20050101, 15:18  #1 
81_{16} Posts 
Pythagorean triples
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 
20050101, 17:38  #2 
Jan 2005
62_{10} Posts 
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. 
20050101, 18:32  #3 
Jan 2005
2×31 Posts 
Correction
Ooops  all possible primitive triples.

20050102, 03:50  #4 
Jun 2003
Pa.,U.S.A.
C4_{16} Posts 
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^21)/2,(n^2+1)/2 Platonic n, any even number n,(n^2)/41.(n^2)4+1 Euclidean x,y any unequal numbers x^2y^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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