\section{Introduction}
  In this paper we will consider two types of $n$-body problems, the first of which being the problem of finding the orbits of  point masses $q_{1}$,...,$q_{n}\in\mathbb{R}^{d}$, $d\in\mathbb{N}$, and respective  masses $m_{1}>0$,...,\textrm{ }$m_{n}>0$ as described by the system of differential equations
  \begin{align}\label{Equations of motion}
    \ddot{q}_{k}=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}(q_{j}-q_{k})f\left(\|q_{j}-q_{k}\|^{2}\right),\textrm{ }k\in\{1,...,n\},
  \end{align}
  where $\|\cdot\|$ is the Euclidean norm, $f:\mathbb{R}_{>0}\rightarrow\mathbb{R}_{>0}$ is a positive-valued scalar function and $\sqrt{x}f(x)$ is a decreasing function. The study of $n$-body problems of this type has applications to, for example, atomic physics, celestial mechanics, chemistry, crystallography,
  differential equations and dynamical systems (see for example \cite{AbrahamMarsden}--\cite{D0}, \cite{DMS}, \cite{DPS4}, \cite{J}--\cite{PV} and the references therein).
  The second $n$-body problem we will investigate is the $n$-body problem in spaces of constant Gaussian curvature, or curved $n$-body problem for short, which can be formulated as follows:
    Let $\sigma=\pm 1$. The $n$-body problem in spaces of constant Gaussian curvature is the problem of finding the dynamics of point masses \begin{align*}q_{1},...,\textrm{ }q_{n}\in\mathbb{M}_{\sigma}^{2}=\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}|x_{1}^{2}+x_{2}^{2}+\sigma x_{3}^{2}=\sigma\},\end{align*} with respective masses $m_{1}>0$,..., $m_{n}>0$, determined by the system of differential equations
  \begin{align}\label{EquationsOfMotion Curved}
   \ddot{q}_{k}=\sum\limits_{\substack{j=1\\j\neq k}}^{n}\frac{m_{j}(q_{j}-\sigma(q_{k}\odot q_{j})q_{k})}{(\sigma -\sigma(q_{k}\odot q_{j})^{2})^{\frac{3}{2}}}-\sigma(\dot{q}_{k}\odot\dot{q}_{k})q_{k},\textrm{ }k\in\{1,...,\textrm{ }n\},
  \end{align}
  where for $x$, $y\in\mathbb{M}_{\sigma}^{2}$  the product $\cdot\odot\cdot$ is defined as
  \begin{align*}
    x\odot y=x_{1}y_{1}+x_{2}y_{2}+\sigma x_{3}y_{3}.
  \end{align*}
  While the curved $n$-body problem for $n=2$ goes back as far as the 1830s, a working model for the $n\geq 2$ case was not found until 2008 by Diacu, P\'erez-Chavela and Santoprete (see \cite{DPS1}, \cite{DPS2} and \cite{DPS3}). This breakthrough then gave rise to further results for the $n\geq 2$ case in \cite{D1}--\cite{DK}, \cite{DPo}, \cite{DSS}, \cite{DT} and \cite{PS}--\cite{ZZ}. See \cite{BM}, \cite{BMK}, \cite{DK}, \cite{DPS1}, \cite{DPS2} and \cite{DPS3} for a historical overview. The study of the curved $n$-body problem has applications to for example geometric mechanics, Lie groups and algebras, non-Euclidean and differential geometry and stability theory, the theory of polytopes and topology (see for example \cite{D6}). A particular use of the curved $n$-body problem is that it may give information about the geometry of the universe. For example: Diacu, P\'erez-Chavela and Santoprete  showed that the configuration of the Sun, Jupiter and the Trojan asteroids cannot exist in curved space (see \cite{DPS1}, \cite{DPS2}). \newline


  By a choreography, or choreographic solution to (\ref{Equations of motion}), or (\ref{EquationsOfMotion Curved}) we mean any solution $q_{1},...,q_{n}$ for which there exists a twice-continuously differentiable periodic vector-valued function $p(t)$ and constants $h_{1},...,h_{n}$ such that $q_{k}(t)=p(t+h_{k})$, $k\in\{1,...,n\}$. For ease of notation, we define $h_{k+Kn}=h_{k}+K\widehat{P}$ for $k\in\{1,...,n\}$, where $\widehat{P}$ is the period of $p$, $K\in\mathbb{Z}$. If $h_{k+1}-h_{k}$ is independent of $k$, then we call a choreographic solution an \textit{equally spaced choreography}, or \textit{equally spaced choreographic solution}. Examples of choreographies are the well-known relative equilibrium where $n$ masses move along a circle, evenly distributed as if they were the vertices of a regular polygon (see for example \cite{AEP}, \cite{CHG}, \cite{T5} and \cite{T7} and the references therein), the famous figure eight, first discovered numerically by Moore (see \cite{Moore}) and independently discovered and proved by Chenciner and Montgomery (see \cite{CM}), which gave rise to the discovery of additional families of choreographies (see for example \cite{CGMS}, \cite{FFO}, \cite{S}, \cite{Mon}, \cite{Y} and the references therein) and numerically discovered choreographies for the curved $n$-body problem (see \cite{MG}, \cite{Mont}), of which the figure eight solution on $\mathbb{S}^{2}$ was analytically proven in \cite{SZ}. Chenciner proved in \cite{C} for general $n$ that for any equally spaced choreographic solution to the classical $n$-body problem, i.e.  $f(x)=x^{-\frac{3}{2}}$ (though it should be noted that his proof does not depend on the choice of $f$), with unequal masses, the same choreographic solution solves the $n$-body problem if all masses are taken equal. He then coined equally spaced choreographic solutions with unequal masses \textit{perverse choreographic solutions} and proved that if $n<6$, such solutions do not exist if the choreography lies in a plane. He additionally stated that choreographies also exist in three-space and that the problem of their perversity is completely open. For the case that a choreographic solution is not assumed to be equally spaced, nothing is known regarding whether the masses are equal even for $n=3$.
  \begin{remark}
    Throughout this paper, when we say that $d$ has a specific value $k\in\mathbb{N}$, we mean that there exists a set $S\subset\left[0,\widehat{P}\right]$ such that the point masses $q_{1}(t)$,...,$q_{n}(t)$ span a $k$-dimensional space for any $t\in S\subset\left[0,\widehat{P}\right]$, where $S$ has Lebesgue measure $\widehat{P}$.
  \end{remark}
  In this paper we will investigate the existence of perverse choreographies in higher-dimensional spaces, in spaces of constant Gaussian curvature and prove that if the curve along which the point masses of a choreography move has an axis of symmetry, the masses have to be equal for $n=3$. Specifically, we will prove the following results:
  \begin{theorem}\label{Main Theorem 1}
    Let $q_{1},...,q_{n}$ be an equally spaced choreographic solution of (\ref{Equations of motion}). If $d=n-1$, $d=n-2$, or if $n$ is odd and $d=n-3$, all masses are equal and if $n$ is even and $d=n-3$, all masses with an odd label are equal and all masses with an even label are equal.
  \end{theorem}
  \begin{corollary}\label{Main Corollary 1}
    Let $q_{1},...,q_{n+1}$ be a solution of (\ref{Equations of motion}), where $q_{1},...,q_{n}$ are an equally spaced choreography and $q_{n+1}=0$. If $d=n-1$, $d=n-2$, or $n$ is odd and $d=n-3$, then the first $n$ masses are equal and if $n$ is even and $d=n-3$, then the first $\frac{n}{2}$ masses with an odd label are equal and the first $\frac{n}{2}$ masses with an even label are equal.
  \end{corollary}
  \begin{theorem}\label{Main Theorem 2}
    Let $q_{1},...,q_{n}$ be a choreographic solution of (\ref{Equations of motion}). If $n=3$ and the curve along which the masses move has an axis of symmetry, then the masses $m_{1}$,$m_{2}$,$m_{3}$ are all equal.
  \end{theorem}
  \begin{corollary}\label{Main Corollary 2}
    Let $q_{1},...,q_{n+1}$ be a solution of (\ref{Equations of motion}), where $q_{1},...,q_{n}$ are a choreography and $q_{n+1}=0$. If $n=3$ and the curve along which the masses move has an axis of symmetry and $q_{n+1}=0$, then the masses $m_{1}$,$m_{2}$,$m_{3}$ are all equal.
