# Unexpected appearance of Pythagorean triples in a homeomorphism between the 1-sphere and the extended real line

I was thinking about various kinds of mappings of prime numbers and wondered in particular what prime numbers would look like when projected from the (extended) real line to the 1-sphere by a homeomorphism linking these two spaces. When I did the calculations I was amazed to find that prime numbers are mapped to a family of Pythagorean triples on the 1-sphere! This came as a complete surprise to me but I later learned that the link between stereographic projection and Pythagorean triples is already well known. Nevertheless, in this note I want to quickly record how I stumbled on this result.

Consider the three points $N$, $Q$ and $P$ in the diagram. Since they are collinear there must be a scalar $t$ such that

$P = N + t(Q - N)$

Writing this equation in vector form we get

$(x, 0) = (0, 1) + t\big((x_1, x_2) - (0, 1) \big)$

$= (x_1 t, t(x_2 - 1) + 1)$

from which we deduce

$x = x_1 t$

$\implies t = \frac{x}{x_1}$

and

$t(x_2 - 1) + 1 = 0$

$\implies t = \frac{1}{1 - x_2}$

Equating these two expressions for $t$ we get

$\frac{x}{x_1} = \frac{1}{1 - x_2}$

$\implies x = \frac{x_1}{1 - x_2}$

The function $\pi(x_1, x_2) = \frac{x_1}{1 - x_2}$ is the homeomorphism which maps points on the 1-sphere to points on the extended real line.

I was more interested in the inverse function $\pi^{-1}(x)$ which maps points on the extended real line to the 1-sphere. To find this, observe that

$x^2 + 1 = \frac{x_1^2}{(1 - x_2)^2} + \frac{(1 - x_2)^2}{(1 - x_2)^2}$

Using $x_1^2 + x_2^2 = 1$ we have $x_1^2 = 1 - x_2^2$ so the above equation can be written as

$x^2 + 1 = \frac{1 - x_2^2}{(1 - x_2)^2} + \frac{(1 - 2x_2 + x_2^2)}{(1 - x_2)^2}$

$= \frac{2 - 2x_2}{(1 - x_2)^2} = \frac{2}{1 - x_2}$

Therefore

$x^2 + 1 = \frac{2}{1 - x_2}$

$\implies x_2 = \frac{x^2 - 1}{x^2 + 1}$

and then

$x = \frac{x_1}{1 - x_2}$

$\implies x_1 = \frac{2x}{x^2 + 1}$

Therefore if $x$ is a prime, the corresponding point on the 1-sphere is

$\bigg( \frac{2x}{x^2 + 1}, \frac{x^2 - 1}{x^2 + 1} \bigg)$

However, the numbers $2x$, $x^2 - 1$ and $x^2 + 1$ are then Pythagorean triples, as can easily be demonstrated by showing that these terms satisfy the identity

$(2x)^2 + (x^2 - 1)^2 \equiv (x^2 + 1)^2$