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pdftitle={Rigidity of Newtonian Centered Co-Circular Central Configurations},
pdfauthor={Chih-Yao Wu}}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
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\title{Rigidity of Newtonian Centered\\Co-Circular Central Configurations}
\author{Chih-Yao Wu\thanks{Correspondence: \href{mailto:andersonwu2000@gmail.com}{andersonwu2000@gmail.com}. AI agents developed the argument and assisted with manuscript preparation under the author's coordination. This manuscript has not received an independent human expert review.}}
\date{October 23, 2026}
\begin{document}
\maketitle
\begin{abstract}
We prove that a Newtonian central configuration of finitely many positive masses on a circle whose center is the center of mass must be a regular polygon with equal masses. The proof selects an extremal mass--gap pair and lifts the circle to a periodic configuration on the line. A positive integral representation of the circular kernel converts the force equations into sharp weighted quadrature bounds. A scalar convexity estimate forces equality at the selected pair, and the equality conditions determine every mass and gap. The argument applies to every finite number of bodies and includes all analytic and rational estimates needed for the conclusion.
\end{abstract}
\section{Introduction}\label{introduction}

A central configuration is an arrangement of point masses whose
gravitational accelerations are proportional to their displacements from
the center of mass. Equal masses at the vertices of a regular polygon
give the simplest circular examples. The converse asks whether a circle
centered at the center of mass can support any other Newtonian central
configuration.

This is the Newtonian case of Problem 12 in the list of Albouy, Cabral
and Santos \cite{ACS}. Cors, Hall and Roberts \cite{CHR} established
uniqueness results for fixed positive masses and, in particular,
regularity when the masses are equal. Hampton \cite{Hampton} proved
local isolation of the equal-mass regular polygon among centered
co-circular relative equilibria. Wang \cite{Wang} obtained a criterion
yielding the Newtonian classification for \(n\leq6\). Tang and Zhu
\cite{TZ} excluded further prescribed patterns of unequal masses. These
results distinguish two issues: the geometry once the masses are equal,
and the exclusion of unequal masses. Here both follow from one equality
argument, for every finite number of bodies.

\begin{theorem}\label{thm:main}
Let $n\geq3$, let $m_1,\ldots,m_n>0$, and let distinct points $r_1,\ldots,r_n$ lie on a circle of radius $R>0$ and center $O$. Suppose that $O$ is their center of mass and that, for some $G>0$ and $\lambda\in\mathbb R$,
\[
G\sum_{j\ne i}m_j\frac{r_j-r_i}{|r_j-r_i|^3}
=-\lambda(r_i-O),\qquad i=1,\ldots,n.
\tag{1}
\]
Then the masses are equal and the points form a regular $n$-gon. Conversely, every equal-mass regular $n$-gon satisfies these conditions.
\end{theorem}

The argument takes place in the plane of the circle. It therefore
applies to circles in any Euclidean ambient space. The two-body case
consists of equal antipodal masses, as follows directly from centering.

The proof begins by selecting a body with maximal mass relative to its
nearest angular gap. After rescaling, that body has mass \(1\) and a
neighbor at distance \(1\) with mass \(b\leq1\). The same selection
gives a separation bound for every pair in the periodic lift of the
circle. A convex quadrature inequality then bounds the contribution of
any exterior distribution by a weighted lattice.

The bridge between this one-dimensional inequality and Newtonian gravity
is a positive integral representation of the circular kernel. It turns
the radial and tangential equations at the selected pair into two
balance identities. Those identities first imply \(b>7/10\); a
mixed-kernel estimate and one exact rational comparison then give
\(b=1\). At that point the quadrature bounds are equalities. Every
change of mass or gap produces a strictly positive loss, so equality
forces the entire lift to be the unit lattice. Its projection is the
regular polygon.

Sections 2--4 establish the normalization, quadrature inequality and
kernel representation. Sections 5--7 force equality of the selected
masses, and Sections 8--9 recover the entire configuration. Appendix A
evaluates the scalar integral used in the periodization. Appendix B
gives a complementary interpretation through a conserved interaction
across cuts of the lifted configuration. All estimates needed for the
theorem are proved in the text, including the finite rational
calculation.

We use angular gaps in radians. A sum over physical bodies omits the
target body; a sum over the periodic lift omits only the particular
target point. This distinction accounts for the self-image term in
Section 4.

\section{An extremal pair and the periodic
lift}\label{an-extremal-pair-and-the-periodic-lift}

Set \(M=\sum_i m_i\), \(p_i=m_i/M\),
\(u_i=(r_i-O)/R=(\cos\theta_i,\sin\theta_i)\), and
\(\lambda_*=\lambda R^3/(GM)\). Put
\(t_i=(-\sin\theta_i,\cos\theta_i)\). For \(i\ne j\), define

\[
K_{ij}=|u_j-u_i|^{-1},\qquad
\Omega_{ij}=\frac{\sin(\theta_j-\theta_i)}{|u_j-u_i|^3},
\]

with zero diagonal entries. Because

\[
(u_j-u_i)\cdot u_i=-\tfrac12|u_j-u_i|^2,
\qquad (u_j-u_i)\cdot t_i=\sin(\theta_j-\theta_i),
\]

the normalized force is exactly

\[
F_i:=\sum_{j\ne i}p_j\frac{u_j-u_i}{|u_j-u_i|^3}
=-\tfrac12(Kp)_i u_i+(\Omega p)_i t_i.
\tag{2}
\]

Thus (1) is equivalent to

\[
Kp=c\mathbf1,\qquad \Omega p=0,\qquad c=2\lambda_*>0.
\tag{3}
\]

Positivity of \(c\) follows from the positive off-diagonal entries of
\(K\). Conversely, (3) gives (1), and pairwise cancellation gives
\(\sum_i p_iF_i=0\), hence \(\sum_i p_iu_i=0\). We will use the given
centering directly.