  \end{corollary}
  \begin{theorem}\label{Main Theorem 3}
    Let $q_{1},...,q_{n}$ be an equally spaced choreographic solution of (\ref{EquationsOfMotion Curved}). If $\sigma=-1$, then if $n<6$, $n\neq 4$, all masses are equal and if $n=4$, then the even labeled masses are equal and the odd labeled masses are equal. If $\sigma=1$, the same holds true, unless the point masses move along a great circle.
  \end{theorem}
  \begin{corollary}\label{Main Corollary 3}
    Let $q_{1},...,q_{n+1}$ be a solution of (\ref{EquationsOfMotion Curved}), where $q_{1},...,q_{n}$ is an equally spaced choreography and $q_{n+1}=(0,0,1)^{T}$. If $\sigma=-1$, then if $n<6$, $n\neq 4$, the masses of the first $n$ point masses are equal and if $n=4$, then $m_{1}=m_{3}$ and $m_{2}=m_{4}$. If $\sigma=1$, the same holds true, unless the first $n$ point masses move along a great circle.
  \end{corollary}
  \begin{remark}
    Solutions of the type discussed in Corollary~\ref{Main Corollary 1}, Corollary~\ref{Main Corollary 2} and Corollary~\ref{Main Corollary 3} exist: It is well-known that one can construct a solution of (\ref{Equations of motion}) and (\ref{EquationsOfMotion Curved}) where the $q_{1}$,...,$q_{n}$ behave as the vertices of a regular polygon rotating around $q_{n+1}=0$ and $q_{n+1}=(0,0,1)^{T}$ respectively.
  \end{remark}
  \begin{remark}
    Choreographic solutions for which $d=n-k$, $k\in\mathbb{Z}_{\geq 1}$, $k\leq n-2$, exist. For example: We know that planar choreographic solutions exist for all $n$ (see \cite{CGMS}). If a choreographic solution is planar, then $d=2$, so choosing $n=k+2$ gives that for any $k$ there is an $n$ such that $d=n-k$. The matter of what choreographies exist for $d\geq 3$ is an open problem, but regardless of that, Theorem~\ref{Main Theorem 1} and Corollary~\ref{Main Corollary 1} are a first step in investigating the perversity of choreographies if $d\geq 3$.
  \end{remark}
  \begin{remark}
    The argument used in the proof of Theorem~\ref{Main Theorem 1} to prove that if $d=n-3$, $n$ odd, all masses are equal, or take on at most two values if $n$ is even, can be applied if $d=n-4$ and $n$ is even as well, (with $n-3$ replaced with $n-4$) under the condition that we can choose a $k\in\{1,...,n-1\}$ in such a way that the number of linearly independent vectors in the linear combinations of (\ref{General equations1}) and (\ref{General equations2}) is the same as in the linear combinations of (\ref{Lead up to general equations1}) and (\ref{Lead up to general equations2}), in which case if $n$ is even and $d=n-4$ all $a_{l}$ in (\ref{Span}), $l\neq\frac{n}{2}$ have to be zero. However, if $n>6$, it is possible that there is no such $k$ and excluding that possibility is nontrivial and likely requires further knowledge on the dynamics of the choreography.
  \end{remark}
  The remainder of this paper is as follows: We will first formulate necessary notation and lemmas in section~\ref{Section Background Theory} and then prove  Theorem~\ref{Main Theorem 1} in section~\ref{Section proof of main theorem 1}, Corollary~\ref{Main Corollary 1} in section~\ref{Section proof of main corollary 1}, Theorem~\ref{Main Theorem 2} in section~\ref{Section proof of main theorem 2}, Corollary~\ref{Main Corollary 2} in section~\ref{Section proof of main corollary 2}, Theorem~\ref{Main Theorem 3} in section~\ref{Section proof of main theorem 3} and Corollary~\ref{Main Corollary 3} in section~\ref{Section proof of main corollary 3}.
  \section{Background theory}\label{Section Background Theory}
  Throughout this paper we will use the notation introduced in the previous section for choreographies. For ease of notation, we will additionally define $m_{k+Kn}=m_{k}$ for $k\in\{1,...,n\}$, $K\in\mathbb{Z}$ and if we deal with equally spaced choreographies, we will choose $h_{k+1}-h_{k}=1$ and $\widehat{P}=n$, as we can always use a rescaling argument if a choreography is equally spaced and $h_{k+1}-h_{k}\neq 1$. Furthermore, we will assume that for any choreography solution of (\ref{Equations of motion}) we have that
      \begin{align}\label{Center of mass zero}
        \sum\limits_{k=1}^{n}m_{k}q_{k}=0.
      \end{align}
      Note that by (\ref{Equations of motion}) we have that
      \begin{align*}
        \sum\limits_{k=1}^{n}m_{k}\ddot{q}_{k}&=\sum\limits_{k=1}^{n}\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{k}m_{j}(q_{j}-q_{k})f\left(\|q_{j}-q_{k}\|^{2}\right)=0,
      \end{align*}
      so there exist constant vectors $A$ and $B$ such that $\sum\limits_{k=1}^{n}m_{k}q_{k}=At+B$. But for the $q_{k}$, $k\in\{1,...,n\}$, to lie on a closed, periodic curve, we need that $A=0$ and if $B\neq 0$, we can replace the $q_{k}$, $k\in\{1,...,n\}$ in (\ref{Equations of motion}) with $\widehat{q}_{k}=q_{k}-\frac{1}{\sum\limits_{j=1}^{n}m_{j}}B$ and work with $\widehat{q}_{k}$ instead. This result is well-known, but was quickly proven to make the paper as self-contained as possible. For ease of notation, we will define $\Delta_{j}(t)=p(t+j)-p(t)$ for all $j\in\{1,...,n-1\}$ and $M=\sum\limits_{j=1}^{n}m_{j}$. Additionally, to prove Theorem~\ref{Main Theorem 1} and Theorem~\ref{Main Theorem 3} the following well-known vectors will be helpful: Let $\widehat{e}_{1}$,..., $\widehat{e}_{n}\in\mathbb{C}^{n}$, where the $j$th component of $\widehat{e}_{l}$ is $\frac{1}{\sqrt{n}}e^{\frac{2\pi li}{n}(j-1)}$, $j\in\{1,...,n\}$, $l\in\mathbb{Z}$. Let $\lambda_{l}=e^{\frac{2\pi li}{n}}$. Note that if $\widehat{B}$ is the $n\times n$ matrix for which the $j$th component of $\widehat{B}v$ is the $j+1$st component of $v$ for all vectors $v\in\mathbb{C}^{n}$, then $\widehat{B}\widehat{e}_{l}=\lambda_{l}\widehat{e}_{l}$. Additionally, note that the Euclidean inner product of $\widehat{e}_{j}$ and $\widehat{e}_{l}$ is zero for $j\neq l$ and $1$ for $j=l$, making the vectors $\widehat{e}_{1}$,..., $\widehat{e}_{n}$ an orthonormal basis of $\mathbb{C}^{n}$. Finally, if the curve with parametrisation $p(t)$ lies in the plane and is axisymmetric, we will write
      \begin{align}\label{reflection identity}
        p(-t)=\begin{pmatrix}
        1 & 0 \\0 & -1
      \end{pmatrix}p(t).
      \end{align}
      Next we will formulate the following lemmas needed to prove Theorem~\ref{Main Theorem 1}, Corollary~\ref{Main Corollary 1}, Theorem~\ref{Main Theorem 2}, Corollary~\ref{Main Corollary 2}, Theorem~\ref{Main Theorem 3} and Corollary~\ref{Main Corollary 3}:
      \begin{lemma}\label{Lemma 1}
        Let $q_{1}$,...,$q_{n}$ be an equally spaced choreography of (\ref{Equations of motion}). Then
        \begin{align}\label{Linear combination f}
          0=\sum\limits_{j=1}^{n-1}\left(m_{k+j}-\frac{M}{n}\right)\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)
        \end{align}
        and
        \begin{align}\label{Linear combination c}
          0=\sum\limits_{j=1}^{n-1}\left(m_{k+j}-\frac{M}{n}\right)\Delta_{j}(t),\textrm{ }k\in\{1,...,n\}.
