1

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

CIJ
  • 3,437
  • 1
  • 12
  • 19
  • 1
    All solutions in positive integers are given by $z=s^2+t^2$ and $x=s^2-t^2$, $y=2st$, or the other way around, where $s\gt t$ and $s,t$ are relatively prime of opposite parity. Wikipedia will have this, among others, in its article on Pythagorean triples. – André Nicolas Dec 21 '14 at 08:07

1 Answers1

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.

Zubin Mukerjee
  • 17,832