# On cubic planar graphs with face boundaries of length divisible by 4

All graphs considered here are finite, simple and undirected.

Let $$\mathscr{G}$$ denote the class of cubic plane graphs for which all face boundaries are of length divisible by four. The 3-cube $$Q_3$$ is a member of $$\mathscr{G}$$ (perhaps, the smallest member of $$\mathscr{G}$$). By an indirect proof, one can prove the following by combining some existing results:

• For every graph $$G\in\mathscr{G}$$, the number of vertices in $$G$$ is divisible by 8.

Can you prove this by an elementary method?
I would like to get a feel of the class $$\mathscr{G}$$; so, indirect proofs are also welcome (as they may show some properties of the class I am not aware of).

By the Euler characteristic and double counting edges with the cubic assumption, we get $$v=2f-4$$. So v is even, and it suffices to prove that f is even too. By the defining property of $$\mathcal{G}$$ and double counting edges, $$\sum_{k \equiv 0 mod 4} k f_k=2e$$, where $$f_k$$ is the number of faces of degree $$k$$. Reducing mod 4, we get that $$2e \equiv 0 mod 4$$, and thus that $$e$$ is even. By the Euler characteristic we deduce that $$f$$ is even which concludes the proof.
• Oops, sorry. I gave the wrong number in the question. It should be "the number of vertices in $G$ is divisible by 8". I have updated the question now (the constant was 4 in the earlier version of the question). Jan 11, 2022 at 13:25
• That $v$ is divisible by 4 can be shown much more easily by double counting: the number of incident vertex-face pairs is, on the one hand, $3v$, on the other hand, the sum of the number of sides of all faces, which is a multiple of 4. (In your notation: $\sum_kkf_k=3v$.) Jan 13, 2022 at 14:25