How many squares can be determined using $n$ points?: revisited

The main observations were:

• In $$\mathbb{R}^2$$ at most $$O(n^2)$$ as every pair of points determines at most 3 squares.
• In $$\mathbb{R^3}$$ at most $$O(n^3)$$ as every triple of points determines at most one square
• This can be improved to $$O(n^{7/3}$$) as it is known that this is sharp for the number of right triangles determined by $$n$$ points in $$\mathbb{R}^3.$$

However it would be more relevant to have a bound on the number of isosceles right triangles.

I would guess that in the long run the optimum arrangement in $$\mathbb{R}^3$$ is a square grid of side $$s=\sqrt{n}.$$

If I calculate correctly, a square grid of $$n=s^2$$ points determines $$\frac{s^4-s^2}{12}$$ squares, so indeed $$O(n^2)$$.

I would also guess that this might be best for $$\mathbb{R}^3$$ and, if not , then a cubic grid of side $$s=\sqrt[3]{n}$$ is best.

These aren't very educated guesses. So my main question (in two parts) is

Suppose $$n=t^6$$,

• Does any arrangement of $$n$$ points in $$\mathbb{R}^2$$ determine more squares that a square grid of side $$t^3?$$

• Does any arrangement of $$n$$ points in $$\mathbb{R}^3$$ determine more squares that a square grid in a plane of side $$t^3?$$

I picked $$n=t^6$$ to allow for a cube of side $$t^2$$ as well.

I think it best to ask one main question, but those seem related enough. Secondary questions are:

• Is anything known about the arrangements in $$\mathbb{R}^3$$ which achieve $$O(n^{7/3})$$ right triangles?
• Is anything known about bounds and arrangements for the number of isosceles right triangles?
• Are formulas or bounds known for the number of squares determined by the points of a cubic grid of $$s^3$$ points?

An answer to that last one might show that the answer to the second part of the main question is NO.

In a cubic grid of $$s^3$$ points one has $$3s$$ squares of side $$s$$ parallel to coordinate planes. So there are $$3s\frac{s^4-s^2}{12}$$ squares in those planes. That is only $$O(n^{5/3})$$. However any two orthogonal vectors of equal length determine a family of planar lattices with a certain number of squares. Determining the number of such squares within a cube may be known but is certainly more delicate that the planar case.

A particularly productive length is $$15:$$

The integer vectors of length $$15$$ are $$(15,0,0),(14,5,2),(12,9,0),(11,10,2),(10,10,5)$$ along with their permutations and changing the sign on some coordinates.

A number of orthogonal pairs are possible such as

• $$(15,0,0),(0,15,0)$$
• $$(10,10,5),(10,-5,-10)$$
• $$(12,9,0),(-9,12,0)$$
• $$(15,0,0),(0,9,12)$$
• $$(14,5,2),(2,-10,11)$$
• $$(14,5,2),(-5,10,10)$$
• $$(11,10,2),(-10,10,5)$$

And again the coordinates can be permuted and signs flipped.