Pythagorean triples and Pell's equations We all know that Pythagorean triples are diophantine equations of order 2. So my question is can the solutions to ALL Pell's equations of the form $X^2-DY^2=1$ be derived from the Pythagorean triples? Is there an algorithm to do it?
P.S: I found an interesting paper (details below) that deals with the solubility of negative Pell's equation in terms of Primitive Pythagorean Triples. I was wondering if an algorithm be derived from it for even the normal Pell's equations.
A. Grytczuk, F. Luca and M. Wójtowicz, “The Negative Pell Equation and Pythagorean Triples"
 A: Pythagorean triples correspond to the special case of rational solutions for $D=-1$ and relate to rational points on the unit circle.
If $X^2+Y^2=1$ then $(X^2-Y^2)^2+(2XY)^2=(X^2+Y^2)^2=1$ and $X^2-Y^2$ and $2XY$ give another solution.
The Pell equation is looking for integer points on $X^2-DY^2=1$. Once one solution is found note that $(X^2+DY^2)^2-D(2XY)^2=(X^2-DY^2)^2=1$ so that $(X^2+DY^2)$ and $2XY$ make a further solution.
Note that there are finitely many integer points on the unit circle - $(0, \pm 1)$ and $(\pm 1, 0)$ while the group of rational points is not finite and not finitely generated (geometrically there are rational points at arbitrarily small angles with the real axis).
On the other hand for Pell's equation, which is looking for integer points, and with positive $D$ (negative $D$ is trivial), the group generated by the solutions is finitely generated (though an infinite abelian group: if solutions exist - we can always generate more as above). The existence of a fundamental unit from which all solutions can be derived is the interesting result in this case - and that is the productive approach.
You know, as you comment, the continued fraction methods here. If you read the literature you will see that fundamental solutions can be surprisingly large in relation to $D$. It would be a real surprise if something as simple as deriving solutions from Pythagoras would work as a general procedure - how would you account for this phenomenon?

Just to note, on the new solutions from old theme:
If $X^2-DY^2=1$ and $W^2-DZ^2=1$ then we can consider $$(X+Y\sqrt D)(W+Z\sqrt D)=(XW+DYZ)+(XZ+YW)\sqrt D$$ and we find that $$(XW+DYZ)^2-D(XZ+YW)^2=(X^2-DY^2)(W^2-DZ^2)=1$$
This works in the Pythagorean case for $D=-1$
