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Is there a way to determine the form of all rational solutions to the equation $a^2+b^2=1$?

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Clear denominators to reduce to the problem of finding all pythagorean triples. – Arturo Magidin Nov 28 '11 at 17:38
Notice that this is the same as integer solutions to $a^2+b^2=c^2$. A full statement of the result and proof can be found on wikipedia: – Eric Naslund Nov 28 '11 at 17:38

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 $b=\dfrac{2mn}{m^2+n^2}$.

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For many more such elegant proofs see Olga Taussky's award-winning 1971 AMM paper Sums of Squares. – Bill Dubuque Nov 28 '11 at 19:07

These are just the Pythagorean triples in disguise; think about clearing the denominators of $a,b$.

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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 $(a,b)$:

$$S\ =\ \left\{\Bigl({1-m^2\over 1+m^2}, {2m\over m^2+1}\Bigr)\ \bigm|\ m\in {\mathbb Q}\cap[0,1]\right\}\ .$$

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+1 This technique shows that the rational points on any rational quadratic curve can be similarly parameterized. – hardmath Nov 28 '11 at 22:02

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 may in this case" (emphasis mine)).


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 $$ (a/c,b/c) = \left( \frac{m^2-n^2}{m^2+n^2}, \frac{2mn}{m^2+n^2} \right) $$ 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.

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Dear Michael Hardy, thank you for the upvote. I don't know if the last comment is directly pointed at me, but I posted my answer after hardmath, Arturo, and Eric had already posted links to the accessible proofs with Pythagorean triples. I always figured that as soon as a satisfactory answer at the "correct" level is posted, the audience is then the whole site/internet, not just the OP. Also, I don't see any indication that Roonie is a secondary-school pupil, or any explicit indication of his/her level of mathematical maturity. – yunone Nov 28 '11 at 19:27

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