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The trivial solutions of the diophantine equation $x^2+y^2=z^2$ are the following: $$x=0, y=n, z=\pm n, n \in \mathbb{Z}$$ $$x= \xi , y=0, z= \pm \xi, \xi \in \mathbb{Z}$$ $$$$ My question is, why is the $\pm$ only at $z$?? Why isn't it for example $x=0, y= \pm n, z=\pm n, n \in \mathbb{Z}$ ??

Mary Star
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Because if you choose $x=0$ and $y=n\in\mathbb{Z}$, then there are only two solutions for $z$: $z=+n$ and $z=-n$. (and of course it works similar for $y=0$ and $x=\xi$)

In other words, every $y=n$ gives you two solutions for $z$.

Kaj
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  • Could we take also $y=-n$?? At the solutions I wrote at my post, there is only $y=n$ and the sign $"-"$ is only at $z$.. Why?? – Mary Star Apr 03 '14 at 20:41
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    $n\in\mathbb{Z}$, which is both the positive integers, zero and the negative. $n$ can be negative. – Kaj Apr 03 '14 at 20:42
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    For example: Let $n=2$, then $z=2$ or $z=-2$. Let's look at another example: let $n=-2$, then $z=2$ or $z=-2$ as well. – Kaj Apr 03 '14 at 20:45
  • A ok!!! I understand!!! Thank you very much!!! – Mary Star Apr 03 '14 at 21:45