Can you arrange a finite number of green and red squares on the plane, sides parallel to the axes, such that:
- Every red square intersects $M$ green squares and no red square;
- Every green square intersects $M$ red squares and no other green square?
For $M=2$, this is easy:
For $M=3$, this is more difficult, and was solved recently by dtldarek:
For $M=5$, this is impossible, because the smallest square necessarily intersects at most 4 squares of the other color.
So, the only remaining case is $M=4$: Is this possible?
It can't be done for $M=4$, for much the same reason as for $M=5$. When two squares intersect, they do so either by each clipping a corner of the other, or by one of them taking a bite from a side of the other. But the latter case requires the biter to be smaller than the bitee, and this leaves only two corners of the smaller (say Green) square available for clipping, hence requires a yet smaller (now Red) square to take a bite from one of its sides, ad infinitum. Hence all the intersections must occur at the corners of the squares. But in any finite arrangement, there'll be squares at the outskirts whose outer corners aren't clipped.
(Side note: In any finite collection of closed sets, there is some minimum distance between non-overlapping pairs. This gives room for the sets to be independently expanded by epsilonish amounts, so that each has a different size while preserving all pairs that do and do not intersect.)