Creating a square with a "nonhypothenus area", under certain restriction

Factual Questions
21

I suppose everyone tires of hearing me work through my thoughts by now, but I always try to find the cleanest, minimal proof of anything, and I realized there’s no need to invoke the line distance formula in proving this:

Let’s say a lattice line is any line which passes through two (and therefore infinitely many regularly spaced) lattice points, a lattice-line square is one whose sides lie along lattice lines, and a lattice-point square is one whose corners lie on lattice points.

Given any lattice line, consider the grid formed by all parallel and perpendicular lattice lines, thus carving up the plane into lots of little cells. Lattice-line squares along this grid consist of, well, squares built out of these cells.

But also, in particular, each cell lies in (in fact, along an edge of) some lattice-point square built out of cells (because each lattice line can be followed far enough in any direction till one ends up at one of its regularly spaced lattice points).

So the area of any lattice-line square is a square multiple of the corresponding cell-area, and also, the area of some lattice-point square (which is by definition a “hypotenuse number”; i.e., our slight misnomer for a sum of two squares) is a square multiple of that same cell-area.

So each lattice-line square’s area is a hypotenuse number divided by a square and multiplied by a square. In other words, a lattice-line square’s area times some square is a lattice-point square’s area times some square.

The last fact we need is that if two whole numbers are in square ratio, and one is a hypotenuse number, then so is the other. One direction of this is trivial, while the other direction requires (as far as I can tell) the less elementary analysis (by, e.g., Gaussian prime factorization) of what kinds of integers are hypotenuse numbers. But with that, we are done.

22

Indeed, even without the last non-elementary step, what this proves is that every lattice-line square has area of the form (a^2 + b^2)/c^2, whether or not this comes out to a whole number. The last, non-elementary thing is the number-theoretic fact that when such an expression comes out to a whole number, it must itself be a sum of two squares.

23

Whoops, ignore this post.

24

More generally, a positive rational area is achievable just in case, in its prime factorization, the exponent of 2 is at least -1, and the exponent of any prime which is 3 mod 4 is even and non-negative.

Thus, we cannot achieve the areas 1/4, 1/8, 1/9, 1/18, etc., even though any square multiple of these which is a whole number IS achievable.

The OP’s example square of area 3.6, however, is achievable, since 3.6 = 2^1 * 3^2 * 5^(-1), where the exponent of 2 is at least -1, the exponent of 3 (which is 3 mod 4) is non-negative and even, and 5 (being 1 mod 4) is unconstrained in exponent.

25

I at least have enjoyed it, even if I spend more time than I should on this trying to visualize and confirm the steps to myself.

26

Ah, I just realized a simpler way of thinking about the original problem:

Any square formed in the desired way has all its corners have rational coordinates. Which means, by choosing a common denominator for these coordinates, it is simply a square with lattice-point corners shrunk down by that denominator in each dimension. Thus, its area is (a^2 + b^2)/c^2 for some a, b, and c. [And then we invoke the number theory.]

Bingo-bango, real simple. But it took a while to achieve this clarity…

27

Put another way, even if we didn’t care about drawing the sides of the square, but just wanted to mark its corners as intersections of lines, the only restriction on square area would be that it would be of the form (a^2 + b^2)/c^2 (equivalently, of the form x^2/(y^2 + z^2) [as multiplying top and bottom by a^2 + b^2 brings one from the former form to the latter, and then symmetrically, one can go back]). And this would be by the super simple “All coordinates are rational” proof.

If we do care about drawing the square-sides, we get a little more restriction. Not just any areas of the form x^2/(y^2 + z^2) will be achievable; rather, precisely those where y and z can be chosen coprime will be achievable. And this would be by the more sophisticated “Consider cell-area by considering line distance…” proof.

In both cases, the achievable whole numbers are precisely the “hypotenuse numbers”, but in the former case, the achievable values in general are precisely those rationals whose lowest-terms numerator and denominator are both hypotenuse numbers, while in the latter case, we add the further constraint that this denominator must further be a “primitive hypotenuse number” (one that can be written as a^2 + b^2 with a/b itself in lowest-terms; equivalently, by the number-theoretic magic we keep using from here, a hypotenuse number neither divisible by 4 nor by any non-hypotenuse factor).

28

Then the area can be anything at all. If they are not lattice points, then I don’t understand the question.

29

The sides of the square must lie on lines which pass through two lattice points. The corners, though, needn’t themselves be lattice points; they can just be intersections of lines which each pass through two other lattice points.

Again, look at the OP’s area 3.6 UVWZ example here.

30

This confused me at first, until I came up with an example. Consider the square (0,0), (0,1/2), (1/2,1/2), (1/2,0). Every one of those points can be the intersection of two geoboard lines (for instance, (0,1/2) is the intersection between the line from (-1,0) to (1,1) with the Y axis), but you can’t get the side from (0,1/2) to (1/2,1/2) on a geoboard, since it never intersects an integer lattice point.