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Suppose $x,y,z$ are mutually coprime positive integers that solve $x^2+y^2=z^2$.Assume that $x$ is odd and exactly one of $y$ or $z$ is even. Show that $z-y$ and $z+y$ are coprime.

So I started of by saying since $x$ is odd $x^2$ is odd $x^2=z^2-y^2=(z-y)(z+y)$ so both $z-y$ and $z+y$ are odd and then tried a few different things but got stuck. Any ideas?

Bill Dubuque
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nikoa
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    If $z-y$ and $z+y$ have a common prime factor $p$, then we have $p\mid 2z$. Since $z-y$ and $z+y$ are obviously odd , we even have $p\mid z$. This implies $p\mid y$ contradicting that $y$ and $z$ are coprime. – Peter Apr 22 '22 at 13:49
  • I didn't understand why you concluded that p|z. And also i don't think its a contradiction question – nikoa Apr 22 '22 at 13:58
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    $(1)$ If $p$ divides $z-y$ (which is odd) , $p$ must be odd. An ODD prime divisor of $2z$ must divide $z$ $(2)$ The obvious approach to show that $z-y$ and $z+y$ are coprime is to assume that they are not and derive a contradiction. – Peter Apr 22 '22 at 14:00
  • It's not necessarily an argument by contradiction. It shows that if p is a positive integer that divides both z-y and z+y then p divides both z and y. Therefore p divides gcd(z,y)=1 and we get p=1. In particular, you can take p=gcd(z-y,z+y). – Jaap Scherphuis Apr 22 '22 at 14:18

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