A few days ago, a peer jokingly posed a puzzle. In essence, the puzzle states that there are some cats, each cat is in a corner, and each cat sees 3 other cats. The puzzle then asks how many cats there are in total. The obvious solution is that there are $4$ cats... However, I have long thought about extensions of the problem. I have found polyhedra that allow for $2n$ "cats", where $n$ is an integer, the "cats" are the vertices, and the edges serve as "hallways" through which cats can see each other.
This naturally made me curious if the problem is possible with an odd number of vertices. I immediately tried to attack the problem and could not solve it... I even failed to attack a restricted form of the problem, requiring a convex polyhedron with an odd number of vertices with each vertex having 3 connected edges. An example of a failing example would be a square pyramid (as one of the vertices has $4$ edges). However, this definition actually discounts the puzzle's intended solution... If we use my "hallway" interpretation of the problem then the "hallways" between cats in opposite corners form an "X" in the center of the square, creating an additional vertex with four connected edges.
Realizing how much harder I had made the puzzle, I decided to pursue the restricted case in two dimensions (which the intended solution lies in). Though I conjecture that an odd number of "cats" is impossible in both two and three dimensions, I thought it would be easier to show in two dimensions, especially for closed, planar graphs. However, I have come up short with a proof here as well.
I realize that my language here is very vague... I lack a decent introduction into topology, and thus my language is incomplete. Should more details be needed I would be glad to elaborate. Nevertheless, I think the most basic form of my question is as such: does there exist a generalization of the cat puzzle to odd numbers of cats in two dimensions? In three dimensions? I conjecture this impossible.