TITLE: planar regular bipartite graph
QUESTION [6 upvotes]: (a) Prove that there exists a 3-regular planar bipartite graph $G$ with $4n$ vertices for all $ n\geq 3$.
(b) Prove that does not exist a 3-regular planar bipartite graph with $10$ vertices.
The only idea had for (a) was that is enough to prove the claim for a 3-regular bipartite graph with $12$ vertices because then we add $2$ vertices. The only theorem I knew about planar graphs is Kuratowski and that if the graph is planar then $ m <3n-6$. I can't use the fact that we are looking for a bipartite graph.
For (b): $2m =\sum deg(v) =3\cdot 10 =30$ so it is $m=15$.

REPLY [4 votes]: Another approach to (a). Start with a 2n-cycle with vertices 1, 2, 3, ..., 2n. Inside this, draw another 2n-cycle with vertices 1', 2', 3', ..., 2n'. Join vertices i and i' together for all i. This is a 3-regular planar bipartite graph of order 4n. (The two colour classes are {1, 3, 5, ..., 2n-1, 2', 4', ..., 2n'} and {2, 4, ..., 2n, 1', 3', ..., (2n-1)'}.)