A parametrization of Heronian triangles

Question

Let $a,b,c$ be integers which are the sides of a triangle with integral area, a so called Heronian triangle. This website attributes to Gauss the result that there must then exist integers $m,n,p,q$ such that

$a = mn(p^2+q^2)$

$b = (mp)^2+(nq)^2$

$c = (m+n)(mp^2-nq^2)$

(where I left out a $4pq$ factor designed to make the radius of the circumscribed circle integral as well). It's not hard to see that the triangle defined by these formulas is indeed Heronian, however I could neither prove nor find a reference for the fact that this parametrization is exhaustive.

Can someone do one of these two things?

Thanks!

(Note: I'm communicating this question on behalf of my dad, who is really the person who looked into that but is not easily capable of asking it himself over here. I may be slow to respond on his behalf if questions come up).

Answer 1

Let your triangle $\triangle{ABC}$ have side lengths $a,b,c \in \mathbb{Q}$ and rational area. Assume WLOG that $c$ is the longest side and drop the altitude from $C$ with length $h\in Q$. The triangle is divided into two right triangles one with hypotenuse $a$ and legs $d,h$, and one with hypotenuse $b$ and legs $e,h$. We have $d+e=c\in \mathbb{Q}$. Also notice that $$d-e=\frac{d^2-e^2}{d+e}=\frac{a^2-b^2}{c}\in \mathbb{Q}$$ so we conclude that $d,e$ are rational. From the pythagorean triples we have relations $$a-d=r,\quad a+d=\frac{h^2}{r},\quad b-e=s,\quad b+e=\frac{h^2}{s}$$ and therefore $$a=\frac{1}{2r}(h^2+r^2),\quad b=\frac{1}{2s}(h^2+s^2),\quad c=\frac{r+s}{2rs}(rs-h^2)$$ This is exactly your parametrization up to scaling.

Answer 2

This paper by Sascha Kurz credits a parametrization much like yours (for triangles with integer sides and rational area) to the seventh-century Indian mathematician Brahmagupta. The paper also gives an algorithm for generating Heronian triangles. It doesn't provide a proof that the parametrization gives all Heronian triangles, but has a reasonable reference list which might be a good place for further searching.