All pythagorean triples appear on the rational unit circle.

Question

Let $a,b$ be positive integers and let $(x,y)$ satisfy $$x^2+y^2=1$$ $$ax-by=0$$

Prove that $(x,y)\in \mathbb{Q}^2$ iff $\exists n\in \mathbb{N}$ such that $a^2+b^2=n^2$.

Answer 1

Let $a,b$ be positive integers and let $(x,y)$ satisfy $$x^m+y^m=1$$ $$ax-by=0$$

Prove that $(x,y)\in \mathbb{Q}^2$ iff $\exists n\in \mathbb{N}$ such that $a^m+b^m=n^m$.

Proof

Multiplying $x^m+y^m=1$ by $a^m$:

$$a^mx^m+a^my^m=a^m$$ Substituting $ax=by$
$$b^my^m+a^my^m=a^m$$

$$(a^m+b^m)y^m=a^m$$

$$y^m=\frac{a^m}{a^m+b^m}$$

$$y=\frac{a}{\sqrt{a^m+b^m}}=\frac{a}{n}$$ Likewise,

$$x=\frac{b}{\sqrt{a^m+b^m}}=\frac{b}{n}$$

If there does not exist any $n$ such that $n=\sqrt{a^m+b^m}$ then we can see that $x$ and $y$ cannot be rational as the numerator is an integer and the denominator is the $m$th root of a non perfect power of $m$.

It looks to me like there is no problem generalizing in this way but this won't work in general for $$x^a+b^y=1$$

Answer 2

Proof

Multiplying $x^2+y^2=1$ by $a^2$:

$$a^2x^2+a^2y^2=a^2$$ Substituting $ax=by$
$$b^2y^2+a^2y^2=a^2$$

$$(a^2+b^2)y^2=a^2$$

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

$$y=\frac{a}{\sqrt{a^2+b^2}}=\frac{a}{n}$$ Likewise,

$$x=\frac{b}{\sqrt{a^2+b^2}}=\frac{b}{n}$$

If there does not exist any $n$ such that $n=\sqrt{a^2+b^2}$ then we can see that $x$ and $y$ cannot be rational as the numerator is an integer and the denominator is the square root of a non perfect square.