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The mapping from the regular solids to $\mathbb{Z}_5$ given by the number of faces in the solid mod 5, interestingly, is a bijection. Any geometers or algebraists know if there is a significant reason for this? I mean, the fact that it is a possibility (i.e. that there are exactly 5 regular solids) is a well known proof, but why is this particular mapping bijective? Just seems too much of a coincidence.

Its been 40 years since I understood the proof in grad school, and it's long since gone from my memory - perhaps this fact is even buried in the proof? Any grad students fresh off understanding the proof?

Following on comments below, more generally in $\mathbb{R}^n$, if there are $k$ regular polytopes, and we look at the mapping from these regular polytopes to $\mathbb{Z}_k$ given by $f|k$ ($f$ modulo $k$) where $f$ is the number of faces in the polytope, then (if comments below are correct), the mapping is a bijection iff $n=3|6$ or $n=5|6$. And so the question is, is there a deep reason for this?

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    $\begingroup$ Maybe it would be more interesting if you had a similar phenomenon for the regular solids in $\mathbb R^n$ for all $n$ ? You get a similar correspondence for $n=2$, but not $n=4$ (where there are six). $\endgroup$
    – Ryan Budney
    Commented Nov 24, 2023 at 20:09
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    $\begingroup$ I think this question is potentially interesting. The classification of regular polyhedra is related to other classifications like of simple Lie algebras, etc., and so it’s not impossible to imagine a somewhat deep reason for this face-counting bijection to work. $\endgroup$
    – Sam Hopkins
    Commented Nov 24, 2023 at 20:13
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    $\begingroup$ The observation holds only in odd dimensions. If $n \ge 5$, there are only three regular polytopes in dimension $n$, the hypercube with $2n$ faces, the hyperoctahedron with $2^n$ faces, the simplex with $n+1$ faces. The numbers $n$, $2n$ and $n+1$ are different modulo 3 if and only if $n$ is odd. $\endgroup$
    – Christophe Leuridan
    Commented Nov 24, 2023 at 20:59
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    $\begingroup$ Kepler tried to match the five Platonic solids to the five spaces between the six planets that were known in his day. It didn't end well. en.wikipedia.org/wiki/Mysterium_Cosmographicum $\endgroup$
    – Gerry Myerson
    Commented Nov 24, 2023 at 21:55
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    $\begingroup$ I'm tempted to say there might be something about the Euler characteristic that helps here. The number of faces is 2 + E - V, and having the mod 5 count of faces be an injective function is the same as saying the difference E - V is an injective function mod 5. This is then a fact about the underlying edge graph, and regularity means the degree D of the vertices to be constant. Then E = DV/2, so that we are left with considering the function V(D/2 - 1). 2 is invertible mod 5, so we are considering the function V(D - 2). We know 1≤D-2≤3 and V≥4 from elementary considerations. $\endgroup$
    – David Roberts ♦
    Commented Nov 25, 2023 at 5:47

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