Order the angles cyclically. Write \(g_i>0\) for the successive angular
gaps, including the seam, so \(\sum_i g_i=2\pi\). Every \(g_i<\pi\).
Indeed, a gap of size at least \(\pi\) places all points in a closed
semicircle. A normal vector pointing toward its midpoint has nonnegative
scalar product with all points. If the containing arc has length less
than \(\pi\), all products are positive. If its length is exactly
\(\pi\), at most its two endpoints have zero product; \(n\geq3\)
supplies an interior point. Positive masses then contradict centering.

\needspace{5\baselineskip}

Define

\[
d_i=\min(g_{i-1},g_i),\qquad \rho=\max_i\frac{p_i}{d_i}.
\]

The maximum exists and is positive. Choose a maximizing atom, relabel it
0, set \(a=p_0\) and \(\delta=d_0=a/\rho\), and choose the orientation
so a nearest neighbor lies in the positive angular direction. Rescale
arcs and masses by

\[
x_i=\theta_i/\delta\pmod L,\qquad q_i=p_i/a,
\qquad L=2\pi/\delta.
\tag{4}
\]

After rotation, the anchor is at 0 with mass 1, its next point is at 1
with mass \(b\in(0,1]\), and \(L>2\). For every atom, maximality of
\(\rho\) gives

\[
q_i\leq\min(g_{i-1}/\delta,g_i/\delta).
\tag{5}
\]

In particular \(b\leq1\). The useful constraint is the local mass--gap
bound (5); other normalized masses may exceed 1.

Let \(X=\{x_i+kL:0\leq i<n,\ k\in\mathbb Z\}\), with the same mass at
every image. Consecutive gaps are the rescaled cyclic gaps. For any
distinct lifted atoms \(x<y\), the first gap to the right of \(x\) and
last gap to the left of \(y\) are at most \(y-x\). Equation (5)
therefore gives

\[
|x-y|\geq\max(q_x,q_y).
\tag{6}
\]

The inequality applies throughout the periodic lift, including different
images of the same body. The lift has a positive minimum gap, a finite
maximum gap, and bounded positive masses, since the base configuration
is finite.

For \(0<t<L\), put

\[
h_L(t)=\frac{\pi}{L}\csc\frac{\pi t}{L}
       =\frac{\delta}{2\sin(\delta t/2)}.
\]

The exact finite radial and derivative rows are

\[
\sum_{j\ne i}q_jh_L(x_j-x_i\pmod L)=c/\rho,
\qquad
\sum_{j\ne i}q_jh_L'(x_j-x_i\pmod L)=0.
\tag{7}
\]

To check the constants, write \(h(\theta)=[2\sin(\theta/2)]^{-1}\) for
\(0<\theta<2\pi\). Then \(h_L(t)=\delta h(\delta t)\), so the first sum
is \(\delta c/a=c/\rho\). Moreover,

\[
h'(\theta)=-\frac{\sin\theta}{[2\sin(\theta/2)]^3},
\qquad h_L'(t)=\delta^2h'(\delta t).
\]

Consequently the second sum is \(-\delta^2(\Omega p)_i/a=0\). Modulo
differences always take their representative in \((0,L)\); self-images
are absent from (7).

We shall prove that these balance equations saturate a sharp bound for
separated weighted points. Its equality case will make the lift a
unit-mass integer lattice.

\needspace{20\baselineskip}

\section{A sharp bound for separated weighted
points}\label{a-sharp-bound-for-separated-weighted-points}

The separation bound (6) limits the mass that can be placed near an
anchor. We need to convert this geometric restriction into a bound for a
weighted sum of kernel values. The comparison distribution will have
equal masses at equally spaced points.

The two sums below describe that comparison: \(V(t,s)\) includes the
point at distance \(t\), whereas \(R(t,s)\) starts one step later.
Inequality (8) allows one spacing to be replaced by the next spacing
while paying for the mass at the intervening point. Iterating this
one-step inequality will prove (9).

\begin{lemma}[Weighted quadrature]\label{lem:quad} Let $\phi$ be continuous, nonnegative, decreasing and convex on $[0,\infty)$, tending to zero. At the arguments in question suppose the following sums are finite, and define

\[
V(t,s)=s\sum_{j\geq0}\phi(t+js),\qquad
R(t,s)=s\sum_{j\geq1}\phi(t+js),\qquad s>0.
\]

Then $V(t,s)$ is nondecreasing and $R(t,s)$ nonincreasing in $s$. Moreover,

\[
\min(s,h)\phi(t)+V(t+h,h)\leq V(t,s).
\tag{8}
\]

If a countable positive source $t_i\geq d>0$ satisfies $q_i\leq t_i$ and $|t_i-t_j|\geq\max(q_i,q_j)$, then

\[
\sum_i q_i\phi(t_i)\leq d\sum_{j\geq1}\phi(jd).
\tag{9}
\]

\end{lemma}

\begin{proof} The nonnegative distributional second derivative $\mu$ of a convex decreasing function tending to zero represents it as

\[
\phi(u)=\int_{(0,\infty)}(z-u)_+\,d\mu(z).
\tag{10}
\]

For completeness, its right derivative is increasing and tends to zero: a strictly negative limiting derivative would force $\phi$ below zero. Thus $-\phi'_+(u)=\mu((u,\infty))$ away from the harmless endpoint convention. Integrating from $u$ to infinity and using Tonelli gives (10); continuity supplies the endpoint value. All applied extensions below are continuous, so no endpoint jump is present.

