Form of rational solutions to $a^2+b^2=1$? Is there a way to determine the form of all rational solutions to the equation $a^2+b^2=1$?
 A: These are just the Pythagorean triples in disguise; think about clearing the denominators of $a,b$.
A: While the pointer to Pythagorean triples is certainly appropriate, falling back to integers rises questions about odds and evens, uniqueness, and so on. The following procedure leads right away to a one-one parametrization of these pairs $(a,b)$:
You may assume $a\geq0$, $b\geq0$. When $(a,b)\in {\mathbb Q}^2$ then the line through the point $(a,b)\in S^1$ and $(-1,0)\in S^1$ has a certain rational slope $m\in [0,1]$. This means that we obtain all such points $(a,b)$ by intersecting lines through $(-1,0)$ of rational slope $m\in [0,1]$ with $S^1$. Such a line has equation
$$y=m(x+1)\ ,$$
and intersecting this with $S^1$ we get $1=x^2+y^2=x^2+m^2(x^2+2x+1)$ or
$$(m^2+1)x^2+ 2m^2 x + m^2-1 =0\ .$$
Now one solution of this equation is $x=-1$, and the other, interesting, one is the solution of
$$\bigl((m^2+1)x^2+ 2m^2 x + m^2-1\bigr):(x+1)=(m^2+1) x + m^2-1 \ =\ 0\ .$$
This gives $x={1-m^2\over 1+m^2}$ and accordingly $y={2m\over 1+m^2}$, so that we obtain the following parametric representation of the indicated set $S$ of pairs $(a,b)$:
$$S\ =\ \left\{\Bigl({1-m^2\over 1+m^2}, {2m\over m^2+1}\Bigr)\ \bigm|\ m\in {\mathbb Q}\cap[0,1]\right\}\ .$$
A: I up-voted yunone's answer, but I notice that it begins by saying "if you know some field theory", and then gets into $N_{\mathbb{Q}(i)/\mathbb{Q}}(a_bi)$, and then talks about Galois groups, and then Hilbert's Theorem 90, and tau functions (where "$\tau$ is just the complex conjugation map in this case" (emphasis mine)).
Sigh.
I would have no hesitation about telling a class of secondary-school pupils that all Pythagorean triples are of the form
$$
(a,b,c) = (m^2-n^2,2mn,m^2+n^2)
$$
and that they are primitive precisely if $m$ and $n$ are coprime and not both odd.
That means rational points on the circle are
$$
\left(\frac a c, \frac b c\right) = \left( \frac{m^2-n^2}{m^2+n^2}, \frac{2mn}{m^2+n^2} \right)
$$
and they're in lowest terms precisely under those circumstances.
But how much of what appears in my first paragraph above would I tell secondary-school pupils?  Maybe it's better not to lose the audience before answering the question.
A: If you know some field theory, it's possible to find the form without much messy algebra. The condition that $a^2+b^2=1$ for rational $a$ and $b$ is equivalent to the fact that $N_{\mathbb{Q}(i)/\mathbb{Q}}(a+bi)=1$.
Since $\text{Gal}(\mathbb{Q}(i)/\mathbb{Q})\cong\mathbb{Z}/2\mathbb{Z}$, the classical statement of Hilbert's Theorem 90 implies that $a+bi=y/\tau(y)$ for some $y\in\mathbb{Q}(i)$, where $\tau$ is just the complex conjugation map in this case. So for some $m+ni\in\mathbb{Q}(i)$,
$$
a+bi=\frac{m+ni}{\tau(m+ni)}=\frac{m+ni}{m-ni}=\frac{(m^2-n^2)+(2mn)i}{m^2+n^2},
$$
which implies $a=\dfrac{m^2-n^2}{m^2+n^2}$ and $b=\dfrac{2mn}{m^2+n^2}$.
