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.