Let $\mathbb{Z}$ be the set of whole numbers and $l,m\in N$. Let's color all elements of $\mathbb{Z}\times\mathbb{Z}$ in $k$ different colors. Prove that we can find two aritmetic progressions $A$ and $B$ of length $l$ and $m$ in $\mathbb{Z}$ such that $A\times B$ contains only elements of the same color.
I know that this has something to do with Van der Waerden's theorem that has to be extended from $\mathbb{Z}^1$ to $\mathbb{Z}^2$ and I have found some pretty neat theorems here. But I could not make a direct connection.
Rather than coloring $\mathbb Z \times \mathbb Z$, color $\mathbb Z \times [n]$ where $[n] = \{1,2,\dots,n\}$ is large enough that any $k$-coloring of $[n]$ contains an $m$-term arithmetic progression.
There are only finitely many possible monochromatic $m$-term arithmetic progressions to be found in $[n]$. One upper bound is $k \cdot \binom n2$, by choosing the color and the first two elements of the progression (which determine the other elements).
For every element $a \in \mathbb Z$, write down which monochromatic $m$-term arithmetic progression we find in $\{a\} \times [n]$. This is itself a $k\cdot\binom n2$-coloring of $\mathbb Z$, so it contains a monochromatic $l$-term arithmetic progression.
By staring at the thing we've found long enough, we realize that it is the thing we wanted to find.