In my previous post I wrote that Euler's formula
implies that there can only be five regular polyhedra and that this can be shown by simply solving a Diophantine equation. In this post I will demonstrate ( read: prove ) this and I will show how to solve Diophantine equations in Mathematica.
We start with Euler's polyhedron formula $$F  E + V = 2,$$
where $F=$ number of faces, $E=$ number of edges and $V=$ number of vertices. What exactly makes a polyhedron regular? Clearly, all the faces are equal ( i.e. all triangles ) of a regular polyhedron. This property alone is not enough though. Also, on a regular polyedron the same number of faces meet at each vertex. Clearly the number of edges remains an unknown for now. These requirements can be related to E, the number of edges, if we introduce two ( new ) variables:
 $A$ : the number of edges surrounding a face. ( i.e. $A = 4$ for a cube ).
 $B$ : the number of edges meeting at a vertex. ( i.e.$B = 3$ for a cube ).
We know the following about a regular polyhedron:
$$FA= 2E $$
$$VB = 2E $$
$$FE+V=2E $$
$$A,B \geq 3 $$
$$F,E,V \geq 1$$
( Note that I have used $FA=2E$ ( instead of $FA=E$ ) because each edge is part of two faces, as well as $VB = 2E$ because each edge connects two vertices. )
By simply algebraically reworking the equations above we get $$\frac{2E}{A}E+\frac{2E}{B}=2.$$
In this particular case there are two approaches to attack this Diophantine equation. ( Named after Diophantus, 3rd century. Diophantine problems have fewer equations than unknown variables and involve finding integers that work correctly for all equations.) We can use trial and error or we can use a tool like Mathematica. I have used Mathematica as follows.
The tool to solve Diophantine equations in Mathematica is
Reduce
. In Mathematica
Reduce[expr, vars, dom]
reduces the statement
expr
by solving equations or inequalities for
vars
and eliminating quantifiers and does this over the domain
dom
. We solve $\frac{2E}{A}E+\frac{2E}{B}=2$ in Mathematica as follows using
Reduce
:
Reduce[{(2 e)/a  e + (2 e)/b == 2, a >= 3, b>= 3, e >= 0}, {a, b, e}, Integers]
Out[1]=
(a == 3 && b == 3 && e == 6) 
(a == 3 && b == 4 && e == 12) 
(a == 3 && b == 5 && e == 30) 
(a == 4 && b == 3 && e == 12) 
(a == 5 &$ b == 3 && e == 30)
So there are indeed five solutions! Let's look at them more closely.
F 
V 
E 
Type 
$\frac{2 \cdot 6}{3} = 4$ triangle 
$\frac{2 \cdot 6}{3}= 4$ 
6 
Tetrahedron 
$\frac{2 \cdot 12}{3} = 8$ triangle 
$\frac{2 \cdot 12}{4}= 6$ 
12 
Octahedron 
$\frac{2 \cdot 30}{3} = 20$ triangle 
$\frac{2 \cdot 30}{5}= 12$ 
30 
Icosahedron 
$\frac{2 \cdot 12}{4} = 6$ square 
$\frac{2 \cdot 12}{3}= 8$ 
12 
Cube 
$\frac{2 \cdot 30}{5} = 12$ pentagon 
$\frac{2 \cdot 30}{3}= 20$ 
30 
Dodecahedron 

A=5 (edges per face), B=3 (edges to a vertice ) = Dodecahedron. 
This proof clearly shows that the various disciplines ( in this case topology, number theory and geometry ) in mathematics are related and thus pointing to mathematics at a deeper layer . It has been said that graph theory, geometry, algebra and number theory are just different manifestations of the same mathematical concepts. I recently read that the Riemann Hypothesis ( analytical number theory ) can be proved by proving an equivalent theorem in graph theory.
Deep stuff, surely.