Minimum needed colors to color the vertices of cube such that no two vertices on the same edge of the cube are the same color.

Question

Full question: All vertices of a cube are colored such that no two vertices on the same edge of the cube are the same color. What is the minimum number of colors that are needed to color the vertices of the cube?

There's 8 vertices in total. If I look at one vertex I see that there are three other vertices directly connected. (Am I right?) Would this mean there are four colors minimum?

Answer 1

place all 8 vertices at points with $x,y,z$ coordinates either 0 or 1. Half the points, such as $(0,0,0),$ have an even sum. Half the points, such as $(0,0,1),$ have an odd sum.

Just two colors, odd and even. Any edge joins a vertex with odd sum to a vertex with even sum

Answer 2

In general, the hypercube graph $Q_n$ is bipartite for all $n$.

This includes the case $n=3$, which is a $3$-dimensional cube, as in your question.

Bipartiteness is equivalent to $2$-colorability, so a cube is $2$-colorable.