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/