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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.

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    $\begingroup$ Presumably you intend to parametrize $(x, y, z)$ in terms of $(s, t)$ using Euclid's formula but you seem to have a typo. $\endgroup$ Dec 27, 2022 at 23:01
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    $\begingroup$ 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 $x^2 + y^2 = z^2$? $\endgroup$ Dec 27, 2022 at 23:04
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    $\begingroup$ Note that if $\gcd(x,z)=d>1$, then $d|y$ as well. Why? $\endgroup$ Dec 27, 2022 at 23:10
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    $\begingroup$ 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 $\gcd(x,y)\gt 1$. $\endgroup$ Dec 27, 2022 at 23:12

2 Answers 2

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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.

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  • $\begingroup$ Thats a relief, the question was in my textbook so I assumed there was an awser too it $\endgroup$
    – cheesewiz
    Dec 27, 2022 at 23:16
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    $\begingroup$ Yes, Zag's answer is correct. $\endgroup$ Dec 27, 2022 at 23:16
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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.

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