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\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item You can prove Gallai with a standard
      focusing + product argument.
    \item \nameref{HJ} for \glsref[dpps]{$2$-point
      parameter set} on $X = \{1, 2\}$ gives a
      rectangle, while \nameref{gallai} can give a
      square.
  \end{enumerate}
\end{remark*}

\newpage
\section{Partition Regular Equations}

Schur's theorem: $x + y = z$ has monochromatic
solutions (if $\Nbb$ is finitely coloured).

Van der Waerden: $x_1, x_2, y_1, \ldots, y_m$ such
that the system
\begin{align*}
  y_1
  &= x_1 + x_2 \\
  y_2
  &= x_1 + 2x_2 \\
  &~\vdots \\
  y_m
  &= x_1 + mx_2
\end{align*}
has a monochromatic solution.

\textbf{Main aim:} decide when a system of
equations is `partition regular'.

\begin{definition*}[Partition regular]
  \glsadjdefn{pr}{partition regular}{matrix}%
  Let $A \neq 0$ be a $m \times n$ matrix over $\Qbb$ and
  we say that $A$ is \emph{partition regular} (PR)
  if whenever $\Nbb$ is finitely coloured, there
  exists a monochromatic $\mathbf{x} \neq 0$ such that $A
  \mathbf{x} = 0$.
\end{definition*}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Schur: says $(1, 1, -1)$ is \gls{pr}.

    \item Van der Waerden: says
      \[
        \begin{pmatrix}
          1 & 1 & -1 & 0 & \cdots & 0 \\
          1 & 2 & 0 & -1 & \cdots & 0 \\
          1 & 3 & 0 & 0 & \cdots & 0 \\
          \vdots & \vdots & \vdots & \vdots & \ddots &
          \vdots \\
          1 & m & 0 & 0 & \cdots & -1
        \end{pmatrix}
      \]
      is \gls{pr}.
    \item $(1, -1)$ is \gls{pr}.
    \item $(3, 4, -7)$ is \gls{pr} (just take all to be
      equal).
    \item $(3, 4, -6)$? Don't know yet.
    \item Non-example: $(2, -1)$. Need $2x = y$.
      Colour $\Nbb$ by setting $n$ to be red if
      the biggest power of $2$ dividing it is
      even, and blue otherwise.
  \end{enumerate}
\end{example*}

\begin{definition*}[Column property]
  \glsadjdefn{cp}{column property}{matrix}%
  We say that a rational matrix
  \[
    A = \begin{pmatrix}
      \uparrow & \uparrow & \cdots & \uparrow \\
      c_1 & c_2 & \cdots & c_n \\
      \downarrow & \downarrow & \cdots &
      \downarrow
    \end{pmatrix}
  \]
  has the column property (CP) if there exists a
  partition of $[n] = B_1 \sqcup B_2 \cdots \sqcup
  B_r$ such that:
  \begin{enumerate}[(1)]
    \item $\sum_{i \in B_1} c_i = 0$
    \item $\sum_{i \in B_t} c_i \in \Span \langle c_j
      : j \in B_1 \cup B_2 \cup \cdots B_{t - 1}
      \rangle$ (note that it doesn't make a
      difference whether the span is the
      $\Rbb$-linear or $\Qbb$-linear span)
  \end{enumerate}
\end{definition*}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $(1, 1, -1)$ can take $B_1 = \{1, 3\}$,
      $B_2 = \{2\}$, hence it does have the
      \gls{cp}.
    \item Van der Waerden matrix from (2) in the
      previous example: take $B_1 = \{1, 3, 4,
      \ldots, n\}$ and $B_2 = \{2\}$, which shows
      that it has the \gls{cp}.
    \item $(3, 4, -7)$, take $B_1 = \{1, 2, 3\}$
      so it has \gls{cp}.
    \item $(2, -1)$ does not have \gls{cp}.
    \item $(\lambda, -1)$ has \gls{cp} if and only
      if $\lambda = 1$.
    \item $(3, 4, -6$ doesn't have the \gls{cp}.
  \end{enumerate}
\end{example*}

\textbf{Aim:}

\begin{theorem}[Rado's Theorem]
  \label{rado}
  % Theorem 2.1
  A matrix over $\Qbb$ is \gls{pr} if and only if
  it has the \gls{cp}.
\end{theorem}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item \gls{pr} is checkable in finite time.
    \item Find $a, b \in \Nbb$ such that
      \[
        \begin{pmatrix}
          1 & -1 & 3 \\
          2 & -2 & a \\
          4 & -4 & b
        \end{pmatrix}
      \]
      is \gls{pr}. Take $(a, b) = (6, 12)$.
    \item Neither direction of \nameref{rado} is
      easy (or obvious)!
  \end{enumerate}
\end{remark*}

Today we will look at a single equation, i.e. a
single row matrix.

If we have $\mathbf{x} = (a_1, \ldots, a_n)$ a $1
\times n$ matrix, then $\mathbf{x}$ is \gls{pr}
if and only if $\lambda \mathbf{x}$ is also
\gls{pr}. So we may assume that $a_i \in \Zbb$.

\textbf{Observation:} $(a_1, \ldots, a_n)$ has the
\gls{cp} if and only if there exists a set
$\{a_{i_1}, \ldots, a_{i_k}\}$ of non-zero
elements such that $\sum_{k = 1}^{n} a_{i_k} = 0$
($*$).

Also note that we may assume that $a_i \neq 0$.

We are going to show that if $\mathbf{x}$ \gls{pr}
then it has the \gls{cp}, which is equivalent to
($*$) in this case.

\begin{remark*}
  Even in this case, neither direction of
  \nameref{rado} is easy.
\end{remark*}

\begin{definition*}
  \glssymboldefn{e}%
  Let $x \in \Nbb$ and $p$ a prime. Then we can
  write
  \[
    x = a_k p^k + \cdots + a_1 p + a_0
  ,\]
  with $0 \le a_i < p$. Denote by $e(x) = \{a_t :
  t = \min,cbi : a_i \neq 0\}$.
\end{definition*}

\begin{example*}
  \phantom{}
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/25a545a4f604456e.png}
  \end{center}
\end{example*}

\begin{proposition}
  \label{prop:2.1}
  % Propositon 2.1
  % Propositon 2.2
  If $A = (a_1, \ldots, a_n)$, $a_i \neq 0$ is
  \gls{pr} then it has the \gls{cp}.
\end{proposition}

\begin{proof}
  Let $p$ be a huge prime, $p > \sum |a_i|$. I give
  to the number $x$ the colour $\e(x) \in \{0,
  \ldots, p - 1\}$. Then by assumption, there
  exists $x_1, \ldots, x_n$ of the same colour
  such that $\sum a_i x_i = 0$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/b6e1574502ca4b0c.png}
  \end{center}
  In symbols, let $t = \min \{t(x_1), \ldots,
  t(x_n)\}$, $I = \{i : t(x_i) = t\}$. When we sum
  $\sum_{i \in [n]} a_i x_i = 0$ and we look at the last digit
  $\bmod p$, we get $\sum_{i \in I} a_i d = 0$,
  where $d$is the colour of our $x_i$s.
  Then $d \left( \sum_{i \in I} a_i \right)
  \equiv 0 \pmod p$. Then $\sum_{i \in I} a_i =
  0$ (and note $I$ is non-empty).
\end{proof}

\begin{remark*}
  To this day, there are no other known proofs of
  this proposition.
\end{remark*}
