac{\sum_{j=1}^{3}m_{j}}{\sum_{j=1}^{4}m_{j}}w_{14}\succ\epsilon.$

By the equation $w_{4}=\sum_{j=1}^{3}m_{j}W_{j4}$, it follows that

Case 1). If $z_{14}\prec\epsilon$, then $w_{4}\sim
m_{1}W_{14}\sim\sum_{j=1}^{3}m_{j}w_{14}$. Then $r_{14}\approx 1$. Thus

$r_{14}^{2}\approx 1\prec r^{2}_{24}\approx r^{2}_{34}\prec\epsilon^{-1},\
4|m_{1}|=|m_{2}|=|m_{3}|;$

Case 2). If $z_{14}\approx\epsilon$, then $w_{14}\preceq W_{14}$. Thus

$\epsilon^{2}\prec r^{2}_{14}\approx r^{2}_{24}\approx r^{2}_{34}\preceq 1;$

Case 3). If $z_{14}\succ\epsilon$, then $w_{14}\preceq W_{14}$. Thus

$\epsilon^{2}\prec r^{2}_{14}\approx r^{2}_{24}\approx r^{2}_{34}\preceq 1.$

####  5.3.3 When $\sum_{j=1}^{3}m_{j}=0$

Then it is necessary that

$\frac{1}{{|m_{1}|^{\frac{1}{2}}}}+\frac{1}{{|m_{2}|^{\frac{1}{2}}}}+\frac{1}{{|m_{3}|^{\frac{1}{2}}}}=m_{1}+m_{2}+m_{3}=0.$
(19)

The above system has six solutions of $(m_{1},m_{2},m_{3})$. It is not
necessary to find them for our purpose. By
$\sum_{j=1}^{4}m_{j}w_{4}=\sum_{j=1}^{3}m_{j}w_{j4}$ it follows that
$w_{4}\prec w_{14}.$

Case 1). If $z_{14}\prec\epsilon$, this is impossible since the system by
equations (18) and (19) has no real solution.

Case 2). If $z_{14}\approx\epsilon$, then

$\epsilon^{2}\prec r^{2}_{14}\approx r^{2}_{24}\approx
r^{2}_{34}\prec\epsilon^{-1};$

Case 3). If $z_{14}\succ\epsilon$, then we have

$\epsilon^{2}\prec r^{2}_{14}\approx r^{2}_{24}\approx
r^{2}_{34}\prec\epsilon^{-4}.$

Similarly, Diagram III has three other variants. Two of them are presented in
the third row of Figure 11. We still call them Diagram III if there is no
confusion arise. The corresponding mass sets are denoted by

$\mathcal{V}_{III}[ijk],$

if the fully edged triangle has vertices i, j, and k. For instance, for the
two diagrams in the third row of Figure 11, the corresponding mass sets are
denoted by $\mathcal{V}_{III}[234]$ and $\mathcal{V}_{III}[134]$ respectively.
To distinguish, we may refer to the first one as Diagram III with
$\triangle_{234}$, the second one as Diagram III with $\triangle_{134}$ and so
on.

###  5.4 Diagram IV

The masses belong to the set

$\mathcal{V}_{IV}[123]:~{}~{}~{}~{}~{}~{}~{}~{}~{}\sum_{j=1}^{3}m_{j}=0,$

and we have

$r_{12},r_{23},r_{13}\approx\epsilon,r_{14}\sim\pm r_{24}\sim\pm
r_{34}\approx\epsilon^{-2}.$ (20)

Similarly, Diagram IV has three other variants. One of them is presented in
the fourth row of Figure 11. We still call them Diagram IV if there is no
confusion arise. The corresponding mass sets are denoted by

$\mathcal{V}_{IV}[ijk],$

if the fully edged triangle has vertices i, j, and k. For instance, for the
second diagram in the fourth row of Figure 11, the corresponding mass sets is
denoted by $\mathcal{V}_{IV}[124]$. To distinguish, we may refer to the first
one as Diagram IV with $\triangle_{123}$ and so on.

###  5.5 Diagram V

Following the argument from [1], we have

$\frac{Z_{12}}{m_{3}}\sim\frac{Z_{23}}{m_{1}}\sim\frac{Z_{31}}{m_{1}}\sim
a\epsilon^{-2},\ {\rm and\ set}\ z_{12}=m_{3},\ z_{31}=m_{2},$

with $a$ being some nonzero constant. By $Z_{kl}=r_{kl}^{-3}z_{kl}$, we see

$\frac{m_{1}}{r_{12}^{3}}\sim\frac{-(m_{2}+m_{3})}{r_{23}^{3}}\sim\frac{m_{1}}{r_{31}^{3}}\sim
am_{1}\epsilon^{-2}.$

Then we have

$r_{12}\sim\epsilon^{\frac{2}{3}}c\sqrt[3]{m_{1}},\
r_{23}\sim\epsilon^{\frac{2}{3}}c\sqrt[3]{-(m_{2}+m_{3})},\
r_{31}\sim\epsilon^{\frac{2}{3}}c\sqrt[3]{m_{1}},\ c\neq 0.$

In this subsection and in this subsection only, $\sqrt[3]{x}\
(x\in\mathbb{R})$ is understood as the real cubic root of $x$. Then the
asymptotic relation
$\frac{1}{m_{1}r_{23}}+\frac{1}{m_{2}r_{31}}+\frac{1}{m_{3}r_{12}}\sim 0$
implies

$\frac{1}{m_{1}\sqrt[3]{-(m_{2}+m_{3})}}+\frac{1^{\frac{1}{3}}}{m_{2}\sqrt[3]{m_{1}}}+\frac{1^{\frac{1}{3}}}{m_{3}\sqrt[3]{m_{1}}}=0.$

The above equation holds if and and only if

$m_{1}\sqrt[3]{-(m_{2}+m_{3})}=m_{2}\sqrt[3]{m_{1}}=m_{3}\sqrt[3]{m_{1}},\
{\it or}\
\frac{1}{m_{1}\sqrt[3]{-(m_{2}+m_{3})}}+\frac{1}{m_{2}\sqrt[3]{m_{1}}}+\frac{1}{m_{3}\sqrt[3]{m_{1}}}=0.$

The first case has no real solutions, and the second case reduces to

$\mathcal{V}_{V}[1,23]:~{}~{}~{}~{}~{}~{}~{}~{}~{}m_{1}^{2}(m_{2}+m_{3})^{4}=m_{2}^{3}m_{3}^{3},$

which has real solutions only if $m_{2}m_{3}>0$.

Note that

$\sum_{j=1}^{4}m_{j}w_{1}+m_{4}w_{14}=\sum_{j=1}^{4}m_{j}w_{j}+m_{2}w_{21}+m_{3}w_{31}=m_{2}w_{21}+m_{2}w_{31}\preceq\epsilon,$

then $w_{14}\approx\epsilon^{-2}$ and $z_{14}\approx\epsilon^{2}$. It follows
that

$\sum_{j=1}^{3}m_{j}\neq 0,r_{12},r_{23},r_{13}\approx\epsilon,r_{14}\approx
1,r_{24}\approx r_{34}\approx\sqrt{\epsilon}^{-1}.$ (21)

Similarly, Diagram V has 11 other variants. Some of them are presented in the
fifth row of Figure 11. We still call them Diagram V if there is no confusion
arise. The corresponding mass sets are denoted by

$\mathcal{V}_{V}[i,jk],$

if the fully edged triangle has vertices i, j, k and vertex i is connected to
the fourth vertex by one $w$-stroke. For instance, for the first two diagrams
in the fifth row of Figure 11, the corresponding mass sets are denoted by
$\mathcal{V}_{V}[3,14]$ and $\mathcal{V}_{V}[4,13]$ respectively. Both of the
first two diagrams will be referred as Diagram V with $\triangle_{134}$.

###  5.6 Diagram VI

We will reach our result without discussing the mass polynomial of this
diagram. Here is one remarks about this diagram. Note that the momentum of
inertia $I$ tends to zero by estimation 2. By Lemma 2.2, the momentum of
inertia is constant on a continuum of central configurations, so such a
singular sequence exists only on the subset $I=0$.

######  Proposition 5.1.

For singular sequences corresponding to Diagram V and VI, we have the
following estimates

$r_{jk}r_{lp}\prec 1,\ r_{jk}^{2}r_{lp}\preceq 1,$

where $r_{jk}$ and $r_{lp}$ are any two nonadjacent distances. The same
estimate holds for Diagram III, except for third case of Subsection 5.3.3, for
which it is necessary that the masses satisfy equation (19).

##  6 Finiteness of Central configurations with $m\neq 0$

In this section, we esabilish the finiteness of central configurations in the
situation where the toatl mass $m\neq 0$. There are two cases, whether there
is some partial sum of three masses being zero or not.

The first case, where all partial sum of three masses are nonzero, is
considered in the first two subsections and we show that

######  Theorem 6.1.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero, if
$\sum_{i=1}^{4}m_{i}\neq 0$ and $\prod_{1\leq j<k<l\leq
4}(m_{j}+m_{k}+m_{l})\neq 0$, then system (3), which defines the normalized
central configurations in the complex domain, possesses finitely many
solutions.

The second case, where at least one partial sum of three masses is zero, is
considered in the next two subsections and we show that

######  Theorem 6.2.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero, if
$\sum_{i=1}^{4}m_{i}\neq 0$ and $\prod_{1\leq j<k<l\leq
4}(m_{j}+m_{k}+m_{l})=0$, then system (3), which defines the normalized
central configurations in the complex domain, possesses finitely many
solutions.

Since the momentum of inertia is a constant on a continuum of solution of
system (3), the proof of the above two theorems completes after we establish
Theorem 6.3, Theorem 6.4, Theorem 6.6 and Theorem 6.7.

###  6.1 Finiteness with $m\neq 0$, $\prod_{1\leq k\leq 4}(m-m_{k})\neq 0$ and
$I\neq 0$

######  Theorem 6.3.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero, if
$\sum_{i=1}^{4}m_{i}\neq 0$, $\prod_{1\leq j<k<l\leq 4}(m_{j}+m_{k}+m_{l})\neq
0$ and $I\neq 0$, then system (3), which defines the normalized central
configurations in the complex domain, possesses finitely many solutions.

In this case only Diagram I, Diagram II, Diagram III and Diagram V are
possible. From now on, we postpone the tedious computations of the proofs to
the Appendix, and we will say $\triangle_{jkl}\longrightarrow 0$, if we have
one singular sequence corresponding to one diagram with a fully edged triangle
with vertices j, k, l.

_Proof of Theorem 6.3_: It is easy to see that giving five of $r_{kl}^{2}$’s,
$1\leq k<l\leq 4$, determines only finitely many geometrical configurations up
to rotation. Suppose that there are infinitely many solutions of system (4) in
the complex domain. Then at least two of $r_{kl}^{2}$’s must take infinitely
many values and thus are dominating by Lemma 2.1. Suppose that $r_{kl}^{2}$ is
dominating for some $1\leq k<l\leq 4$. There must exist a singular sequence of
central configurations with $r_{kl}^{2}\longrightarrow 0$, which happens only
in Diagram III or Diagram V. In either case, by Proposition 5.1,

$r_{kl}^{2}r_{ij}^{2}\longrightarrow 0,\ r_{ki}^{2}r_{jl}^{2}\longrightarrow
0,\ r_{il}^{2}r_{kj}^{2}\longrightarrow 0,$

along the singular sequence. Then all the three polynomials

$r_{13}^{2}r_{24}^{2},\ r_{14}^{2}r_{23}^{2},\ r_{12}^{2}r_{34}^{2},$

are dominating. It is easy to see that there exist
$\triangle_{ijk}\longrightarrow 0$ (say $\triangle_{123}\longrightarrow 0$),
so $r^{2}_{12},r^{2}_{23}$ and $r^{2}_{13}$ are dominating.

There also exist singular sequences with
$r_{12}^{2}r_{34}^{2}\longrightarrow\infty$, singular sequences with
$r_{13}^{2}r_{24}^{2}\longrightarrow\infty$ and singular sequences with
$r_{14}^{2}r_{23}^{2}\longrightarrow\infty$. These sequences must correspond
to Diagram I or Diagram II.

