Diophantine problems, pythagorean equation

Question

It can be shown that that for $x,y,z \in \mathbb{N}_{>0}$ the Pythagorean equation $x^2 + y^2 = z^2$ has the general solution $x = 2grs$,$y = g(r^2 - s^2)$ and $z = g(r^2+s^2)$ with $g>0, r>s>0$ and $(r,s) = 1$ and $r,s$ not both odd.

I am trying to prove now that the integer solutions to $x^2+y^2=z^2$ with $(x,y,z) =1$ are in $1$-to-$1$ correspondence with the rational solutions $u,v$ to $u^2+v^2 = 1$. I cannot really see where to start from here ...

Answer 1

"To start from here", consider a solution of $x^2+y^2=z^2$. If $z=0$, then $x=y=0$, which is not in positive integers. Hence we can divide by $z^2$ and obtain $$ \left(\frac{x}{z}\right)^2+\left(\frac{y}{z}\right)^2=1. $$ This gives $u^2+v^2=1$. Conversely, suppose that there are $r,s\in \Bbb Q$ with $r^2+s^2=1$. Then writing $r=x/z$ and $s=y/w$ we have $$ \left(\frac{x}{z}\right)^2+\left(\frac{y}{w}\right)^2=1. $$ Multiplying with $(zw)^2$ we obtain $$ (wx)^2+(yz)^2=(zw)^2. $$ Hence $(wx,yz,zw)$ is a Pythagorean triple.

If we start with a primitive Pythagorean triple, and pass through the rational solution then we obtain back the triple $(zx,zy,z^2)$, which we have to rescale to obtain the bijection.

Answer 2

Hint: If $x^2+y^2=z^2$ where $x,y,z$ are integers and $z\ne0$, then $\dfrac{x^2}{z^2}+\dfrac{y^2}{z^2}=\dfrac{z^2}{z^2}=1$.

Answer 3

Part of this depends on something called class number, in that, given a rational solution to $a^2 + 6 b^2 = 1,$ clearing the denominators gives rise to more than one possibility. One family of primitive integer solutions to $x^2 + 6 y^2 = z^2$ is (using absolute values to save space), and might as well demand $u,v \geq 0 \; : \;$ $$ x = | u^2 - 6 v^2 |, \; \; y = 2uv, \; \; z = u^2 + 6 v^2 \; \; , $$ when $u$ is not divisible by $2$ or $3,$ along with $\gcd(u,v)=1.$ We get a second family $$ x = | 2u^2 - 3 v^2 |, \; \; y = 2uv, \; \; z = 2u^2 + 3 v^2 \; \; , $$ when $u$ is not divisible by $3,$ then $v$ is not divisible by $2,$ along with $\gcd(u,v)=1.$

Try a few small $(u,v)$ pairs. The first family has $z \equiv 1 \pmod 6,$ when $u \neq 0 \; . \; $ The second family, when $u,v \neq 0, $ has $z \equiv 5 \pmod 6.$