TITLE: 7 Color Theorem
QUESTION [6 upvotes]: I studied and understood the proof of the 4 color theorem, which states that given any separation of a plane into contiguous regions, the regions can be colored using at most four colors so that no two adjacent regions have the same color; now, I encountered with the following statement 

For any separation of a torus into contiguous regions, the regions can be colored using at most seven colors so that no two adjacent regions have the same color.

I have been looking for a proof of such statement; which I cannot find, I was hoping if anyone could give me some reference of where could I find such proof, are maybe give some hint to prove it myself. Thanks in advance

REPLY [2 votes]: It can be shown that for any orientable surface, the minimum number of colors is given by 
$$
\left \lfloor \frac{7+\sqrt{1+48g}}{2} \right \rfloor
$$
where $g$ is the genus of the surface (the "number of holes"). Since for a torus $g=1$, it follows that the minimum number of colors equals $7$.
Note that the formula does not hold for $g=0$, otherwise it would be an elegant and short proof of the $4$ color theorem in the plane.