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 $m>3$.
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 $m>3$.