Can I narrow my search range in seeking solutions for equations?

Question

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?

Answer 1

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.

Answer 2

So you are indeed trying to find integer solutions to the equation $$mn^3 - m^3n = D$$ It looks like you've flipped the sign of $n_1$, presumably because the associated root is always negative (if $m$ and $D$ are both positive, this equation will always have exactly one negative root, with the other two roots either positive or complex conjugates). Since you are only interested in positive roots, it does not matter whether $n_1$ is integer or not. The actual root $-n_1$ is negative. Thus you are actually only interested in $n_0$ and $n_2$.

For your purposes, there is a better expression for $n_2$. First, to simplify the notation, define $$A := \frac {2m}{\sqrt 3}$$ and $$\theta := \frac 13\arccos\left(-\frac {3\sqrt 3 D}{2m^4}\right)$$

then $$n_0 = A\cos (\theta)\\n_2 = A\cos\left(\theta - \frac{2\pi}3\right)$$ This way, you only have to take the arccosine of one value.

That is fairly minor. I think the biggest help for you comes from the observation that $m$ divides $D$, since $m(n^3 - m^2n) = D$. So while you have a wide range for $m$, most of the values can be quickly eliminated.

This points out that your final entry above is incorrect: $48$ does not divide $1878120$, so it cannot be a valid value of $m$.