A Pythagorean triple is given by $(x,y,z)=(p(m^2-n^2),p(2mn),p(m^2+n^2))$. Is there a way to show that $m=m'$, $n=n'$ and $p=p'$ or that there's possibly a counterexample where this isn't the case?
2026-03-26 17:38:23.1774546703
If $m,n,p$ and $m',n',p'$ produce the same Pythagorean triple, does the following have to hold? $m=m'$, $n=n'$ and $p=p'$.
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It is not Always valid : take for example any $p,m,n$ with $ p\bmod 4\equiv 0$ and then take $p' = p/4\:,m' = 2m\:, n' = 2n$