For a triangular kernel $(z-u)_+$, set $A_0=(z-t)_+$ and $m=\lfloor A_0/s\rfloor$. Finite summation gives

\[
R=s m A_0-\tfrac12s^2m(m+1),\qquad
V=s(m+1)A_0-\tfrac12s^2m(m+1).
\]

On each interval where $m$ is fixed,

\[
\partial_sR=m[A_0-(m+1)s]\leq0,
\qquad \partial_sV=(m+1)[A_0-ms]\geq0.
\]

The formulas agree at breakpoints, including the zero terms at integer $A_0/s$. Integration against $\mu$, justified by positivity, proves the two monotonicities. If $h\geq s$, the left side of (8) is $s\phi(t)+R(t,h)\leq s\phi(t)+R(t,s)$. If $h\leq s$, it is $V(t,h)\leq V(t,s)$.

For a finite selected subset $t_1<\cdots<t_m$, put $s_1=t_1\geq d$ and $s_i=t_i-t_{i-1}$ for $i\geq2$. The assumptions give $q_i\leq\min(s_i,s_{i+1})$ for $i<m$, and $q_m\leq s_m$. The last point satisfies $q_m\phi(t_m)\leq V(t_m,s_m)$. Suppose that the points after $t_i$ have already been bounded by $V(t_{i+1},s_{i+1})$. Since $t_{i+1}=t_i+s_{i+1}$, inequality (8), with $t=t_i$, $s=s_i$ and $h=s_{i+1}$, gives
\[
\begin{aligned}
\sum_{r=i}^{m}q_r\phi(t_r)
&\leq q_i\phi(t_i)+V(t_{i+1},s_{i+1})\\
&\leq \min(s_i,s_{i+1})\phi(t_i)
       +V(t_i+s_{i+1},s_{i+1})\\
&\leq V(t_i,s_i).
\end{aligned}
\]
Backward induction therefore yields

\[
\sum_{i=1}^m q_i\phi(t_i)\leq V(t_1,s_1)=R(0,s_1)\leq R(0,d).
\]

Exhausting the countable source by finite subsets proves (9) by positivity. If a differentiable kernel has the required shape only on $[d,\infty)$, extend it below $d$ by its tangent at $d$. This extension is continuous, nonnegative, decreasing and convex; all sampled values in (9) remain unchanged. \end{proof}

For strictness we will use a more specialized elementary observation. If
\(\phi(t)=e^{-ut}\), \(u>0\), then for \(t>0\)

\[
V(t,s)=e^{-ut}\frac{s}{1-e^{-us}},\qquad
R(t,s)=e^{-ut}\frac{s}{e^{us}-1}.
\tag{11}
\]

The first is strictly increasing and the second strictly decreasing in
\(s>0\). Indeed, their derivative numerators are \(1-(1+us)e^{-us}>0\)
and \(e^{us}-1-us e^{us}<0\), respectively; these signs follow by
differentiating from zero. The same strictness holds for integrals
against any positive Laplace density nonzero on a set of positive
measure. At \(t=0\) only \(R\) is needed and remains finite for the
kernels used here.

\section{The periodic kernel and the two balance
identities}\label{the-periodic-kernel-and-the-two-balance-identities}

Fix \(L\geq2\). For \(t>0\), define

\[
w(s)=\tanh(Ls/2),\quad
k(t)=\int_0^\infty w(s)e^{-st}\,ds,\quad U(t)=-k'(t),
\] \[
C(t)=k(t)-k(t+1),\quad
A(t)=C(t)-U(t+1),\quad B(t)=U(t)-C(t).
\tag{12}
\]

Here \(C\) is the change in potential over a unit step, while \(A\) and
\(B\) compare this change with the force at the two endpoints. All
differentiations at positive \(t\) are justified by exponential
domination. Since \(0<w(s)\leq Ls/2\),

\[
k(t)\leq L/(2t^2),\qquad U(t)\leq L/t^3.
\tag{13}
\]

For \(0<t<L\), geometric summation and Tonelli give

\[
\sum_{j\in\mathbb Z}k(|t+jL|)
=\int_0^\infty\frac{e^{-st}+e^{-s(L-t)}}{1+e^{-Ls}}\,ds
=\frac{\pi}{L}\csc\frac{\pi t}{L}=h_L(t).
\tag{14}
\]

Substituting \(v=e^{-Ls}\) and inverting the variable in the second term
reduces the integral to \[
\frac1L\int_0^\infty\frac{v^{t/L-1}}{1+v}\,dv.
\] Appendix A evaluates this integral and completes the proof of (14).

