Finding an example of a Pythagorean triple, such that $\gcd(x,y,z)=1$ but $\gcd(x,z)>1$, $\gcd(x,y)>1$, and $\gcd(y,z)>1$

Question

Finding an example of a Pythagorean triple $x^2+y^2=z^2$, with the $\gcd(x,y,z)=1$ but $\gcd(x,z)>1$, $\gcd(x,y)>1$, and $\gcd(y,z)>1$.

In order for this to be the case then I need to have an $x,y,z$ such that $x$ and $y$ have a common divisor that is greater than $1$, $y$ and $z$ to have a common divisor that is greater than $1$ but also not a common divisor of $x$ and $y$, and so on.

I attempted just to try some numbers with the formula \begin{align} x &= 2st, \\ y &= t^2 - s^2 \\ z &= t^2 + s^2 \end{align} but to no avail. I've tried different methods to no avail, and I feel like the most promising method might be using the fundamental theorem of arithmetic, but I don't know how to progress from there.

Answer 1

Such an example doesn't exist.

If $x, y, z \in \mathbb Z$ are such that $x^2 + y^2 = z^2$ and if $p$ is a prime number that divides say both $x$ and $z$, then $p$ also divides $y^2 = z^2 - x^2$, hence $p$ divides $y$. So $\gcd(x, z) \neq 1 \implies \gcd(x, y, z) \neq 1$. The same implication holds for the premices $\gcd(x, y)$ and $\gcd(y, z)$ by the same kind of arguments.

Answer 2

Let $\,d\,$ be a common factor of $x$ and $y$

\begin{align*} z^2=&x^2+y^2\\ =&(dq)^2+(dr)^2\\ =&d^2(q^2+r^2)\\ =&d^2s^2=z^2\\ \implies &d^2|z^2\\ \end{align*} Then $\,d\,$ is a common factor of $z$ as well and there is no such triple as the one you seek.