Connected simple cubic graph

Question

I am trying to understand this problem and yes this is from my assignment and I should be doing it myself, but I have been staring at it for 2 hours and not getting anywhere, so decided to post it here.

Let G be a connected cubic simple graph that contains 2 edge-disjoint spanning trees show that |G| = 4.

Answer 1

Let $G$ be a simple cubic graph, with two edge-disjoint spanning trees, that has $n$ vertices and $m$ edges. Since $G$ is regular of degree 3, we have

$3n = 2m.$

Now any spanning tree has $n-1$ edges, and $G$ has two disjoint spanning trees. So we have

$m \ge 2n - 2.$

Combining the first equation and the second inequality, we get

$n \le 4$.

Now there are no cubic simple graphs on 1, 2 or 3 vertices, so the result follows. (It is debatable whether there is one on 0 vertices.)

Answer 2

Express the number of edges in a connected cubic simple graph in terms of the number of vertices. Then do the same for a graph that contains 2 edge-disjoint spanning trees. Solve this equation for the number of vertices.