        \end{align}
      \end{lemma}
      \begin{proof}
        Substituting $q_{k}(t)$ with $p(t+k)$ and $q_{j}(t)$ with $p(t+j)$ in (\ref{Equations of motion}) and subsequently replacing $t+k$ with $t$ gives
        \begin{align*}
          \ddot{p}(t)=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}(p(t+j-k)-p(t))f\left(\|p(t+j-k)-p(t)\|^{2}\right),
        \end{align*}
        which can be rewritten as
        \begin{align}\label{Sum 1f}
          \ddot{p}(t)=\sum\limits_{j=1}^{n-1}m_{j+k}(p(t+j)-p(t))f\left(\|p(t+j)-p(t)\|^{2}\right)=\sum\limits_{j=1}^{n-1}m_{j+k}\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)
        \end{align}
        for any fixed value $k$. Summing both sides of (\ref{Sum 1f}) from $1$ to $n$ with respect to $k$ then gives
        \begin{align}\label{Sum 2f}
          n\ddot{p}(t)&=\sum\limits_{j=1}^{n-1}\left(\sum\limits_{k=1}^{n}m_{j+k}\right)\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)=\sum\limits_{j=1}^{n-1}M\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right).
        \end{align}
        Dividing both sides of (\ref{Sum 2f}) by $n$ and subtracting the resulting equation from (\ref{Sum 1f}) then gives (\ref{Linear combination f}).
        Next, we will prove (\ref{Linear combination c}) by using that $\sum\limits_{k=1}^{n}m_{k}q_{k}=0$: Note that
        \begin{align*}
          Mq_{k}=\sum\limits_{j=1}^{n}m_{j}q_{k}-0=\sum\limits_{j=1}^{n}m_{j}q_{k}-\sum\limits_{j=1}^{n}m_{j}q_{j}=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}(q_{k}-q_{j}),
        \end{align*}
        which, writing $q_{k}(t)=p(t+k)$, $q_{j}(t)=p(t+j)$ and replacing $t+k$ with $t$, can be rewritten as
        \begin{align*}
          -Mp(t)=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}(p(t+j-k)-p(t)),
        \end{align*}
        which, for any fixed $k$, can be rewritten as
        \begin{align}\label{Sum 2c}
          -Mp(t)=\sum\limits_{j=1}^{n-1}m_{j+k}(p(t+j)-p(t))=\sum\limits_{j=1}^{n-1}m_{j+k}\Delta_{j}(t).
        \end{align}
        Summing both sides of (\ref{Sum 2c}) with respect to $k$ from $1$ to $n$ and dividing the resulting equation on both sides by $n$ then gives
        \begin{align}\label{Sum 3c}
          -Mp(t)=\sum\limits_{j=1}^{n-1}\frac{M}{n}\Delta_{j}(t).
        \end{align}
        Subtracting (\ref{Sum 3c}) from (\ref{Sum 2c}) then finally gives (\ref{Linear combination c}). This completes the proof.
      \end{proof}
      \begin{remark}
        Lemma~\ref{Lemma 1} was proven by Chenciner in \cite{C} for $f(x)=x^{-\frac{3}{2}}$ and his proof works for general $f$ as well. As technically speaking the proof in \cite{C} was written down only for $f(x)=x^{-\frac{3}{2}}$ and to make the paper self-contained, we have included the proof here.
      \end{remark}
  \begin{lemma}\label{Lemma 2}
    Consider any choreography of (\ref{Equations of motion}) for which the curve given by $p(t)$ has an axis of symmetry. Then for all $l$, $k\in\{1,...,n\}$, $l\neq k$ we have that
    \begin{align}\label{Equations of motion choreography result 1}
      &(m_{k}-m_{l})\bigg(p(t+h_{l}-h_{k})-p(t)\bigg)f\left(\|p(t+h_{l}-h_{k})-p(t)\|^{2}\right)\nonumber\\
      &=\sum\limits_{\substack{j=1\\j\neq k,l}}^{n}m_{j}\Bigg(\bigg(p(t+h_{j}-h_{k})-p(t)\bigg)f\left(\|p(t+h_{j}-h_{k})-p(t)\|^{2}\right)\nonumber\\
      &-\bigg(p(t-(h_{j}-h_{l}))-p(t)\bigg)f\left(\|p(t-(h_{j}-h_{l}))-p(t)\|^{2}\right)\Bigg),
    \end{align}
    we have
    \begin{align}\label{Equations of motion choreography result 2}
      &(m_{k}-m_{l})\bigg(p(t+(h_{l}-h_{k}))-p(t)\bigg)f\left(\|p(t+(h_{l}-h_{k}))-p(t)\|^{2}\right)\nonumber\\
      &=\sum\limits_{\substack{j=1\\j\neq k,l}}^{n}m_{j}\Bigg(\bigg(p(t+(h_{j}-h_{k}))-p(t+(h_{l}-h_{k}))\bigg)f\left(\|p(t+(h_{j}-h_{k}))-p(t+(h_{l}-h_{k}))\|^{2}\right)\nonumber\\
      &-\bigg(p(t+(h_{l}-h_{j}))-p(t+(h_{l}-h_{k}))\bigg)f\left(\|p(t+(h_{l}-h_{j}))-p(t+(h_{l}-h_{k}))\|^{2}\right)\Bigg),
    \end{align}
    we have
    \begin{align}\label{Equations of motion choreography result 3}
      &(m_{k}-m_{l})\bigg(p(t+h_{l}-h_{k})-p(t)\bigg)\nonumber\\
      &=\sum\limits_{\substack{j=1\\j\neq k,l}}^{n}m_{j}\Bigg(\bigg(p(t+h_{j}-h_{k})-p(t)\bigg)-\bigg(p(t-(h_{j}-h_{l}))-p(t)\bigg)\Bigg),
    \end{align}
    and we have
    \begin{align}\label{Equations of motion choreography result 4}
      &(m_{k}-m_{l})\bigg(p(t+(h_{l}-h_{k}))-p(t)\bigg)\nonumber\\
      &=\sum\limits_{\substack{j=1\\j\neq k,l}}^{n}m_{j}\Bigg(\bigg(p(t+(h_{j}-h_{k}))-p(t+(h_{l}-h_{k}))\bigg)-\bigg(p(t+(h_{l}-h_{j}))-p(t+(h_{l}-h_{k}))\bigg)\Bigg).
    \end{align}
  \end{lemma}
  \begin{proof}
  Substituting the $q_{k}$ and $q_{j}$ with $p(t+h_{k})$ and $p(t+h_{j})$ respectively in (\ref{Equations of motion}) we get that
  \begin{align*}
    \ddot{p}(t+h_{k})=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}\bigg(p(t+h_{j})-p(t+h_{k})\bigg)f\left(\|p(t+h_{j})-p(t+h_{k})\|^{2}\right),
  \end{align*}
  which holds for all $t\in\mathbb{R}$. As such we may replace $t+h_{k}$ with $t$ to obtain
  \begin{align}\label{Equations of motion choreography 1}
    \ddot{p}(t)=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}\bigg(p(t+h_{j}-h_{k})-p(t)\bigg)f\left(\|p(t+h_{j}-h_{k})-p(t)\|^{2}\right).
  \end{align}
  Let $s=-t$. Substituting $t$ with $-s$ in (\ref{Equations of motion choreography 1}) and using (\ref{reflection identity}) we get
  \begin{align}\label{Equations of motion choreography 2}
    &\begin{pmatrix}
      1 & 0 \\
      0 & -1
    \end{pmatrix}(-1)^{2}\frac{d^{2}}{ds^{2}}{p}(s)\nonumber\\
    &=\begin{pmatrix}
      1 & 0 \\
      0 & -1
    \end{pmatrix}\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}\bigg(p(s-h_{j}+h_{k})-p(s)\bigg)f\left(\|p(s-h_{j}+h_{k})-p(s)\|^{2}\right)
  \end{align}
  and multiplying both sides of (\ref{Equations of motion choreography 2}) from the left with $\begin{pmatrix}
      1 & 0 \\
      0 & -1
    \end{pmatrix}^{-1}$ gives
  \begin{align}\label{Equations of motion choreography 3}
    \ddot{p}(s)=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}\bigg(p(s-h_{j}+h_{k})-p(s)\bigg)f\left(\|p(s-h_{j}+h_{k})-p(s)\|^{2}\right).
