For a cubic map:

`4C`_{2} + 3C_{3} + 2C_{4} + C_{5} - C_{7} - 2C_{8} - 3C_{9} ... = 12

Each `C`_{n}

term
represents the number of countries, or faces, in a cubic map that have n-sides. So, `C`_{4}

is the number
of squares, `C`_{5}

the number of pentagons, etc...

This is an important consequence of Euler's Polyhedra formula.

To derive this result, begin by noting that the total number of countries is:

`F = `_{2} + C_{3} + C_{4} + C_{5} + C_{6} + C_{7} + ... + C_{n}

The next step in deriving the formula is to count the number of vertices in a map. For each triangle, there
are 3, each square has 4, and in general `C`_{n}

has `n`

vertices. So:

`V = 2C`_{2} + 3C_{3} + 4C_{4} + 5C_{5} + 6C_{6} + 7C_{7} + ... + nC_{n}

But, as we are considering cubic maps, each vertex is shared by three countries. Therefore, the proper expression is:

`3V = 2C`_{2} + 3C_{3} + 4C_{4} + 5C_{5} + 6C_{6} + 7C_{7} + ... + nC_{n}

` V = `^{2}/_{3}C_{2} + C_{3} + ^{4}/_{3}C_{4} + ^{5}/_{3}C_{5} + 2C_{6} + ^{7}/_{3}C_{7} + ... + ^{n}/_{3}C_{n}

The same can be done with the number of edges. However, this time, each edge is only shared between two countries.

`2E = 2C`_{2} + 3C_{3} + 4C_{4} + 5C_{5} + 6C_{6} + 7C_{7} + ... + nC_{n}

` E = C`_{2} + ^{3}/_{2}C_{3} + 2C_{4} + ^{5}/_{2}C_{5} + 3C_{6} + ^{7}/_{2}C_{7} + ... + ^{n}/_{2}C_{n}

With these expressions for `F`

, `E`

and `V`

, we can make use of Euler's Polyhedra formula:

`2 = F - E + V`

` = (C`_{2} + C_{3} + C_{4} + C_{5} + C_{6} + C_{7} + ... + C_{n})

` - (C`_{2} + ^{3}/_{2}C_{3} + 2C_{4} + ^{5}/_{2}C_{5} + 3C_{6} + ^{7}/_{2}C_{7} + ... + ^{n}/_{2}C_{n})

` + (`^{2}/_{3}C_{2} + C_{3} + ^{4}/_{3}C_{4} + ^{5}/_{3}C_{5} + 2C_{6} + ^{7}/_{3}C_{7} + ... + ^{n}/_{3}C_{n})

` = `^{2}/_{3}C_{2} + ^{1}/_{2}C_{3} + ^{1}/_{3}C_{4} + ^{1}/_{6}C_{5} + 0C_{6} - ^{1}/_{6}C_{7} - ...

Although not immediately clear, this is the desired result. Multiplying through by 6 makes things clearer:

`12 = 4C`_{2} + 3C_{3} + 2C_{4} + C_{5} - C_{7} - 2C_{8} - 3C_{9} ...

This result does provide some interesting insight to the realm of cubic maps, and polyhedra.

Notice that only `C`_{2}

through `C`_{5}

have positive coefficients. This means that
in any cubic map one of these shapes must appear. This is the Five neighbour theorem.