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:::: '''[9, 12, 15] and [15, 20, 25] can be obtained just as easily from Euclid using ''m'' = 2, ''n'' = 1, ''k'' = 3 for the first one , and ''m'' = 2, ''n'' = 1, ''k'' = 5 for the second. This is clearly explained and sourced in the Wikipedia article on [[Pythagorean triple|Pythagorean triples]], along with an explanation for the the need to introduce parameter ''k'' when generating ALL triples as opposed to just the primitives. But as Euclid well knew, it is enough to consider only the set of ''primitive'' triples (all of which are generated by the equation you cite), since ALL ''non-primitive'' solutions can be generated trivially from the primitive ones. In addition to all of the primitives, the version of the equation using only parameters ''m'' and ''n'' produces an infinite number of non-primitive triples of the form [''ak'',''bk'',''ck''] where ''k'' is an "even square (or half-square) integer". To get only the primitives, ''m'' and ''n'' must be coprime, with ''m'' > ''n'', and one of ''m'',''n'' must be even.'''
::::'''I assume you agree that Dickson's equations produce all the primitives. This is true when ''s'' and ''t'' are coprime as in line 1 below. (If they were not coprime, then we could remove common factor ''k'' as in the equation on line 2). But if instead we apply common factor ''k'' to our primitive triple(s) [ ''x, y, z''], we get the non-primitive triple(s) ['' x', y', z'''] as in line 3 where ''r', s', t' '' share a common factor k>1. '''
:::::<math>\begin{align}
& x=\text{ }(r+s),\text{ }y=(r+t),\text{ }z=(r+s+t) \\
& {x}'=(r+s)k,\text{ }{y}'=(r+t)k,\text{ }{z}'=(r+s+t)k \\
& {x}'=({r}'+{s}'),\text{ }{y}'=({r}'+{t}'),\text{ }{z}'=({r}'+{s}'+{t}') \\
\end{align}</math>
::::'''Thus given an arbitrary primitive [ ''x, y, z''] with ''s'','' t'' coprime, we get all of its multiples too. The latter have the form ['' x', y', z' '' ] where ''r', s', t''' share common factor k>1.'''
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