$3$ primitive pythagorean triples from 6 integers.

Question

Do there exist $3$ different primitive Pythagorean triples $(a,d,w), (a,b,z)$ and $(c,d,z)$?

Explicitly, we want $6$ different integers $a,b,c,d,w,z$ such that...

(1) $a^2 + d^2 = w^2$

(2) $a^2 + b^2 = c^2 + d^2 = z^2$

Answer 1

Unless I've made a mistake with my arithmetic these 6 will do...

a = 7015 b = 46248 c = 46535 d = 4752 w = 8473 z = 46777

Answer 2

This is surprisingly simple given a function derived from Euclid's formula: $$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2$$

To find a triple matching side-C, $$C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor$$ The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$, for example $$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$ $$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$

$$33^2+56^2=63^2+16^2=65^2$$

The larger the value of $C$ the greater the chances there are more solutions, for example $C=1105$:

$f(24,23)=(47,1104,1105)\\ f(31,12)=(817,744,1105)\\ f(32,9)=(943,576,1105)\\ f(33,4)=(1073,264,1105)$