Show that an unrestricted $3$-coloring of a plane triangulation $G$ has an even number of $3$-colored faces Show that an unrestricted $3$-coloring of a plane triangulation $G$ has an even number of $3$-colored faces.
Here an unrestricted $3$-coloring means that adjacent vertices can receive the same color. And a $3$-colored face is a face where its frontier has distinct colors in each vertex.
 A: I think the idea should be something like this below.
Because I've seen somewhere 25+ years ago a competitive math problem of this kind whose solution impressed me quite a lot back then.
Let $s$ be any segment. Let also
$f(s) = 1$ if the two ends/vertices of the segment are colored differently;
$f(s) = 0$ if the two ends are of the same color.
Now for each triangular face $a$ denote $G(a)$ to be the sum of $f(s_{ai})$ for its three sides $s_{ai}$ where $i=1,2,3$.
Obviously $G$ can take only the values $0, 2, 3$.
You can make a few drawings and convince yourself in this.
Now sum $G(a)$ over all triangular faces $a$ and denote this sum by $S$.
On one side: $$S = n_0 \cdot 0 + n_2 \cdot 2 + n_3 \cdot 3$$
where $n_i$ is the number of triangular faces for which $G$ takes the value $i$.
On the other side each segment takes part in exactly two triangles. So in $S$ we  in fact sum up the $f(s)$ values for all segments, but actually each segment $s$ contributes two times its own $f(s)$ to $S$.
Therefore $S$ is even.
But then $n_3$ has to be even (from the above equation).
And this is what we wanted to prove.
Side note: I don't quite understand what you mean by triangulation of the plane.
Is this infinite? If so, how can we talk about even or odd number of the 3-colored faces?
Also I see in Wikipedia that different types of triangulations can be defined.
Triangulation (geometry)
But I am sure this problem (which you state) has something to do with this idea.