Consider a singular sequence with $r_{13}^{2}r_{24}^{2}\longrightarrow\infty$,
thus, the masses must be belong to

$\mathcal{V}_{IA}[12,34]\bigcup\mathcal{V}_{IA}[14,23]\bigcup\mathcal{V}_{IB}[12,34]\bigcup\mathcal{V}_{IB}[14,23]\bigcup\mathcal{V}_{II}[13,24].$

Note that
$\mathcal{V}_{IA}[12,34]\bigcup\mathcal{V}_{IA}[14,23]\subset\mathcal{V}_{II}[13,24]$.
Thus we the above set is the union of

$\begin{array}[]{lc}1).~{}~{}~{}\mathcal{V}_{II}[13,24];&\\\
2).~{}~{}~{}\mathcal{V}_{IB}[12,34]\bigcup\mathcal{V}_{IB}[14,23].&\end{array}$
(22)

Repeat the argument with $r_{14}^{2}r_{23}^{2}$. Then the masses must belong
to one of

$\begin{array}[]{lc}1).~{}~{}~{}\mathcal{V}_{II}[14,23];&\\\
2).~{}~{}~{}\mathcal{V}_{IB}[12,34]\bigcup\mathcal{V}_{IB}[13,24].&\end{array}$
(23)

Repeat the argument with $r_{12}^{2}r_{34}^{2}$. Then the masses must belong
to one of

$\begin{array}[]{lc}1).~{}~{}~{}\mathcal{V}_{II}[12,34];&\\\
2).~{}~{}~{}\mathcal{V}_{IB}[13,24]\bigcup\mathcal{V}_{IB}[14,23].&\end{array}$
(24)

Our strategy is to show that the three constraints (22), (23) and (24),
together with other available constraints, can not be satisfied
simultaneously.

Define

$\displaystyle V[123]\triangleq$
$\displaystyle\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[1,23]\bigcup\mathcal{V}_{V}[2,13]\bigcup\mathcal{V}_{V}[3,12];$
$\displaystyle V[124]\triangleq$
$\displaystyle\mathcal{V}_{III}[124]\bigcup\mathcal{V}_{V}[1,24]\bigcup\mathcal{V}_{V}[2,14]\bigcup\mathcal{V}_{V}[4,12];$
$\displaystyle V[134]\triangleq$
$\displaystyle\mathcal{V}_{III}[134]\bigcup\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\bigcup\mathcal{V}_{V}[4,13];$
$\displaystyle V[234]\triangleq$
$\displaystyle\mathcal{V}_{III}[234]\bigcup\mathcal{V}_{V}[2,34]\bigcup\mathcal{V}_{V}[3,24]\bigcup\mathcal{V}_{V}[4,23].$

Case 1: Three of $1)$ are satisfied.  That is,
$m_{1}m_{3}=m_{2}m_{4},m_{1}m_{4}=m_{2}m_{3},m_{1}m_{2}=m_{3}m_{4}$. Then
$m_{1}^{2}=m_{2}^{2}=m_{3}^{2}=m_{4}^{2}$. Then $m_{1}=m_{2}=m_{3}=m_{4}$.
However, recall that $\triangle_{123}\longrightarrow 0$, consequently, the
masses belong to the set $V[123]$, a contradiction.

Case 2: Two of $1)$ are satisfied. Without lose of generality, assume that the
masses belong to the set $\mathcal{V}_{II}[13,24]\mathcal{V}_{II}[14,23]$,
i.e., $m_{1}m_{3}=m_{2}m_{4},m_{1}m_{4}=m_{2}m_{3}$. Then
$m_{1}^{2}=m_{2}^{2},m_{3}^{2}=m_{4}^{2}$. Then $m_{1}=m_{2},m_{3}=m_{4}$. The
masses also belong to the set

$\mathcal{V}_{IB}[13,24]\bigcup\mathcal{V}_{IB}[14,23].$

However, recall that $\triangle_{123}\longrightarrow 0$, consequently, the
masses belong to the set $V[123]$, a contradiction.

Case 3: One of $1)$ is satisfied. Without lose of generality, assume that the
masses belong to the set $\mathcal{V}_{II}[13,24]$, i.e.,
$m_{1}m_{3}=m_{2}m_{4}$. Then the masses also belong to the set

$\begin{array}[]{c}(\mathcal{V}_{IB}[12,34]\bigcup\mathcal{V}_{IB}[13,24])(\mathcal{V}_{IB}[13,24]\bigcup\mathcal{V}_{IB}[14,23])\\\
=(\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[14,23])\bigcup\mathcal{V}_{IB}[13,24].\end{array}$

Subcase 1: $\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[14,23]$:

Note that $\triangle_{123}\longrightarrow 0$. It is easy to check that
$\mathcal{V}_{II}[13,24]\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[14,23]V[123]$
is empty, a contradiction.

Figure 10:

Subcase 2: $\mathcal{V}_{IB}[13,24]$:

Except that $\triangle_{123}\longrightarrow 0$, note that in this case one of
the two diagrams in Figure 10 occurs, thus $r_{14}^{2}$ and $r_{34}^{2}$ are
dominating polynomials. Consider a singular sequence with
$r^{2}_{14}\longrightarrow 0$, then the sequence corresponds to Diagram III or
V. In Diagram III, the distances not in the fully edged triangle $\succeq 1$
since $I\neq 0$. Thus, we have

$\triangle_{124}\longrightarrow 0,\ {\it or}\ \triangle_{134}\longrightarrow
0.$

Repeat the argument with $r_{34}^{2}$, we have

$\triangle_{134}\longrightarrow 0,\ {\it or}\ \triangle_{234}\longrightarrow
0.$

Thus, we conclude that we have

$\triangle_{134}\longrightarrow 0,\ {\it or\ }\ \triangle_{124}\longrightarrow
0,\triangle_{234}\longrightarrow 0.$

However, it is easy to check that both of the following two sets

$\displaystyle\mathcal{V}_{II}[13,24]\mathcal{V}_{IB}[13,24]V[123]V[134],\
\mathcal{V}_{II}[13,24]\mathcal{V}_{IB}[13,24]V[123]V[124]V[234],$

are empty, a contradiction.

Case 4: No $1)$ is satisfied.Then the masses belong to the set

$\begin{array}[]{c}(\mathcal{V}_{IB}[12,34]\bigcup\mathcal{V}_{IB}[14,23])(\mathcal{V}_{IB}[12,34]\bigcup\mathcal{V}_{IB}[13,24])(\mathcal{V}_{IB}[13,24]\bigcup\mathcal{V}_{IB}[14,23])\\\
=(\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[14,23])\bigcup(\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[13,24])\bigcup(\mathcal{V}_{IB}[13,24]\mathcal{V}_{IB}[14,23]).\end{array}$

Without lose of generality, assume that the masses belong to the set
$\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[13,24]$. Except that
$\triangle_{123}\longrightarrow 0$, note that:

By $\mathcal{V}_{IB}[12,34]$, repeat the argument utilized above, we have two
dominating polynomials, $r_{14}^{2},r_{24}^{2}$. Then let
$r_{14}^{2}\longrightarrow 0$, we have $\triangle_{124}\longrightarrow 0,\
{\it or}\ \triangle_{134}\longrightarrow 0.$ Let $r_{24}^{2}\longrightarrow
0$, we have $\triangle_{124}\longrightarrow 0,\ {\it or}\
\triangle_{234}\longrightarrow 0.$ Thus, we conclude that we have

$\triangle_{124}\longrightarrow 0,\ {\it or\ }\ \triangle_{134}\longrightarrow
0,\triangle_{234}\longrightarrow 0.$

Repeat the argument for $\mathcal{V}_{IB}[13,24]$ and the two polynomials
$r_{14}^{2},r_{34}^{2}$. We have

$\triangle_{134}\longrightarrow 0,\ {\it or\ }\ \triangle_{124}\longrightarrow
0,\triangle_{234}\longrightarrow 0.$

In short, at least two of $\triangle_{124}\longrightarrow 0$,
$\triangle_{134}\longrightarrow 0$ and $\triangle_{234}\longrightarrow 0$
occur.

Subcase 1: $\triangle_{124}\longrightarrow 0\ {\rm and}\
\triangle_{134}\longrightarrow 0$:

However, it is easy to check that

$\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[13,24]V[123]V[124]V[134]$

is empty, a contradiction.

Subcase 2: $\triangle_{124}\longrightarrow 0\ {\rm and}\
\triangle_{234}\longrightarrow 0$:

However, it is easy to check that

$\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[13,24]V[123]V[124]V[234]$

is empty, a contradiction.

Subcase 3: $\triangle_{134}\longrightarrow 0\ {\rm and}\
\triangle_{234}\longrightarrow 0$:

However, it is easy to check that

$\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[13,24]V[123]V[134]V[234]$

is empty, a contradiction.

To summarize, we proved that the system (3) possesses finitely many solutions,
if $\sum_{i=1}^{4}m_{i}\neq 0$, $\prod_{1\leq j<k<l\leq
4}(m_{j}+m_{k}+m_{l})\neq 0$ and $I\neq 0$.

$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\Box$

###  6.2 Finiteness with $m\neq 0$, $\prod_{1\leq k\leq 4}(m-m_{k})\neq 0$ and
$I=0$

######  Theorem 6.4.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero, if
$\sum_{i=1}^{4}m_{i}\neq 0$, $\prod_{1\leq j<k<l\leq 4}(m_{j}+m_{k}+m_{l})\neq
0$ and $I=0$, then system (3), which defines the normalized central
configurations in the complex domain, possesses finitely many solutions.

In this case only Diagram I, Diagram II, Diagram III, Diagram V and Diagram VI
are possible.

_Proof of Theorem 6.4_: It is easy to see that giving five of $r_{kl}^{2}$’s,
$1\leq k<l\leq 4$, determines only finitely many geometrical configurations up
to rotation. Suppose that there are infinitely many solutions of system (4) in
the complex domain. Then at least two of $\\{r_{kl}^{2}\\}$’s must take
infinitely many values and thus are dominating by Lemma 2.1. Suppose that
$r_{kl}^{2}$ is dominating for some $1\leq k<l\leq 4$. There must exist a
singular sequence of central configurations with $r_{kl}^{2}\longrightarrow
0$, which happens only in Diagram III, Diagram V or Diagram VI. In either
case, by Proposition 5.1,

$r_{kl}^{2}r_{ij}^{2}\longrightarrow 0,\ r_{ki}^{2}r_{jl}^{2}\longrightarrow
0,\ r_{il}^{2}r_{kj}^{2}\longrightarrow 0,$

along the singular sequence. Then all the three polynomials

$r_{13}^{2}r_{24}^{2},\ r_{14}^{2}r_{23}^{2},\ r_{12}^{2}r_{34}^{2},$

are dominating. It is easy to see that there exist
$\triangle_{ijk}\longrightarrow 0$ (say $\triangle_{123}\longrightarrow 0$),
so $r^{2}_{12},r^{2}_{23}$ and $r^{2}_{13}$ are dominating.

Then there also exist singular sequences with
$r_{12}^{2}r_{34}^{2}\longrightarrow\infty$, singular sequences with
$r_{13}^{2}r_{24}^{2}\longrightarrow\infty$ and singular sequences with
$r_{14}^{2}r_{23}^{2}\longrightarrow\infty$. These sequences must correspond
to Diagram I or Diagram II.

Consider a singular sequence with $r_{13}^{2}r_{24}^{2}\longrightarrow\infty$,
thus, the masses must belong to the set

$\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{I0}[14,23]\bigcup\mathcal{V}_{II0}[13,24],$

which is the union of

$\begin{array}[]{l}1).~{}~{}~{}\mathcal{V}_{II0}[13,24];\\\
2).~{}~{}~{}\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{I0}[14,23].\end{array}$
(25)

Repeat the argument with $r_{14}^{2}r_{23}^{2}$. Then the masses must belong
to one of the sets

$\begin{array}[]{l}1).~{}~{}~{}\mathcal{V}_{II0}[14,23];\\\
2).~{}~{}~{}\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{I0}[13,24].\end{array}$
(26)

Repeat the argument with $r_{12}^{2}r_{34}^{2}$. Then the masses belong to one
of the sets

$\begin{array}[]{l}1).~{}~{}~{}\mathcal{V}_{II0}[12,34];\\\
2).~{}~{}~{}\mathcal{V}_{I0}[13,24]\bigcup\mathcal{V}_{I0}[14,23].\end{array}$
(27)

Our strategy is to show that the three constraints (25), (26) and (27) are
incompatible.