  \end{align}
  Replacing $s$ with $t$ and $k$ with $l$, $l\in\{1,...,n\}$ in (\ref{Equations of motion choreography 3}) and subtracting the resulting equation from (\ref{Equations of motion choreography 1}) gives
  \begin{align}\label{Equations of motion choreography 4}
    &0=(m_{l}-m_{k})\bigg(p(t+h_{l}-h_{k})-p(t)\bigg)f\left(\|p(t+h_{l}-h_{k})-p(t)\|^{2}\right)\nonumber\\
    &+\sum\limits_{\substack{j=1\\j\neq k,l}}^{n}m_{j}\Bigg(\bigg(p(t+h_{j}-h_{k})-p(t)\bigg)f\left(\|p(t+h_{j}-h_{k})-p(t)\|^{2}\right)\nonumber\\
    &-\bigg(p(t-(h_{j}-h_{l}))-p(t)\bigg)f\left(\|p(t-(h_{j}-h_{l}))-p(t)\|^{2}\right)\Bigg),
  \end{align}
  which proves (\ref{Equations of motion choreography result 1}).
  Replacing $t$ with $-s-(h_{l}-h_{k})$ in (\ref{Equations of motion choreography 4}), multiplying both sides of the resulting equation from the left with $-\begin{pmatrix}
    1 & 0 \\0 & -1
  \end{pmatrix}^{-1}$ and using (\ref{reflection identity}) gives
  \begin{align}\label{Equations of motion choreography 5}
    &0=(m_{l}-m_{k})\bigg(p(s+(h_{l}-h_{k}))-p(s)\bigg)f\left(\|p(s+(h_{l}-h_{k}))-p(s)\|^{2}\right)\nonumber\\
    &+\sum\limits_{\substack{j=1\\j\neq k,l}}^{n}m_{j}\Bigg(\bigg(p(s+(h_{j}-h_{k}))-p(s+(h_{l}-h_{k}))\bigg)f\left(\|p(s+(h_{j}-h_{k}))-p(s+(h_{l}-h_{k}))\|^{2}\right)\nonumber\\
    &-\bigg(p(s+(h_{l}-h_{j}))-p(s+(h_{l}-h_{k}))\bigg)f\left(\|p(s+(h_{l}-h_{j}))-p(s+(h_{l}-h_{k}))\|^{2}\right)\Bigg).
  \end{align}
  Replacing $s$ with $t$ in (\ref{Equations of motion choreography 5}) then proves (\ref{Equations of motion choreography result 2}). \\
  Note that by (\ref{Center of mass zero}) we have that
  \begin{align}\label{Center of mass zero 1}
    -Mq_{k}=\sum\limits_{j=1}^{n}m_{j}(-q_{k})+0=-\sum\limits_{j=1}^{n}m_{j}q_{k}+\sum\limits_{j=1}^{n}m_{j}q_{j}=\sum\limits_{j=1}^{n}m_{j}(q_{j}-q_{k})=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}(q_{j}-q_{k})
  \end{align}
  and if we replace $q_{k}$ with $p(t+h_{k})$ for all $k\in\{1,...,n\}$ in (\ref{Center of mass zero 1}) and replace $t+h_{k}$ with $t$ in the resulting equation, then we get
  \begin{align}\label{Equations of motion choreography 1 center of mass}
    -Mp(t)=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}\bigg(p(t+h_{j}-h_{k})-p(t)\bigg).
  \end{align}
  Note that the right-hand side of (\ref{Equations of motion choreography 1 center of mass}) is exactly (\ref{Equations of motion choreography 1}) with $f$ replaced by $1$ and $\ddot{p}(t)$ replaced by $-Mp(t)$. Repeating the steps that led from (\ref{Equations of motion choreography 1}) to (\ref{Equations of motion choreography 4}) and (\ref{Equations of motion choreography 5}) but starting with (\ref{Equations of motion choreography 1 center of mass}) instead then proves (\ref{Equations of motion choreography result 3}) and (\ref{Equations of motion choreography result 4}). This completes the proof.
  \end{proof}
  \begin{lemma}\label{Lemma 3}
    Let $q_{1}$,..., $q_{n}$ be an equally spaced choreographic solution of (\ref{EquationsOfMotion Curved}). Then
    \begin{align}\label{CurvedHelp}
       0=\sum\limits_{j=1}^{n-1}\frac{\left(m_{j+k}-\frac{M}{n}\right)\Bigg(p(t+j)-\sigma\bigg(p(t+j)\odot p(t)\bigg)p(t)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+j)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}
    \end{align}
  for all $k\in\{1,...,n\}$.
  \end{lemma}
  \begin{proof}
    Let $q_{1}$,..., $q_{n}$ be an equally spaced choreographic solution of (\ref{EquationsOfMotion Curved}). Then by (\ref{EquationsOfMotion Curved}) we have that
    \begin{align}\label{Step1 Curved}
   \ddot{p}(t+k)&=\sum\limits_{j=1,\textrm{ }j\neq k}^{n}\frac{m_{j}\Bigg(p(t+j)-\sigma\bigg(p(t+j)\odot p(t+k)\bigg)p(t+k)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+j)\odot p(t+k)\bigg)^{2}\Bigg)^{\frac{3}{2}}}\nonumber\\&-\sigma\bigg(\dot{p}(t+k)\odot\dot{p}(t+k)\bigg)p(t+k),\textrm{ }k\in\{1,...,\textrm{ }n\}.
  \end{align}
  Let $s=t+k$. Then we can rewrite (\ref{Step1 Curved}) as
  \begin{align}\label{Step2 Curved}
   \ddot{p}(s)&=\sum\limits_{j=1,\textrm{ }j\neq k}^{n}\frac{m_{j}\Bigg(p(s+j-k)-\sigma\bigg(p(s+j-k)\odot p(s)\bigg)p(s)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(s+j-k)\odot p(s)\bigg)^{2}\Bigg)^{\frac{3}{2}}}-\sigma\bigg(\dot{p}(s)\odot\dot{p}(s)\bigg)p(s)\nonumber\\
   &=\sum\limits_{j=1}^{n-1}\frac{m_{j+k}\Bigg(p(s+j)-\sigma\bigg(p(s+j)\odot p(s)\bigg)p(s)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(s+j)\odot p(s)\bigg)^{2}\Bigg)^{\frac{3}{2}}}-\sigma\bigg(\dot{p}(s)\odot\dot{p}(s)\bigg)p(s).
  \end{align}
  Summing both sides of (\ref{Step2 Curved}) from $1$ to $n$ with respect to $k$ and subsequently dividing both sides by $n$ gives
  \begin{align}\label{Step3 Curved}
   \ddot{p}(s)&=\sum\limits_{j=1}^{n-1}\frac{\frac{M}{n}\Bigg(p(s+j)-\sigma\bigg(p(s+j)\odot p(s)\bigg)p(s)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(s+j)\odot p(s)\bigg)^{2}\Bigg)^{\frac{3}{2}}}-\sigma\bigg(\dot{p}(s)\odot\dot{p}(s)\bigg)p(s).
  \end{align}
  Subtracting (\ref{Step3 Curved}) from (\ref{Step2 Curved}) and replacing $s$ with $t$ then finally gives
  \begin{align*}
       0=\sum\limits_{j=1}^{n-1}\frac{\left(m_{j+k}-\frac{M}{n}\right)\Bigg(p(t+j)-\sigma\bigg(p(t+j)\odot p(t)\bigg)p(t)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+j)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}
    \end{align*}
  for all $k\in\{1,...,n\}$. This completes the proof.
  \end{proof}
  \section{Proof of Theorem~\ref{Main Theorem 1}}\label{Section proof of main theorem 1}
    By Lemma~\ref{Lemma 1}, we have that
    \begin{align}\label{number 1}
          0=\sum\limits_{j=1}^{n-1}\left(m_{k+j}-\frac{M}{n}\right)\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)
    \end{align}
    and
    \begin{align}\label{number 2}
          0=\sum\limits_{j=1}^{n-1}\left(m_{k+j}-\frac{M}{n}\right)\Delta_{j}(t),\textrm{ }k\in\mathbb{Z}.
