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.

  • $\begingroup$ Could you explain what a Fibonacci box is and how the first box creates three more boxes? I think this is not standard. $\endgroup$ – Thomas Aug 26 '15 at 7:32
  • $\begingroup$ I did it like stated here en.wikipedia.org/wiki/… placing the numbers on the right on different places and fill the rest of the matrix so that the properties are fulfilled. $\endgroup$ – sch0rschi Aug 26 '15 at 7:48
  • $\begingroup$ Did you check out the arXiv paper provided as a reference in the Wikipedia article? If you haven't, you should. If you have, you should point out which part of it you don't understand or disagree with; otherwise your question is basically a request to reproduce that paper here. $\endgroup$ – joriki Aug 26 '15 at 9:08

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