Finding solutions of $z^2=x^2+y^2$ where $\gcd(z, y) =1$

Question

Is there an easy way to find the solutions of $$z^2=x^2+y^2$$ where $\gcd(z,y)=1$?

I apologize if this is a duplicate

Answer 1

These triples $(x,y,z)$ are called "primitive Pythagorean triples" because they are the integer side lengths of right triangles, and they have no common factors (note that $\gcd(z,y) = 1$ implies $\gcd(z,x)=1$ ... why?).

The triples have a general form that André Nicolas mentioned in a comment. This Wikipedia article on some structure that these triples have is very well-written for learning about the subject for the first time, unlike most Wikipedia articles on math topics.