    \end{align}
    If $d=n-1$, then there exists a subset $S$ of $[0,n]$ of Lebesgue measure $n$ for which the vectors $\Delta_{1}(t),...,\Delta_{n-1}(t)$ are linearly independent for all $t\in S$, which means that by both (\ref{number 1}) and (\ref{number 2}) we have that $m_{j+k}=\frac{M}{n}$ for all $k$, $j\in\mathbb{Z}$, which proves that if $d=n-1$ all masses are equal. If $d=n-2$, there exists an $S\subset[0,n]$ of Lebesgue measure $n$, such that for all $t\in S$ we have that the vectors $\Delta_{1}(t),...,\Delta_{n-1}(t)$ span an $n-2$-dimensional space. For any $t\in S$ there then exists an $r\in\{1,...,n-1\}$ (which may depend on $t$), such that by (\ref{number 1}) and (\ref{number 2}) we have that
    \begin{align}\label{number 3}
          -\left(m_{k+r}-\frac{M}{n}\right)\Delta_{r}(t)f\left(\|\Delta_{r}(t)\|^{2}\right)=\sum\limits_{\substack{j=1\\j\neq r}}^{n-1}\left(m_{k+j}-\frac{M}{n}\right)\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)
    \end{align}
    and
    \begin{align}\label{number 4}
          -\left(m_{k+r}-\frac{M}{n}\right)\Delta_{r}(t)=\sum\limits_{\substack{j=1\\j\neq r}}^{n-1}\left(m_{k+j}-\frac{M}{n}\right)\Delta_{j}(t),
    \end{align}
    where the vectors $\Delta_{j}(t)$, $j\in\{1,...,n-1\}$, $j\neq r$, span an $n-2$-dimensional space. The $\Delta_{j}(t)$ are linearly independent for $j\in\{1,...,n-1\}$, $j\neq r$, so for those $j$ we have that if we multiply both sides of (\ref{number 4}) with $f\left(\|\Delta_{r}(t)\|^{2}\right)$ and compare the coefficients of the $\Delta_{j}(t)$ of the resulting identity with the coefficients of the $\Delta_{j}(t)$ in (\ref{number 3}), we find that  \begin{align*}\left(m_{k+j}-\frac{M}{n}\right)f\left(\|\Delta_{r}(t)\|^{2}\right)=\left(m_{k+j}-\frac{M}{n}\right)f\left(\|\Delta_{j}(t)\|^{2}\right)\end{align*} for all $k\in\mathbb{Z}$, $j\in\{1,...,n-1\}$. So if there is at least one $t\in S$ for which there are an $r$ and a $j$ for which  $f\left(\|\Delta_{j}(t)\|^{2}\right)\neq f\left(\|\Delta_{r}(t)\|^{2}\right)$, then all masses are equal. The alternative is that $f\left(\|\Delta_{j_{1}}(t)\|^{2}\right)=f\left(\|\Delta_{j_{2}}(t)\|^{2}\right)$ for all $j_{1}$, $j_{2}\in\{1,...,n-1\}$ and all $t\in S$. But if that is the case, then because $f$ is a strictly decreasing and therefore bijective function, we have that $\|\Delta_{j_{1}}(t)\|=\|\Delta_{j_{2}}(t)\|$ for all $j_{1}$, $j_{2}\in\{1,...,n-1\}$, for all $t\in[0,n]$, which means that, by using a suitable change of variables, we have that $\|q_{j_{1}}(t)-q_{j_{2}}(t)\|=\|q_{l_{1}}(t)-q_{l_{2}}(t)\|$ for all $j_{1},j_{2},l_{1},l_{2}\in\{1,...,n\}$, $j_{1}\neq j_{2}$ and $l_{1}\neq l_{2}$, for all $t\in [0,n]$, which means that the $q_{1}$,...,$q_{n}$ represent the vertices of an $n-1$-dimensional simplex, which contradicts that $d=n-2$. So all masses are equal.

    If $d=n-r$, $r\geq 3$, we have that there exists a set $S\subset[0,n]$ with Lebesgue measure $n$, for which the vectors $\Delta_{j}(t)$, $j\in\{1,...,n-1\}$ span an $n-r$-dimensional space for all $t\in S$. Additionally, note that because $\widehat{e}_{1}$,..., $\widehat{e}_{n}$ span $\mathbb{C}^{n}$, there exist constants $a_{1},...,a_{n-1}$ such that
    \begin{align}\label{Span}
      \begin{pmatrix}
        m_{1+k}-\frac{M}{n}\\
        m_{2+k}-\frac{M}{n}\\
        \vdots\\
        m_{n+k}-\frac{M}{n}
      \end{pmatrix}=\sum\limits_{l=1}^{n-1}a_{l}\lambda_{l}^{k}\widehat{e}_{l}.
    \end{align}
    Note that by construction $\widehat{e}_{n}$ is orthogonal to $\left(m_{1+k}-\frac{M}{n},m_{2+k}-\frac{M}{n},...,m_{n+k}-\frac{M}{n}\right)^{T}$, so we may exclude $\widehat{e}_{n}$ in the linear combination in (\ref{Span}).
    Again by Lemma~\ref{Lemma 1}, this means that
    \begin{align*}
          0=\sum\limits_{j=1}^{n-1}\left(m_{k+j}-\frac{M}{n}\right)\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)=\frac{1}{\sqrt{n}}\sum\limits_{j=1}^{n-1}\left(\sum\limits_{l=1}^{n-1}a_{l}\lambda_{l}^{k+j-1}\right)\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)
    \end{align*}
    and
    \begin{align*}
          0=\sum\limits_{j=1}^{n-1}\left(m_{k+j}-\frac{M}{n}\right)\Delta_{j}(t)=\frac{1}{\sqrt{n}}\sum\limits_{j=1}^{n-1}\left(\sum\limits_{l=1}^{n-1}a_{l}\lambda_{l}^{k+j-1}\right)\Delta_{j}(t),
    \end{align*}
    so
    \begin{align*}
          0=\sum\limits_{l=1}^{n-1}a_{l}\lambda_{l}^{k}\sum\limits_{j=1}^{n-1}\lambda_{l}^{j-1}\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)
    \textrm{ and }
          0=\sum\limits_{l=1}^{n-1}a_{l}\lambda_{l}^{k}\sum\limits_{j=1}^{n-1}\lambda_{l}^{j-1}\Delta_{j}(t),
    \end{align*}
    which means by the linear independence of the $\lambda_{l}^{k}$ as functions of $k$ that for all $l\in\{1,...,n-1\}$ we have that $a_{l}=0$, or
    \begin{align}
          0=\sum\limits_{j=1}^{n-1}\lambda_{l}^{j-1}\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)\textrm{ and }\label{DeltaForce1}\\
          0=\sum\limits_{j=1}^{n-1}\lambda_{l}^{j-1}\Delta_{j}(t).\label{DeltaForce2}
    \end{align}
    If there are any $l$ for which $a_{l}\neq 0$, $l\neq\frac{n}{2}\textrm{ }mod\textrm{ }n$ if $n$ is even, then taking complex conjugates on both sides of (\ref{DeltaForce1}) and (\ref{DeltaForce2}) gives
    \begin{align}
          0=\sum\limits_{j=1}^{n-1}\lambda_{\pm l}^{j-1}\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)\textrm{ and }\label{Lead up to general equations1}\\
          0=\sum\limits_{j=1}^{n-1}\lambda_{\pm l}^{j-1}\Delta_{j}(t),\label{Lead up to general equations2}
    \end{align}
    so for all $k\in\{1,...,n\}$ we have that
    \begin{align*}
          &0=\lambda_{l}^{k-1}\sum\limits_{j=1}^{n-1}\lambda_{-l}^{j-1}\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)-\lambda_{-l}^{k-1}\sum\limits_{j=1}^{n-1}\lambda_{l}^{j-1}\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)\textrm{ and }\\
          &0=\lambda_{l}^{k-1}\sum\limits_{j=1}^{n-1}\lambda_{-l}^{j-1}\Delta_{j}(t)-\lambda_{-l}^{k-1}\sum\limits_{j=1}^{n-1}\lambda_{l}^{j-1}\Delta_{j}(t),
    \end{align*}
    so as $\lambda_{l}^{k-1}\lambda_{-l}^{j-1}-\lambda_{-l}^{k-1}\lambda_{-l}^{k}\lambda_{l}^{j-1}=2i\sin{\frac{2\pi l}{n}(k-j)}$ that means that
    \begin{align*}
          0=\sum\limits_{\substack{j=1\\j\neq k}}^{n-1}\left(\sin{\frac{2\pi l}{n}(k-j)}\right)\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)\textrm{ and }\\
          0=\sum\limits_{\substack{j=1\\j\neq k}}^{n-1}\left(\sin{\frac{2\pi l}{n}(k-j)}\right)\Delta_{j}(t)
    \end{align*}
    for all $k\in\{1,...,n\}$. Particularly, if $n$ is even, this means that
    \begin{align}
          0=\sum\limits_{\substack{j=1\\j\neq k,k+\frac{n}{2}}}^{n-1}\left(\sin{\frac{2\pi l}{n}(k-j)}\right)\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)\textrm{ and }\label{General equations1}\\
          0=\sum\limits_{\substack{j=1\\j\neq k,k+\frac{n}{2}}}^{n-1}\left(\sin{\frac{2\pi l}{n}(k-j)}\right)\Delta_{j}(t)\label{General equations2}
    \end{align}
    for all $k\in\{1,...,n\}$. \\
    So if $d=n-3$, we have, reusing the argument we used for the $d=n-2$ case, that there are two ways we can write a vector as a linear combination of $n-3$ linearly independent vectors, unless $\|\Delta_{j}(t)\|$ is independent of $j$, in which case the point masses are the vertices of an $n$-simplex, which is a contradiction.   This means that if $d=n-1$, $d=n-2$, or if $n$ is odd and $d=n-3$, all masses are equal (as the $a_{l}$ are all zero) and if $n$ is even and $d=n-3$, all masses with an odd label are equal and all masses with an even label are equal (as we have not excluded the possibility that $a_{\frac{n}{2}}\neq 0$). This completes the proof.
