TITLE: Can I finitely color Z^2 such that (x,a) and (a,y) are different for every x,y,a?
QUESTION [6 upvotes]: I ran into this obstacle in a harmonic analysis problem; I know epsilon about coloring problems.
Is it possible to finitely color Z^2 so that the points (x,a) and (a,y) are differently colored for every x, y and a in the integers (excepting, of course, the trivial cases x=y=a)?

REPLY [11 votes]: Nope.
Suppose this were possible. For z \in Z, let Rz be the colors that appear in the row {(x,z) : x≠z \in Z}, and let Cz be the colors that appear in the column {(z,y) : y≠z \in Z}. The coloring condition is that each Cz and Rz is disjoint. Since there are finitely many possibilities for each set, find distinct z and z' where Rz = Rz', and Cz = Cz'. What color is (z,z')? Well, it's in Rz' (and not Cz'), and it's in Cz (but not Rz). So it's both in Rz and not: a contradiction.
(Ninja'd. Drat. Well, different proof at least.)