Problem source

A polyhedron is a solid bounded by $F$ plane faces, which meet in $E$ edges and $V$ vertices. You may assume \textit{Euler's
formula}, that $V-E+F=2$.
In a regular polyhedron the faces are equal regular $m$-sided
polygons, $n$ of which meet at each vertex. Show that
$$
F={4n\over h}  \,,
$$
where $h=4-(n-2)(m-2)$.
By considering the possible values of $h$, or otherwise, prove that there are only five regular polyhedra, and find $V$, $E$ and $F$ for each.

Solution source

Note that each of the $F$ faces have $m$ edges. If we count edges from each face we will get $mF$ edges, but this counts each edge twice, so $E = \frac{m}{2}F$. Similarly each edge goes into $2$ vertices, but this counts each vertex $n$ times, therefore $V = \frac{2E}{n} = \frac{m}{n}F$

Therefore \begin{align*}
&& 2 &= V - E + F \\
&&&= \frac{m}{n}F - \frac{m}{2}F + F \\
&&&= F \left ( \frac{2m-nm+2n}{2n} \right) \\
&&&= F \left ( \frac{4-(n-2)(m-2)}{2n} \right) \\
\Rightarrow && F &= \frac{4n}{4-(n-2)(m-2)} = \frac{4n}{h}
\end{align*}

Notice that $1 \leq h \leq 4$, so

If $h = 4$, then $n = 2$ or $m = 2$ but we can't have two-sided polygons or polyhedra with two faces making a vertex so this is not possible.

If $h = 3$ then $(n-2)(m-2) = 1$ and $3 \mid n$, so we have $n = 3, m = 3$. ie we have triangular faces meeting at $3$ per point. We also have $F = \frac{4 \cdot 3 }3$ which is $4$ faces so this is a tetrahedron and $(V,E, F) = (4, 6, 4)$

If $h = 2$ then $(n-2)(m-2) = 2$:
Case 1: $n = 4, m = 3$ which is four triangles meeting at each vertex with $(V,E,F) = (6, 12, 8)$, ie an octohedron.
Case 2: $n = 3, m = 4$ so we have three squares meeting at each vertex, with $(V,E,F) = (8,12,6)$ which is a cube.

If $h = 1$ then $(n-2)(m-2) = 3$
Case 1: $n = 5, m = 3$ which is five triangles meeting at each vertex, with $(V,E,F) = ( 12,30, 20)$ ie an icosahedron.
Case 2: $n = 3, m = 5$ which is three pentagons meeting at each vertex, with $(V,E,F) = (20, 30,12)$ which is a dodecahedron.

These are all the possible cases and hence we have found the five platonic solids.