Case 1: Three of $1)$ are satisfied.  A straightforward computation shows that
all the conditions are impossible.

Case 2: Two of $1)$ are satisfied. Without lose of generality, assume that the
masses belong to the set

$\mathcal{V}_{II0}[13,24]\mathcal{V}_{II0}[14,23](\mathcal{V}_{I0}[13,24]\bigcup\mathcal{V}_{I0}[14,23]).$

However, one checks that the set is empty.

Case 3: One of $1)$ is satisfied. Without lose of generality, assume that the
masses belong to the set

$\begin{array}[]{c}\mathcal{V}_{II0}[13,24](\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{I0}[13,24])(\mathcal{V}_{I0}[13,24]\bigcup\mathcal{V}_{I0}[14,23])\\\
=\mathcal{V}_{II0}[13,24]\mathcal{V}_{I0}[12,34]\mathcal{V}_{I0}[14,23]\bigcup\mathcal{V}_{II0}[13,24]\mathcal{V}_{I0}[13,24].\end{array}$

However, a straightforward computation shows that that the set is empty.

Case 4: No $1)$ is satisfied. Then the masses belong to the set

$\begin{array}[]{c}(\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{I0}[14,23])(\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{I0}[13,24])(\mathcal{V}_{I0}[13,24]\bigcup\mathcal{V}_{I0}[14,23])\\\
=(\mathcal{V}_{I0}[12,34]\mathcal{V}_{I0}[14,23])\bigcup(\mathcal{V}_{I0}[12,34]\mathcal{V}_{I0}[13,24])\bigcup(\mathcal{V}_{I0}[13,24]\mathcal{V}_{I0}[14,23]).\end{array}$

Without lose of generality, assume that the masses belong to the set
$\mathcal{V}_{I0}[12,34]\mathcal{V}_{I0}[13,24]$. A straightforward
computation shows that the set is empty.

To summarize, we proved that the system (3) possesses finitely many solutions,
if $\sum_{i=1}^{4}m_{i}\neq 0$, $\prod_{1\leq j<k<l\leq
4}(m_{j}+m_{k}+m_{l})\neq 0$ and $I=0$.

$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\Box$

###  6.3 Finiteness of Central configurations with $m\neq 0$, $\prod_{1\leq
k\leq 4}(m-m_{k})=0$ and $I\neq 0$

In this case Diagram I, Diagram II, Diagram III, Diagram IV , Diagram V and
Diagram VI are all possible. First, we establish a result without regard to
the value of $I$.

######  Lemma 6.5.

If some product $r_{jk}^{4}r_{lm}^{2}$ is not dominating on the closed
algebraic subset $\mathcal{A}$. Then system (3) possesses finitely many
solutions.

_Proof of Lemma 6.5_:

If system (3) possesses infinitely many solutions, without lose of generality,
assume that $r_{12}^{4}r_{34}^{2}$ is not dominating on the closed algebraic
subset $\mathcal{A}$. Then some level set $r_{12}^{4}r_{34}^{2}\equiv
const\neq 0$ also includes infinitely many solutions of system (3).

In this level set, all the corresponding possible diagrams are presented in
Figure 11. In particular, for Diagram III, we are in the third case of
Subsection 5.3.3. More precisely, if the fully edged triangle is
$\triangle_{jkl}$ and the other vertex is p, then $r^{2}_{pj}\sim
r^{2}_{pk}\sim r^{2}_{pl}\approx\epsilon^{-1}$, and the masses belong to the
set $\mathcal{V}_{III}[jkl]\mathcal{V}_{IV}[jkl]$.

Figure 11: All possible diagrams in the level set $r_{12}^{4}r_{34}^{2}=const$

Similarly, at least two of $\\{r_{kl}^{2}\\}$’s must take infinitely many
values and thus are dominating. Suppose that $r_{kl}^{2}$ is dominating for
some $1\leq k<l\leq 4$. There must exist a singular sequence of central
configurations with $r_{kl}^{2}\longrightarrow 0$, which happens only in
Diagram III, Diagram IV or Diagram V. By Proposition 5.1,
$r_{12}^{2}r_{34}^{2}\longrightarrow 0$ along the singular sequence. Then the
polynomial $r_{12}^{2}r_{34}^{2}$ is dominating.

Consider a singular sequence with $r_{12}^{2}r_{34}^{2}\longrightarrow\infty$,
then it is easy to see that this happens only in Diagram IV. Without lose of
generality, assume that Diagram IV with $\triangle_{123}$ occurs. Note that
$r^{4}_{12}r^{2}_{14}r^{2}_{24}\longrightarrow\infty$ along a singular
sequence corresponding to Diagram IV with $\triangle_{123}$, so
$r^{4}_{12}r^{2}_{14}r^{2}_{24}$ is dominating. When
$r^{4}_{12}r^{2}_{14}r^{2}_{24}\longrightarrow 0$, it is easy to see that this
happens only in Diagram IV with $\triangle_{124}$. Thus

$m_{1}+m_{2}+m_{3}=m_{1}+m_{2}+m_{4}=0,$ (28)

and $r^{2}_{13}r^{2}_{14}$ is a dominating polynomial.

Consider a singular sequence with $r^{2}_{13}r^{2}_{14}\longrightarrow 0$. It
is easy to see that this happens only in Diagram III with $\triangle_{134}$,
and in the first two of Diagram V. However, it is easy to check that equation
(28) is not consistent with any one of the following three sets

$\mathcal{V}_{III}[134]\mathcal{V}_{IV}[134],\ \mathcal{V}_{V}[3,14],\
\mathcal{V}_{V}[4,13].$

This is a contradiction.
$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\Box$

Hence, we assume that all $r_{jk}^{4}r_{lm}^{2}$’s are dominating on the
closed algebraic subset $\mathcal{A}$ from now on.

If Diagram IV and Diagram III (the third case of Subsection 5.3.3 with
$r_{lp}^{2}\succeq\epsilon^{-2}$, where vertex p is not in the fully edged
triangle) do not occur, then $r_{jk}r_{lp}\prec 1$ holds for Diagram III, and
hence the proof of Theorem 6.2 reduces to those of Theorem 6.1. Thus, we
assume that either Diagram IV or Diagram III (the third case of Subsection
5.3.3 with $r_{lp}^{2}\succeq\epsilon^{-2}$, where vertex p is not in the
fully edged triangle) occurs. Then there exists some
$\triangle_{ijk}\longrightarrow 0$. Without lose of generality, we assume that
$\triangle_{123}\longrightarrow 0$ below. Then

$m_{1}+m_{2}+m_{3}=0$ (29)

and it is easy to see that all of $r_{kl}^{2}$’s, $1\leq k<l\leq 4$, are
dominating.

Now we prove the finiteness in the case of $I\neq 0$.

######  Theorem 6.6.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero, if
$\sum_{i=1}^{4}m_{i}\neq 0$, $\prod_{1\leq j<k<l\leq 4}(m_{j}+m_{k}+m_{l})=0$
and $I\neq 0$, then system (3), which defines the normalized central
configurations in the complex domain, possesses finitely many solutions.

In this case only Diagram I, Diagram II, Diagram III, Diagram IV and Diagram V
are possible.

_Proof of Theorem 6.6_:

Consider a singular sequence with $r_{14}^{2}\longrightarrow 0$, then the
sequence corresponds to $\triangle_{124}\longrightarrow 0$,
$\triangle_{134}\longrightarrow 0$ in Diagram III, IV or V. Repeat the
argument with $r_{24}^{2}\longrightarrow 0$ and $r_{34}^{2}\longrightarrow 0$,
then it is easy to see that at least two of $\triangle_{124}\longrightarrow
0$, $\triangle_{134}\longrightarrow 0$ and $\triangle_{234}\longrightarrow 0$
occur. Without lose of generality, assume that $\triangle_{124}\longrightarrow
0$ and $\triangle_{134}\longrightarrow 0$ occur.

Corresponding to $\triangle_{124}\longrightarrow 0$ we have

$\begin{array}[]{c}(\frac{1}{|m_{1}|^{\frac{1}{2}}}+\frac{1}{|m_{2}|^{\frac{1}{2}}}+\frac{1}{|m_{4}|^{\frac{1}{2}}})(m_{1}+m_{2}+m_{4})(m_{1}^{2}(m_{2}+m_{4})^{4}-m_{2}^{3}m_{4}^{3})\\\
(m_{2}^{2}(m_{1}+m_{4})^{4}-m_{1}^{3}m_{4}^{3})(m_{4}^{2}(m_{2}+m_{1})^{4}-m_{2}^{3}m_{1}^{3})=0\end{array}$
(30)

Corresponding to $\triangle_{134}\longrightarrow 0$ we have

$\begin{array}[]{c}(\frac{1}{|m_{1}|^{\frac{1}{2}}}+\frac{1}{|m_{3}|^{\frac{1}{2}}}+\frac{1}{|m_{4}|^{\frac{1}{2}}})(m_{1}+m_{3}+m_{4})(m_{1}^{2}(m_{3}+m_{4})^{4}-m_{3}^{3}m_{4}^{3})\\\
(m_{3}^{2}(m_{1}+m_{4})^{4}-m_{1}^{3}m_{4}^{3})(m_{4}^{2}(m_{3}+m_{1})^{4}-m_{3}^{3}m_{1}^{3})=0\end{array}$
(31)

Let us further consider $r_{12}^{4}r_{34}^{2}\longrightarrow\infty$, then we
are in Diagram I, Diagram II, Diagram III or Diagram IV.

If we are in Diagram I of Diagram II.  Then the masses belong to the set

$\mathcal{V}_{IA}[13,24]\bigcup\mathcal{V}_{IA}[14,23]\bigcup\mathcal{V}_{IB}[13,24]\bigcup\mathcal{V}_{IB}[14,23]\bigcup\mathcal{V}_{II}[12,34].$
(32)

However, a straightforward computation shows that masses of the above set are
not consistent with equations (29), (30) and (31).

If we are in Diagram III. Then we have

$\left\\{\begin{array}[]{lr}\frac{1}{|m_{1}|^{\frac{1}{2}}}+\frac{1}{|m_{3}|^{\frac{1}{2}}}+\frac{1}{|m_{4}|^{\frac{1}{2}}}=0&\\\
m_{1}+m_{3}+m_{4}=0&\end{array}\right.$ (33)

or

$\left\\{\begin{array}[]{lr}\frac{1}{|m_{2}|^{\frac{1}{2}}}+\frac{1}{|m_{3}|^{\frac{1}{2}}}+\frac{1}{|m_{4}|^{\frac{1}{2}}}=0&\\\
m_{2}+m_{3}+m_{4}=0&\end{array}\right.$ (34)

However, a straightforward computation shows that all of these relations are
not consistent with equations (29), (30) and (31).

If we are in Diagram IV. Then it is Diagram IV with $\triangle_{234}$ or
$\triangle_{134}$.

Case 1: If we are in Diagram IV with $\triangle_{234}$: Then

$m_{2}+m_{3}+m_{4}=0.$ (35)

A straightforward computation shows that the above equation is not consistent
with equations (29), (30) and (31).

Case 2: If we are in Diagram IV with $\triangle_{134}$ : Then

$m_{1}+m_{3}+m_{4}=0.$ (36)

In this case, equation (36) is consistent with equations (29), (30) and (31)
and it is necessary to include more polynomials. Let us further consider
$r_{13}^{4}r_{24}^{2}\longrightarrow\infty$, we are in Diagram I, Diagram II,
Diagram III and Diagram IV.

Subcase 1: If in Diagram I or Diagram II. Then

$\mathcal{V}_{IA}[12,34]\bigcup\mathcal{V}_{IA}[14,23]\bigcup\mathcal{V}_{IB}[12,34]\bigcup\mathcal{V}_{IB}[14,23]\bigcup\mathcal{V}_{II}[13,24].$
(37)

However, a straightforward computation shows that masses of the above set are
not consistent with equations (29), (30) and (36).

Subcase 2: If in Diagram III. We have

$\left\\{\begin{array}[]{lr}\frac{1}{|m_{1}|^{\frac{1}{2}}}+\frac{1}{|m_{2}|^{\frac{1}{2}}}+\frac{1}{|m_{4}|^{\frac{1}{2}}}=0&\\\
m_{1}+m_{2}+m_{4}=0&\end{array}\right.$ (38)

or equation (34). However, a straightforward computation shows that neither
equation (38) nor equation (34) are consistent with equations (29), (30) and
(36).

