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\begin{proposition}
  \label{prop:3.3}
  % Proposition 3.3
  $X = \{0, 1, 2\}$ is not \gls{eram}.
\end{proposition}

\begin{proof}
  Recall in $\Rbb^n$ we have
  \[
    \|x + y\|_2^2 + \|x - y\|_2^2
    = 2(\|x\|_2^2 + \|y\|_2^2)
  .\]
  Every copy of $\{0, 1, 2\}$ is $\{x - y, x, x +
  y\}$ with $\|y\|_2 = 1$ (in any $\Rbb^n$). We
  have
  \[
    \|x + y\|_2^2 + \|x - y\|_2^2
    = 2 \|x\|_2^2 + 2
  .\]
  If we can find a colouring of $\Rbb_{\ge 0}$
  such that there does not exist a monochromatic
  solution to $a + b = 2c + 2$. Then use $c(x) \to
  \phi(\|x\|_2^2)$.

  We $4$-colour $\Rbb_{\ge 0}$ by $\phi(x) =
  \lfloor x \rfloor \pmod{4}$.

  Suppose $a, b, c$ all have colour $n \in \{0, 1,
  2, 3\}$. Then $a + b = c + 2$ implies
  \[
    a + b - 2c = M_4 + \{a\} + \{b\} - 2 \{c\} = 2
  \]
  where $M_4$ is a multiple of $4$. This is
  impossible as $-2 < \{a\} + \{b\} - 2 \{c\} <
  2$.
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item For all $n$, there is a $4$-colouring of
      $\Rbb^n$ that stops every copy of $X = \{0,
      1, 2\}$ from being monochromatic.
    \item Very important in $a + b = 2c + 2$, we
      got this $2$ to not be $0$.
    \item It will turn out that the property that
      made $\{0, 1, 2\}$ not \gls{eram} is ``it
      does not lie on a sphere''.
  \end{enumerate}
\end{remark*}

\begin{definition*}[Spherical]
  \glsadjdefn{spher}{spherical}{set}
  A set $X \subseteq \Rbb^n$ is called
  \emph{spherical} if it lies on the surfcace of a
  sphere.
\end{definition*}

For example, a triangle, a rectangle, any simplex
(non-degenerate).

\begin{definition*}[Simplex]
  \glsnoundefn{simp}{simplex}{simplexes}%
  Let $x_1, \ldots, x_d$ be some points. They form
  a \emph{simplex} if $x_1 - x_d, x_2 - x_d,
  \ldots, x_{d - 1} - x_d$ are linearly
  independent. In other words, they do not lie in
  a $(d - 2)$-dimensional affine space.
\end{definition*}

\textbf{Aim:} If $X$ is \gls{eram}, then $X$ is
spherical.

To do so, we will use a ``generalised
parallelogram law''.

\begin{lemma}
  \label{lemma:3.4}
  % Lemma 3.4
  \begin{iffc}
    \lhs
    $x_1, \ldots, x_d$ in $\Rbb^n$ are not
    \gls{spher}
    \rhs
    there exists $c_i \in \Rbb$, not all $0$, such
    that:
    \begin{enumerate}[(1)]
      \item $\sum_i c_i = 0$
      \item $\sum_i c_i x_i = 0$
      \item $\sum_i c_i \|x_i\|_2^2 \neq 0$
    \end{enumerate}
  \end{iffc}
\end{lemma}

In the previous proof, we took $(1, 1, -2)$ and
$2$ being the value in (3).

\begin{iffproof}
  \rightimpl
  $x_1, \ldots, x_d$ are not \gls{spher}. The
  first two conditions say that $x_1, \ldots, x_d$
  is not a \gls{simp}: so there exists $c_1,
  \ldots, c_{d - 1}$ (not all $0$) such that
  $\sum_i c_i(x_i - x_d) = 0$ or $\sum_{i = 1}^{N -
  1} c_i x_i + (-c_1 - c_2 - \cdots - c_{d -
  1})x_d = 0$.

