The question is in the title:



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> **Question:** Is there any 4-dimensional polytope without 3-gonal and 4-gonal faces (of dimension two), other than the [120-cell](https://en.wikipedia.org/wiki/120-cell)?
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I consider only *convex* polytopes (convex hull of finitely many points) that are full-dimensional (not contained in a proper subspace).
And I consider a polytope to be distinct from the 120-cell if it has a non-isomorphic face-lattice.


It is known that any 4-polytope must have a 3-gonal, 4-gonal or 5-gonal face of dimension two.
The 120-cell has only 5-gonal faces of dimension two.



Is there a 4-polytope without 3-gonal and 4-gonal faces, other than the 120-cell?
There are other polytopes. To construct one let's do the following. Remember first that in the hyperbolic $4$-space there exists a regular compact *right-angled* 120-cell. Here, right-angled means that any two adjacent faces intersect under angle $\frac{\pi}{2}$. Regular means, that all the faces are isomeric, and the polytope has the same group of self-isometries as the Euclidean 120-cell. This polytope is discussed, for example, in 


<https://pdfs.semanticscholar.org/a0eb/ccbed0687d966a9aaaac2f370bc930a556be.pdf>


at the bottom of page 65. The references to more classical articles are given there.


Now, if we double it in one face then we get a new convex polytope, and it is not hard to see, that it doesn't have 2-faces that are triangles and quadrilaterals. But any convex hyperbolic polytope is also combinatorially equivalent to a Euclidean one.


More generally, you can take any compact right-angled hyperbolic polytope in $\mathbb H^4$. Since it is hyperbolic and right-angled, it can not have $2$-faces that are triangles of quadrilaterals. And there is a infinite number of such polytopes in dimension 4. Each of them gives a Euclidean one as well.


