rational points on one dimensional sphere

Let

$${𝕊}^1=\{(x,y)\in {ℝ}^2|x^2+y^2=1\}$$

be a one dimensional sphere.

We will denote by

$${𝕊}_{\mathbb{Q}}^1=\{(x,y)\in {\mathbb{Q}}^2|(x,y)\in {𝕊}^1\}$$

the rational sphere. We shall try to describe ${𝕊}_{\mathbb{Q}}^1$ in terms of Pythagorean triplets.

Theorem. Let $(x,y)\in {ℝ}^2$. Then $(x,y)\in {𝕊}_{\mathbb{Q}}^1$ if and only if there exists a Pythagorean triplet $a,b,c\in \mathbb{Z}$ (i.e. $|a|,|b|,|c|\in \mathbb{N}$ is a Pythagorean triplet) such that $x,y$ are of the form $\frac{a}{c}$ and $\frac{b}{c}$.

Proof. ,,$\Leftarrow$” If (for example) $x=\frac{a}{c}$ and $y=\frac{b}{c}$ for a Pythagorean triplet $a,b,c\in \mathbb{Z}$, then we have

$$x^2+y^2=\frac{a^2}{c^2}+\frac{b^2}{c^2}=\frac{a^2+b^2}{c^2}=\frac{c^2}{c^2}=1$$

and thus $(x,y)\in {𝕊}_{\mathbb{Q}}^1$.

,,$\Rightarrow$” Assume that $(x,y)\in {𝕊}_{\mathbb{Q}}^1$. Then $x=\frac{p}{q}$ for some $p,q\in \mathbb{Z}$. It follows, that

$$1=x^2+y^2=\frac{p^2}{q^2}+y^2$$

and this is if and only if

$$y=\frac{\sqrt{q^2-p^2}}{q}$$

(up to a sign of course). Therefore $y\in \mathbb{Q}$ if and only if $\sqrt{q^2-p^2}=n$ is an integer. In this case we have

$$x=\frac{p}{q},y=\frac{n}{q}.$$

Note, that

$$q^2-p^2=n^2$$

and thus

$$q^2=n^2+p^2,$$

so $n,p,q\in \mathbb{Z}$ is a Pythagorean triplet, which completes the proof. $\mathrm{\square }$

Corollary. The rational sphere ${𝕊}_{\mathbb{Q}}^1$ is an infinite set.

Proof. Let $m,n\in \mathbb{N}$ be natural numbers such that $n$ is fixed and even. Let $m$ run through primes. Then (due to the theorem in parent entry)

$$2mn,m^2-n^2,m^2+n^2\in \mathbb{N}$$

is a Pythagorean triplet.Let

$$x_m=\frac{2mn}{m^2+n^2}=\frac{2n}{m+\frac{n^2}{m}}.$$

Ir follows from the theorem, that there exists $y_m\in \mathbb{Q}$ such that $(x_m,y_m)\in {𝕊}_{\mathbb{Q}}^1$. It is easy to see, that $x_m=x_{m^{\prime }}$ if and only if $m=m^{\prime }$ and thus we generated infinitely many rational points on sphere. This completes the proof. $\mathrm{\square }$