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#1 |
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1,553 Posts |
Hi.
Could anyone explain to me what formulae are used to obtain Pythagorean triples? I've heard there're three or four of them. Thanks |
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#2 |
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Jan 2005
2×31 Posts |
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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#3 |
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Jan 2005
2×31 Posts |
Ooops - all possible primitive triples.
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#4 |
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Jun 2003
Pa.,U.S.A.
110001002 Posts |
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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