A Platonic solid is a convex polyhedron that has faces that are regular polygons and has the same number of regular polygons around each vertex.
Let's show that any Platonic solid has to be either a tetrahedron, a cube, an octahedron, a dodecahedron, or an icosahedron.
Let V, E, and F be the number of vertices, edges, and faces of the Platonic solid. Let N be the number of edges of each face, and let M be the number of faces (and hence also edges) around each vertex.
Then we have M x V = 2E and N x F = 2E (in each equation, we're noting that each edge has been counted twice).
Now we have to use Euler's formula: for a convex polyhedron, V - E + F = 2. I'll prove that later on. Substituting V = 2E/M and F = 2E/N, we get
2E/M - E + 2E/N = 2.
Divide both sides by 2E, and we get 1/M - 1/2 + 1/N = 1/E, or 1/M + 1/N = 1/E + 1/2.
Since E is positive, we must have 1/M + 1/N > 1/2, and also M and N have to be at least 3 (do you see why?). The only pairs (M,N) that satisfy these inequalities are (3,3) (tetrahedron), (3,4) (cube), (4,3) (octahedron), (3,5) (dodecahedron), and (5,3) (icosahedron). End of proof.
Now, why is Euler's formula true? There are lots of proofs, but here is one of the easier ones to understand:
Remove one face from your Platonic solid, thinking of it as just the outer skin of vertices, edges, and faces, not the stuff inside. If we can show V - E + F = 1 for that thing, then we will have shown that V - E + F = 2 for the polyhedron.
Imagine flattening that ball with a hole onto a table. That doesn't change V - E + F. Now, one at a time, remove an edge that is on the border and also remove the face that it is next to. E goes down 1, and so does F, so that's a net change of 0 to V - E + F. Continue doing that until you have one polygon left. That polygon satisfies V - E + F = 1, and so the original polyhedron satisfies V - E + F = 2. That proves Euler's formula.