Date: 18 May 1995 19:31:20 -0400 From: Roger Gillies Subject: equation for pythagorean triples Dear Dr. Math: Myself and another teacher have been tearing out our hair over this: There is a relation between Pythagorean triples we have been working with. It is as follows: let n, x and x+1 represent the numbers of a pythagorean triple where n is always an odd number and x is always an even number. The first triple is: 3, 4 and 5 The next is: 5, 12 and 13 Followed by: 7, 24 and 25 and so on......... For each triple, the first odd digit (n) is multiplied by its order in the sequence (7's order is 3rd, so 7x3), then the number 3 is added to the product, giving x. The question is this: How can this relationship be expressed as a formula? Hope you can help, and thanks in advance for trying. Sincerely, Roger Gillies and John Reid
Date: 20 May 1995 10:19:18 -0400 From: Dr. Ken Subject: Re: equation for pythagorean triples Hello there! Let me first introduce (you may have seen this before) the formula for Pythagorean triples: a = (m^2 - n^2), b = 2mn, c = (m^2 + n^2). You can verify that these three expressions will always form a P-triple for any positive m and n, and in fact I believe that all P-triples are expressible in this form. This will be a handy tool. In your case, you say that b is one less than c. Then we have 1 + 2mn = m^2 + n^2 1 = m^2 - 2mn + n^2 1 = (m - n)^2 Assuming that m is bigger than n (which it is by examining our first term m^2 - n^2, which is positive), this gives us m = n+1. Let's plug that back into our original formulae: a = (n+1)^2 - n^2 = 2n + 1 b = 2(n+1)n = 2n^2 + 2n c = (n+1)^2 + n^2 = 2n^2 + 2n + 1 Now try plugging in n=1,2,3,... into these formulae. This is really neat stuff, thanks for writing us about it!! -K
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