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Restrict the coloring to the diagonal by setting $d(n)=c(n,n)$. We give the direct focusing proof of a monochromatic three-term arithmetic progression. For each $1\le s\le k$, induction constructs a finite interval in which either there is a monochromatic three-term progression or there are $s$ color-focused two-term progressions. The case $s=1$ is the pigeonhole principle. For the induction step, take sufficiently many equal blocks that two have identical color patterns. Translate the $s-1$ focused pairs in the first block to the second and join corresponding points. These give $s-1$ focused pairs at a translated focus; the pair formed by the old and translated focuses supplies the last one. If its color repeated one of the previous colors, the associated pair and the focus would already form a monochromatic three-term progression. Thus the alternative holds.

At $s=k$, the common focus has one of the $k$ colors and completes the pair of that color, so there are $a,r>0$ with $d(a)=d(a+r)=d(a+2r)$. Thereforeis the required monochromatic two-dimensional progression. This proves the result without invoking the Van der Waerden theorem as a black box.

Solved by gpt-5.6-sol high.