\needspace{5\baselineskip}

The nonzero images of a body at its own target contribute

\[
S:=2\sum_{j\geq1}k(jL)
=2\int_0^\infty\frac{ds}{e^{Ls}+1}
=\frac{2\log2}{L}.
\tag{15}
\]

Consequently the full lifted radial row at atom \(i\), excluding only
that particular lifted atom, is

\[
\sum_{y\in X\setminus\{x_i\}}q_yk(|y-x_i|)=c/\rho+S q_i.
\tag{16}
\]

The extra term \(Sq_i\) records the images of the target body and must
therefore be kept when two rows are subtracted. Opposite self-images
cancel in the derivative row. Equation (13), finiteness per period, and
bounded masses justify absolute summation and differentiation of the
image series away from their singularities, so (7) gives exactly zero
lifted derivative row as well.

\needspace{5\baselineskip}

Define left and right exterior sums on this one actual source by

\[
\mathcal L_K=\sum_{x<0}q_xK(-x),\qquad
\mathcal R_K=\sum_{x>1}q_xK(x-1).
\]

The left distances satisfy \(t\geq1\) and \(q\leq t\); the right
distances satisfy \(t\geq b\) and \(q\leq t\), by (6). Both inherit
pairwise exclusion. Let \(D=k(1)+S\), distinct from the earlier density
\(\rho\). Subtracting (16) at 0 and 1 gives

\[
(b-1)k(1)+\mathcal L_C-\mathcal R_C=S(1-b),
\qquad \mathcal L_C-\mathcal R_C=(1-b)D.
\tag{17}
\]

At 0 the direct right force is \(bU(1)\), and at 1 the direct left force
is \(U(1)\). The two torque balances therefore give

\[
\mathcal L_U-\mathcal R_{U(\,\cdot+1)}=bU(1),\qquad
\mathcal R_U-\mathcal L_{U(\,\cdot+1)}=U(1).
\tag{18}
\]

The kernels \(A\) and \(B\) were chosen to combine these radial and
tangential balances. Indeed, their definitions give \[
\begin{aligned}
\mathcal L_A+\mathcal R_B
 &=\bigl(\mathcal R_U-\mathcal L_{U(\,\cdot+1)}\bigr)
   +\bigl(\mathcal L_C-\mathcal R_C\bigr),\\
\mathcal L_B+\mathcal R_A
 &=\bigl(\mathcal L_U-\mathcal R_{U(\,\cdot+1)}\bigr)
   -\bigl(\mathcal L_C-\mathcal R_C\bigr).
\end{aligned}
\] The first parentheses in each line are supplied by (18), and the
remaining difference by (17). We obtain

\[
\mathcal L_A+\mathcal R_B=U(1)+(1-b)D,
\qquad
\mathcal L_B+\mathcal R_A=bU(1)-(1-b)D.
\tag{19}
\]

These identities couple the two exterior distributions. The positive
kernels \(A\) and \(B\) will allow their contributions to be bounded by
weighted lattices.

\section{A strict lower bound for the neighboring
mass}\label{a-strict-lower-bound-for-the-neighboring-mass}

Before using the sharper mixed estimate, we first keep the neighboring
mass away from zero. The two positive balance identities already provide
such a bound. Write

\[
p_A(s)=1-(1+s)e^{-s},\qquad p_B(s)=s-1+e^{-s}.
\]

Then \(A\) and \(B\) have Laplace densities \(wp_A\) and \(wp_B\). Both
densities are positive for \(s>0\), since \(p_A'(s)=s e^{-s}>0\),
\(p_B'(s)=1-e^{-s}>0\), and both vanish at zero. Further,

\[
p_B-p_A=s-2+(s+2)e^{-s},\qquad (p_B-p_A)'=p_A>0,
\]

so \(B>A>0\). The ratio \(r=p_A/p_B\) strictly decreases because

\[
p_A'p_B-p_Ap_B'=e^{-s}[s^2-4\sinh^2(s/2)]<0.
\tag{20}
\]

The last inequality follows from \(2\sinh(s/2)>s\), by differentiating.
Also \(v(s)=w(s)/s\) strictly decreases: for \(z>0\),
\(\tanh z-z\operatorname{sech}^2z>0\), since its derivative is
\(2z\operatorname{sech}^2z\tanh z>0\) and its limit at zero is zero.

If \(r,v\) are decreasing and \(\nu\) is a probability measure, then

\[
\int rv\,d\nu-\int r\,d\nu\int v\,d\nu
=\tfrac12\iint(r(s)-r(t))(v(s)-v(t))\,d\nu(s)d\nu(t)\geq0.
\tag{21}
\]

Apply this to the normalized finite measure with density
\(s p_B(s)e^{-ts}\,ds\). Dividing by its positive integrals gives

\[
\frac{A(t)}{B(t)}\geq
\frac{\int_0^\infty s p_A(s)e^{-ts}\,ds}
     {\int_0^\infty s p_B(s)e^{-ts}\,ds}
=\frac{t(3t+1)}{(t+1)(3t+2)}=:c(t).
\tag{22}
\]

The numerator integral is \(t^{-2}-(t+1)^{-2}-2(t+1)^{-3}\), and the
denominator is \(2t^{-3}-t^{-2}+(t+1)^{-2}\), which verify the displayed
ratio directly. Moreover

\[
c'(t)=\frac{2(6t^2+6t+1)}{(t+1)^2(3t+2)^2}>0,\qquad 0<c(t)<1.
\]

