A simple graph $G$ is a quadrangulation of the plane if it can be embedded in the plane in such a way that all faces are quadrangles. We say an edge $e=(uv)$ is an $(a,b)$-edge if the degrees of the two end-vertices, $u$ and $v$, are $a$ and $b$, respectively.
I am thinking about the following question: Does a quadrangulation with minimum degree $3$ must contain either a $(3,3)$ edge or a $(3,4)$ edge?
I know the discharging method might work, but I am not yet able to master it proficiently.
Here is my initial attempt: First, we set the charge of each vertex to be $d_G(v)-4$ and the charge of each face to be $0$. In this case, the total charge, as calculated, is -8.
Suppose that a quadrangulation does not contain (3,3) and (3,4) edges, and then apply the following rules:
Rule 1: Each vertex with a degree of 5 or higher redistributes its initial charge uniformly to each incident face.
Rule 2: Each face redistributes its charge uniformly to incident 3-degree vertex.
Our goal is to obtain the charge of each vertex and face be greater than or equal to 0 through charge transfer, and then derive a contradiction. However, it seems that the two rules above do not work. I haven't found any contradictions. I might need a more clever discharging method.
Perhaps there are other non-discharging methods. Or a counterexample can be founded.