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

Question

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).

Answer 1

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.