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Let $(x,y)$ be the non-hypotenuse sides of a Pythagorean triple. Does there exist another such pair $(x,a)$, where $a \ne y$?

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    $\begingroup$ From $(3,4,5)$ we obtain $(12,16,20)$ and $(9,12,15)$ $\endgroup$ – Hagen von Eitzen Jun 26 '17 at 16:17
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$(32,126)$ and $(32,24)$, where $24=4!$.

There are infinitely many these pairs such that $(x,y,z)$ and $(x,p,q)$ are Pythagorean triples

with $z=\max\{x,y,z\}$ ,$q=\max\{x,p,q\}$, $y\neq p$ and $p=a!$.

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Sure: $(9, \color{red}{12}, 15)$ and $(5, \color{red}{12}, 13)$.

If you want two triples where all three numbers have no common factors, then take $(11, \color{red}{60}, 61)$ and $(\color{red}{60}, 91, 109)$.

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$(9,12,15)$ and $(9,40,41)$ is a solution.

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