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\begin{document}

\title[]{Every planar graph admits a six coloring}

\author[]{Stu D. Ent}
%    Address of record for the research reported here
\address{Department of Mathematics, Massachusetts Institute of Technology}

%    Information for third author

%\thanks{}
\date{\today}


\begin{abstract}
  We show that the vertices of every planar graph can be colored by
  six colors with no two adjacent vertices sharing the same color.
\end{abstract}

\maketitle
%\tableofcontents

\section{Introduction}
Let $G = (V,E)$ be a finite, undirected, planar graph. A $k$-coloring
of $G$ is a function $f \colon V \to \{1,\ldots,k\}$ such that $f(u)
\neq f(w)$ for all $(u,w) \in E$. We say that $G$ is $k$-colorable if
it admits a $k$-coloring.

\begin{theorem}
  \label{thm:six-coloring}
  Every planar graph admits a $6$-coloring.
\end{theorem}

To prove this theorem we will use the following lemma, which was
proved in class.
\begin{lemma}
  \label{lemma:degree-five}
  In every planar graph there exists a vertex with degree at most five.
\end{lemma}


\begin{proof}[Proof of Theorem~\ref{thm:six-coloring}]
  {\bf Your proof goes here.}
\end{proof}


\end{document}