Van der Waerden's theorem, strengthened so the progression and its common difference have one colour

Ordinary van der Waerden existence follows by induction on $m$. Length $1$ is immediate. Assuming finite witnesses for length $m-1$ for every number of colours, apply [L1] with $r=c$; if its first alternative occurs, it gives length $m$, while in the second alternative the focus has one of the $c$ colours and extends the focused progression of that colour to length $m$.

We now prove the strengthened statement. If $m=1$, take $a=d=1$, so assume $m\ge 2$. Induct on $c$. The assertion is immediate for $c=1$. Assume it for $c-1$ colours and let $n$ be a finite witness for the same target length $m$ with $c-1$ colours. By step 1.1, choose an ordinary monochromatic progression $a,a+d,\dots ,a+n(m-1)d$ in a sufficiently long $c$-coloured interval.

If one of $d,2d,\dots ,nd$ has the progression's colour, say $td$ does, then $a,a+td,\dots ,a+(m-1)td$ together with its difference $td$ has one colour. Otherwise the colouring $t\mapsto \mathrm{colour}(td)$ on $\{1,\dots ,n\}$ uses at most $c-1$ colours. The induction hypothesis gives $u,u+q,\dots ,u+(m-1)q,q$ of one colour there, and multiplication by $d$ gives the required progression and difference in the original colouring.

The colour induction proves the strengthened theorem for every finite $c$, while step 1.1 supplies the ordinary finite witnesses used in its construction.