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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.

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  • $\begingroup$ Is it given that $x,y$ are even? $\endgroup$ Commented Jul 24, 2022 at 12:03
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    $\begingroup$ Yes, $x$ and $y$ are even. $\endgroup$ Commented Jul 24, 2022 at 12:03

2 Answers 2

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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 $f_1^2\,|\,\frac{m+z}2$.

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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.

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