On the solutions of $x^2+y^2=z^2$

Question

Let $x,y,z \in \mathbb{N}$ where x is even; x,y are relatively prime and $x^{2}+y^{2}=z^{2}$. It will be tried to show that there exist $u,v \in \mathbb{N}$ relatively prime and $u> v$ and

$x=2uv, y=u^{2}-v^{2}, z=u^{2}+v^{2}$

We look at : $x^{2} = z^{2}-y^{2} = (z-y)(z+y)$ so we get the conditions:

$x= \sqrt{z-y)(z+y)} ; y= \sqrt{z^{2}-x^{2}} ; z=\sqrt{(x+y)(x+y)} $

then plug one of them into the conditions for u,v :

$x = 2uv ; \sqrt{(z-x)}\sqrt{(z+x)} = u^{2}-v^{2}; z=u^{2}+v^{2} $

Original idea was to plug x into

Now there are 4 variables and 3 equations! So this seems to be wrong.

Does anybody see the right way.

Tell me. Please.

Answer 1

The easiest way to see this is to know that unique factorization exists in $\mathbb Z[i]$. Then, if $x$ an $y$ are relatively prime, and $x$ is even, you can show that $x+yi$ and $x-yi$ must be relatively prime in $\mathbb Z[i]$. But $(x+yi)(x-yi)=z^2$ is a perfect square, so, since the factors are relatively prime, $x+yi=e(u+vi)^2$ where $e$ is a unit (so $e\in \{ 1,-1,i,-i \}$.)

But $(u+vi)^2 = (u^2-v^2) + 2uv i$, so, since $y$ is necessarily odd, $e$ must be $i$ or $-i$, and you get that $x+yi = \pm 2uv \mp (u^2-v^2)i$

In particular, if $x,y>0$, we can see that $x=2|u||v|$ and $y=|u|^2-|v|^2$ or $|v|^2-|u|^2$.

Answer 2

It's not true. In the first place, if $x$ is odd, then $x\ne2uv$. But even if you fix that, the assertion the $u$ and $v$ are relatively prime implies that $x$, $y$, and $z$ are (nearly) relatively prime, which doesn't have to be the case.

Answer 3

It is trivial to show that, with those expressions for u and v, that $u^2,v^2 \in \mathbb{N}$ as you are simply choosing $u^2$ to be the value halfway between $b^2$ and $c^2$. Also note that $a/2 = uv \in \mathbb{N}$, because a is even by assumption. $u^2$ and $v^2$ are coprime because otherwise (a,b,c) is not a primitive triple. Now, if $u=d\sqrt{e}$ and $v=f\sqrt{g}$ then the only way for $a=2uv$ to be an integer is if e = g = 1, since otherwise $u^2$ and $v^2$ would have a common factor. Therefore, we see that $u,v\in\mathbb{N}$.

Funny that I just spent time figuring this out last week or so in an effort to understand Fermat's descent proof for $n=4$. I'm interested if someone can simplify it.

Answer 4

We know that $x,y,z$ can not be all even if $x$ is even and $x,y$ are relatively prime. If x is even, then z can not be even. now with $\frac{1}{2}(z-y)(z+y)=(\frac{x}{2})^{2}$. Assume the factors are not relative primes, and there is a number p dividing them. Then p divides y and z. Then the factors also have to be squares due to prime factorisation, of the form : $\frac{1}{2}(z+y)=u^{2}$ and $(\frac{1}{2}(z-y)=v^{2}$ for $u,v \in \mathbb{N}$. These both factors are relatively prime so u and v must also be relatively prime. The equations can be solved into the conditions: $x=2uv , y=u^{2}-v^{2}, z=u^{2}+v^{2}$

For $y \in \mathbb{N}$, it also holds that with $\frac{1}{2}(z+y)=u^{2}, \frac{1}{2}(z-y)=v^{2}$ $u>v$.

Is this proof correct?