Pythagorean TripleDate: 8/28/96 at 17:6:55 From: Patrick Nethercot Subject: Pythagorean Triple Does anybody know the formula for finding the three lengths in a Pythagorean triple? Patrick Date: 8/30/96 at 10:58:38 From: Doctor Jerry Subject: Re: Pythagorean Triple I can give a "constructive" method of deriving the formulas a = p^2+q^2 b = p^2-q^2 c = 2pq for generating the Pythagorean triple (a,b,c) - so that a^2 = b^2+c^2 -- from integers p and q. Seek to express a and b in terms of integers x and y, so that a = x+y b = x-y To satisfy a^2=b^2+c^2, c^2=(x+y)^2-(x-y)^2=4xy. If x and y are integers, a and b will be integers. The number c will be an integer only if sqrt(pq) is an integer. So, take x = p^2 and y = q^2. Then c = 2pq. I note that c is even but it's not necessarily the shortest side. Take p = 4 and q = 3, for example. Then a = p^2+q^2 = 25 b = p^2-q^2 = 7 c = 2pq = 24. As an side note, the ancient Babylonians had a table of Pythagorean triples, generated, so it argued, by the above formulas. See Otto Neugebauer's The Exact Sciences in Antiquity, where I saw the above material for the first time. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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