= Solution
Write $\chi$ for the red-blue <finite colouring> of unordered pairs. The two alternatives concern one <arithmetic progression> or a pair of disjoint <arithmetic progressions>; they need not be mutually exclusive. For $m=1$, a singleton already has no non-blue edge, so the first alternative holds. Assume henceforth $m\ge2$.
If the first alternative fails, every $m$-term <arithmetic progression> contains a red pair. Label each $(a,d)\in\mathbb N^2$ by one such pair's positions $0\le r<s<m$, choosing the first in a fixed ordering when there are several. This is a <finite colouring> with at most $\binom m2$ colours. The following two branches establish the <arithmetic-progression bipartite Ramsey dichotomy>.
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