What's the connection between hypotenuse-squared being integers and simple fractions of $\pi$ or whole circles?

Question

I noticed something interesting when I think of Pythagoras theorem as addition of two integers to get another integer. For example, if the hypotenuse-squared is equal to 4 then it leads to the following 5 triangles:

$\sqrt(0)^2 + \sqrt(4)^2 = \sqrt(4)^2$

$\sqrt(1)^2 + \sqrt(3)^2 = \sqrt(4)^2$

$\sqrt(2)^2 + \sqrt(2)^2 = \sqrt(4)^2$

$\sqrt(3)^2 + \sqrt(1)^2 = \sqrt(4)^2$

$\sqrt(4)^2 + \sqrt(0)^2 = \sqrt(4)^2$

The angles corresponding to these turn out to be nice, simple fractions of the circle: $0, \pi/6, \pi/4, \pi/3, \pi/2$. If we normalize the denominator to 12, then these angles are: $0\pi/12, 2\pi/12, 3\pi/12, 4\pi/12, 6\pi/12$ ). These are not only "nice", but also very useful as they show up in squares, equilateral triangles (and their halves), hexagons and more.

But doing the same with hypotenuse-squared = 3 or 5, leads to angles like $arctan(1/\sqrt2)$ or $arctan(1/\sqrt 4)$ and $arctan(\sqrt 2/\sqrt 3)$. Sure, these are interesting angles. $1/\sqrt2$ is the Lichtenberg Ratio and is the aspect ratio of metric paper size standard. But these do not seem to be connected to the circle or $\pi$ in any interesting way. I am also not aware of their connections to other common shapes.

My question is this: What is so special about the number 4 which leads to those angles turning out to be simple parts of the whole circle? If it's about 4 being an even number, should we expect similar "nice" angles when the hypotenuse-squared is 6 and the squares of the other two sides are integers (eg: $arctan(1/\sqrt 5)$)? Or is it the fact that 3 and 5 are primes, somehow making them less nice/useful? Is there a theorem that exposes the link between the hypotenuse-squared being an integer and fraction of circles those angles represent?

Answer 1

I wouldn't say there's anything special about 4 per se. There are nice properties to it, the most striking one being that it is a square, and thus the hypotenuse is an integer, and thus the two special right triangles (45-45-90, 30-60-90) can be scaled up nicely to fit it. If you consider 16, and look at the angles formed by the points

• (0,0), (4,0), ($\sqrt{12}, 2$)
• (0,0), (4,0), ($\sqrt8$, $\sqrt8$)
• (0,0), (4,0), (2, $\sqrt{12}$)

they are also the nice angles $\dfrac \pi6$, $\dfrac \pi4$, $\dfrac \pi3$ that were observed with using 4. These same angles can be achieved with any starting hypotenuse length, but the side lengths corresponding with the "nice" angles will be integers only when the hypotenuse lengths are even integers. Thus, your example of 6 would not work since it results in a non-integer hypotenuse length of $\sqrt6$.