Re: [math-fun] Futility Closet

You can strengthen that to three colors, and there's still a nice solution: Show that for every possible coloring of the plane with three colors there exit, for every distance d, two points of distance d with the same color. Tom Joerg Arndt writes:

Here's one that I find delightful: Show that for every possible coloring of the plane with two colors there exist, for every distance d, two points of distance d with the same color.