# Diophantine problems, pythagorean equation

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 ...

"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.

• Thanks for replying Dietrich Burde. Can you explain again what is meant with the last sentence: "If we start with a primitive Pythagorean triple, and pass through the rational solution then we obtain back the triple (𝑧𝑥,𝑧𝑦,𝑧2), which we have to rescale to obtain the bijection."? – JustusK Nov 17 '19 at 20:14
• Triples with $\operatorname{gcd}(x,y,z)=1$ are called primitive. You have asked for a bijection of primitive triples, not just of triples. The proof gives both. – Dietrich Burde Nov 17 '19 at 20:31
• The definition is clear to me. I just cannot follow the last step. What is meant with rescaling? – JustusK Nov 17 '19 at 21:20
• Rescaling means to rescale $(zx,zy,z^2)$ to $(x,y,z)$. In terms of equations, $(zx)^2+(zy)^2=(z^2)^2$ is rescaled to $x^2+y^2=z^2$ by dividing out $z^2$. – Dietrich Burde Nov 18 '19 at 8:43

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

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 $$z \equiv 5 \pmod 6.$$