Pythagorean Quadruplets

Date: 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/