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).


1 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.

  • 1
    $\begingroup$ 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). $\endgroup$ Commented Jan 11, 2022 at 13:25
  • 3
    $\begingroup$ 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: $\sum_kkf_k=3v$.) $\endgroup$ Commented Jan 13, 2022 at 14:25

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