Subcase 3: If in Diagram IV. We have $\triangle_{234}\longrightarrow 0$ or
$\triangle_{124}\longrightarrow 0$. If $\triangle_{234}\longrightarrow 0$,
then we have

$m_{2}+m_{3}+m_{4}=0,$

not consistent with equations (29), (30) and (36).

If $\triangle_{124}\longrightarrow 0$, then $m_{1}+m_{2}+m_{4}=0.$ By (29) and
(36), the masses are

$m_{1}=-2m_{2},m_{3}=m_{2},m_{4}=m_{2}.$ (39)

Note that $r_{23}^{2}r_{24}^{2}r_{34}^{2}\longrightarrow\infty$, it is
dominating. Let $r_{23}^{2}r_{24}^{2}r_{34}^{2}\longrightarrow 0$, we are in
Diagram III, IV and V. The masses satisfying equation (39) can not admit
Diagram III and V. In Diagram IV, the fully edged triangle can only be
$\triangle_{234}$, so we have

$m_{2}+m_{3}+m_{4}=0,$

which contradicts with (39).

To summarize, we proved that the system (3) possesses finitely many solutions,
if $\sum_{i=1}^{4}m_{i}\neq 0$, $\prod_{1\leq j<k<l\leq
4}(m_{j}+m_{k}+m_{l})=0$ and $I\neq 0$.

$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\Box$

###  6.4 Finiteness with $m\neq 0$, $\prod_{1\leq k\leq 4}(m-m_{k})=0$ and
$I=0$

######  Theorem 6.7.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero, if
$\sum_{i=1}^{4}m_{i}\neq 0$, $\prod_{1\leq j<k<l\leq 4}(m_{j}+m_{k}+m_{l})\neq
0$ and $I=0$, then system (3), which defines the normalized central
configurations in the complex domain, possesses finitely many solutions.

In this case all of Diagram I, Diagram II, Diagram III, Diagram IV, Diagram V
and Diagram VI are possible.

_Proof of Theorem 6.7_:

Let us first consider $r_{12}^{4}r_{34}^{2}\longrightarrow\infty$, then we are
in Diagram I, Diagram II, Diagram III or Diagram IV.

Then the masses belong to the set

$\begin{array}[]{c}\mathcal{V}_{I0}[13,24]\bigcup\mathcal{V}_{I0}[14,23]\bigcup\mathcal{V}_{II0}[12,34]\\\
\bigcup(\mathcal{V}_{III}[134]\mathcal{V}_{IV}[134])\bigcup(\mathcal{V}_{III}[234]\mathcal{V}_{IV}[234])\bigcup\mathcal{V}_{IV}[134]\bigcup\mathcal{V}_{IV}[234]\triangleq
V[12,34].\end{array}$ (40)

Similarly, when considering $r_{13}^{4}r_{24}^{2}\longrightarrow\infty$ the
masses belong to the set

$\begin{array}[]{c}\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{I0}[14,23]\bigcup\mathcal{V}_{II0}[13,24]\\\
\bigcup(\mathcal{V}_{III}[124]\mathcal{V}_{IV}[124])\bigcup(\mathcal{V}_{III}[234]\mathcal{V}_{IV}[234])\bigcup\mathcal{V}_{IV}[124]\bigcup\mathcal{V}_{IV}[234]\triangleq
V[13,24];\end{array}$ (41)

when considering $r_{23}^{4}r_{14}^{2}\longrightarrow\infty$ the masses belong
to the set

$\begin{array}[]{c}\mathcal{V}_{I0}[13,24]\bigcup\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{II0}[14,23]\\\
\bigcup(\mathcal{V}_{III}[124]\mathcal{V}_{IV}[124])\bigcup(\mathcal{V}_{III}[134]\mathcal{V}_{IV}[134])\bigcup\mathcal{V}_{IV}[124]\bigcup\mathcal{V}_{IV}[134]\triangleq
V[23,14].\end{array}$ (42)

Or

$\begin{array}[]{c}V[12,34]=(\mathcal{V}_{I0}[13,24])\bigcup(\mathcal{V}_{I0}[14,23])\bigcup(\mathcal{V}_{II0}[12,34])\bigcup\mathcal{V}_{IV}[134]\bigcup\mathcal{V}_{IV}[234];\\\
V[13,24]=(\mathcal{V}_{I0}[12,34])\bigcup(\mathcal{V}_{I0}[14,23])\bigcup(\mathcal{V}_{II0}[13,24])\bigcup\mathcal{V}_{IV}[124]\bigcup\mathcal{V}_{IV}[234];\\\
V[23,14]=(\mathcal{V}_{I0}[13,24])\bigcup(\mathcal{V}_{I0}[12,34])\bigcup(\mathcal{V}_{II0}[14,23])\bigcup\mathcal{V}_{IV}[124]\bigcup\mathcal{V}_{IV}[134].\end{array}$
(43)

Other than the above relations, recall that we have assumed that the masses
belong to $\mathcal{V}_{IV}[123]$.

Case 1: If Diagram I occurs. Without lose of generality, assume that the
masses belong to the set $\mathcal{V}_{I0}[13,24]$, then we claim that

$(\mathcal{V}_{I0}[13,24])\bigcap
V[13,24]\bigcap\mathcal{V}_{IV}[123]=\emptyset.$ (44)

A straightforward computation shows the claim.

Case 2: If Diagram II occurs. Without lose of generality, assume that the
masses belong to the set $\mathcal{V}_{II0}[12,34]$, then we claim that

$(\mathcal{V}_{II0}[12,34])\bigcap V[13,24]\bigcap\mathcal{V}_{IV}[123]\bigcap
V[23,14]=\emptyset.$ (45)

A straightforward computation shows the claim.

Therefore, Diagram I and Diagram II can not occur and the masses belong to the
following set

$\mathcal{V}_{IV}[123]\left(\mathcal{V}_{IV}[124]\mathcal{V}_{IV}[134]\bigcup\mathcal{V}_{IV}[124]\mathcal{V}_{IV}[234]\bigcup\mathcal{V}_{IV}[134]\mathcal{V}_{IV}[234]\right),$

which contains only three solutions

$m_{2}=m_{3}=m_{4}=-\frac{1}{2}m_{1},\ m_{1}=m_{3}=m_{4}=-\frac{1}{2}m_{2},\
m_{1}=m_{2}=m_{4}=-\frac{1}{2}m_{3}.$

It is easy to see that the first solution is not consistent with the
constraints corresponding to Diagram I, Diagram II, Diagram III or Diagram V,
therefore, only Diagram IV and Diagram VI are possible now. We consider the
function $r_{12}^{2}r_{23}^{2}r_{13}^{2}r_{14}^{2}r_{24}^{2}r_{34}^{2}$, then
it is easy to see that it is dominating. When considering
$r_{12}^{2}r_{23}^{2}r_{13}^{2}r_{14}^{2}r_{24}^{2}r_{34}^{2}\rightarrow 0$,
we are in Diagram VI, thus the function $r_{12}^{2}r_{13}^{2}r_{14}^{2}$ is
dominating. When considering
$r_{12}^{2}r_{13}^{2}r_{14}^{2}\rightarrow\infty$, we are in Diagram IV with
$\triangle_{234}\longrightarrow 0$. As a result, we have
$m_{2}+m_{3}+m_{4}=0$, which is a contradiction.

Similarly, the other two solutions do not admit infinite solutions neither.

To summarize, we proved that the system (3) possesses finitely many solutions,
if $\sum_{j=1}^{4}m_{j}\neq 0$, $\prod_{1\leq j<k<l\leq
4}(m_{j}+m_{k}+m_{l})=0$ and $I=0$.

$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\Box$

##  7 Finiteness of Central configurations with $m=0$

###  7.1 Problematic diagrams with vanishing total mass

We prove that the first and third diagram in Figure 5 and all diagrams in
Figure 6 and Figure 7 are impossible.

Case 1: The First and third diagram in Figure 5. First, it is easy to see that

$\begin{array}[]{c}m_{1}+m_{2}=0,\\\ m_{3}+m_{4}=0,\\\ w_{1}\sim w_{2}\sim
w_{3}\sim w_{4}\sim a\epsilon^{-2},\\\ z_{12}\approx z_{34}\approx
w_{12}\approx w_{34}\approx\epsilon.\end{array}$ (46)

It follows that

$w_{12}=m_{3}(W_{32}-W_{31})+m_{4}(W_{42}-W_{41}),$

we claim that $z_{1}\sim z_{2}\sim z_{3}\sim z_{4}$ for the first three
diagrams in Figure 5. Otherwise, we have $z_{jk}\approx\epsilon^{-2}$ and then
$W_{jk}=\frac{1}{z_{jk}^{\frac{3}{2}}w_{jk}^{\frac{1}{2}}}\prec\epsilon^{2}$
for $j=3,4,k=1,2$. This contradicts with $w_{12}\approx\epsilon$.

For the third diagram, we have $z_{2},z_{4}\prec z_{1},z_{2}$. This is a
contradiction, so the third diagram does not exist.

For the first diagram, we have

$z_{1}\sim m_{2}Z_{21},\ z_{2}\sim m_{1}Z_{12}+m_{3}Z_{32}.$

This is a contradiction, so the first diagram does not exist.

Case 2: The Diagrams in Figure 6. First, it is easy to see that the second and
third diagram are impossible.

For the first diagram, note that $r_{14},r_{24},r_{34}\succ\epsilon$. Without
loss of generality, assume that $w_{14}\succ\epsilon$, then we have
$w_{24}\sim w_{34}\sim w_{14}\succ\epsilon$. By

$0=\sum_{j=1}^{4}m_{j}w_{4}=\sum_{j=1}^{3}m_{j}w_{j4},$

it follows that $\sum_{j=1}^{3}m_{j}=0$, a contradiction.

Case 3: The Diagram in Figure 7. First, it is easy to see that
$w_{14}\succ\epsilon$, then we have $w_{24}\sim w_{34}\sim
w_{14}\succ\epsilon$. Repeat the argument in the previous case. We then arrive
at $\sum_{j=1}^{3}m_{j}=0$, a contradiction.

We could not eliminate the diagrams in Figure 12. Some singular sequence could
still exist and approach any of these diagrams.

Figure 12: Problematic diagrams with $m=0$

####  7.1.1 Diagram I

Note that the hypothesis that $m\neq 0$ is not utilized in Subsection 5.1,
thus the estimates on the distances and mass polynomials derived there still
holds for the present case. In particular, we have

$(m_{1}+m_{2})(m_{3}+m_{4})\neq 0,$

$r_{12}^{3}\sim m_{1}+m_{2},\ r_{34}^{3}\sim m_{3}+m_{4},\
r_{jk}\approx\epsilon^{-2},j=1,2,k=3,4.$ (47)

The masses corresponding to Diagram I belong to one of the following two sets

$\left\\{\begin{array}[]{ll}\mathcal{V}_{IA0}[12,34]:,&\hbox{$m_{1}=m_{2},\
m_{3}=m_{4},\ m_{1}+m_{3}=0;$}\\\
\mathcal{V}_{IB}[12,34]:,&\hbox{$\mu_{1}^{3}\mu_{3}^{3}+\mu_{2}^{3}\mu_{3}^{3}+\mu_{1}^{3}\mu_{4}^{3}+\mu_{2}^{3}\mu_{4}^{3}=0,\
m_{1}m_{2}<0,\ m_{3}m_{4}<0,$}\\\ &\hbox{$m_{1}+m_{2}\neq 0,\ m_{3}+m_{4}\neq
0.$}\\\ \end{array}\right.$

####  7.1.2 Diagram II

Following the argument from [1], the masses corresponding to Diagram II
satisfy the following equation

$m_{1}m_{3}=m_{2}m_{4}.$

Furthermore, we claim that

$m_{1}+m_{2}=m_{2}+m_{3}=m_{3}+m_{4}=m_{1}+m_{4}=0.$ (48)

Indeed, if $m_{1}+m_{2}\neq 0$, then we have

$z_{12}\approx\epsilon^{-2},\ m_{3}+m_{4}\neq 0,\ z_{34}\approx\epsilon^{-2}.$

By the fact
$z_{12}=(m_{1}+m_{2})Z_{12}+m_{3}(Z_{32}-Z_{31})+m_{4}(Z_{42}-Z_{41})$, we
have $z_{12}\sim(m_{1}+m_{2})Z_{12}$ and then $w_{12}\approx\epsilon^{2}$.
Hence, $W_{12}=w_{12}^{-1/2}z_{12}^{-3/2}\approx\epsilon^{2}$. Similarly,
$W_{34}\approx\epsilon^{2}$. We also would have $z_{13}\approx\epsilon^{-2}$
and $w_{13}\succ\epsilon$, then $W_{13}\prec\epsilon^{2}$. Similarly,
$W_{24}\prec\epsilon^{2}$.