On the right, \(A\geq c(b)B\). On the nonempty left, \(B>A>c(b)A\). The
left is nonempty by periodicity, or directly by the first balance in
(18). Therefore (19) gives the strict inequality

\[
bU(1)-(1-b)D>c(b)[U(1)+(1-b)D].
\tag{23}
\]

Since \(w=1-2/(e^{Ls}+1)\),

\[
D=1+2\int_0^\infty\frac{1-e^{-s}}{e^{Ls}+1}\,ds>1,
\qquad 0<U(1)<1.
\tag{24}
\]

Dividing (23) by \(U(1)\), and weakening \(D/U(1)>1\) to 1 with its
nonnegative coefficient \(1-b\), gives

\[
2b-1>c(b)(2-b).
\tag{25}
\]

For \(b\leq1/2\) this is impossible. Its numerator after multiplying by
the positive denominator \((b+1)(3b+2)\) is

\[
P(b)=9b^3+2b^2-3b-2.
\]

On \([1/2,7/10]\), \(P'(b)=27b^2+4b-3>0\), but \(P(7/10)=-33/1000<0\).
Hence

\[
b>7/10.
\tag{26}
\]

\section{A mixed-kernel bound}\label{a-mixed-kernel-bound}

The preliminary estimate leaves \(7/10<b\leq1\). To exclude its open
part, we subtract one quarter of the second balance in (19) from the
first. Before making that subtraction, we must check that the resulting
kernels satisfy Lemma 1. The coefficient \(1/4\) is chosen for two
purposes: the left kernel has the required shape on \(t\geq1\), and the
right kernel has a positive Laplace density. The endpoint \(7/10\) and
the eight-term truncation in Section 7 will then give a strict scalar
bound. Only these properties of the constants are used.

\needspace{5\baselineskip}

The resulting kernels are

\[
f_L=A-\tfrac14B,\qquad g_L=B-\tfrac14A.
\tag{27}
\]

The density of \(g_L\) is \(w(p_B-p_A/4)>0\); consequently \(g_L\) is
positive, strictly decreasing, and strictly convex for all \(t>0\).

To handle \(f_L\), introduce only the explicit comparison kernels
obtained by replacing \(w\) by 1:

\[
A_\infty(t)=\frac1{t(t+1)^2},\qquad
B_\infty(t)=\frac1{t^2(t+1)},
\] \[
f(t)=\frac{3t-1}{4t^2(t+1)^2},\qquad
 g(t)=\frac{3t+4}{4t^2(t+1)^2}.
\tag{28}
\]

These follow by integrating \(p_Ae^{-st}\), \(p_Be^{-st}\). This
comparison replaces only the kernel density; the configuration remains
fixed. Direct differentiation yields

\[
f'(t)=-\frac{9t^2-t-2}{4t^3(t+1)^3},
\quad
f''(t)=\frac{18t^3+2t^2-7t-3}{2t^4(t+1)^4},
\] \[
-f'''(t)=\frac{3(30t^4+10t^3-15t^2-15t-4)}{2t^5(t+1)^5}.
\tag{29}
\]

At \(t=1+u\), \(u\geq0\), the three sign-controlling numerators are
respectively

\[
6+17u+9u^2,\quad
10+51u+56u^2+18u^3,\quad
6+105u+195u^2+130u^3+30u^4.
\]

Thus \(-f',f'',-f'''>0\) for \(t\geq1\). The ratio \(p_A/p_B\) decreases
from 1 to 0, by (20) and its elementary endpoint limits. Hence
\(p_f=p_A-p_B/4\) changes sign exactly once, from positive to negative,
at some \(s_*>0\). For \(j=0,1,2\),

\[
\int_0^\infty s^{j+1}p_f(s)e^{-ts}\,ds
=(-1)^{j+1}f^{(j+1)}(t)>0\quad(t\geq1).
\]

To pass from these comparison integrals to the actual kernel, we use the
following elementary observation.

\begin{lemma}[One sign change]\label{lem:signchange}
Suppose $\psi$ is nonnegative on $(0,s_*)$ and nonpositive on $(s_*,\infty)$, and $v$ is positive and nonincreasing. If $\psi$ and $v\psi$ are absolutely integrable and $\int_0^\infty\psi(s)\,ds>0$, then
\[
\int_0^\infty v(s)\psi(s)\,ds
\geq v(s_*)\int_0^\infty\psi(s)\,ds>0.
\]
\end{lemma}
\begin{proof}
On each side of $s_*$, the two factors in
$(v(s)-v(s_*))\psi(s)$ have the same sign. Integrating this nonnegative product proves the inequality.
\end{proof}

Apply the lemma with \(\psi(s)=s^{j+1}p_f(s)e^{-ts}\) and
\(v(s)=w(s)/s\), whose decrease was proved in Section 5. The exponential
factor gives absolute integrability at infinity; near zero,
\(p_f(s)=O(s^2)\) and \(w(s)/s\) is bounded. The preceding integral
calculation supplies \(\int\psi>0\). Hence, for \(j=0,1,2\), \[
(-1)^j f_L^{(j)}(t)
=\int_0^\infty s^jw(s)p_f(s)e^{-ts}\,ds>0
\qquad(t\geq1).
\] These three signs say exactly that \(f_L\) is positive, strictly
decreasing and strictly convex on the domain of the left exterior
distances.

