Only 5 platonic hedrons, connectivity only proof? I am reading a course on discrete differential geometry and found this neat problem:

After thinking about it for 15 minutes curiosity got the better of me and I cheated. And here's one possible proof:
https://www.mathsisfun.com/geometry/platonic-solids-why-five.html
However, that used angles, and the hint implies you can do it only with the connectivity (I assume Euler's formula?).
I was wondering if someone could prove this with only connectivity information.
 A: I'll show here that the cube is the only Platonic solid with square faces.
We start with a genus 0 polyhedron with square faces, and three edges and three faces to each vertex. Using only this information and the Euler characteristic, what can we say?
Well, we have $V-E+F = 2$, for starters (where, as is conventional, $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number of faces). Next, each edge is the edge of two faces, and each face has four edges, so we get that $4F = 2E$. Finally, from each edge going between two vertices, we get $2E = 3V$. This is three equations in three unknowns, and solving them is rather straight-forward: We get $V = 8, E = 12, F = 6$. This necessarily gives us a cube.
What about four edges and four faces to each vertex? Setting up the same equations, we get
$$
\cases{V-E+F = 2\\
4F = 2E\\
2E = 4V}
$$
Trying to solve these, we get $0 = 2$, so this is impossible. Trying with more than four will give negative solutions, which is absurd. So we conclude that the cube is the only Platonic solid with square faces.
The three solids with triangular faces and the one solid with pentagonal faces may be solved completely analogously.
