proof of Pythagorean triples

Suppose that a2+b2=1 where a,b. a2+b2=N(i)/(a+bi) (here N is the norm), so a2+b2=1 if and only if N(i)/(a+bi)=1. (i) is cyclic over with Galois group isomorphicPlanetmathPlanetmathPlanetmath to /2, so by Hilbert’s Theorem 90, there is some element s+ti(i) such that


so that


Now, given any integer right triangleMathworldPlanetmath p,q,r with p2+q2=r2, we have


where p/r,q/r, so for some s,t,


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


for m,n. Thus for some d,

Title proof of Pythagorean triplesPlanetmathPlanetmath
Canonical name ProofOfPythagoreanTriples1
Date of creation 2013-03-22 17:44:34
Last modified on 2013-03-22 17:44:34
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 5
Author rm50 (10146)
Entry type Proof
Classification msc 11-00