The kernels have summable tails. For example \(p_A(s),p_B(s)=O(s^2)\) at
zero, while \(w(s)=O(s)\); splitting the integral at \(s=1\) shows
\(A,B,f_L,g_L=O(t^{-4})\) at infinity. Alternatively, for their positive
combinations \(A,B\), the bound \(w\leq1\) and (28) already gives
\(O(t^{-3})\). Either bound suffices.

Apply Lemma 1, with tangent extension below 1 for \(f_L\) and below
\(b\) for \(g_L\), to the actual left and right sources. This gives

\[
\mathcal L_{f_L}\leq\sum_{j\geq1}f_L(j),\qquad
\mathcal R_{g_L}\leq b\sum_{j\geq1}g_L(jb).
\tag{30}
\]

Subtract one quarter of the second equality in (19) from the first.
Together with (30), it implies

\[
F_L(b):=\sum_{j\geq1}f_L(j)+b\sum_{j\geq1}g_L(jb)
 -(1-b/4)U(1)-\tfrac54(1-b)D\geq0.
\tag{31}
\]

Because \(f_L+g_L=\tfrac34[U(t)-U(t+1)]\), telescoping gives

\[
F_L(1)=0
\tag{32}
\]

for every real \(L>0\). The identity holds before any restriction on the
period has been deduced.

\section{The neighboring masses are
equal}\label{the-neighboring-masses-are-equal}

Section 6 gives \(F_L(b)\geq0\) for a configuration satisfying the force
equations. We will prove the opposite strict inequality whenever
\(7/10\leq b<1\). There are two estimates to establish: \[
F_L(b)\leq F(b)+\frac{1-b}{18},\qquad
F(b)<-\frac{1-b}{18}.
\] The first controls the difference between the periodic kernel and an
explicit comparison kernel. For the second, convexity will reduce the
entire interval to a calculation at \(b=7/10\), because both functions
vanish at \(b=1\).

\paragraph{The convex comparison.}

Define

\[
F(b)=\sum_{j\geq1}f(j)+b\sum_{j\geq1}g(jb)-\tfrac94+\tfrac32 b.
\tag{33}
\]

Since \(A_\infty+B_\infty=t^{-2}-(t+1)^{-2}\), \(F(1)=0\). Each summand
\(h_j(b)=b g(jb)\) is convex, since

\[
h_j''(b)=\frac{9b^3j^3+24b^2j^2+16bj+4}
 {2b^3j^2(bj+1)^4}>0.
\tag{34}
\]

The series is locally uniformly convergent on \(b>0\), with
\(O(j^{-3})\) tails on every compact positive \(b\) interval. Summing
convexity inequalities proves convexity of \(F\); the corresponding
derivative series also converge locally uniformly, directly from the
rational formulas.

\paragraph{The finite-period correction.}

We next compare the two functions, keeping the self-image term. Let

\[
\delta_U=1-U(1)=2\int_0^\infty\frac{s e^{-s}}{e^{Ls}+1}\,ds,
\quad
\delta_D=D-1=2\int_0^\infty\frac{1-e^{-s}}{e^{Ls}+1}\,ds.
\tag{35}
\]

Both are positive, and \(\delta_U\leq2/(L+1)^2\). Set
\(p_g=p_B-p_A/4>0\). The difference between the right comparison sum and
the right finite-period sum is

\[
H(b)=b\sum_{j\geq1}[g(jb)-g_L(jb)]
=2\int_0^\infty\frac{p_g(s)}{e^{Ls}+1}
                  \frac{b}{e^{bs}-1}\,ds.
\tag{36}
\]

Tonelli justifies the identity. For each \(s>0\), the factor
\(b/(e^{bs}-1)\) strictly decreases in \(b\): its derivative numerator
is \(e^{bs}-1-bs e^{bs}<0\). Thus \(H'(b)\leq0\). Differentiation is
locally dominated: \(p_g=O(s^2)\) near zero, the factor is \(O(s^{-1})\)
and its \(b\)-derivative is \(O(1)\) there, uniformly on compact
positive \(b\) intervals, and exponential decay controls infinity.

Define the absolutely convergent, \(b\)-independent sum \[
J_L=\sum_{j\geq1}\bigl[f_L(j)-f(j)\bigr].
\] Subtracting (33) from (31), with (35)--(36), gives the exact identity
\[
F_L(b)-F(b)=J_L-H(b)+(1-b/4)\delta_U
                 -\tfrac54(1-b)\delta_D.
\] The established differentiability of \(F\) and \(H\) therefore also
gives that of \(F_L\), and

\[
F_L'(b)-F'(b)=-H'(b)-\tfrac14\delta_U+\tfrac54\delta_D
\geq-\tfrac14\delta_U.
\]

Set \(D_L(b)=F_L(b)-F(b)\). Since \(D_L(1)=0\), the derivative bound
gives \[
D_L(b)=-\int_b^1D_L'(u)\,du
\leq\frac{1-b}{4}\delta_U.
\] Thus the lower bound on the derivative gives the required upper bound
on the function. Using \(L\geq2\), we obtain

\[
F_L(b)\leq F(b)+\tfrac14(1-b)\delta_U
\leq F(b)+\frac{1-b}{2(L+1)^2}
\leq F(b)+\frac{1-b}{18}
\quad(0<b\leq1,\ L\geq2).
\tag{37}
\]