On the other hand, note $z_{13}\sim z_{12},z_{14}\prec\epsilon.$ The the
identity

$0=\sum_{j=1}^{4}m_{j}z_{1}=\sum_{j=2}^{4}m_{j}z_{j1}$

implies that $m_{2}+m_{3}=0$, then $m_{1}+m_{4}=0$.

Hence, we have
$w_{14}=(m_{1}+m_{4})W_{14}+m_{2}(W_{24}-W_{21})+m_{3}(W_{34}-W_{31})\preceq\epsilon^{2}$.
By Estimate 1, $w_{14}\approx\epsilon^{2}$. Note that
$W_{14}\approx\epsilon^{-2}$. It follows that $r_{14}\approx\epsilon^{4/3}$,
which contradicts with Estimate 1. Thus equation (48) follows.

Note that

$\begin{array}[]{c}r_{12},r_{23},r_{34},r_{14}\prec 1,\\\
r_{13}\prec\epsilon^{-2},r_{24}\prec\epsilon^{-2}.\end{array}$ (49)

####  7.1.3 Diagrams with two $zw$-edges

Obviously, we have $m_{1}+m_{2}=m_{3}+m_{4}=0$. Then the identities
$0=\sum_{i=1}^{4}m_{i}z_{i}=\sum_{i=1}^{4}m_{i}w_{i}$ imply

$m_{1}z_{12}=-m_{3}z_{34},\ m_{1}w_{12}=-m_{3}w_{34}.$ (50)

For the last two diagrams with two $zw$-edges in Figure 12, we have

$w_{23}=(m_{2}+m_{3})W_{23}+m_{1}(W_{13}-W_{12})+m_{4}(W_{43}-W_{42})\prec\epsilon^{-2},$

which leads to

$m_{1}W_{12}\sim m_{3}W_{34}.$ (51)

With (50), we have $m_{1}^{2}=m_{3}^{2}$.

For the first diagram with two $zw$-edges in Figure 12, we claim that at least
one of $z_{23,}w_{23}$ shall have order less than $\epsilon^{-2}$. Otherwise,
$W_{23}\approx W_{24}\approx W_{13}\approx W_{14}\approx\epsilon^{4}$. This
contradicts with the equation

$w_{12}=(m_{1}+m_{2})W_{12}+m_{3}(W_{32}-W_{31})+m_{4}(W_{42}-W_{41}).$

Then we have either

$m_{1}W_{12}\sim m_{3}W_{34},\ {\it or}\ m_{1}Z_{12}\sim m_{3}Z_{34}.$ (52)

With (50), we have $m_{1}^{2}=m_{3}^{2}$.

Therefore, the masses corresponding to Diagrams with two $zw$-edges satisfy

$m_{1}+m_{2}=m_{3}+m_{4}=0,\ m_{1}^{2}=m_{3}^{2}.$ (53)

####  7.1.4 Fully edged diagrams

We will reach our result without discussing the mass polynomial of this
diagram.

###  7.2 Finiteness of Central configurations with $m=0$

######  Theorem 7.1.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero with
$\sum_{i=1}^{4}m_{j}=0$, then system (3), which defines the normalized central
configurations in the complex domain, possesses finitely many solutions except
perhaps for, up to renumeration of the bodies, two groups of masses:

$m_{1}=m_{2}=-m_{3}=-m_{4},\
m_{1}=m_{2},m_{3}=-\alpha^{2}m_{1},m_{4}=-(2-\alpha^{2})m_{1},$

where $\alpha$ is the unique positive root of equation (54).

Numerically, $\alpha\approx 1.2407$.

Note that if Diagram I with $\mathcal{V}_{IA0}$, Diagram II, or Diagrams with
two $zw$-edges occurs, then $m_{k}=\pm m_{1},k=2,3,4$. Thus we assume that
Diagram I with $\mathcal{V}_{IA0}$, Diagram II, and Diagrams with two
$zw$-edges do not occur in the following discussion.

_Proof of Theorem 7.1_: It is easy to see that giving five of $r_{kl}^{2}$’s,
$1\leq k<l\leq 4$, determines only finitely many geometrical configurations up
to rotation. Suppose that there are infinitely many solutions of system (4) in
the complex domain. Then at least two of $\\{r_{kl}^{2}\\}$’s must take
infinitely many values and thus are dominating by Lemma 2.1. Suppose that
$r_{kl}^{2}$ is dominating for some $1\leq k<l\leq 4$. There must exist a
singular sequence of central configurations with $r_{kl}^{2}\longrightarrow
0$, which happens only in the fully edged diagrams. In each case, it is easy
to see that all of $r_{kl}^{2}$’s, $1\leq k<l\leq 4$, and
$r_{13}^{2}r_{24}^{2},\ r_{14}^{2}r_{23}^{2},\ r_{12}^{2}r_{34}^{2}$ are
dominating.

Consider a singular sequence with $r_{12}^{2}\longrightarrow\infty$, then the
masses must belong to $\mathcal{V}_{IB}[13,24]\bigcup\mathcal{V}_{IB}[14,23]$.
Consider a singular sequence with $r_{13}^{2}\longrightarrow\infty$, then the
masses must belong to $\mathcal{V}_{IB}[12,34]\bigcup\mathcal{V}_{IB}[14,23]$.
Consider a singular sequence with $r_{14}^{2}\longrightarrow\infty$, then the
masses must belong to $\mathcal{V}_{IB}[12,34]\bigcup\mathcal{V}_{IB}[13,24]$.
Hence, at least two of $\mathcal{V}_{IB}[12,34]$, $\mathcal{V}_{IB}[13,24]$
and $\mathcal{V}_{IB}[14,23]$ occur. Without lose of generality, assume that
$\mathcal{V}_{IB}[13,24]$ and $\mathcal{V}_{IB}[14,23]$ occur. That is, the
masses belong to

$\mathcal{V}_{IB}[13,24]\bigcap\mathcal{V}_{IB}[14,23]\bigcap\mathcal{V}_{m0},$

where $\mathcal{V}_{m0}$ denotes the set $\sum_{j=1}^{4}m_{j}=0$.

It is easy to see $m_{1}m_{2}>0$ and $m_{3}m_{4}>0$. Without loss of
generality, assume that $m_{1},m_{2}>0,m_{3},m_{4}<0$. Then a straightforward
computation shows that the set
$\mathcal{V}_{IB}[13,24]\bigcap\mathcal{V}_{IB}[14,23]\bigcap\mathcal{V}_{m0}$
is empty, provided that $\prod_{i\neq j}(|m_{i}|-|m_{j}|)\neq 0.$ Note that

$(m_{1}+m_{3})(m_{2}+m_{4})\neq 0,\ (m_{1}+m_{4})(m_{2}+m_{3})\neq 0,$

Hence, assume that $m_{1}=m_{2}$, or $m_{3}=m_{4}$. A straightforward
computation shows the set
$\mathcal{V}_{IB}[13,24]\bigcap\mathcal{V}_{IB}[14,23]\bigcap\mathcal{V}_{m0}$
reduces to

$\displaystyle\\{m_{1}=m_{2},m_{3}=-\alpha^{2}m_{1},m_{4}=-(2-\alpha^{2})m_{1}\\}\bigcup\\{m_{1}=m_{2},m_{3}=-(2-\alpha^{2})m_{1},m_{4}=-\alpha^{2}m_{1}\\}$
$\displaystyle\bigcup\\{m_{1}=-\alpha^{2}m_{3},m_{2}=-(2-\alpha^{2})m_{1},m_{3}=m_{4}\\}\bigcup\\{m_{1}=-(2-\alpha^{2})m_{3},m_{2}=-\alpha^{2}m_{1},m_{3}=m_{4}\\}$

where $\alpha$ is the unique positive root of the following equation:

$(1-\alpha^{3})^{2}-(1+\alpha^{3})^{2}(2-\alpha^{2})^{3}=0.$ (54)

$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\Box$

###  7.3 Finiteness with $m_{1}=m_{2}=-m_{3}=-m_{4}$, or
$m_{1}=m_{2},m_{3}=-\alpha^{2}m_{1},m_{4}=-(2-\alpha^{2})m_{1}$.

######  Theorem 7.2.

Suppose that $m_{1},m_{2},m_{3},m_{4}$ are real and nonzero, if
$m_{1}=m_{2}=-m_{3}=-m_{4}$, or
$m_{1}=m_{2},m_{3}=-\alpha^{2}m_{1},m_{4}=-(2-\alpha^{2})m_{1}$, then system
(2) possesses finitely many solutions.

We divide the proof into the collinear case and planar case.

####  7.3.1 The collinear case

_Proof of Theorem 7.2, Part 1_:

Assume that the masses are on the $x$-axis, and $x_{i}$ is the coordinate of
$m_{i}$. Then system (2) becomes

$\displaystyle\sigma x_{1}$
$\displaystyle=m_{2}\frac{x_{1}-x_{2}}{|x_{1}-x_{2}|^{3}}+m_{3}\frac{x_{1}-x_{3}}{|x_{1}-x_{3}|^{3}}+m_{4}\frac{x_{1}-x_{4}}{|x_{1}-x_{4}|^{3}}$
$\displaystyle\cdots$ (55) $\displaystyle\sigma x_{4}$
$\displaystyle=m_{1}\frac{x_{4}-x_{1}}{|x_{4}-x_{1}|^{3}}+m_{2}\frac{x_{4}-x_{2}}{|x_{4}-x_{2}|^{3}}+m_{3}\frac{x_{4}-x_{3}}{|x_{4}-x_{3}|^{3}},$

where $\sigma=\pm 1$.

If $m_{1}=m_{2}=-m_{3}=-m_{4}$, without lose of generality, assume
$x_{1}<x_{i}$, $i=2,3,4$. Since $\sum m_{i}x_{i}=0$, it suffices to assume
that the ordering is $x_{1}<x_{3}<x_{4}<x_{2}$. A straightforward computation
shows that system (55) has finitely many solutions.

If $m_{1}=m_{2},m_{3}=-\alpha^{2}m_{1},m_{4}=-(2-\alpha^{2})m_{1}$, by the
reflection symmetry and the fact that $m_{1}=m_{2}$, it suffices to consider
the following six orderings: $x_{1}<x_{2}<x_{3}<x_{4}$,
$x_{1}<x_{2}<x_{4}<x_{3}$, $x_{1}<x_{4}<x_{2}<x_{3}$,
$x_{4}<x_{1}<x_{2}<x_{3}$, $x_{1}<x_{3}<x_{2}<x_{4}$,
$x_{1}<x_{4}<x_{2}<x_{3}$. Note that the first two orderings are impossible
since $\frac{x_{1}+x_{2}}{2}\in(x_{1},x_{2})$, and
$\frac{-m_{3}x_{3}-m_{4}x_{4}}{2}>x_{2}$. The third ordering is also
impossible since

$x_{4}=\frac{x_{1}+x_{2}+m_{3}x_{3}}{-m_{4}}=\frac{(2+m_{3})x_{1}+(1+m_{3})(x_{2}-x_{1})-m_{3}(x_{2}-x_{3})}{-m_{4}}<x_{1}.$

For each of the last three orderings, a straightforward computation shows that
system (55) has finitely many solutions. For instance, let
$x_{4}<x_{1}<x_{2}<x_{3}$ and $\sigma=1$. The system becomes

$\displaystyle
x_{1}+\frac{1}{d_{12}^{2}}+\frac{m_{3}}{d_{13}^{2}}-\frac{(-2-m_{3})}{d_{14}^{2}}$
$\displaystyle=0$ $\displaystyle
x_{2}-\frac{1}{d_{12}^{2}}+\frac{m_{3}}{d_{23}^{2}}-\frac{(-2-m_{3})}{d_{24}^{2}}$
$\displaystyle=0$ (56) $\displaystyle
x_{3}-\frac{1}{d_{13}^{2}}-\frac{m_{2}}{d_{23}^{2}}-\frac{(-2-m_{3})}{d_{34}^{2}}$
$\displaystyle=0,$

where $x_{2}=x_{1}+d_{12},x_{3}=x_{1}+d_{13},x_{4}=x_{1}-d_{14}$,
$d_{12}=-m_{3}d_{13}-(2+m_{3})d_{14},d_{23}=d_{13}-d_{12},d_{24}=d_{12}+d_{14},d_{34}=d_{13}+d_{14}$.
After clearing the denominators, the above system consists of three
polynomials in $x_{1},d_{13},d_{14}$. Denote them by $p_{1},p_{2},p_{3}$
respectively. Note that each of them is degree one in $x_{1}$. We compute the
resultants $r_{1}(d_{13},d_{14})=Res(p_{1},p_{2},x_{1})$ and
$r_{2}(d_{13},d_{14})=Res(p_{1},p_{3},x_{1})$, and then compute
$r_{3}(d_{14})=Res(r_{1},r_{3},d_{13})$, which is a nonzero polynomial in
$d_{14}$ with degree 135. Thus, we conclude that system (56) has finitely many
solutions.

