= Solution
The <Strengthened Van der Waerden theorem> says that every finite coloring contains, for each prescribed $m$, a monochromatic set
$$
\{d,a,a+d,\ldots,a+(m-1)d\}.
$$
We prove the finite form by induction on the number $k$ of colors. The case $k=1$ is immediate. Let $n$ work for $m$ and $k-1$ colors, and apply the ordinary <Van der Waerden theorem> to obtain a monochromatic progression
$$
a,a+d,\ldots,a+n(m-1)d.
$$
If one of $d,2d,\ldots,nd$ has the progression's color, say $rd$, then
$$
rd,a,a+rd,\ldots,a+(m-1)rd
$$
works. Otherwise $d,2d,\ldots,nd$ use at most $k-1$ colors. By the induction hypothesis their indices contain $b,b+r,\ldots,b+(m-1)r$ together with $r$ in one color. Multiplying by $d$ yields
$$
rd,bd,(b+r)d,\ldots,(b+(m-1)r)d,
$$
which is the required progression together with its <common difference>.
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