  \section{Proof of Corollary~\ref{Main Corollary 1}}\label{Section proof of main corollary 1}
  If $q_{1}$,..., $q_{n}$ move along a curve like an equally spaced choreography and have masses $m_{1},...,m_{n}$ and $q_{n+1}=0$ and has mass $m$, then by (\ref{Equations of motion}) we have for $k\neq n+1$ that
  \begin{align*}
    \ddot{q}_{k}=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}(q_{j}-q_{k})f\left(\|q_{j}-q_{k}\|^{2}\right)-m q_{k}f\left(\|q_{k}\|^{2}\right),
  \end{align*}
  which means that the same argument that gave (\ref{Sum 1f}) gives
  \begin{align}\label{the n+1 case 1}
     \ddot{p}(t)+m p(t)f\left(\|p(t)\|^{2}\right)&=\sum\limits_{j=1}^{n-1}m_{j+k}(p(t+j)-p(t))f\left(\|p(t+j)-p(t)\|^{2}\right)\nonumber\\
     &=\sum\limits_{j=1}^{n-1}m_{j+k}\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right),
  \end{align}
  if we use our definition of $m_{j+k}$, $j$, $k\in\mathbb{Z}$ for the masses $m_{1}$,...,$m_{n}$.

  Summing both sides of (\ref{the n+1 case 1}) with respect to $k$ from $1$ to $n$ and dividing both sides of the resulting equation by $n$ then gives
  \begin{align}\label{the n+1 case 2}
     \ddot{p}(t)+m p(t)f\left(\|p(t)\|^{2}\right)=\sum\limits_{j=1}^{n-1}\frac{M}{n}\Delta_{j}(t)f\left(\|\Delta_{j}(t)\|^{2}\right)
  \end{align}
  and subtracting (\ref{the n+1 case 2}) from (\ref{the n+1 case 1}) then gives (\ref{Linear combination f}) again. As the formula for the center of mass is exactly the same as in the proof of Lemma~\ref{Lemma 1}, (\ref{Linear combination c}) holds true as well and as (\ref{Linear combination f}) and (\ref{Linear combination c}) generated the proof of Theorem~\ref{Main Theorem 1}, the proof of Theorem~\ref{Main Theorem 1} proves Corollary~\ref{Main Corollary 1} as well.
  \section{Proof of Theorem~\ref{Main Theorem 2}}\label{Section proof of main theorem 2}
  Consider a choreography solution of (\ref{Equations of motion}) for $n=3$ for which $p$ has an axis of symmetry. Then by Lemma~\ref{Lemma 2} we have that for any $i$, $j$, $k\in\{1,2,3\}$, $i\neq j$, $i\neq k$, $j\neq k$,
  \begin{align}\label{Boom1}
      &(m_{i}-m_{k})\bigg(p(t+h_{k}-h_{i})-p(t)\bigg)f\left(\|p(t+h_{k}-h_{i})-p(t)\|^{2}\right)\nonumber\\
      &=m_{j}\Bigg(\bigg(p(t+h_{j}-h_{i})-p(t)\bigg)f\left(\|p(t+h_{j}-h_{i})-p(t)\|^{2}\right)\nonumber\\
      &-\bigg(p(t-(h_{j}-h_{k}))-p(t)\bigg)f\left(\|p(t-(h_{j}-h_{k}))-p(t)\|^{2}\right)\Bigg),
    \end{align}
    \begin{align}\label{Boom2}
      &(m_{i}-m_{k})\bigg(p(t+(h_{k}-h_{i}))-p(t)\bigg)f\left(\|p(t+(h_{k}-h_{i}))-p(t)\|^{2}\right)\nonumber\\
      &=m_{j}\Bigg(\bigg(p(t+(h_{j}-h_{i}))-p(t+(h_{k}-h_{i}))\bigg)f\left(\|p(t+(h_{j}-h_{i}))-p(t+(h_{k}-h_{i}))\|^{2}\right)\nonumber\\
      &-\bigg(p(t-(h_{j}-h_{k}))-p(t+(h_{k}-h_{i}))\bigg)f\left(\|p(t-(h_{j}-h_{k}))-p(t+(h_{k}-h_{i}))\|^{2}\right)\Bigg),
    \end{align}
    \begin{align}\label{Boom3}
      &(m_{i}-m_{k})\bigg(p(t+h_{k}-h_{i})-p(t)\bigg)\nonumber\\
      &=m_{j}\Bigg(\bigg(p(t+h_{j}-h_{i})-p(t)\bigg)-\bigg(p(t-(h_{j}-h_{k}))-p(t)\bigg)\Bigg),
    \end{align}
    and
    \begin{align}\label{Boom4}
      &(m_{i}-m_{k})\bigg(p(t+(h_{k}-h_{i}))-p(t)\bigg)\nonumber\\
      &=m_{j}\Bigg(\bigg(p(t+(h_{j}-h_{i}))-p(t+(h_{k}-h_{i}))\bigg)-\bigg(p(t-(h_{j}-h_{k}))-p(t+(h_{k}-h_{i}))\bigg)\Bigg).
    \end{align}
    If $p(t+h_{j}-h_{i})-p(t)$ and $p(t-(h_{j}-h_{k}))-p(t)$ are linearly independent, or pointing in the same direction, then for (\ref{Boom1}) and (\ref{Boom3}) to both be true, because $xf(x^{2})$ is a decreasing function, we need that these vectors are the same length, as otherwise $(m_{i}-m_{k})(p(t+h_{k}-h_{i})-p(t))$ has two different directions. By the same argument, working with (\ref{Boom2}) and (\ref{Boom4}) instead, we need that $p(t+(h_{j}-h_{i}))-p(t+(h_{k}-h_{i}))$ and $p(t-(h_{j}-h_{k}))-p(t+(h_{k}-h_{i}))$ have the same length if they are linearly independent, or pointing in the same direction. If for all $t$ at least one of these conditions is not met, then it is easy to see that using the continuity of $p$ all point masses lie on a straight line, which contradicts that they lie on a closed curve by \S 331 of \cite{W}. Thus we find that $p(t+(h_{j}-h_{i}))$ and $p(t-(h_{j}-h_{k}))$ lie on a circle of radius $\|p(t+(h_{k}-h_{i}))-p(t)\|$ around $p(t)$ and we find that $p(t+(h_{j}-h_{i}))$ and $p(t-(h_{j}-h_{k}))$ lie on a circle of radius $\|p(t+(h_{k}-h_{i}))-p(t)\|$ around $p(t+(h_{k}-h_{i}))$. This is only possible if the vectors on the right-hand sides of (\ref{Boom1}), (\ref{Boom2}), (\ref{Boom3}) and (\ref{Boom4}) cancel out against each other, which means that as $p(t+(h_{k}-h_{i}))-p(t)\neq 0$, we have that $m_{i}-m_{k}=0$ for all $i$, $k\in\{1,2,3\}$. This completes the proof.