\paragraph{One endpoint controls the interval.}

It remains to prove the strict bound for \(F\). Set \(\beta=7/10\). For
\(t\geq1\), \(f(t)\leq A_\infty(t)\leq t^{-3}\); for every \(t>0\),
\(g(t)\leq B_\infty(t)\leq t^{-3}\). Since

\[
\sum_{j>8}j^{-3}\leq\int_8^\infty x^{-3}\,dx=1/128,
\]

we have

\[
F(\beta)\leq E:=\sum_{j=1}^{8}[f(j)+\beta g(j\beta)]
 +\frac{1+\beta^{-2}}{128}-\frac94+\frac32\beta
=-\frac{1794985965991645696553}{84600393758389463834880}.
\tag{38}
\]

In particular,

\[
E+\frac{1-\beta}{18}
=-\frac{76995880670364259861}{16920078751677892766976}<0.
\tag{39}
\]

Both values follow by substituting the rational functions (28) into the
eight-term sum in (38). Thus (39) is an exact rational inequality.

For \(\beta\leq b<1\), set \(\eta=(1-b)/(1-\beta)>0\). Then
\(b=\eta\beta+(1-\eta)1\). Convexity and \(F(1)=0\) give

\[
F(b)\leq\eta F(\beta)<-\eta\frac{1-\beta}{18}
=-\frac{1-b}{18}.
\tag{40}
\]

Equations (37), (40) imply \(F_L(b)<0\), contradicting (31). In view of
(26), the only possibility is

\[
b=1.
\tag{41}
\]

\section{Equality and regularity}\label{equality-and-regularity}

At \(b=1\), the first equality of (19) is

\[
\mathcal L_A+\mathcal R_B=U(1).
\tag{42}
\]

Both sources now have initial distance at least 1. Lemma 1 applied to
the positive Laplace kernels \(A,B\), with tangent extensions if needed,
gives

\[
\mathcal L_A\leq\sum_{j\geq1}A(j),\qquad
\mathcal R_B\leq\sum_{j\geq1}B(j).
\]

Their bounds sum to \(U(1)\), since \(A+B=U(t)-U(t+1)\). Therefore both
inequalities are equalities.

To identify the equality case, we write the loss in quadrature as a sum
of nonnegative local deficits. On either half-line, list its distances
increasingly as \(t_1<t_2<\cdots\), and let \(K\) be its kernel, \(A\)
or \(B\). The list exists because the base source is finite and
periodic. Define \(s_1=t_1\geq1\), \(s_i=t_i-t_{i-1}\) for \(i\geq2\),
and let \(V_K,R_K\) be the sums of Lemma 1 formed from the original
Laplace kernel \(K\). All sampled arguments in the deficit calculation
below are at least 1, so the tangent extension used to apply the lemma
changes none of these values. All \(s_i\) are bounded above and bounded
below away from zero. Exclusion gives

\[
q_i\leq\min(s_i,s_{i+1}).
\]

For each \(i\geq1\), define nonnegative transition and mass deficits

\[
T_i=V_K(t_i,s_i)-\min(s_i,s_{i+1})K(t_i)
                       -V_K(t_{i+1},s_{i+1}),
\] \[
M_i=[\min(s_i,s_{i+1})-q_i]K(t_i).
\]

Equation (8) gives \(T_i\geq0\), and \(M_i\geq0\). The boundary deficit
is

\[
B_0=R_K(0,1)-R_K(0,s_1)\geq0.
\]

For every \(m\geq1\), telescoping gives the exact identity

\[
R_K(0,1)-\sum_{i=1}^{m}q_iK(t_i)
=B_0+\sum_{i=1}^{m}(T_i+M_i)+V_K(t_{m+1},s_{m+1}).
\tag{43}
\]

The terminal term tends to zero. For example \(K(t)=O(t^{-3})\) and
\(s_i\in[\sigma,H]\) imply

\[
V_K(t_i,s_i)\leq H C\sum_{j\geq0}(t_i+j\sigma)^{-3}\longrightarrow0.
\]

Equality in quadrature makes the left side of (43) tend to zero. Every
term on its right is nonnegative. In particular, \(B_0\) is bounded
above by that left side for every \(m\), so \(B_0=0\). For any fixed
\(i\), take \(m\geq i\); then \(T_i\) and \(M_i\) are each bounded above
by the same quantity. Letting \(m\to\infty\) gives \(T_i=M_i=0\). This
applies to each fixed index, and therefore to every gap and mass in the
ray.

The positive Laplace densities \(wp_A,wp_B\) and (11) imply strict
decrease of \(R_K(0,s)\). Hence \(B_0=0\) forces \(s_1=1\). In the proof
of (8), if \(s_{i+1}>s_i\), its deficit is
\(R_K(t_i,s_i)-R_K(t_i,s_{i+1})>0\); if \(s_{i+1}<s_i\), it is
\(V_K(t_i,s_i)-V_K(t_i,s_{i+1})>0\). Thus \(T_i=0\) forces
\(s_{i+1}=s_i\). Inductively every gap is 1. Finally \(K(t_i)>0\), so
\(M_i=0\) forces \(q_i=1\).

Applied to both sides of (42), the exterior atoms are exactly the
negative integers and \(2,3,\ldots\), each of mass 1. Together with the
anchors 0 and 1, the full lift is exactly the unit-mass integer lattice.

