Mathologer's secret of row 10, tetrahedral version Mathologer put out this great video some time ago and it has to be viewed for you to understand this question.
Here, I'm talking about the tetrahedral version (refer to timestamp 28:37 onward) of the puzzle, presented at the end of the video:
Each layer consists of spheres arranged in the shape of an equilateral triangle. The bottom layer is the biggest, with side length $n$, and is colored arbitrarily, with each sphere being one of four colors. Each subsequent layer is built up as follows:

*

*If the three adjacent spheres on one layer are all different colors, then the sphere on top gets the fourth color.

*If the three adjacent spheres are all the same color, then the sphere on top also gets that same color.

*If two of the three adjacent spheres are the same color but one is different, then the sphere on top gets the color that only appears once.

There are some values of $n$ where the four vertices will always follow the rules given above, regardless of the initial coloring. It's given that the smallest non-trivial size where this happens is $n=3$. After knowing that, it's easy to prove that $n=2^{k}+1$ is sufficient for it to work.
How do I prove that $3$ is the smallest case of this without having to check all 4096 initial colorings?
 A: Think of the four colors as $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$, and define an "addition" operation on the colors: add both components modulo $2$ (a result of $2$ in a component becomes $0$). For example, $(1,0) + (1,1) = (1+1, 0+1) = (2,1) = (0,1)$. Notably:

*

*Any color plus $(0,0)$ is itself;

*Any color plus itself is $(0,0)$;

*Any three colors together sum to the fourth.

Then your coloring rule can be restated as: color the fourth sphere with the sum of the colors of the first three. (If all three spheres are different, their sum is the fourth color; otherwise, cancel out two spheres of the same color and use the color of the third.)
Suppose that the six colors on the bottom have the colors
\begin{array}{ccccc}
  & & A \\
  & B && C \\
  D && E && F
\end{array}
Then the next layer will be
\begin{array}{ccc}
     & A+B+C \\
   B+D+E && C+E+F
\end{array}
and the sum of those three colors is $(A+B+C)+(B+D+E)+(C+E+F)$. But $B+B$, $C+C$, and $E+E$ cancel out, and we are just left with $A+D+F$: the sum of the three corners of the bottom triangle. Therefore the fourth corner obeys the coloring rule with the first three corners.