  \section{Proof of Corollary~\ref{Main Corollary 2}}\label{Section proof of main corollary 2}
  If $q_{1}$,..., $q_{n}$ move along a curve like a choreography and $q_{n+1}=0$ with mass $m$, then, as in the proof of Corollary~\ref{Main Corollary 1}, by (\ref{Equations of motion}) we have for $k\neq n+1$ that
  \begin{align}\label{Corollary 2 Identity 1}
    \ddot{q}_{k}=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}(q_{j}-q_{k})f\left(\|q_{j}-q_{k}\|^{2}\right)-m q_{k}f\left(\|q_{k}\|^{2}\right).
  \end{align}
  Using that $q_{j}(t)=p(t+h_{j})$ for $j<n+1$ and defining $s=t+h_{k}$, (\ref{Corollary 2 Identity 1}) can be rewritten as
  \begin{align}\label{Corollary 2 Identity 2}
     \ddot{p}(s)+m p(s)f\left(\|p(s)\|^{2}\right)=\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}\bigg(p(s+h_{j}-h_{k})-p(s)\bigg)f\left(\|p(s+h_{j}-h_{k})-p(s)\|^{2}\right).
  \end{align}
  Replacing $s$ with $-t$  in (\ref{Corollary 2 Identity 2}) and subtracting the resulting equation from (\ref{Corollary 2 Identity 2}), using that $p(-u)=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}p(u)$ for any $u\in\mathbb{R}$ gives
  \begin{align}\label{Corollary 2 Identity 3}
     &\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\bigg(\ddot{p}(t)+m p(t)f\left(\|p(t)\|^{2}\right)\bigg)\nonumber\\
     &=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\sum\limits_{\substack{j=1\\j\neq k}}^{n}m_{j}\bigg(p(t-(h_{j}-h_{k}))-p(t)\bigg)f\left(\|p(t-(h_{j}-h_{k}))-p(t)\|^{2}\right).
  \end{align}
  Multiplying (\ref{Corollary 2 Identity 3}) on both sides with $\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}^{-1}$ from the left and subtracting (\ref{Corollary 2 Identity 3}) from (\ref{Corollary 2 Identity 2}) (having replaced $s$ in (\ref{Corollary 2 Identity 2}) with $t$) gives (\ref{Equations of motion choreography result 1}), after which the proof of Lemma~\ref{Lemma 2} can be followed to the letter to obtain (\ref{Equations of motion choreography result 2}), (\ref{Equations of motion choreography result 3}) and (\ref{Equations of motion choreography result 4}), after which we can repeat the proof of Theorem~\ref{Main Theorem 2} to obtain our result. This completes the proof.
  \section{Proof of Theorem~\ref{Main Theorem 3}}\label{Section proof of main theorem 3}
  Let $q_{1}$,...,$q_{n}$ be an equally spaced choreographic solution of (\ref{EquationsOfMotion Curved}). Then by Lemma~\ref{Lemma 3} we have that
  \begin{align*}
       0&=\sum\limits_{j=1}^{n-1}\frac{\left(m_{j+k}-\frac{M}{n}\right)\Bigg(p(t+j)-\sigma\bigg(p(t+j)\odot p(t)\bigg)p(t)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+j)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}
  \end{align*}
  for all $k\in\mathbb{Z}$.
  We again note, as in the proof of Theorem~\ref{Main Theorem 1}, that because $\widehat{e}_{1}$,..., $\widehat{e}_{n}$ span $\mathbb{C}^{n}$, there exist constants $a_{1},...,a_{n-1}$ such that
    \begin{align}\label{Span curved}
      \begin{pmatrix}
        m_{1+k}-\frac{M}{n}\\
        m_{2+k}-\frac{M}{n}\\
        \vdots\\
        m_{n+k}-\frac{M}{n}
      \end{pmatrix}=\sum\limits_{l=1}^{n-1}a_{l}\lambda_{l}^{k}\widehat{e}_{l}.
    \end{align}
    Again, note that by construction $\widehat{e}_{n}$ is orthogonal to $\left(m_{1+k}-\frac{M}{n},m_{2+k}-\frac{M}{n},...,m_{n+k}-\frac{M}{n}\right)^{T}$, so we may exclude $\widehat{e}_{n}$ in the linear combination in (\ref{Span curved}). Finally, analogous to the proof of Theorem~\ref{Main Theorem 1}, we find by (\ref{Span curved}) that proving that $m_{j+k}=\frac{M}{n}$ for all $j$, $k\in\mathbb{Z}$ is equivalent to proving that $a_{l}=0$ for all $l\in\{1,...,n-1\}$, which can be done by proving that
    \begin{align}\label{The L}
       0\neq\sum\limits_{j=1}^{n-1}\lambda_{l}^{j-1}\frac{\Bigg(p(t+j)-\sigma\bigg(p(t+j)\odot p(t)\bigg)p(t)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+j)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}
  \end{align}
  for all $l\in\{1,...,n-1\}$. \\
  If $n=3$, we immediately get that the right-hand side of (\ref{The L}) cannot be real-valued and therefore cannot be zero.
  If $n=4$, we have that $\lambda_{l}=e^{\frac{2\pi i}{4}}$, $\lambda_{l}=e^{\frac{4\pi i}{4}}$, or $\lambda_{l}=e^{\frac{6\pi i}{4}}$. If $\lambda_{l}=e^{\frac{2\pi i}{4}}$, or $\lambda_{l}=e^{\frac{6\pi i}{4}}$, then the imaginary part of the right-hand side of (\ref{The L}) is \begin{align*}\pm \frac{p(t+2)-\sigma\bigg(p(t+2)\odot p(t)\bigg)p(t)}{\Bigg(\sigma -\sigma\bigg(p(t+2)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}},\end{align*} which cannot be zero. However, if $\lambda_{l}=e^{\frac{4\pi i}{4}}=-1$, we cannot use a similar argument to prove that the right-hand side of (\ref{The L}) is not zero, which means that $a_{2}$ is not necessarily zero, which translates into the possibility of $m_{1}=m_{3}$ and  $m_{2}=m_{4}$, but not all masses being equal.
  For $n=5$ we have that if (\ref{The L}) does not hold for a certain $l<5$, then
  \begin{align}\label{The NOT L1}
       0=\sum\limits_{j=1}^{4}\lambda_{l}^{j-1}\frac{p(t+j)-\sigma\bigg(p(t+j)\odot p(t)\bigg)p(t)}{\Bigg(\sigma -\sigma\bigg(p(t+j)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}
  \end{align}
  and by Lemma~\ref{Lemma 3}
  \begin{align}\label{Help5}
       0&=\sum\limits_{j=1}^{4}\frac{\left(m_{j+k}-\frac{M}{5}\right)\Bigg(p(t+j)-\sigma\bigg(p(t+j)\odot p(t)\bigg)p(t)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+j)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}.
  \end{align}
  Multiplying (\ref{The NOT L1}) on both sides with $m_{1+k}-\frac{M}{5}$ and subtracting the resulting equation from (\ref{Help5}) gives
  \begin{align}\label{Help5 1}
       0&=\sum\limits_{j=2}^{4}\frac{\left(m_{j+k}-\frac{M}{5}-\lambda_{l}^{j-1}\left(m_{1+k}-\frac{M}{5}\right)\right)\Bigg(p(t+j)-\sigma\bigg(p(t+j)\odot p(t)\bigg)p(t)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+j)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}
  \end{align}
  and taking complex conjugates on both sides of (\ref{Help5 1}) gives
  \begin{align}\label{Help5 2}
       0&=\sum\limits_{j=2}^{4}\frac{\left(m_{j+k}-\frac{M}{5}-\lambda_{5-l}^{j-1}\left(m_{1+k}-\frac{M}{5}\right)\right)\Bigg(p(t+j)-\sigma\bigg(p(t+j)\odot p(t)\bigg)p(t)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+j)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}.