$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\Box$

####  7.3.2 The planar case

Recall that solving system (2) is equivalent to finding critical points of the
potential restricted on the set $I=c$. For these two groups of masses,
following Celli, [2], we use $r_{jk}$ as coordinates. For system with zero
total mass, there are extra constraints, namely

$\sum_{k\neq 1}m_{k}r^{2}_{1k}=\sum_{k\neq 2}m_{k}r^{2}_{2k}=\sum_{k\neq
3}m_{k}r^{2}_{3k}=\sum_{k\neq 4}m_{4}r^{2}_{4k}=c_{0}.$

Then system (2) is equivalent to the following system

$\begin{array}[]{cc}m_{j}m_{k}+2(\lambda_{j}m_{k}+\lambda_{k}m_{j})r_{jk}^{3}=0,&1\leq
j<k\leq 4\\\ \sum_{j\neq k}m_{j}r_{jk}^{2}=I,&1\leq k\leq 4,\end{array}$ (57)

where $\lambda_{j}$, $1\leq j\leq 4$, are the Lagrange multiplier.

_Proof of Theorem 7.2, Part 2_:

Consider system (57).

Case 1: $m_{1}=m_{2}=-m_{3}=-m_{4}.$

In this case, it is easy to see that

$r_{12}=r_{14}=r_{23}=r_{24}$

By imposing the normalization $r_{24}=1$ and setting $r_{34}=t$, system (57)
reduces to

$r_{12}^{2}=4-t^{2},\ \frac{1}{(4-t^{2})^{3/2}}+\frac{1}{t^{3}}=2.$

Thus, $t\approx 0.816713,1.825645$ and the system has only two solutions.

Case 2: $m_{1}=m_{2},m_{3}=-\alpha^{2}m_{1},m_{4}=-(2-\alpha^{2})m_{1}$.  In
this case, after eliminating the $\lambda_{j}$ and $I$ we have the relations:

$\begin{array}[]{cc}\frac{1}{r_{12}^{3}}+\frac{1}{{r_{34}}^{3}}=\frac{1}{r_{13}^{3}}+\frac{1}{{r_{24}}^{3}}=\frac{1}{r_{14}^{3}}+\frac{1}{{r_{23}}^{3}},&\\\
\alpha^{2}\left(r_{13}^{2}-r_{23}^{2}\right)=(2-\alpha^{2})\left(r_{24}^{2}-r_{14}^{2}\right).&\end{array}$

It follows that

$r_{13}-r_{23}=r_{14}-r_{24}=0,$

then by imposing the normalization $r_{34}=1$, system (57) reduces to

$\begin{array}[]{c}\frac{1}{r_{12}^{3}}=\frac{1}{r_{13}^{3}}+\frac{1}{r_{14}^{3}}-1,\\\
r_{12}^{2}-\left(r_{13}^{2}-1\right)\alpha^{2}-r_{14}^{2}\left(4-\alpha^{2}\right)=0,\\\
r_{13}^{2}-r_{14}^{2}+\alpha^{2}-1=0.\end{array}$ (58)

The last two equations of system (58) imply
$r_{12}^{2}=4r_{14}^{2}-\alpha^{4}$. Then $r_{14}$ satisfies

$\frac{1}{(4r_{14}^{2}-\alpha^{4})^{3/2}}=\frac{1}{(r_{14}^{2}+1-\alpha^{2})^{3/2}}+\frac{1}{r_{14}^{3}}-1.$

which has finitely many solutions. Thus, system (58) has finitely many
solutions.

$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\Box$

##  8 Appendix

Some tedious computations in the proofs of Theorems in Section 6 and Section 7
are omitted there. We now provide them in this appendix.

We denote $\pm 1$ by $\sigma_{i}$, and $\sqrt{|m_{i}|}>0$ by $\mu_{i}$,
$i=1,2,3,4$. Without lose of generality, let $\mu_{1}=1$. Then those algebraic
constraints can be written as polynomials of the three real variable
$\mu_{2},\mu_{3},\mu_{4}$. For instance, the constraints in
$\mathcal{V}_{IB}[12,34]$ are

$\displaystyle(\mu_{3}^{3}+\sigma_{1}\mu_{4}^{3})+\sigma_{2}\mu_{2}^{3}(\mu_{3}^{3}+\sigma_{3}\mu_{4}^{3})=0,$
$\displaystyle m_{2}<0,\ m_{3}m_{4}<0,\ 1+m_{2}\neq 0,\ m_{3}+m_{4}\neq 0.$

We then see that $\sigma_{1}\sigma_{3}=-1$ and the polynomial reduces to

$(\mu_{3}^{3}+\sigma_{1}\mu_{4}^{3})+\sigma_{2}\mu_{2}^{3}(\mu_{3}^{3}-\sigma_{1}\mu_{4}^{3})=0.$

Note that $\sigma_{1}=1$ if and only if $1<\mu_{2}$. The constraint in
$\mathcal{V}_{III}[124]$ is

$(\mu_{2}-\frac{\mu_{4}}{\mu_{4}-1})(\mu_{2}-\frac{\mu_{4}}{1-\mu_{4}})(\mu_{2}-\frac{\mu_{4}}{1+\mu_{4}})=0.$

_Computations in the proof of Theorem 6.3_:

Case 2: The masses satisfy $m_{1}=m_{2}=1,m_{3}=m_{4}$, and it belongs to

$\mathcal{V}_{IB}[13,24]\bigcup\mathcal{V}_{IB}[14,23]\bigcap\left(\mathcal{V}_{III}[123]\bigcup(\mathcal{V}_{V}[1,23]\bigcup\mathcal{V}_{V}[2,13]\bigcup\mathcal{V}_{V}[3,12])\right).$

Then we have $m_{1}m_{3}<0$, so it can not belong to
$\mathcal{V}_{V}[1,23]\bigcup\mathcal{V}_{V}[2,13]$.

Thus, the masses are in $\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[3,12]$,
then $m_{3}=m_{4}=-\frac{1}{4}$. Obviously, the masses is not in
$\mathcal{V}_{IB}[13,24]\bigcup\mathcal{V}_{IB}[14,23]$.

Case 3, subcase 1: It is easy to see that the masses are in

$\mathcal{V}_{II}[13,24]\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[14,23](\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[2,13]).$

Then $\mu_{3}=\mu_{2}\mu_{4}$. By the fact that the masses are in
$\mathcal{V}_{IB}[12,34]$, we find

$\left(\mu_{2}^{6}+1\right)^{2}\left(\mu_{2}^{12}-6\mu_{2}^{6}+1\right)\mu_{4}^{12}=0.$

Since $m_{1}+m_{2}\neq 0$, we obtain $\mu_{2}=(\sqrt{2}\pm 1)^{1/3}$.
Similarly, we have $\mu_{4}=(\sqrt{2}\pm 1)^{1/3}$. It is easy to check that
the masses are not in $\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[2,13]$.

Case 3, subcase 2: It is easy to see that the masses are in

$\mathcal{V}_{II}[13,24]\mathcal{V}_{IB}[13,24](\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[1,23]\bigcup\mathcal{V}_{V}[3,12])(\mathcal{V}_{III}[134]\bigcup\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]).$

By the fact that the masses are in $\mathcal{V}_{II}[13,24]$, we have
$\mu_{3}=\mu_{2}\mu_{4}$. Substituting $\mu_{3}=\mu_{2}\mu_{4}$ into the
polynomials corresponding to $\mathcal{V}_{IB}[13,24]$,
$\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[1,23]\bigcup\mathcal{V}_{V}[3,12]$,
and
$\mathcal{V}_{III}[134]\bigcup\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14])$,
we obtain three polynomials in terms of $(\mu_{2},\mu_{4})$, with degree 36,
17, and 17 respectively. We will not write them explicitly here.
Straightforward computation, for instance, by using Gröbner basis, shows that
they have no common positive root. Hence, the set is empty.

Case 3, subcase 3: It is easy to see that the masses are in

$\displaystyle\mathcal{V}_{II}[13,24]\mathcal{V}_{IB}[13,24]\bigcap(\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[1,23]\bigcup\mathcal{V}_{V}[3,12])\bigcap$
$\displaystyle(\mathcal{V}_{III}[124]\bigcup\mathcal{V}_{V}[2,14]\bigcup\mathcal{V}_{V}[4,12])\bigcap(\mathcal{V}_{III}[234]\bigcup\mathcal{V}_{V}[2,34]\bigcup\mathcal{V}_{V}[4,23]).$

This set is also empty. The proof is similar to the above subcase and is
omitted.

Case 4, subcase 1: It is easy to see that the masses are in

$\displaystyle\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[13,24]\bigcap(\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[1,23])\bigcap$
$\displaystyle(\mathcal{V}_{III}[124]\bigcup\mathcal{V}_{V}[2,14])\bigcap(\mathcal{V}_{III}[134]\bigcup\mathcal{V}_{V}[3,14]).$

Note that the relation corresponding to
$\mathcal{V}_{III}[124]\bigcup\mathcal{V}_{V}[2,14]$ can be written as

$0=(\mu_{2}-\frac{\mu_{4}}{\mu_{4}-1})(\mu_{2}-\frac{\mu_{4}}{1-\mu_{4}})(\mu_{2}-\frac{\mu_{4}}{1+\mu_{4}})(\mu_{2}-\frac{\mu_{4}^{3/2}}{1+\mu_{4}^{2}}).$

There are four possibilities.

If $\mu_{2}=\frac{\mu_{4}}{\mu_{4}-1}$, then $\mu_{2}>1$ and the constraints
from $\mathcal{V}_{IB}[12,34]$ is

$\mu_{2}^{6}(\mu_{3}^{3}-\mu_{4}^{3})^{2}-(\mu_{3}^{3}+\mu_{4}^{3})^{2}=0.$

Substituting $\mu_{2}=\frac{\mu_{4}}{\mu_{4}-1}$ into it and clearing the
denominator, we obtain one polynomial in terms $(\mu_{3},\mu_{4})$ with degree
12. Similarly, with the constraints from
$\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[1,23]$ and
$\mathcal{V}_{III}[134]\bigcup\mathcal{V}_{V}[3,14]$, we obtain two
polynomials in terms $(\mu_{3},\mu_{4})$ with degree 12. Straightforward
computation shows that they have no common positive root. Hence, the set is
empty if $\mu_{2}=\frac{\mu_{4}}{\mu_{4}-1}$.

Similarly, we can show that the set is empty if
$\mu_{2}=\frac{\mu_{4}}{1-\mu_{4}}$, $\mu_{2}=\frac{\mu_{4}}{\mu_{4}+1}$, or
$\mu_{2}=\frac{\mu_{4}^{3/2}}{1+\mu_{4}^{2}}$.

To summarize, the set corresponding to this subcase is empty.

Case 4, subcase 2 and subcase 3: It is easy to see that the masses are in

$\displaystyle\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[13,24]\bigcap(\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[1,23])\bigcap$
$\displaystyle(\mathcal{V}_{III}[234]\bigcup\mathcal{V}_{V}[4,23])\bigcap(\mathcal{V}_{III}[124]\bigcup\mathcal{V}_{V}[2,14]),$

and

$\displaystyle\mathcal{V}_{IB}[12,34]\mathcal{V}_{IB}[13,24]\bigcap(\mathcal{V}_{III}[123]\bigcup\mathcal{V}_{V}[1,23])\bigcap$
$\displaystyle(\mathcal{V}_{III}[234]\bigcup\mathcal{V}_{V}[4,23])\bigcap(\mathcal{V}_{III}[134]\bigcup\mathcal{V}_{V}[3,14]),$

respectively. The two sets are also empty. The proof is similar to the above
subcase and is omitted.

