Suppose that $a^2+b^2=1$ where $a,b\in\Rats$ . $a^2+b^2=N_{\Rats(i)/\Rats}(a+bi)$ (here $\N$ is the norm), so $a^2+b^2=1$ if and only if $N_{\Rats(i)/\Rats}(a+bi)=1$ . $\Rats(i)$ is cyclic over $\Rats$ with Galois group isomorphic to $\Ints/2\Ints$ , so by Hilbert's Theorem 90, there is some element $s+ti\in\Rats(i)$ 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$$a=\frac{s^2-t^2}{s^2+t^2}, \qquad b=\frac{2st}{s^2+t^2$$

Now, given any integer right triangle $p,q,r$ with $p^2+q^2=r^2$ , we have$$\left(\frac{p}{r}\right)^2+\left(\frac{q}{r}\right)^2=$$ where $p/r, q/r\in\Rats$ , so for some $s,t\in\Rats$ ,$$\frac{p}{r}=\frac{s^2-t^2}{s^2+t^2}, \qquad \frac{q}{r}=\frac{2st}{s^2+t^2$$ 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$$\frac{p}{r}=\frac{m^2-n^2}{m^2+n^2}, \qquad \frac{q}{r}=\frac{2mn}{m^2+n^2$$ for $m,n\in\Ints$ . Thus for some $d\in\Ints$ ,$$p=d(m^2-n^2), \qquad q=2mnd, \qquad r=d(m^2+n^2$$