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Lul Thyme
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Quebec, Canada
Aug 2000 time: 05:16
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If I remember correctly there are two infinite families of irreducible integer pythagorean triple (meaning no common divisor to each...)
the
3,4,5
5,12,13
7,24,25
etc...
is one
let me check if I can find my reference
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civman2000
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of the Martian Empire
Jun 2001 time: 23:16
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quote: Originally posted by KrazyHorse
In fact, it's quite easy to show that up to a scale factor all pythagorean triplets are equivalent to a = (2n + 1), b = (2n^2 + 2n), c = (2n^2 + 2n + 1) |
This may be true (I would conjecture it is; it wouldn't take long to check but I'm feeling lazy) but this:
quote: It is the case up to a scale factor, as I've said. For all intents and purposes 3 4 5, 5 12 13, 7 24 25 etc. are the only pythagorean triplets. All others are direct multiples of these. |
...is not. Take, say, 20-21-29. To get c-b=1 (where b is the even one), you would have to divide by 9, giving 20/9-7/3-29/9. Thus it reduces to the above equations with n=2/3, but cannot be generated by your cases with n an integer.
I'm surprised that no one else here knows (or knew until snwflake went hunting) of the very simple formula for generating all of the triplets: a=u^2-v^2, b=2uv, c=u^2+v^2 for integers u>v>0. That's how I came up with my counterexample 20-21-29; I just tested values of u and v until fidning one (5 and 2) that worked.
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