Everybody knows what a Pythagorean triplet is, it is the tuple (a,b,c) such that a2 + b2 = c2. It turns out that if we restrict ourselves to the set of primitive integral pythagorean triplets, every such triplet where a, b and c are coprime can be represented by integers m, n where
a = m2 - n2
b = 2mn
c = m2 + n2
Besides being of mathematical interest, this result also gives us an efficient method to enumerate integral pythagorean triplets.
Visit this link for the proof.
UPDATE: The link seems to imply that the relation holds for all a,b,c. But I think there is a flaw in the proof when the author assumes that since a2 is square, c-b and c+b must also be square. It seems to me that this is not necessarily true unless c-b and c+b are coprime. This is why I added the qualification that a, b and c must be coprime.
All feedback is welcome. :-)
Wednesday, May 28, 2008
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