proof of Pythagorean triples

Suppose that $a^2+b^2=1$ where $a,b\in \mathbb{Q}$. $a^2+b^2=N_{\mathbb{Q}(i)/\mathbb{Q}}(a+bi)$ (here $\mathrm{N}$ is the norm), so $a^2+b^2=1$ if and only if $N_{\mathbb{Q}(i)/\mathbb{Q}}(a+bi)=1$. $\mathbb{Q}(i)$ is cyclic over $\mathbb{Q}$ with Galois group isomorphic to $\mathbb{Z}/2\mathbb{Z}$, so by Hilbert’s Theorem 90, there is some element $s+ti\in \mathbb{Q}(i)$ such that

$$a+bi=\frac{s+ti}{\sigma (s+ti)}=\frac{s+ti}{s-ti}=\frac{s^2-t^2+2sti}{s^2+t^2}$$

so that

$$a=\frac{s^2-t^2}{s^2+t^2},b=\frac{2st}{s^2+t^2}$$

Now, given any integer right triangle $p,q,r$ with $p^2+q^2=r^2$, we have

$${\left(\frac{p}{r}\right)}^2+{\left(\frac{q}{r}\right)}^2=1$$

where $p/r,q/r\in \mathbb{Q}$, so for some $s,t\in \mathbb{Q}$,

$$\frac{p}{r}=\frac{s^2-t^2}{s^2+t^2},\frac{q}{r}=\frac{2st}{s^2+t^2}$$

Clearing fractions on the right hand side of these equations by multiplying numerator and denominator by the square of the least common multiple of the denominators of $s,t$, we get

$$\frac{p}{r}=\frac{m^2-n^2}{m^2+n^2},\frac{q}{r}=\frac{2mn}{m^2+n^2}$$

for $m,n\in \mathbb{Z}$. Thus for some $d\in \mathbb{Z}$,

$$p=d(m^2-n^2),q=2mnd,r=d(m^2+n^2)$$