I'm sorry, I'm not that familiar with methods of generating Pythagorean Triples.
You are saying that if n is odd, then . Okay, I verified that, simple enough. Since this works for all odd there are an infinite number of triples of this form. For odd, is even and the other two values are odd. Since it is an even perfect square.
Now you have to show that there exist an infinite number of n's for which , thus making the triple primitive.
Well, since , . But since they are both odd, their . But for if , then adding 8 will certainly mean . Therefore, for all n odd,
Suppose for some odd prime . Then . So , so . So .
Ergo, for n odd, this triple is primitive.