Pythagorean QuadrupletsDate: 12/28/98 at 21:48:12 From: John Basias Subject: Pythagorean quadruplets I am trying to find a formula that would generate Pythagorean quadruplets. For example: 3, 4, 12, 13 and 1, 2, 2, 3 are quadruplets since the sum of the squares of the first three natural numbers equals the square of the last number. I think I have found a formula that generates Pythagorean quadruplets where the sum of the first and third number in the set is the last number in that set. For example: 1, 2, 2, 3; 1, 4, 8, 9; or 4, 12, 18, 22. The formula I was able to come up with is: p^2, 2pq, 2q^2, p^2+2q^2 where p,q are elements of N. This formula does not work for quadruplets such as 3, 4, 12, 13, or 8, 9, 12, 17. Is there a formula that would generate any set of quadruplets given any natural number? Please advise how I can go about deriving such a formula. Is there any research in this area that could help me? Thank you. Sincerely, John Date: 12/29/98 at 09:51:59 From: Doctor Rob Subject: Re: Pythagorean quadruplets Here is a different set of formulas that give all such answers: p^2 + q^2 - r^2 2*p*r 2*q*r p^2 + q^2 + r^2 where p, q, and r are all positive. Your formulas above are a special case, with r = q. It is worth noting that p, q, and r do not necessarily have to be rational. If: p = 1/sqrt(2) q = 3/sqrt(2) r = 2/sqrt(2) you get the quadruple 3, 2, 6, and 7, for example. They do have to all be integral multiples of the reciprocal of the square root of some fixed integer, in this case 1/sqrt(2). If you have a quadruple, you can find the values of p, q, and r by subtracting the first number from the fourth, which gives you 2*r^2. That allows you to find r, and then you can find p and q from the second and third formulas. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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