I know of Van der Waerden's theorem which says that given $k, m$, there is $n \in \mathbb{N}$ such that for any $k$-colouring of $[n]$, there is a monochromatic arithmetic progression of length $m$. I'm trying to prove the following version – that there is a monochromatic AP of length $m$ such that even the common difference $d$ gets the same color.
I think the $k, m$ in the original statement may have to be manipulated to somehow involve $d$, but I'm not sure how to. Some help will be appreciated.
Best Answer
Proceed by induction on $k$. If $k=1$ the statement is trivial. If $k>1$ assume by inductive hypothesis that there is $n$ such that every $(k-1)$-colouring of $n$ contains a monochromatic AP of length $m$, with common difference also of the same colour.
Now let $c$ be big enough so that any $k$-colouring of $c$ contains a monochromatic AP of length $n(m-1)+1$, with common difference $d$. Write it as $$a,a+d,\ldots,a+n(m-1)d$$If there is $i\leq n$ such that $id$ has the same colour as $a$, then $$a,a+id,\ldots,a+i(m-1)d$$ is a monochromatic AP of length $m$ with common difference $id$ of the same colour. If there is no such $i$, then $\{id\mid i\leq n\}$ is a $(k-1)$-coloured set of size $n$, on which you can use the inductive hypothesis