There are two parts to the proof:

- Some map requires at least four colors.
- No map requires more than four colors.

But part 1 is easy. Here is one such map.

One of the assumptions of this theorem is that the map is on a flat plane. But what if we have a map on a donut's surface? It turns out that we need seven colors. Can you find a map on a donut that requires at least seven colors?

You can send solutions to skepticsplay at gmail dot com. A tip: you can represent a donut with a flat square in which each edge wraps around to the opposite edge.

solution posted

## 6 comments:

It is difficult to not find immediatly an answer in Google. So I didn't have a chance to find myself. An oblique tiling with hexagons does the job.

True, this is one puzzle which is easy to google. If you want more, you can also try to prove that a mobius requires six colors (but don't google it).

I tried, but then I googled. There is much less info about the six colored mobius. I miss especialy a 3D turnable model (as the NON-6-colored mobius on the wolfram page).

The 6-colored mobius in tissue is not satisfying.

I just sent you a map which I think meets the criteria. Then I came here and see that a hex map works. Looking at mine again, I think it's topologically similar but I drew it in Paint so it's all R angles.

I'll have a look at the Moebius later.

I haven't received any e-mails from you. You might have to resend.

I made a 6-color-Mobius model with paper. I had to print "both sides". One with the mirror image. Because a mobius strip is not orientabel his thickness must be zero and "both sides" must locally have the same color. I send a photo of my model.

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