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

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.

• Presumably you intend to parametrize $(x, y, z)$ in terms of $(s, t)$ using Euclid's formula but you seem to have a typo. Dec 27, 2022 at 23:01
• Also, to clarify, what roles do $x$, $y$, and $z$ play in the Pythagorean equation? Can you edit to make this explicit? Is it meant to by $x^2 + y^2 = z^2$? Dec 27, 2022 at 23:04
• Note that if $\gcd(x,z)=d>1$, then $d|y$ as well. Why? Dec 27, 2022 at 23:10
• What you want is impossible. If $p$ is a prime that divides both $x$ and $y$, then it must divide $x^2+y^2=z^2$, hence $z$; if it divides both $x$ and $z$, then it must divide $y$, If it divides both $y$ and $z$, then it must divide $x$. You cannot have $\gcd(x,y,z)=1$ and also have $\gcd(x,y)\gt 1$. Dec 27, 2022 at 23:12

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.
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.