Pythagorean Quadruples (Integer Question) We have $x^2+y^2+z^2=m^2$ with $x,y,z,m$ integers, $\gcd(x,y,z)=1$, $z$ is odd, $x$ and $y$ are even.
Set $x_1:=\frac{x}{2}$ und $y_1:=\frac{y}{2}$. We get
\begin{gather*}
x_1^2+y_1^2=\frac{1}{4}(x^2+y^2)=\frac{1}{4}(m^2-z^2)=\left(\frac{m+z}{2}\right)\left(\frac{m-z}{2}\right).
\end{gather*}
Set $f:=\gcd(x_1,y_1)$, $f_1:=\gcd(f,\frac{m+z}{2})$, $f_2:=\gcd(f,\frac{m-z}{2})$.
We can proof, that $\gcd(f_1,f_2)=1$ is true.
If $d=\gcd(f_1,f_2)>1$, then we have $d\mid f$, $d\mid \frac{m+z}{2}$, $d\mid \frac{m-z}{2}$ and consequently also $d\mid \frac{m+z}{2}-\frac{m-z}{2}=z$. Moreover we get $d\mid \frac{x}{2}$, especially $d\mid x$. Analog $d\mid y$, which creates a contradiction to $\gcd(x,y,z)=1$.
Set $x_2:=\frac{x_1}{f}$, $y_2:=\frac{y_1}{f}$, $z_1:=\frac{m+z}{2f_1^2}$ and $z_2:=\frac{m-z}{2f_2^2}$.
Can someone explain to me why $z_1$ and $z_2$ are integers?
I unterstand that for example $\frac{m+z}{2f_1}$ is an integer, but I don't understand why $z_1$ is an integer.
Thanks for your help.
 A: Since $f_1|x_1,y_1$, we have
$$f_1^2\,|\,x_1^2+y_1^2\ =\ \left(\frac{m+z}2\right)\left(\frac{m-z}2\right)$$
and $f_1$ is coprime to $f_2$, hence also to $\frac{m-z}2$, so $f_1^2\,|\,\frac{m+z}2$.
A: A Pythagorean quadruple is the combination of two Pythagorean triples where the left two digits are the legs of a triple whose hypotenuse (which must be odd) matches the "odd" leg of a triple containing the right two digits. i,e $\space(3,4,5)\land(5,12,13)\longrightarrow
(x,y,z,m)=(3,4,12,13).\quad$ As such, your statement that $\space x,y\space$ are both even cannot be true and
$x^2+y^2\ne m^2-z^2.\quad$ The former would require that the number
$\space 5\space$ be even because the sum of two even numbers is even.
The even leg(s) of a triple or quadruple is/are always a multiple of $\space4\space$ so, if we let
$\space y^2+z^2=m^2-x^2=(m+x)(m-x),\space$ we find a multiple of
$\space8\space$ and, division by $\space2\space$ or $\space4\space$ still leaves a GCD that is even.
Also, the $\space x\space$ and $\space m\space$ values are always odd so their sums and their differences are always even and divisible by multiples of
$\space2\space$ that correspond to the size of the quadruple elements.
