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I want to prove it isn't possible to make a football (a convex polyhedron such that at least 3 edges meet at each vertex) out of exactly 9 squares and m octagons where $m>3$.

Raul
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The original poster hasn't supplied the proof mentioned in his comment. Here's one for the benefit of anyone interested:

The 9 squares and $m$ octagons give $f = 9 + m$ faces. Counting 4 edges for each square and 8 for each octagon gives $2e = 36 + 8m$. The Euler characteristic is $f - e + v = 2$, or $v = 2 + e - f = 2 + 18 + 4m - 9 - m = 11 + 3m$. Exactly 3 edges meet at each vertex (min 3 for a polyhedron, max 3 when squares and/or octagons meet at a vertex to retain convexity), so $3v = 2e$, i.e. $3(11+3m) = 36+8m$, or $m=3$.

I don't know whether a polyhedron of 9 squares and 3 octagons exists, but the calculations rule out $m>3$.

Frentos
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    Convexity does take a part in the argument. There are non-convex polyhedron which contains vertex with more than 3 square/octahedron attached to it. The statement that "exactly 3 edges meet at vertex" is only justified when the polyhedron is convex. – achille hui Dec 18 '15 at 13:15
  • Thanks - will amend my comment! I've carried around other characteristics of convexity but never seen the (obvious once mentioned) connection with sum-of-angles-around-a-vertex. – Frentos Dec 19 '15 at 01:05