Number Fields - Pythagorean Triples

Pythagorean Triples

We wish to solve $x^{2} + y^{2} = z^{2}$ over the integers. Without loss of generality assume $x, y, z$ have no common factor. This implies that $z$ is odd (consider the equation modulo 4).

First we move to $ℤ [i]$ :

(x + y i) (x - y i) = z^{2}

Claim: $x + y i = u a^{2}$ where $u$ is a unit and $a \in ℤ [i]$ .

Proof: Suppose a prime $p$ divides $x + y i$ . Then $p$ divides $z$ , thus $p^{2}$ must divide both sides. If $p$ also divides $x - y i$ , then it must also divide their sum $2 x$ . But $2 x$ and $z$ are coprime, implying $p$ is a unit, which is a contradiction.

Hence we may write $x + y i = u (m + n i)^{2}$ , from which we obtain all solutions

{x, y, z} = {\pm (m^{2} - n^{2}), \pm 2 m n, \pm (m^{2} + n^{2})}