If I have 3 Pythagorean triples $(X,R;L),(S,2R;Y),(2R,Z;T)$, do they imply the existence of three Pythagorean triples which are primitive? 
If I have $3$ Pythagorean triples $(X,R;L),(S,2R;Y),(2R,Z;T)$, do they imply the existence of three Pythagorean triples which are primitive, and also have the same 'interlinking'?

The notation I use is $(A,B;C)$ which means we have $A^2 + B^2 = C^2$. Moreover, $A,B,C$ are positive integers.
For context, this was used to solve a problem on Quora: The Problem on Quora.
I guess it boils down to the properties of double Pythagorean numbers? Would very much appreciate a rather in-depth look at the whole subject, I guess. Most of what I've found has been rather low level, for school kids. I want something at the level of a university course.
One thing that's linked on Quora as well is: Berggrens's tree. It seems like it would help, but my "math sense" is a bit faulty :P
 A: If we maintain that $A$ and $C$ are always odd, there are no primitive solutions. If, however, we permit $A$ and $B$ to trade places, one of the smallest solutions is
$$(7,24,25)\qquad (140,48,148)\qquad (280,351,449)$$
To explore these further, we can use functions derived from Euclid's formula  shown here as $F(m,k)$ in a form excerpted from a paper––there will be fewer mistakes in presentation if the form is left unchanged.
$$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2$$
To find a triple matching side-A,
$$A=m^2-k^2\implies k=\sqrt{m^2-A}\qquad\text{for}\qquad  \sqrt{A+1} \le m \le \frac{A+1}{2}$$
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$
for example
$$A=15\implies \sqrt{15+1}=4\le m \le \frac{15+1}{2} =8\quad\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\qquad \qquad F(8,7)=(15,112,113) $$
To find a triple matching side-B,
$$B=2mk\implies k=\frac{B}{2m}\qquad\text{for}\qquad \bigg\lfloor \frac{1+\sqrt{2B+1}}{2}\bigg\rfloor \le m \le \frac{B}{2}$$
The lower limit ensures $m>k$ and the upper limit ensures $m\ge 2$ for example
$$B=44\implies\qquad \bigg\lfloor \frac{1+\sqrt{88+1}}{2}\bigg\rfloor =5 \le m \le \frac{44}{2}=22\quad \land \quad m\in\{11,22\}\implies k\in\{2,1\}$$
$$F(11,2)=(117,44,125)\qquad\qquad F(22,1)=(483,44,485)$$
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
$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$
