right, (ii) will always generate a triple, but not all of them:

Suppose $\displaystyle r=nd $ and $\displaystyle s=md $, then can you see that $\displaystyle d^2\mid x,y,z $? So we'll always have a square dividing $\displaystyle x,y,z $ but this isn't always the case in (i).

Therefore using (ii), we can't get all possible Pythagorean triples.