_Computations in the proof of Theorem 6.4_：

Case 1: The masses are in
$\mathcal{V}_{II0}[13,24]\mathcal{V}_{II0}[14,23]\mathcal{V}_{II0}[12,34]$. We
have $m_{2}^{2}=m_{3}^{2}=m_{4}^{2}=1=m_{1}$ and $m_{i}+m_{j}\neq 0$. Then
$m_{1}=m_{2}=m_{3}=m_{4}=1$, which does not belong to
$\mathcal{I}_{II}[13,24]$, a contradiction.

Case 2: Without lose of generality, consider only the case
$\mathcal{V}_{II0}[13,24]\mathcal{V}_{II0}[14,23]\mathcal{V}_{I0}[13,24]$.
First we have

$m_{1}=m_{2}^{2}=1,\ m_{3}^{2}=m_{4}^{2},\
(m_{1}+m_{2})(m_{2}+m_{3})(m_{3}+m_{4})\neq 0,\ m_{1}m_{3}<0,\ m_{2}m_{4}<0.$

Thus, we set the masses are $m_{1}=m_{2}=1,m_{3}=m_{4}=s<0,s\neq-1$, which
does not belong to $\mathcal{I}_{I}[13,24]$, a contradiction.

Case 3: The masses are in either
$\mathcal{V}_{II0}[13,24]\mathcal{V}_{I0}[12,34]\mathcal{V}_{I0}[14,23]$ or
$\mathcal{V}_{II0}[13,24]\mathcal{V}_{I0}[13,24]$.

In the first subcase, the signs of the masses are $(+,-,+,-)$. Set
$m_{2}=s<0,m_{4}=t<0$. Then $m_{3}=st$. Then that the masses are in
$\mathcal{I}_{I}[12,34]$ implies

$\frac{s}{\sqrt[3]{1+s}}+\frac{st^{2}}{\sqrt[3]{t}\sqrt[3]{1+s}}=0.$

Then $t=-1$ and $m_{1}+m_{4}=0$, hence the masses are not in
$\mathcal{V}_{IB}[14,23]$, a contradiction.

In the second subcase, we have $m_{1}m_{3}=m_{2}m_{4}$. Then the masses are in
$\mathcal{I}_{I}[13,24]$ implies

$m_{1}+m_{3}=-(m_{2}+m_{4}),$

which contradicts with the hypothesis that the total mass $m\neq 0$.

Case 4: The masses are in $\mathcal{V}_{I0}[12,34]\mathcal{V}_{I0}[13,24]$.
The masses are $(1,-\mu_{2}^{2},-\mu_{3}^{2},\mu_{4}^{2})$. Then the masses
satisfy the system

$\displaystyle
f_{1}=(\mu_{3}^{3}+\sigma_{1}\mu_{4}^{3})+\sigma_{2}\mu_{2}^{3}(\mu_{3}^{3}-\sigma_{1}\mu_{4}^{3})=0,$
$\displaystyle
f_{2}=(\mu_{2}^{3}+\sigma_{3}\mu_{4}^{3})+\sigma_{4}\mu_{3}^{3}(\mu_{2}^{3}-\sigma_{3}\mu_{4}^{3})=0,$
$\displaystyle\mu_{2}^{6}(\mu_{3}^{2}-\mu_{4}^{2})-\mu_{3}^{6}\mu_{4}^{6}(1-\mu_{2}^{2})=0,$
$\displaystyle\mu_{3}^{6}(\mu_{2}^{2}-\mu_{4}^{2})-\mu_{2}^{6}\mu_{4}^{6}(1-\mu_{3}^{2})=0.$

The last two equations imply

$\mu_{3}>\mu_{4},\Leftrightarrow\mu_{2}<1,\ {\rm and\
}\mu_{2}>\mu_{4},\Leftrightarrow\mu_{3}<1.$

Hence, there are four possibilities: $\mu_{2}<1,\mu_{3}<1$;
$\mu_{2}<1,\mu_{3}>1$; $\mu_{2}>1,\mu_{3}>1$; $\mu_{2}>1,\mu_{3}<1$. Without
lose of generality, we consider only the first two subcases.

If $\mu_{2}<1,\mu_{3}<1$, then we have
$\sigma_{1}=\sigma_{2}=\sigma_{3}=\sigma_{4}=-1$, i.e.,

$f_{1}=(\mu_{3}^{3}-\mu_{4}^{3})-\mu_{2}^{3}(\mu_{3}^{3}+\mu_{4}^{3})=0,\
f_{2}=(\mu_{2}^{3}-\mu_{4}^{3})-\mu_{3}^{3}(\mu_{2}^{3}+\mu_{4}^{3})=0.$

Then we have

$0=f_{1}-f_{2}=(1+\mu_{4}^{3})(\mu_{3}^{3}-\mu_{2}^{3}),$

thus $\mu_{2}=\mu_{3}$. By the equation the third equation, we get
$\mu_{3}=\mu_{4}\sqrt{\frac{1+\mu_{4}^{4}}{1+\mu_{4}^{6}}}$. Then

$0=f_{1}=\mu_{4}^{3}\left(\frac{\left(\mu_{4}^{5}+\mu_{4}\right)^{3}}{\left(\mu_{4}^{6}+1\right)^{3}}+\left(\mu_{4}^{3}-1\right)\left(\frac{\mu_{4}^{4}+1}{\mu_{4}^{6}+1}\right)^{3/2}+1\right).$

It is easy to verify that the above equation has no root for
$\mu_{4}\in(0,1)$. Thus, the system has no positive solutions.

If $\mu_{2}<1,\mu_{3}>1$, then we have $\sigma_{1}=\sigma_{2}=-1$, and
$\sigma_{3}=\sigma_{4}=1$, i.e.,

$f_{1}=(\mu_{3}^{3}-\mu_{4}^{3})-\mu_{2}^{3}(\mu_{3}^{3}+\mu_{4}^{3})=0,\
f_{2}=(\mu_{2}^{3}+\mu_{4}^{3})+\mu_{3}^{3}(\mu_{2}^{3}-\mu_{4}^{3})=0.$

Then

$0=f_{1}+f_{2}=\left(1-\mu_{4}^{3}\right)\left(\mu_{2}^{3}+\mu_{3}^{3}\right),$

thus $\mu_{4}=1$. By the equation $f_{1}=0$, we get
$\mu_{2}^{3}=\frac{\mu_{3}^{3}-1}{\mu_{3}^{3}+1}$. By the third equation, we
have

$0=\frac{\left(\mu_{3}^{3}-1\right)^{2}\mu_{3}^{18}}{\left(\mu_{3}^{3}+1\right)^{2}}+\left(\frac{\left(\mu_{3}^{2}-1\right)\left(\mu_{3}^{3}-1\right)^{2}}{\left(\mu_{3}^{3}+1\right)^{2}}-\mu_{3}^{6}\right)^{3}.$

It is easy to verify that the above equation has no root for
$\mu_{3}\in(1,\infty)$. Thus, the system has no positive solutions.

_Computations in the proof of Theorem 6.6_：

By assumption, the masses satisfy (29) (30) (31), i.e., belong to the set

$\displaystyle\mathcal{V}_{IV}[123]\bigcap$
$\displaystyle\left(\mathcal{V}_{III}[124]\bigcup\mathcal{V}_{IV}[124]\bigcup\mathcal{V}_{V}[1,24]\bigcup\mathcal{V}_{V}[2,14]\bigcup\mathcal{V}_{V}[4,12]\right)$
$\displaystyle\bigcap$
$\displaystyle\left(\mathcal{V}_{III}[134]\bigcup\mathcal{V}_{IV}[134]\bigcup\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\bigcup\mathcal{V}_{V}[4,13]\right).$

We need to show that any masses of the above set is not consistent with any
one of

$\eqref{DiaIandII},\ \eqref{III134},\ \eqref{III234},\ \eqref{IV234},$

and if the relation (36) is assumed, the masses are not consistent with any
one of

$\eqref{DiaIandIIagain},\ \eqref{III124}.$

We separate the masses satisfying (29) (30) (31) into the union of the
following five subsets:

$\displaystyle{\it Subset\ 1},\ $
$\displaystyle\mathcal{V}_{IV}[123]\bigcap\left(\mathcal{V}_{IV}[124]\bigcup\mathcal{V}_{V}[1,24]\bigcup\mathcal{V}_{V}[2,14]\bigcup\mathcal{V}_{V}[4,12]\right)$
$\displaystyle\bigcap\left(\mathcal{V}_{IV}[134]\bigcup\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\bigcup\mathcal{V}_{V}[4,13]\right)$
$\displaystyle{\it Subset\ 2},\ $
$\displaystyle(\mathcal{V}_{IV}[123]\mathcal{V}_{III}[124]\mathcal{V}_{IV}[134])\bigcup(\mathcal{V}_{IV}[123]\mathcal{V}_{IV}[124]\mathcal{V}_{III}[134])$
$\displaystyle{\it Subset\ 3},\ $
$\displaystyle\mathcal{V}_{IV}[123]\mathcal{V}_{III}[124]\mathcal{V}_{III}[134]$
$\displaystyle{\it Subset\ 4},\ $
$\displaystyle\mathcal{V}_{IV}[123]\mathcal{V}_{III}[124]\left(\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\bigcup\mathcal{V}_{V}[4,13]\right)$
$\displaystyle{\it Subset\ 5},\ $
$\displaystyle\mathcal{V}_{IV}[123]\mathcal{V}_{III}[134]\left(\mathcal{V}_{V}[1,24]\bigcup\mathcal{V}_{V}[2,14]\bigcup\mathcal{V}_{V}[4,12]\right).$

For the first two subsets, straightforward computation show that there are
only five positive solutions

$\displaystyle\left\\{\mu_{1}=1,\mu_{2}=2,\mu_{3}=\sqrt{5},\mu_{4}=2\right\\},$
$\displaystyle\left\\{\mu_{1}=1,\mu_{2}=\sqrt{5},\mu_{3}=2,\mu_{4}=2\right\\},$
$\displaystyle\left\\{\mu_{1}=1,\mu_{2}=2,\mu_{3}=\sqrt{3},\mu_{4}=2\right\\},$
$\displaystyle\left\\{\mu_{1}=1,\mu_{2}=\sqrt{3},\mu_{3}=2,\mu_{4}=2\right\\},$
$\displaystyle\left\\{\mu_{1}=1,\mu_{2}=1/\sqrt{2},\mu_{3}=1/\sqrt{2},\mu_{4}=1/\sqrt{2}\right\\}.$

It is easy to check that none of the above five solutions of $\mu_{i}$’s are
consistent with any one of
$\eqref{DiaIandII},\eqref{III134},\eqref{III234},\eqref{IV234},$ and that
those satisfy equation (36) do not satisfy any one of
$\eqref{DiaIandIIagain},\eqref{III124}.$

For the third subset,
$\mathcal{V}_{IV}[123]\mathcal{V}_{III}[124]\mathcal{V}_{III}[134]$, the
masses satisfy the system

$1+\sigma_{1}\mu_{2}^{2}+\sigma_{2}\mu_{3}^{2}=0,\
1+\sigma_{3}/\mu_{2}+\sigma_{4}/\mu_{4}=0,\
1+\sigma_{5}/\mu_{3}+\sigma_{6}/\mu_{4}=0.$

The above system reduces to 27 quartic equations of $\mu_{4}$, of which we can
find their exact solutions. There are in total eight positive solutions. We
will not write them explicitly here. It is easy to check that the
corresponding masses are not consistent with (32), and satisfy neither
$m_{1}+m_{3}+m_{4}=0,$ nor $m_{2}+m_{3}+m_{4}=0$.

The last two subsets are similar. We consider only the fourth one. Firstly, we
claim that the fourth subset,
$\mathcal{V}_{IV}[123]\mathcal{V}_{III}[124]\left(\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\bigcup\mathcal{V}_{V}[4,13]\right)$,
of masses are not consistent with (32).

