**Proof

There are 3 formulas describing such polyhedra...

Euler's formula for polyhedra:

F = E-V+2

Another one saying that the number of Faces times the number of sides on each face equals twice the number of edges, because every edge connects two faces.

F*S = 2*E

And one more saying that the number of vertices times the number of edges connecting to it equals twice the number of edges, because every edge connects two vertices.

V*P = 2*E

These solve in terms of S and P as the formula for E, F, and V given.

For example if we want triangles so S=3 meeting 5 at each vertex:

It says there must be 30 edges, 20 faces and 12 vertices, which is the regular icosahedron.

I find it interesting that the assumptions are quite general, Euler's formula for polyhedra can be proven even over topologies with holes in them like a donut, and the other two would certainly still be true. But it says you could never have say a 1000 vertex donut polyhedra where the whole surface is divided into triangles and 5 meet at every vertex, because the only solution is for there to be 12 vertices for that configuration of S and P.**Note**

I was thinking about how a toroid can be broken into almost any number of 4 sided figures with 4 edges meeting at each vertex.. there is a caveat that these formulas end up dividing by 0 for S=4 and P=4...

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