Proof

Recall that all the faces of a regular polyhedron are identical regular polygons, and that each vertex has the same degree. Consider four cases, depending on the type of regular polygon.

Case 1: Each face is a triangle. Let \(f\) be the number of faces. There are then \(3f/2\) edges. Using Euler's formula we have \(v - 3f/2 + f = 2\) so \(v = 2 + f/2\text{.}\) Now each vertex has the same degree, say \(k\text{.}\) So the number of edges is also \(kv/2\text{.}\) Putting this together gives

\begin{equation*} e = \frac{3f}{2} = \frac{k(2+f/2)}{2}\text{,} \end{equation*}

which says

\begin{equation*} k = \frac{6f}{4+f}\text{.} \end{equation*}

Both \(k\) and \(f\) must be positive integers. Note that \(\frac{6f}{4+f}\) is an increasing function for positive \(f\text{,}\) bounded above by a horizontal asymptote at \(k=6\text{.}\) Thus the only possible values for \(k\) are 3, 4, and 5. Each of these are possible. To get \(k = 3\text{,}\) we need \(f = 4\) (this is the tetrahedron). For \(k = 4\) we take \(f = 8\) (the octahedron). For \(k = 5\) take \(f = 20\) (the icosahedron). Thus there are exactly three regular polyhedra with triangles for faces.

Case 2: Each face is a square. Now we have \(e = 4f/2 = 2f\text{.}\) Using Euler's formula we get \(v = 2 + f\text{,}\) and counting edges using the degree \(k\) of each vertex gives us

\begin{equation*} e = 2f = \frac{k(2+f)}{2}\text{.} \end{equation*}

Solving for \(k\) gives

\begin{equation*} k = \frac{4f}{2+f} = \frac{8f}{4+2f}\text{.} \end{equation*}

This is again an increasing function, but this time the horizontal asymptote is at \(k = 4\text{,}\) so the only possible value that \(k\) could take is 3. This produces 6 faces, and we have a cube. There is only one regular polyhedron with square faces.

Case 3: Each face is a pentagon. We perform the same calculation as above, this time getting \(e = 5f/2\) so \(v = 2 + 3f/2\text{.}\) Then

\begin{equation*} e = \frac{5f}{2} = \frac{k(2+3f/2)}{2}\text{,} \end{equation*}

\begin{equation*} k = \frac{10f}{4+3f}\text{.} \end{equation*}

Now the horizontal asymptote is at \(\frac{10}{3}\text{.}\) This is less than 4, so we can only hope of making \(k = 3\text{.}\) We can do so by using 12 pentagons, getting the dodecahedron. This is the only regular polyhedron with pentagons as faces.

Case 4: Each face is an \(n\)-gon with \(n \ge 6\text{.}\) Following the same procedure as above, we deduce that

\begin{equation*} k = \frac{2nf}{4+(n-2)f}\text{,} \end{equation*}

which will be increasing to a horizontal asymptote of \(\frac{2n}{n-2}\text{.}\) When \(n = 6\text{,}\) this asymptote is at \(k = 3\text{.}\) Any larger value of \(n\) will give an even smaller asymptote. Therefore no regular polyhedra exist with faces larger than pentagons.⁸