<p>By the Euler characteristic and double counting edges with the cubic assumption, we get <span class="math-container">$v=2f-4$</span>. So v is even, and it suffices to prove that f is even too. By the defining property of <span class="math-container">$\mathcal{G}$</span> and double counting edges, <span class="math-container">$\sum_{k \equiv 0 mod 4} k f_k=2e$</span>, where <span class="math-container">$f_k$</span> is the number of faces of degree <span class="math-container">$k$</span>. Reducing mod 4, we get that <span class="math-container">$2e \equiv 0 mod 4$</span>, and thus that <span class="math-container">$e$</span> is even. By the Euler characteristic we deduce that <span class="math-container">$f$</span> is even which concludes the proof.</p>

CC BY-SA 4.0
2022-01-11T13:25:20.217Z
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).
Cyriac Antony
CC BY-SA 4.0
2022-01-13T14:25:37.863Z
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: $\sum_kkf_k=3v$.)
Emil Jeřábek
CC BY-SA 4.0
2022-01-11T10:55:32.093Z
2022-01-11T10:55:32.093Z
Arnaud
<p>All graphs considered here are finite, simple and undirected.</p>
<p>Let <span class="math-container">$\mathscr{G}$</span> denote the class of cubic plane graphs for which all face boundaries are of length divisible by four. The 3-cube <span class="math-container">$Q_3$</span> is a member of <span class="math-container">$\mathscr{G}$</span> (perhaps, <em>the</em> smallest member of <span class="math-container">$\mathscr{G}$</span>). By an indirect proof, one can prove the following by combining some existing results:</p>
<ul>
<li><em>For every graph <span class="math-container">$G\in\mathscr{G}$</span>, the number of vertices in <span class="math-container">$G$</span> is divisible by 8.</em></li>
</ul>
<p><strong>Can you prove this by an elementary method?</strong><br />
I would like to get a feel of the class <span class="math-container">$\mathscr{G}$</span>; so, indirect proofs are also welcome (as they may show some properties of the class I am not aware of).</p>

<graph-theory><co.combinatorics><planar-graphs>
On cubic planar graphs with face boundaries of length divisible by 4
Cyriac Antony
# On cubic planar graphs with face boundaries of length divisible by 4

(Asked by: username_0 on 2022-01-11)

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).

**Tags:** graph-theory, co.combinatorics, planar-graphs


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## 1 answers

### Answer by username_1 on 2022-01-11. Score: 2

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.

> **Comments:**
> **username_0**: 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).  
> **username_2**: 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: $\sum_kkf_k=3v$.)  

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