Every Pythagorean triple is a multiple of a *primitive* Pythagorean triple,
where a PPT is a Pythagorean triple (a,b,c) with GCD(a,b,c) = 1 (Proof: if
there is a GCD, factor it out of each; the resultant numbers are still a
Pythagorean triple, and are relatively prime). Note that if any two of the
triple have a common factor, the third must also have that as a factor.
The PPTs [since at least one of a and b has to be even, choose it to be b]
can be parametrized by the following:
a = m^2 - n^2,
b = 2mn,
c = m^2 + n^2
where m and n are relatively prime integers of opposite parity (if m is
even, n is odd, and vice-versa). That is, every PPT is of this form, and
every a,b,c of this form is a PPT. Since there are an infinite number of
choices for m and n, there are an infinite number of PPTs. So the infinity
of Pythagorean triples is not due to the existence of multiples of PPTs,
but to the infinity of PPTs.
Note also that not every Pythagorean triple can be parametrized in the form
above, since (9,12,15) is a Pythagorean triple (= 3 times the PPT (3,4,5)),
but 15 cannot be written as m^2+n^2 for any m,n. Thus, only PPTs can be
parametrized in this way.
mark snyder
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