Let $H$ be three regular and planar. If the regions of $H$ can be properly colored using at most four colors, then the edges of $H$ can be properly colored using at most three colors.
I'm not sure how to go about this. So the regions are properly colored with four or less colors and then we want to color all the edges using no more than three colors? Is this like showing the Four Color Theorem is equivalent to something else? I was given the hint to use $\mathbb{Z_2}$ $\times$ $\mathbb{Z_2}$. Any solutions or help is greatly appreciated.
If you also have the condition that the graph is bridgless then you can let the colors be $0,1,2,3$ and color an edge with the xor value of the two faces around the edge.
If you don't have the bridgeless condition I believe there are counterexamples, for example the following graph.