Can every congruent number be written as $\dfrac{nm(n-m)(m+n)}{\delta}$

Question

I want to show every congruent number $\alpha $ be written as $$\alpha = \dfrac{nm(n-m)(m+n)}{\delta}$$ for some $m,n$ where $\gcd(m,n)=1$, $m\not \equiv n\mod 2 $, $m>n$ and $\delta$ a divisor of $mn$ or $(m-n)(m+n)$. I would also like to show that whenever $\delta$ is a divisor of the square part of $nm(n-m)(m+n)$ then $$\alpha = \dfrac{nm(m-n)(m+n)}{\delta}$$ is a congruent number.

Answer 1

Using Euclid's formula $\quad A=m^2-n^2\quad B=2mn\quad C=m^2+n^2\quad$ area is by definition $\quad\alpha=\dfrac{2mn(m^2-n^2)}{2} =mn(m-n)(m+n).\quad$ Most congruent numbers can be written with $\,m,n\in\mathbb{N}\quad$ e.g $\quad\alpha=6\implies(m,n)=(2,1)\,$ but some $\quad \bigg(\dfrac{20}{3}\bigg)^2 + \bigg(\dfrac{3}{2}\bigg)^2 =\bigg(\dfrac{41}{6}\bigg)^2 \implies\alpha=5\,$ are elusive: $\quad \alpha=5\implies m\ge2 \sqrt{\dfrac{5}{3}},\, n = \pm \sqrt{m^2 - \dfrac{20}{3}}.\quad$ This means that, while $\,\alpha=5\,$ is congruent number, being an integer area of a right triangle, it is not the area of any Pythagorean triple which requires integers for all sides.

Therefore, Euclid's formula and the area-derived formula you supplied cannot be used to describe all congruent numbers.