What are the constructible angles and regular/uniform/semiregular/etc polytopes in n-dimensional integer lattices? In a 2-dimensional regular square lattice, using only lattice points, lines connecting lattice points, and points where such lines intersect, I am aware it is impossible to make a 30-degree angle between two lines, and hence, impossible to make an equilateral triangle. Using a regular triangular lattice, however, you can make such angles and triangles, but you lose the ability to construct 45-degree angles, and hence, no squares.
In 3-dimensional space, however, using a cubical lattice with the same principles and extending to constructible planes given three non-colinear points, you can construct both 30-degree and 45-degree angles. For example, the points (0,0,0), (1,1,0), (1,0,1), and (0,1,1) together make a tetrahedron, which includes equilateral triangles.
One thing, iirc, you apparently cannot construct in 2-d or 3-d lattices of this type, despite being able to make 45-degree angles, are regular octagons. You can make equiangular ones, but not with equal side lengths as well.
So I’m wondering, does using regular lattices in four (or more) dimensions allow constructions of any new similar angles, polygons, or polyhedra, that can not be constructed in 3 dimensions? Also, just like how square and triangular lattices in 2-d allows different constructions, are there different regular lattices in 3-d that differ? It seems to me that everything that can be constructed in a 3d cubic lattice can be constructed by a 3d tetrahedral-octahedral one, and vice-versa, though tilted at a different angle from the axes (or am I incorrect in thinking that?). Finally, I know any lines with rational endpoints can only intersect at rational points in 2d Cartesian space, but does this hold in 3 dimensions? Anyone have any insights or resources on this topic?
 A: Somewhat surprisingly, arbitrarily many dimensions in a Cartesian lattice will not get you all angles.  For instance, no Cartesian lattice enables construction of equiangular planar pentagons.
How is this so?  Draw any two vectors $u$ and $v$ from the origin to other lattice points.  Then$|u|^2$ and $|v|^2$ are whole numbers.  Meanwhile
$(u\cdot v) = \Sigma(u_iv_i) = |u||v|\cos\theta$
where $\theta$ is the angle between the vectors.  This forces the cosine of the angle to be a square root of a nonnegative rational number, so for example you can never get $90°-18°=72°$ whose cosine is $(\sqrt{5}-1)/4$.
A: Whilst traditional geometric construction relates to the plane with ruler and compass, iterative application of these simple methods reaches all constructible lengths and the 2d lattice can be generalised.  Viewed algebraically these finitely constructible lengths sit in a tower of field extensions as the roots of quadratics whose coefficients are the level below.  Pythagoras works in any number of dimensions because any three points in space are coplanar and form a triangle.
