> 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
>
Could this parameterization also be used to prove the following:
If a, b, and c are a PT, then 3 is a factor of a, b, or c, AND 4 is a
factor of a, b or c, AND 5 is a factor of a, b, or c?
Paul Miller
Parkersburg WV
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