Finding Pythagorean Triplets Algebraically
Date: 04/13/2002 at 16:53:37 From: Krista Subject: Pythagorean triplet of type I Call the triplet of the natural numbers (a, b, h) a Pythagorean triplet of type I if a^2 + b^2 = h^2 and h = b + 1. Thus (3, 4, 5) is of type I, while (6, 8, 10) is not. Then the number of Pythagorean triplets of type I with a+b+h < 200 is equal to what? The only way I could figure out this question was by looking up a chart of Pythagorean triplets. Is there a way to find these triplets algebraically? Thanks.
Date: 04/13/2002 at 18:35:59 From: Doctor Jubal Subject: Re: Pythagorean triplet of type I ?? Hi Krista, Thanks for writing to Dr. Math. There is absolutely an easier way. Since h = b+1, we can write a^2 + b^2 = (b+1)^2 a^2 + b^2 = b^2 + 2b + 1 a^2 = 2b + 1 That is to say, a^2 is an odd number, so a is also an odd number. Rewriting this in terms of b gives b = (1/2)(a^2 - 1) So you can systematically generate the Pythagorean triplets of type I by taking an odd number for a, calculating b as shown above, and then finding h = b+1. The first few are a b h ---------- 3 4 5 5 12 13 7 24 25 ... et cetera, and you could get carried away with yourself and find triples like (17, 144, 145) or (31, 480, 481), but these are obviously larger than a+b+h = 200. You might notice that I skipped a = 1. I could pick a = 1 and I'd calculate b = 0, h = 1, which does satisfy the Pythagorean formula a^2 + b^2 = h^2, but really isn't a Pythagorean triple because you can't have a triangle with zero as a side length. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/
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