<s>[INST] Proof of Pythagorean quadruple Given $x^2 + y^2 + z^2 = m^2$ with integers $x, y, z$ and $m > 0$, z odd and $(x,y,z)=1$.
Set $x_{1}=\frac{1}{2}x$ and $y_{1}=\frac{1}{2}y$.
Then $x_{1}^2+y_{1}^2=\frac{m+z}{2}\frac{m-z}{2}$.
Set $f=(x_{1},y_{1})$, $f_{1}=(f,\frac{1}{2}(m+z))$, $f_{2}=(f,\frac{1}{2}(m-z))$.
The prof says, that with a simple argument you can see, that $(f_{1},f_{2})=1$.
Can you help me with this argument? I tried to understand it, but I failed.
Thanks for your help!
 [/INST] Suppose, for a contradiction, that $(f_{1}, f_{2}) = d > 1.$ Since $f_{1} = (f, \frac{1}{2}(m+z))$, this implies that $d$ divides both $f$ and $\frac{1}{2}(m+z).$ Similarly, by the definition of $f_{2},$ it follows that $d$ divides $\frac{1}{2}(m-z).$
Since $d$ divides $\frac{1}{2}(m+z)$ and $\frac{1}{2}(m-z),$ $d$ divides $z = \frac{1}{2}(m+z) - \frac{1}{2}(m-z).$
Then, since $d$ divides $f = (x_{1}, y_{1}),$ it follows that $d$ divides $x_{1} = \frac{1}{2}x.$ So, $d$ must also divide $x.$ By the same argument, $d$ divides $y.$
We see that $d > 1$ divides $x, y$ and $z,$ contradicting the assumption that $(x, y, z) = 1.$
So, our initial assumption was wrong, and $(f_{1}, f_{2})$ must be $1.$
</s>