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 \(\mathbb{Z}[i]\):

\[(x+y i)(x-y i) = z^2\]

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

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 \(2x\). But \(2x\) 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

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

Ben Lynn 💡