2^{2k}$. For a general $m$, choose $k$ so that $2^k\leq m<2^{k+1}$. Then $w(3;m)\geq w(3;2^k)>2^{2k}>\frac14m^2$.\end{proof} \section{Remarks} \begin{enumerate} \item The lower bound in Theorem \ref{t3} is not the best possible. Indeed, in the standard reference Ramsey Theory \cite{graham+rothschild+spencer1990}, the authors show with more elaborate techniques that for some positive constant $c$, $w(3;m)>m^{c\log m}$. \item Corollary \ref{c1} shows that a constant/1-1 canonical version of Theorem \ref{t1} is not true. We also know by the Bergelson/Hindman/McCutcheon example that a density version of Theorem \ref{t1} is not true. The following three simple examples, most involving only 3-element sets, illustrate various combinations of the truth or falsity of the ``constant/1-1 version'' and the ``density version''. \begin{enumerate} \item The simplest non-trivial case of van der Waerden's theorem says that every finite coloring of the positive integers produces a monochromatic 3-term arithmetic progression. The constant/1-1 version of the result holds by the Erd\H{o}s-Graham theorem, and the density version holds by Szemer\'edi's theorem. \item Schur's theorem says that if the positive integers are finitely colored, then there is a monochromatic solution of $x+y=z$. The density version does not hold by taking all the odd integers. The constant/1-1 version does not hold by coloring each $x$ with the highest power of 2 dividing $x$. \item Kevin O'Bryant showed me this example: If the positive integers are finitely colored, then there is a monochromatic 3-term geometric progression (a set of the form $\{a,ad,ad^2\}$). To get the constant/1-1 version, let a coloring $g$ of the positive integers be given. Define a new coloring $h$ by setting $h(x) = g(2^x)$. Then, by the Erd\H{o}s-Graham theorem, there is a set $\{a,a+d,a+2d\}$ on which the coloring $h$ is either constant or 1-1, so the coloring $g$ is either constant or 1-1 on the set $\{2^a,2^a2^d,2^a(2^d)^2\}$. The density version does not hold, since the set of square-free numbers has positive density. \item It seems natural to ask for a collection $P$ of 3-element sets (if such a collection exists!) for which: \begin{enumerate} \item Every set of positive integers with positive upper density contains an element of $P$. \item It's not the case that for every coloring of the positive integers, there is an element of $P$ on which the coloring is either constant or 1-1. Allen R. Freedman communicated the following example to me, involving infinite sets. Here instead of considering those collections of triples $\{x,y,z\}$ for which $x+z=2y$, or $x+y=z$, or $xz = y^2$, one considers the collection $P$ of all subsets of $\omega$ which have positive density. Then trivially every set of positive density contains an element of $P$. However, any coloring which is constant on each interval $[2^{n-1},2^n]$, with different colors for different $n$, shows that the constant/1-1 version does not hold. \end{enumerate} \end{enumerate} \item We have used a particular partition of $\omega$ into infinitely many translates of an infinite set. Perhaps it's possible to describe \emph{all} partitions of $\omega$ into infinitely many translates of an infinite set. \end{enumerate} \bibliographystyle{amsplain} \bibliography{tom-all} \end{document}