Every map can be coloured with six colours such that no neighbouring countries have the same colour.
This is a theorem that is closly related to the more famous four color theorem, although is notably
weaker. The proof of this theorem is much easier than for its fourcoloured cousin.
Using proof by contradiction, we are going to assume that there exists a minimal map that with the fewest possible
countries, requires 7 colours; all maps with fewer countries can be coloured using 6 colours.
Using the Five Neighbour theorem, we know that any such minimal map must contain a country with at most
five neighbours. In other words, it must contain either a twosided contry (a 'diagon'?), triangle, square or pentagon.
We need to consider each of these cases and show that because they exist in a map, that map cannot be a minimal map  and as
such, a minimal map cannot possibly exist at all.
To begin with, we shall consider the pentagon.


1 o 2
/ \
/ \
o 6 o
\ /
5 oo 3
/ 4 \
/ \
If we remove one of the boundaries of this country we are left with a map that has one country
less than the minimal map. By our original statement, we must be able to colour this new map using six colours.


R o G
\
\
o o
\ /
P oo B
/ \
/ Y \
Once this is done, if the boundary that was removed is reinstated, we have a coloured map with a blank country.
This blank country has five neighbours and therefore it has five colours neighbouring it. This
leaves us with the sixth colour to complete the map.


R o G
/ \
/ \
o I o
\ /
P oo B
/ \
/ Y \
It has not been necessary to use the seventh colour at all.
This means that the map that we claimed to be minimal, which contains a pentagon,
is not infact a minimal map.
Using the same argument, we can also show that none of the other 'required' shapes can exist in the minimal map. Therefore,
a minimal map that requires 7 colours cannot exist.
The next step from here is to show that no more than 5 colours are required.