Suppose that a2+b2=1 where a,b∈ℚ. a2+b2=Nℚ(i)/ℚ(a+bi) (here N is the norm), so a2+b2=1 if and only if Nℚ(i)/ℚ(a+bi)=1. ℚ(i) is cyclic over ℚ with Galois group isomorphic

to ℤ/2ℤ, so by Hilbert’s Theorem 90, there is some element s+ti∈ℚ(i) such that
|
a+bi=s+tiσ(s+ti)=s+tis-ti=s2-t2+2stis2+t2 |
|
so that
Now, given any integer right triangle
p,q,r with p2+q2=r2, we have
where p/r,q/r∈ℚ, so for some s,t∈ℚ,
|
pr=s2-t2s2+t2,qr=2sts2+t2 |
|
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
|
pr=m2-n2m2+n2,qr=2mnm2+n2 |
|
for m,n∈ℤ. Thus for some d∈ℤ,
|
p=d(m2-n2),q=2mnd,r=d(m2+n2) |
|