When I was young, I proved the famous 4-color map theorem!
Or thought I did, *for just a brief moment*.
It was a few days before I looked
at it clearly enough to spot the flaw, but I knew there had to be one.
Let me give that simple proof here. See if you can spot the flaw!
Remember that a planar graph has Vertices, Edges and Faces, and it's the
Vertices that we are coloring.

**Lemma:** Any finite planar graph will have a vertex of
valence five or less, i.e. a vertex connected via edges to five
or fewer other vertices.

Proof: Recall Euler's V + F = E + 2. Suppose the average valence is k.
Then the number of Edges is E = kV/2. (Divide by 2 since each Edge has two
Vertices.) Similarly E = mF/2 where m is the average number of edges per face.
Combine these three equations to solve for V in terms of m and k and get
V = 4m / (2m + 2k - km). The denominator must be positive; so k < 2m/(m-2).
But we know m ≥ 3 since no faces have fewer edges than a triangle.
These two inequalities imply that k < 6. The average valence is less
than 6; all we needed was the weaker result that *some* valence
is less than 6.

Now suppose that G is the smallest finite planar graph that
cannot be 4-colored. We will remove a vertex ( In the first figure, we show part of the graph
{G- The plan is to manipulate the colors until the pentagon uses only
three colors instead of four. If we can do that, For example, we could change the pentagon's green vertex to yellow.
To maintain a valid 4-coloring for {G- In the diagram we show that the pentagon's yellow vertex is part of
that subgraph. So let's not change that Green to Yellow after all; let's look for something else. |

A similar attempt can be made to recolor the green vertex blue. And again it must fail, for the same reason. In the second diagram we show that the pentagon's green and blue vertices are in the same connected green-blue subgraph. Foiled again! But not so fast ... Look at the red vertex at the top of our pentagon
We can change its color to yellow (and its adjacent yellows to reds, and
their adjacent reds to yellows and so on) without fear of encountering the
yellow vertex at the pentagon's lower left.
planar graph: a yellow-red edge cannot
pass a blue-green fence without tunnelling under it, or leaping over it;
these are forbidden in planar graphs.
To avoid cluttering the diagram, we don't show it, but the Red vertex in the lower right of the pentagon can be similarly changed to Blue. And again the remainder of the pentagon is unaffected: the red-blue subgraph cannot tunnel through the green-yellow fence we've proven to exist. So, we change one of the pentagon's red vertices to yellow, and the other
red vertex to blue, without affecting the rest of the pentagon.
The color Red is now available! Restore the vertex |

**This proof is fallacious!**
I do know where the fallacy is but I'm tired
of typing for now. As an experiment to see how many people enjoy my pages,
please feel free to e-mail me if you want to tell me what the fallacy is!