Pythagorean triple ($x^2+y^2=z^2$)

A Pythagorean triple consists of three positive integers satisfying $x^2+y^2=z^2$. It is primitive when their greatest common divisor is one. Reducing their square numbers modulo $3$ shows that at least one leg is divisible by $3$; reducing modulo $4$ shows that both legs cannot be odd.

The Berggren tree is a rooted tree of primitive Pythagorean triples, with three children at every node. Fix the convention $a$ odd, $b$ even and $c>0$. Starting from the column vector $(3,4,5{)}^{\mathsf{T}}$, multiply by any ofThe first children are $(5,12,13)$, $(21,20,29)$ and $(15,8,17)$. These unimodular matrices preserve the quadratic form $a^2+b^2-c^2$, integer parity and greatest common divisor. Every primitive Pythagorean triple occurs exactly once. F. J. M. Barning's report gives this matrix-generation theorem with the even leg first.^[ref]

An infinite descent argument explains completeness. Except at the root, the unique parent isIts hypotenuse is positive and smaller than $c$, since $c<a+b\le \sqrt{2}c$. The first two unmodified entries cannot both be negative; their signs select the inverse of $A$, $B$ or $C$. The odd entry cannot vanish; the even entry vanishes only at $(3,4,5)$. Repeated descent therefore reaches that root. Multiplying all three coordinates by a positive integer supplies every nonprimitive Pythagorean triple.

A Pythagorean triple $(a,b,c)$ whose coordinates have greatest common divisor 1. Its two legs have opposite integer parity, and every Pythagorean triple is a unique positive integer multiple of a primitive Pythagorean triple up to the ordering of its legs.

The formulas $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$ give a Pythagorean triple for integers $m>n>0$, because $(m^2-n^2{)}^2+(2mn{)}^2=(m^2+n^2{)}^2$. They give a primitive Pythagorean triple exactly when $m,n$ are coprime and have opposite integer parity. Every primitive Pythagorean triple arises this way after possibly exchanging the legs; scaling gives the nonprimitive triples.

Every primitive Pythagorean triple, after exchanging its legs, has this form with $m>n>0$, $\gcd (m,n)=1$ and opposite parity. Its area is $mn(m-n)(m+n)$. The factors $m,n,m-n,m+n$ are pairwise coprime, because $m\pm n$ are odd. Conversely these conditions give a primitive triple. The parametrization follows by factoring the relatively prime integers $(c+b)/2$ and $(c-b)/2$ when $a$ is the even leg.