For a cubic map:
`4C2 + 3C3 + 2C4 + C5 - C7 - 2C8 - 3C9 ... = 12`

Each `Cn` term represents the number of countries, or faces, in a cubic map that have n-sides. So, `C4` is the number of squares, `C5` 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 + C3 + C4 + C5 + C6 + C7 + ... + Cn`

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 `Cn` has `n` vertices. So:
`V = 2C2 + 3C3 + 4C4 + 5C5 + 6C6 + 7C7 + ... + nCn`

But, as we are considering cubic maps, each vertex is shared by three countries. Therefore, the proper expression is:
`3V = 2C2 + 3C3 + 4C4 + 5C5 + 6C6 + 7C7 + ... + nCn`
` V = 2/3C2 + C3 + 4/3C4 + 5/3C5 + 2C6 + 7/3C7 + ... + n/3Cn`

The same can be done with the number of edges. However, this time, each edge is only shared between two countries.
`2E = 2C2 + 3C3 + 4C4 + 5C5 + 6C6 + 7C7 + ... + nCn`
` E = C2 + 3/2C3 + 2C4 + 5/2C5 + 3C6 + 7/2C7 + ... + n/2Cn`

With these expressions for `F`, `E` and `V`, we can make use of Euler's Polyhedra formula:
`2 = F - E + V`
`  = (C2 + C3 + C4 + C5 + C6 + C7 + ... + Cn)`
`    - (C2 + 3/2C3 + 2C4 + 5/2C5 + 3C6 + 7/2C7 + ... + n/2Cn)`
`    + (2/3C2 + C3 + 4/3C4 + 5/3C5 + 2C6 + 7/3C7 + ... + n/3Cn)`
`  = 2/3C2 + 1/2C3 + 1/3C4 + 1/6C5 + 0C6 - 1/6C7 - ... `

Although not immediately clear, this is the desired result. Multiplying through by 6 makes things clearer:
`12 = 4C2 + 3C3 + 2C4 + C5 - C7 - 2C8 - 3C9 ...`

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

Notice that only `C2` through `C5` have positive coefficients. This means that in any cubic map one of these shapes must appear. This is the Five neighbour theorem.