I got help here with a formula to find Pythagorean triples given only area. Using Euclid's formula, $Area = D=m^3 n-mn^3\quad$ the final equations are:
$$n_0=2\sqrt{\frac{m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\arccos{\biggl(-\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}$$
$$n_1=2\sqrt{\frac{m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\arccos{\biggl(\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}$$
$$n_2=n_1-n_0$$
where $$\lfloor\sqrt[4]{D}\rfloor\le m\le \lceil\sqrt[3]{D}\space \rceil$$
Given an area $D$, any value of $m$ that yields an integer for one-or-more of $n_0, n_1, n_2$ provides the $m,n$ values that identify a triple that has area $D$. The limits I show are heuristic; I found them by experimentation in a spreadsheet. I would like to justify them with something more than, "they work", but I can't and I would also like to do better if possible.
Are there logical reasons for my limits? and Is there a way to narrow the search or is this as good as it gets?
Best Answer
The value $m$ needed to create a triple with a given area magnitude is higher when
$m^3(1)-m(1)^3-D=0\quad$ than when $\quad m^3 (m - 1) - m (m - 1)^3 - D = 0$
but it is smallest when $n\approx\frac{m}{2}.\quad $ The limits found by solving with $(m,m/2)$ and then with $(m,1)$ are
$$\bigg\lfloor\sqrt[4]{\frac{8P}{3}}\bigg\rfloor\le m \le \frac{\sqrt[3]{\sqrt{3} \sqrt{27 D^2 - 4} + 9 D}}{\sqrt[3]{2} \sqrt[3]{3^2}} + \frac{\sqrt[3]{\frac{2}{3}}}{\sqrt[3]{\sqrt{3} \sqrt{27 D^2 - 4} + 9 D}}$$
and the search can be limited, as indicated by one of the other posters, by using only the factors of $D$ within this range.