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The following property has been stated without proof in a problem solving book (not as a problem, hence no solution). I also looked at the number theory text I have, and I cannot find it.

All pairwise prime triples of integers satisfying $x^2+y^2=z^2$ are given by $x=|u^2-v^2|\;,\;y=2uv\;,\; z=u^2+v^2\;,\;\text{gcd}(u,v)=1\;,\; u\neq v \;\text{mod} \;2$

Particularly, why $u,v$ should be coprime and $u\neq v\; \bmod \;2$ as otherwise this is trivial algebra. A reference or an intuitive explanation of this result would be appreciated.

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    I think one can find $\text{a proof of this fact}$ in $\text{ Silverman's : Rational points on elliptic curves}$ book.2011-06-29

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The reason for the $u$ and $v$ coprime and opposite parity is that you have the expression for primitive Pythagorean triples, i.e. those where $x$, $y$ and $z$ have no common factor apart from $1$. If $u$ and $v$ have a common factor $k$ then $x$, $y$ and $z$ have a common factor $k^2$, while if $u$ and $v$ are both odd then $x$, $y$ and $z$ have a common factor $2$.

To extend these to capture all Pythagorean triples, merely multiply the expressions for $x$, $y$ and $z$ terms by an arbitrary positive integer.