I'm trying to solve my own question. I almost got the answer (which I'll post in a few days), but there are two things I'm not able to prove but that I know are true (On Polyhedra with Specified Types of Face):
- There does not exists a convex heptahedron with only quadrilateral faces
- There does not exists a convex $14$-hedron with exactly $14$ pentagonal faces
The heptahedron if it exists would have $9$ vertices, $14$ edges and $7$ faces. I can prove that it has $8$ vertices of degree $3$ and one of degree $4$.
The $14$-hedron if it exists would have $23$ vertices, $35$ edges and $14$ faces. I can prove that it has $22$ vertices of degree $3$ and one of degree $4$.
Question: Why do such polyhedra not exist?
Remark: An argument of the non-existence of only one polyhedron is good enough for an answer. I could extrapolate it to the other case.