  \end{align}
  Note that
  \begin{align}\label{MassVector1}
    \begin{pmatrix}
      m_{2+k}-\frac{M}{5}-\lambda_{l}\left(m_{1+k}-\frac{M}{5}\right)\\
      m_{3+k}-\frac{M}{5}-\lambda_{l}^{2}\left(m_{1+k}-\frac{M}{5}\right)\\
      m_{4+k}-\frac{M}{5}-\lambda_{l}^{3}\left(m_{1+k}-\frac{M}{5}\right)
    \end{pmatrix}
    \end{align}
    and
    \begin{align}
    \begin{pmatrix}\label{MassVector2}
      m_{2+k}-\frac{M}{5}-\lambda_{5-l}\left(m_{1+k}-\frac{M}{5}\right)\\
      m_{3+k}-\frac{M}{5}-\lambda_{5-l}^{2}\left(m_{1+k}-\frac{M}{5}\right)\\
      m_{4+k}-\frac{M}{5}-\lambda_{5-l}^{3}\left(m_{1+k}-\frac{M}{5}\right)
    \end{pmatrix}
  \end{align}
  are linearly independent if and only if $m_{1+k}-\frac{M}{5}\neq 0$. Additionally, note that using the Euclidean inner product, the vectors
  \begin{align*}
    &\frac{p(t+2)-\sigma\bigg(p(t+2)\odot p(t)\bigg)p(t)}{\Bigg(\sigma -\sigma\bigg(p(t+2)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}},\textrm{ }\frac{p(t+3)-\sigma\bigg(p(t+3)\odot p(t)\bigg)p(t)}{\Bigg(\sigma -\sigma\bigg(p(t+3)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}\\
    &\textrm{and }\frac{p(t+4)-\sigma\bigg(p(t+4)\odot p(t)\bigg)p(t)}{\Bigg(\sigma -\sigma\bigg(p(t+4)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}
  \end{align*}
  in (\ref{Help5 1}) and (\ref{Help5 2}) are orthogonal to
  \begin{align*}
    v(t):=\begin{pmatrix}
      1 & 0 & 0 \\
      0 & 1 & 0 \\
      0 & 0 & \sigma
    \end{pmatrix}p(t)
  \end{align*}
  and span a two-dimensional space if and only if $v(t)$, $p(t+2)$, $p(t+3)$ and $p(t+4)$ do not lie in the same plane. For $\sigma=-1$, by the geometry of the hyperboloid, we directly see that $v(t)$, $p(t+2)$, $p(t+3)$ and $p(t+4)$ do not lie in the same plane. For $\sigma=1$ we have that $v(t)=p(t)$ and $p(t)$, $p(t+2)$, $p(t+3)$ and $p(t+4)$ may lie in the same plane if that plane goes through the center of the sphere, in which case the point masses $p(t)$, $p(t+2)$, $p(t+3)$ and $p(t+4)$ lie on a great circle. Because $p(t)$, $p(t+2)$, $p(t+3)$ and $p(t+4)$ then lie on a great circle and lie in the same plane, we can write $p(t)=a(t)p(t+2)+b(t)p(t+3)$, where $a$ and $b$ are scalar functions on $\mathbb{R}$, which means that $p(t+1)=a(t+1)p(t+3)+b(t+1)p(t+4)$, so because $p(t+3)$ and $p(t+4)$ lie in the same plane, $p(t+1)$ lies in that plane and all the point masses lie on a great circle. So that means that unless the point masses lie on a great circle, we have that
  \begin{align*}
    &\frac{p(t+2)-\sigma\bigg(p(t+2)\odot p(t)\bigg)p(t)}{\Bigg(\sigma -\sigma\bigg(p(t+2)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}},\textrm{ }\frac{p(t+3)-\sigma\bigg(p(t+3)\odot p(t)\bigg)p(t)}{\Bigg(\sigma -\sigma\bigg(p(t+3)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}\\
    &\textrm{and }\frac{p(t+4)-\sigma\bigg(p(t+4)\odot p(t)\bigg)p(t)}{\Bigg(\sigma -\sigma\bigg(p(t+4)\odot p(t)\bigg)^{2}\Bigg)^{\frac{3}{2}}}
  \end{align*}
  are three vectors that span a two-dimensional space (if the point masses do not lie on a great circle), which means that if (\ref{MassVector1}) and (\ref{MassVector2}) are linearly independent, we have a contradiction by (\ref{Help5 1}) and (\ref{Help5 2}). So this means that $m_{1+k}-\frac{M}{5}=0$, which proves that all masses are equal if $n=5$.
  This then finally proves that if $\sigma=-1$, if $n<6$ and $n\neq 4$, all masses are equal and that if $n=4$ the odd labeled masses are equal and the even labeled masses are equal and that the same holds true if $\sigma=1$, unless the point masses all move along a great circle.
  \section{Proof of Corollary~\ref{Main Corollary 3}}\label{Section proof of main corollary 3}
  Let $q_{1},...,q_{n+1}$ be a solution of (\ref{EquationsOfMotion Curved}), where $q_{1},...,q_{n}$ is an equally spaced choreography and $q_{n+1}=(0,0,1)^{T}$ with mass $m$. Let $\widehat{e}=(0,0,1)^{T}$. Then by (\ref{EquationsOfMotion Curved}) we have for $k\in\{1,...,n\}$ that
  \begin{align*}
   &\ddot{p}(t+k)+\sigma\bigg(\dot{p}(t+k)\odot\dot{p}(t+k)\bigg)p(t+k)\\
   &=\sum\limits_{j=1,\textrm{ }j\neq k}^{n}\frac{m_{j}\Bigg(p(t+j)-\sigma\bigg(p(t+k)\odot p(t+j)\bigg)p(t+k)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+k)\odot p(t+j)\bigg)^{2}\Bigg)^{\frac{3}{2}}}\\
   &+\frac{m\Bigg(\widehat{e}-\sigma\bigg(p(t+k)\odot \widehat{e}\bigg)p(t+k)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(t+k)\odot \widehat{e}\bigg)^{2}\Bigg)^{\frac{3}{2}}},
  \end{align*}
  which, writing $s=t+k$, can be rewritten as
  \begin{align}\label{Final}
   &\ddot{p}(s)+\sigma\bigg(\dot{p}(s)\odot\dot{p}(s)\bigg)p(s)\nonumber\\
   &=\sum\limits_{j=1}^{n-1}\frac{m_{j+k}\Bigg(p(s+j)-\sigma\bigg(p(s)\odot p(s+j)\bigg)p(s)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(s)\odot p(s+j)\bigg)^{2}\Bigg)^{\frac{3}{2}}}+\frac{m\Bigg(\widehat{e}-\sigma\bigg(p(s)\odot \widehat{e}\bigg)p(s)\Bigg)}{\Bigg(\sigma -\sigma\bigg(p(s)\odot \widehat{e}\bigg)^{2}\Bigg)^{\frac{3}{2}}}.
  \end{align}
  Summing both sides of (\ref{Final}) with respect to $k$ from $1$ to $n$, dividing both sides of the resulting equation by $n$ and subtracting that equation from (\ref{Final}) then gives (\ref{CurvedHelp}), which is the result on which the proof of Theorem~\ref{Main Theorem 3} was based.
   So if $\sigma=-1$, if $n<6$ and $n\neq 4$, then the masses $m_{j}$, $j\in\{1,...,n\}$ are equal and if $n=4$ the odd labeled masses $m_{j}$ are equal and the even labeled masses $m_{j}$ are equal, $j\in\{1,...,n\}$ and the same holds true for $\sigma=1$, unless the point masses all move along a great circle.
   \section{Acknowledgements}
   The author would like to thank Dr. Pascal Grange and Dr. Yinna Ye of Xi'an Jiaotong-Liverpool University for their useful feedback during the final stage of the writing of this paper.
 
