Pythagoras using Hilbert 90

Dummit and Foote has a fun way of classifying Pythagorean triples. It is a classic fact that if \((a, b, c)\) is a reduced Pythagorean triple then there exists integers \(m, n\) such that

\[ \begin{align*} a &= m^2 - n^2, \\ b &= 2mn, \\ c &= m^2 + n^2. \end{align*} \]

To prove this using Hilbert’s 90, first we change the problem to rational numbers i.e. let \(a, b\in \mathbb{Q}\) be such that \(a^2 + b^2 = 1\). We want to show that there exists \(m, n \in \mathbb{Q}\) such that \[ \begin{align*} a &= \frac{m^2 - n^2}{m^2 + n^2}, \\ b &= \frac{2mn}{m^2 + n^2}. \end{align*} \]

Consider the cyclic extensions \(\mathbb{Q}(i)/\mathbb{Q}\). The norm map is given by \(N(a + bi) = a^2 + b^2\). So, \(a ^ 2 + b ^ 2 = 1\) implies that \(N(a + bi) = 1\). The Galois group \(\mathbb{Z}/2\mathbb{Z}\) is generated by complex conjugation \(\sigma : i \mapsto -i\). So by Hilbert’s 90, \[ a + bi = \frac{m + ni}{\sigma(m + ni)} = \frac{m + ni}{m - ni} \] for some \(m, n \in \mathbb{Q}\). We can clear out the denominators to assume that \(m, n \in \mathbb{Z}\) so that \[ a + bi = \frac{m + ni}{m - ni} = \frac{m^2 - n^2 + 2mni}{m^2 + n^2} = \frac{m^2 - n^2}{m^2 + n^2} + \frac{2mn}{m^2 + n^2}i \] as desired.

Now my math history isn’t the best, but I have a feeling that this isn’t how Pythagoras originally proved his theorem.