  First we note that (1)--(3) are invariant under
  translation by $v \in \Rbb^n$:
  \begin{itemize}
    \item $\sum_i c_i(x_i + v) = 0$ since $\sum_i
      c_i = 0$
    \item $\sum_i c_i \|x_i + v\|_2^2 = \sum_i c_i
      \|x_i\|_2^2 + 2c_i (x_i, v) + c_i \|v\|_2^2
      = \sum_i c_i \|x_i\|_2^2 + 2 \left( \sum_i
      c_i x_i, v \right) + \|v\|_2^2 \sum_i c_i =
      \sum_i c_i \|x_i\|_2^2$
  \end{itemize}
  Let us look at a minimal subset of $x_1, \ldots,
  x_d$ that is not \gls{spher}. If we can show
  this for without loss of generality $x_1,
  \ldots, x_k$ ($k \le d$) then take $c_i = 0$ for
  all $i \in [k + 1, d]$. Let $\{x_2, \ldots,
  x_k\}$. This is \gls{spher} by minimality.
  Suppose the sphere radius is $r$, and centred at
  $y$.

  By the above, translate the set such that
  $\{x_2, \ldots, x_k\}$ is centred at $0$. Since
  $\{x_1, \ldots, x_k\}$ are not \gls{spher}, it
  is not a \gls{simp}. Hence there exists $c_i$
  such that $\sum_i c_i(x_i - x_k) = 0$. Then
  \[
    c_1 x_1 + c_2 x_2 + \cdots + c_{k - 1} x_{k -1
    } + (-c_1 - c_2 - \cdots - c_{k - 1}) x_k = 0
  .\]
  Without loss of generality $c_i \neq 0$. This is
  fine because the same $c_i$'s work after
  translations ((1)--(3) is totally invariant
  under translations). Then
  \[
    \sum_{i = 1}^k c_i \|x_i\|_2^2
    = c_1 \|x_1\|_2^2 + r^2 \left( \sum_{i =
    2}^{k} c_i \right) \neq 0
  \]
  as $\|x_i\| \neq r$.

  \leftimpl
  Suppose there exists $c_i$ as in the statement,
  and assume $(x_1, \ldots, x_d)$ are \gls{spher},
  centred at $r$, radius $r$.

  Translate the set so that they are centred at
  the origin (this preserves all conditions and
  does not vhange the value of $\sum_I c_i
  \|x_i\|^2 \neq 0$).

  Let $\|x_i\|^2 = r^2$. Then
  \[
    \sum_i c_i \|x_i\|^2 = r^2 \sum_i c_i = 0
  \]
  so $(x_1, \ldots, x_d)$ are not \gls{spher}.
\end{iffproof}

\begin{corollary}[Generalised parallelogram law]
  \label{coro:3.5}
  % Corollary 3.5
  Let $X = \{x_1, \ldots, x_n\}$ be non-spherical.
  Then there exists $c_1, \ldots, c_n$ not all $0$
  with $\sum c_i = 0$ and there exists $b \neq 0$
  such that $\sum_i c_i \|x_i\|^2 = b$.
\end{corollary}

\textbf{Very important:} This is tru for
\emph{every} copy $X'$ of $X$ (with the same $c_i$
and $b$!).

Choosing the $c_i$ as in \cref{lemma:3.4}. If
$\phi(X)$ is a copy of $X$
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/5ad066944ee343dd.png}
\end{center}
then as we have seen we can translate and the
$c_i$ and $b$ are unaffected, and $\phi(0) = 0$.

After that apply $A$ that corresponds to rotation
/ reflection, and $\|Ax\|_2 = \|x\|_2$, so (3)
holds.

\begin{fcthm}[]
  \label{thm:3.6}
  % Theorem 3.6
  Assuming:
  - $X$ is \gls{eram}
  Then:
  $X$ is \gls{spher}.
\end{fcthm}

\begin{proof}
  Suppose $X$ is not \gls{spher}. Then there
  exists $c_i$ (not all $0$) such that $\sum_i
  \|x_i\|^2 = b$ and $\sum_i c_i = 0$.

  Also true for any copy of $X'$.

  Going to split $[0, 1)$ into $[0, \delta),
  [\delta, 2\delta), \ldots$ for small $\delta$.
  Then colour depending on where $c_i \|x\|^2$
  lies.