Case I, in $\mathcal{V}_{IA}[13,24]\bigcup\mathcal{V}_{IA}[14,23]$.  If the
masses are also in $\mathcal{V}_{IA}[13,24]$, then we have $m_{1}=m_{3}=1$,
$m_{2}=m_{4}=-2$. Hence, the masses are not in $\mathcal{V}_{III}[124]$, a
contradiction. Similarly, the masses can not belong to
$\mathcal{V}_{IA}[14,23]$.

Case II, in $\mathcal{V}_{IB}[13,24]$ or in $\mathcal{V}_{IB}[14,23]$.  We
consider only the first one, i.e., the masses are also in
$\mathcal{V}_{IB}[13,24]$. Then $m_{1}m_{3}<0$, so the subset reduces to

$\mathcal{V}_{IV}[123]\mathcal{V}_{III}[124]\left(\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\right)\mathcal{V}_{IB}[13,24].$

Subcase I, in $\mathcal{V}_{V}[1,34]$.  In this case, $m_{3}m_{4}>0$, so
$m_{1},m_{2}>0$, $m_{3},m_{4}<0$ since $m_{1}+m_{2}+m_{3}=0$. Then we have

$1+\mu_{2}^{2}-\mu_{3}^{2}=0.$ (59)

So $1<\mu_{3},\mu_{2}<\mu_{3}$. The polynomial corresponding to
$\mathcal{V}_{V}[1,34]$ is

$(\mu_{3}^{2}+\mu_{4}^{2})^{2}-\mu_{3}^{3}\mu_{4}^{3}=0.$ (60)

Note that the relation corresponding to $\mathcal{V}_{III}[124]$ can be
written as

$(\mu_{2}-\frac{\mu_{4}}{1-\mu_{4}})(\mu_{2}-\frac{\mu_{4}}{1+\mu_{4}})(\mu_{2}-\frac{\mu_{4}}{\mu_{4}-1}).$

There are three possibilities.

If $\mu_{2}=\frac{\mu_{4}}{1-\mu_{4}}$, then $\mu_{4}<\mu_{2}$. Note that the
polynomial (59) becomes

$-\mu_{3}^{2}\mu_{4}^{2}+2\mu_{3}^{2}\mu_{4}-\mu_{3}^{2}+2\mu_{4}^{2}-2\mu_{4}+1=0,$
(61)

and the polynomial corresponding to $\mathcal{V}_{IB}[13,24]$ is
$-\mu_{2}^{3}(1-\mu_{3}^{3})=\mu_{4}^{3}(1+\mu_{3}^{3})$, or

$\left(\mu_{3}^{3}-1\right)-\left(\mu_{3}^{3}+1\right)(1-\mu_{4})^{3}=0.$ (62)

However, straightforward computation shows that (60), (61), (62) has no common
root. Hence, the set is empty if $\mu_{2}=\frac{\mu_{4}}{1-\mu_{4}}$.

Similarly, we can show that the set is empty if
$\mu_{2}=\frac{\mu_{4}}{1+\mu_{4}}$, or $\mu_{2}=\frac{\mu_{4}}{\mu_{4}-1}$.

Subcase II, in $\mathcal{V}_{V}[3,14]$.  The set
$\mathcal{V}_{IV}[123]\mathcal{V}_{III}[124]\mathcal{V}_{V}[3,14]\mathcal{V}_{IB}[13,24]$
is also empty. The proof is similar to the above case and is omitted.

To summarize, the set
$\mathcal{V}_{IV}[123]\mathcal{V}_{III}[124]\left(\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\right)\mathcal{V}_{IB}[13,24]$
is empty.

Case III, in $\mathcal{V}_{II}[12,34]$.  If the masses are also in
$\mathcal{V}_{II}[12,34]$, then $\mu_{2}=\mu_{3}\mu_{4}.$ Then after the
substitution $\mu_{2}=\mu_{3}\mu_{4}$, the constraints corresponding to

$\mathcal{V}_{IV}[123],\ \mathcal{V}_{III}[124],\
\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\bigcup\mathcal{V}_{V}[4,13]$

become three polynomials in terms of $(\mu_{3},\mu_{4})$ with degree 12, 6 and
18. Straightforward computation shows that the three polynomials have no
common root, a contradiction.

Secondly, the fourth subset,
$\mathcal{V}_{IV}[123]\mathcal{V}_{III}[124]\left(\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\bigcup\mathcal{V}_{V}[4,13]\right)$,
of masses satisfy neither $m_{1}+m_{3}+m_{4}=0,$ nor $m_{2}+m_{3}+m_{4}=0$.
For instance, if $m_{1}+m_{3}+m_{4}=0$, then $m_{2}=m_{4}$. The fact that the
masses are in $\mathcal{V}_{III}[124]\mathcal{V}_{IV}[123]$ implies that the
masses are

$\left\\{m_{1}=1,m_{2}=4,m_{3}=-5,m_{4}=4\right\\}\mbox{ or
}\left\\{m_{1}=1,m_{2}=-4,m_{3}=3,m_{4}=-4\right\\}.$

It is easy to check that both of them are not in the set
$\mathcal{V}_{V}[1,34]\bigcup\mathcal{V}_{V}[3,14]\bigcup\mathcal{V}_{V}[4,13]$.

_Computations in the proof of Theorem 6.7_：

Case 1: The masses are in
$\mathcal{V}_{I0}[13,24]\mathcal{V}[13,24]\mathcal{V}_{IV}[123]$. Recall that

$V[13,24]=(\mathcal{V}_{I0}[12,34])\bigcup(\mathcal{V}_{I0}[14,23])\bigcup(\mathcal{V}_{II0}[13,24])\bigcup\mathcal{V}_{IV}[124]\bigcup\mathcal{V}_{IV}[234].$

We ignore the first two choices
$\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{I0}[14,23]$, since we have showed

$\mathcal{V}_{I0}[13,24](\mathcal{V}_{I0}[12,34]\bigcup\mathcal{V}_{I0}[14,23])=\emptyset,$

see Case 4 of the proof of Theorem 6.4. Hence, there are three possibilities.

If the masses belong to
$\mathcal{V}_{I0}[13,24]\mathcal{V}_{II0}[13,24]\mathcal{V}_{IV}[123]$. Then
$m_{1}m_{3}=m_{2}m_{4}$ and the masses are in $\mathcal{I}_{I}[13,24]$. Then
we have $m=0$, a contradiction.

If the masses belong to
$\mathcal{V}_{I0}[13,24]\mathcal{V}_{IV}[124]\mathcal{V}_{IV}[123]$, then
$m_{3}=m_{4}$. Set $m_{3}=s$, then $m_{2}=-(s+1)$. The fact that the masses
are in $\mathcal{I}_{I}[13,24]$ implies

$\frac{s}{\sqrt[3]{s+1}}+s(s+1)=0,$

which is impossible. Similarly, the masses can not belong to
$\mathcal{V}_{I0}[13,24]\mathcal{V}_{IV}[234]\mathcal{V}_{IV}[123]$.

Case 2: The masses are in
$\mathcal{V}_{II0}[12,34]\mathcal{V}[13,24]\mathcal{V}_{IV}[123]\mathcal{V}[23,14]$.
In fact, the set
$\mathcal{V}_{II0}[12,34]\mathcal{V}[13,24]\mathcal{V}_{IV}[123]$ is empty.
Since $m_{2}=m_{3}m_{4}$ and $1+m_{2}+m_{3}=0$, we have

$m_{1}=1,\ m_{2}=\frac{-t}{1+t},\ m_{3}=\frac{-1}{1+t},\ m_{4}=t.$

By Case 1, Diagram I can not occur, then the masses are in
$\mathcal{V}_{II0}[13,24]\bigcup\mathcal{V}_{IV}[124]\bigcup\mathcal{V}_{IV}[234].$

If the masses belong to $\mathcal{V}_{II0}[13,24]$, then $m_{3}=m_{2}m_{4}$,
so we have $t=1$ (since $m_{1}+m_{4}\neq 0$). Thus,
$m_{2}=m_{3}=-\frac{1}{2}$. This contradicts with the fact that the masses are
in $\mathcal{I}_{II}[13,24]$, i.e.,

$0=\frac{m_{1}+m_{3}}{\sqrt[3]{m_{1}+m_{3}}}+\frac{m_{3}+m_{2}}{\sqrt[3]{m_{3}+m_{2}}}+\frac{m_{2}+m_{4}}{\sqrt[3]{m_{2}+m_{4}}}+\frac{m_{4}+m_{1}}{\sqrt[3]{m_{4}+m_{1}}}.$

If the masses belong to $\mathcal{V}_{IV}[124]$, then $m_{3}=m_{4}$, or
$(t+1)t+1=0$, which has no real solution. Similarly, the masses can not belong
to $\mathcal{V}_{IV}[234]$.

_Computations in the proof of Theorem 7.1_:

We want to show that the system

$\displaystyle\mu_{2}^{3}(1+\sigma_{1}\mu_{3}^{3})=\sigma_{2}\mu_{4}^{3}(1-\sigma_{1}\mu_{3}^{3})$
(63)
$\displaystyle\mu_{2}^{3}(1+\sigma_{3}\mu_{4}^{3})=\sigma_{4}\mu_{3}^{3}(1-\sigma_{3}\mu_{4}^{3})$
$\displaystyle 1+\mu_{2}^{2}=\mu_{3}^{2}+\mu_{4}^{2}$

has no solution, if we assume
$(\mu_{2}-1)(\mu_{3}-1)(\mu_{4}-1)(\mu_{2}-\mu_{3})(\mu_{2}-\mu_{4})(\mu_{3}-\mu_{4})\neq
0.$

Without lose of generality, assume $\mu_{3}>\mu_{4}$. There are three cases:
$\mu_{2}>\mu_{3}>\mu_{4}$, $\mu_{3}>\mu_{4}>\mu_{2}$, and
$\mu_{3}>\mu_{2}>\mu_{4}$.

Case 1: $\mu_{2}>\mu_{3}>\mu_{4}$. Then $\mu_{4}>1,\
\sigma_{1}=\sigma_{3}=-1,\ \sigma_{2}=\sigma_{4}=-1.$ Therefore, we have

$\mu_{3}^{3}\frac{1-\mu_{3}^{3}}{1+\mu_{3}^{3}}=\mu_{4}^{3}\frac{1-\mu_{4}^{3}}{1+\mu_{4}^{3}},$

which has no solution since $t^{3}\frac{1-t^{3}}{1+t^{3}}$ is an increasing
function on $(1,\infty)$.

Case 2: $\mu_{3}>\mu_{4}>\mu_{2}$. System (63) has no solution. The proof is
similar to the above case and is omitted.

Case 3: $\mu_{3}>\mu_{2}>\mu_{4}$. Note that $\mu_{4}<1<\mu_{3}$. Rewrite the
first two equation as

$\displaystyle(\mu_{2}^{3}+\sigma_{1}\mu_{4}^{3})=\sigma_{2}\mu_{3}^{3}(\mu_{2}^{3}-\sigma_{1}\mu_{4}^{3}),\
\mu_{2}^{3}(1+\sigma_{3}\mu_{4}^{3})=\sigma_{4}\mu_{3}^{3}(1-\sigma_{3}\mu_{4}^{3}).$

Then it is easy to see that $\sigma_{1}=\sigma_{2}=\sigma_{3}=\sigma_{4}=1.$
Therefore, we have

$\frac{\mu_{2}^{3}+\mu_{4}^{3}}{\mu_{2}^{3}(\mu_{2}^{3}-\mu_{4}^{3})}=\frac{1+(\mu_{4}/\mu_{2})^{3}}{\mu_{2}^{3}(1-(\mu_{4}/\mu_{2})^{3})}=\frac{1+\mu_{4}^{3}}{1-\mu_{4}^{3}}.$
(64)

Note that $\mu_{4},\mu_{4}/\mu_{2}\in(0,1)$ and $\frac{1+t^{3}}{1-t^{3}}$ is
an increasing function on $(0,1)$. If $\mu_{2}<1$, then
$\mu_{4}<\mu_{4}/\mu_{2}<1$, and

$\frac{1+(\mu_{4}/\mu_{2})^{3}}{\mu_{2}^{3}(1-(\mu_{4}/\mu_{2})^{3})}>\frac{1+(\mu_{4}/\mu_{2})^{3}}{(1-(\mu_{4}/\mu_{2})^{3})}>\frac{1+\mu_{4}^{3}}{1-\mu_{4}^{3}},$

a contradiction. Similarly, equation (64) can not hold if $\mu_{2}>1$.

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