It is also true that the automorphism group of the quadratic form is known. See Is there a topograph for Pythagorean triples? and the three matrices. If you have any particular $x^2 + y^2 - z^2 = n,$ write $(x,y,z)$ as a column vector. Multiply by any of the three square matrices or its inverse and you get another solution for $n.$ Multiply again you get another, and so on for any combination of group elements.

I see, for your ordering you need to switch first and last elements to use these three matrices.

It is also true that the automorphism group of the quadratic form is known. See Is there a topograph for Pythagorean triples? and the three matrices. If you have any particular $x^2 + y^2 - z^2 = n,$ write $(x,y,z)$ as a column vector. Multiply by any of the three square matrices or its inverse and you get another solution for $n.$ Multiply again you get another, and so on for any combination of group elements.

I see, for your ordering you need to switch first and last elements to use these three matrices.

It is also true that the automorphism group of the quadratic form is known. See Is there a topograph for Pythagorean triples? and the three matrices. If you have any particular $x^2 + y^2 - z^2 = n,$ write $(x,y,z)$ as a column vector. Multiply by any of the three square matrices or its inverse and you get another solution for $n.$ Multiply again you get another, and so on for any combination of group elements.

It is also true that the automorphism group of the quadratic form is known. See Is there a topograph for Pythagorean triples? and the three matrices. If you have any particular $x^2 + y^2 - z^2 = n,$ write $(x,y,z)$ as a column vector. Multiply by any of the three square matrices or its inverse and you get another solution for $n.$ Multiply again you get another, and so on for any combination of group elements.

I see, for your ordering you need to switch first and last elements to use these three matrices.