When following the approach to generate the ternary tree of Pythagorean triples with Fibonacci boxes, one has the root box \begin{bmatrix}1&1\\2&3\end{bmatrix} which corresponds to the Pythagorean Triple \begin{bmatrix}3&4&5\end{bmatrix} the box then produces 3 more boxes \begin{bmatrix}2&1\\3&5\end{bmatrix} \begin{bmatrix}1&3\\4&5\end{bmatrix} \begin{bmatrix}3&1\\4&7\end{bmatrix} those seem to be right since they are the same as here. When you produce the Pythagorean triples out of the boxes they are \begin{bmatrix}12&5&13\end{bmatrix} \begin{bmatrix}8&15&7\end{bmatrix} \begin{bmatrix}24&7&25\end{bmatrix} but when you look at the image of the ternary tree one can see that two of the produced triples are in layer 2 and the third one is in layer 3.
My question now is, if this approach really finds all Pythagorean triples uniquely, like stated on the first link. And how to produce the tree, like shown on the image on the second link.
Very good Mathologer video about the ternary tree, which tells you how to generate the tree: https://www.youtube.com/watch?v=94mV7Fmbx88&t=38s