0
$\begingroup$

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 ...

$\endgroup$
2
$\begingroup$

"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.

$\endgroup$
  • $\begingroup$ 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."? $\endgroup$ – JustusK Nov 17 '19 at 20:14
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Nov 17 '19 at 20:31
  • $\begingroup$ The definition is clear to me. I just cannot follow the last step. What is meant with rescaling? $\endgroup$ – JustusK Nov 17 '19 at 21:20
  • $\begingroup$ 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 $z^2$. $\endgroup$ – Dietrich Burde Nov 18 '19 at 8:43
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $z \equiv 5 \pmod 6.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.