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Is it possible to have a planar graph with a chromatic number of $4$ such that all vertices have degree $4$?

Every time I try to make the degree condition to work on a graph, it loses its planarity.

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Here is an example.

enter image description here

The graph is 4 regular and 4 colorable. However, it's not possible to color its vertices with 3 colors. The picture shows a partial coloring that cannot be extended to the whole graph.

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    This graph is more symmetric than it looks, by the way. It's the graph of a square antiprism, and is vertex-transitive.2018-07-13
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Gerhard Koester in 1985 constructed infinitely many planar 4-regular graphs that not only are 4-chromatic, but also 4-critical, that is, all their proper subgraphs are 3-colorable.

The only planar 3-regular 4-critical graph is the complete graph of order 4, by Brooks's Theorem. Curiously it is known that no 5-regular planar graph is 4-critical. But I am not aware of any proof that every 5-regular planar graph is 4-colorable, without using the Four Color Theorem.