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Old 2005-01-01, 15:18   #1

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Default Pythagorean triples

Could anyone explain to me what formulae are used to obtain Pythagorean triples? I've heard there're three or four of them.

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Old 2005-01-01, 17:38   #2
PBMcL's Avatar
Jan 2005

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Cool reply - Pythagorean triples

Easiest formula I know is:

Choose integers m, n with m > n > 0. Let x = m^2 - n^2, y = 2mn, and z = m^2 + n^2. Then (x, y, z) satisfies x^2 + y^2 = z^2. For primitive triples (x, y, and z having no common factor), add the conditions gcd(m, n) = 1 and m != n mod 2. I believe all possible triples can be generated this way, but i'm going from memory here.
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Old 2005-01-01, 18:32   #3
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Jan 2005

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Red face Correction

Ooops - all possible primitive triples.
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Old 2005-01-02, 03:50   #4
David John Hill Jr
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Jun 2003

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Default one way to see 3 methods.

If I may refer to Heilbron's 'geometry civilized' p151,
as one of the cleanest summaries of pythagorean triplets I've run across.

Set: Generation Triplet
Pythagorean n, any odd number n,n^2-1)/2,(n^2+1)/2
Platonic n, any even number n,(n^2)/4-1.(n^2)4+1
Euclidean x,y any unequal numbers x^2-y^2,2xy,x^2+y^2

I personally found the odd case is , with some work, obvious as simply building squares on a plane. Rather enlightening an approach and exercise.
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