\section{Completion of the proof}\label{completion-of-the-proof}

The anchor image at \(L\) is an occupied lattice point, so \(L\) is an
integer. The half-open interval \([0,L)\) contains exactly \(L\) lattice
atoms, but it also contains exactly the \(n\) distinct physical atoms.
Hence \(L=n\). The count uses the given period \(L\), whether or not it
is primitive.

All \(q_i=1\), so all \(p_i=a\), and \(\sum_i p_i=1\) gives \(a=1/n\).
Every angular gap equals \(\delta=2\pi/L=2\pi/n\). Restoring \(M,R,O\)
proves the theorem's forward implication.

Conversely, at the vertices of a regular \(n\)-gon with equal mass
\(m>0\), tangential forces cancel in offset pairs \(j,n-j\), and the
radial row is constant by cyclic symmetry. The center of mass is the
circle center, because the sum of the \(n\)-th roots of unity is zero.
The inward coefficient is

\[
\lambda=\frac{Gm}{4R^3}\sum_{j=1}^{n-1}\csc\frac{\pi j}{n}>0.
\]

This proves the converse directly. For \(n=2\), centering says
\(m_1u_1=-m_2u_2\), whose norms give \(m_1=m_2\) and then \(u_2=-u_1\).
The antipodal pair has \(\lambda=Gm/(4R^3)\). For \(n=1\), its sole
point is its center of mass and cannot lie at positive distance from it.

On a fixed circle, for a fixed total mass and oriented cyclic label
order, the configuration is therefore unique modulo rotations of that
circle.

\appendix

\section{The periodization
integral}\label{the-periodization-integral}

The only special-function evaluation in (14) is

\[
I(\beta)=\int_0^\infty\frac{x^{\beta-1}}{1+x}\,dx
=\frac{\pi}{\sin\pi\beta},\qquad 0<\beta<1.
\]

We evaluate it by a keyhole contour. Take the branch
\(z^{\beta-1}=\exp((\beta-1)(\log|z|+i\arg z))\), \(0<\arg z<2\pi\),
with a slit on the positive real axis. Integrate \(z^{\beta-1}/(1+z)\)
on the positively oriented keyhole contour of radii \(\varepsilon<1<R\).
The upper ray contributes the truncated real integral, and the lower
ray, traversed backwards, contributes \(-e^{2\pi i\beta}\) times it. The
small circle is \(O(\varepsilon^\beta)\), the large circle
\(O(R^{\beta-1})\), both tending to zero. The sole pole is at \(-1\),
with residue \(e^{i\pi(\beta-1)}\). Therefore

\[
(1-e^{2\pi i\beta})I(\beta)=2\pi i e^{i\pi(\beta-1)}.
\]

Using \(1-e^{2\pi i\beta}=-2i e^{i\pi\beta}\sin\pi\beta\) gives the
stated value. This scalar evaluation completes the derivation of (14).

\section{A conserved interaction across
cuts}\label{a-conserved-interaction-across-cuts}

The two balance identities in Section 4 arise from adjacent bodies. A
conserved quantity also expresses the compatibility of the entire
periodic configuration. We record it here to clarify the relation
between local balance and the exterior distributions used in the proof.

Enumerate the lifted points increasingly as \((x_j)_{j\in\mathbb Z}\),
with masses \(q_j\). With \(U=-k'\) as in (12), define the interaction
across the cut after \(x_\ell\) by \[
\Pi_\ell=\sum_{j\leq\ell<k}q_jq_kU(x_k-x_j).
\] The lift has a minimum gap \(\sigma>0\) and a maximum mass
\(Q<\infty\). For each index difference \(r\geq1\), exactly \(r\) pairs
cross the cut, and their distances are at least \(r\sigma\). The bound
\(U(t)\leq L/t^3\) therefore gives \[
0<\Pi_\ell\leq Q^2L\sigma^{-3}\sum_{r\geq1}r^{-2}<\infty.
\] Moving the cut past \(x_{\ell+1}\) removes its interactions with
points on the left and adds its interactions with points on the right.
Absolute convergence justifies this subtraction, yielding \[
\begin{split}
\Pi_{\ell+1}-\Pi_\ell=q_{\ell+1}\bigg(&\sum_{k>\ell+1}q_kU(x_k-x_{\ell+1})\\
&-\sum_{j<\ell+1}q_jU(x_{\ell+1}-x_j)\bigg)=0.
\end{split}
\] The last equality is the lifted tangential balance. Thus \(\Pi_\ell\)
is independent of the cut.

For the selected anchors \(0\) and \(1\), define the exterior
exponential sums \[
X(s)=\sum_{x<0}q_xe^{sx},\qquad
Y(s)=\sum_{x>1}q_xe^{-s(x-1)}.
\] Tonelli's theorem factors the interaction across the gap as \[
\Pi=\int_0^\infty s\tanh(Ls/2)e^{-s}(1+X(s))(b+Y(s))\,ds.
\] The separated-point inequality applied to the exponential kernel
gives \(X(s)\leq(e^s-1)^{-1}\) and \(Y(s)\leq b(e^{bs}-1)^{-1}\). These
bounds and the conserved interaction concern the same two exterior
distributions. In particular, balance couples the two sides even when
each is bounded separately. The main proof uses the linear identities
(19) to obtain the stronger conclusion that both quadrature bounds are
saturated.

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\end{document}