Barnette's conjecture

Let $G$ be a finite undirected graph with the following four properties:

$G$ is planar: it can be embedded in the Euclidean plane with no two edges crossing. 2. $G$ is $3$-vertex-connected: $G$ has at least $4$ vertices and, for every set $S$ of at most two vertices, the graph $G - S$ obtained by deleting $S$ and all incident edges is connected. (A planar $3$-connected graph is called polyhedral.) 3. $G$ is bipartite: its vertex set admits a partition into two classes such that every edge of $G$ has one endpoint in each class. 4. $G$ is cubic ($3$-regular): every vertex of $G$ is an endpoint of exactly three edges.

Then $G$ is Hamiltonian, i.e. $G$ contains a cycle passing through each of its vertices exactly once.

Equivalently, in the language of polytopes: if $P$ is a simple $3$-dimensional convex polytope (every vertex of $P$ lies on exactly three edges) such that every face of $P$ has an even number of edges, then the graph formed by the vertices and edges of $P$ has a Hamiltonian cycle.