Generating Pythagorean Triples S.T. $b = a+1$ I am looking for a method to generate Pythagorean Triples $(a,b,c)$. There are many methods listed on Wikipedia but I have a unique constraint that I can't seem to integrate into any of the listed methods.
I need to generate Pythagorean Triples $(a,b,c)$ such that:
$$a^2 + b^2 = c^2$$
$$a\lt b\lt c \,; \quad a,b,c \in \Bbb Z^+$$
$$and $$
$$b=a+1$$
Is there a way to modify one of the listed methods to include this constraint?

Actually, it's there in the special cases section as Almost-isosceles Pythagorean triple, or when the non-hypotenuse sides differ by 1. The complete formula is,
$$\Big(\frac{x-1}{2}\Big)^2+\Big(\frac{x+1}{2}\Big)^2=y^2\tag{1}$$
where $x,y$ are solutions to the Pell equation,
$$x^2-2y^2 = -1\tag{2}$$
Since from $(2)$ one can tell that $x$ is always odd, then the terms of $(1)$ are integers.
