Is it possible to draw a regular pentagon on a regular 3D grid by only connecting the intersection points? Think of it as infinite rows of cubes side by side and infinite rows on top of each other.
Like this, but as many cubes as needed
Let's say the distance between the points (the edges of the cubes) are of 10 units.
Is it possible to draw a pentagon by only connecting the vertices of the cubes?
If so, is it possible to draw it without using millions of cubes?
As an example, you can make an equilateral triangle by connecting 3 vertices of a single cube, see: 

 A: No. 
Suppose you have found such a regular pentagon $ABCDE$
then, since the lattice points are well.. on a lattice, the points $A' = A + \vec{BC}, B' = B + \vec{CD}, \ldots , E' = E + \vec{AB}$ are also on the lattice.
And they also form a regular pentagon, but one whose side is smaller than the side of the original pentagon by some factor $K < 1$.
Iterating this procedure you can find regular pentagons with lattice points, that get smaller and smaller and smaller.
However, you can't get two points on your lattice that are closer than the side of one of the original cube (your $10$ units)  so there can't be any regular pentagon on lattice points whose side is smaller than $10$ units, which is a contradiction.
A: No. Computing the scalar product between two consecutive sides of a regular pentagon and dividing by their squared norm we should obtain $\cos 108=\frac{1-\sqrt{5}}{4}$. However for two segments $(a,b)$ and $(c,d)$ on the lattice with the same length this quantity is
$$\frac{ac+bd}{a^2+b^2}$$
which is rational.
