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Pythagorean Triples

Date: 05/22/99 at 06:04:39
From: Charlie Smith
Subject: The general formula for pythagorean triples


I am British and I have been doing some gcse coursework. The project 
is called "Beyond Pythagoras". I have already found the formula for 
the shortest, middle, and longest sides of both odd and even triples. 
I have also found the formulas for the area and perimeter of both the 
odds and evens, but I am stuck on trying to get the general formula 
for all sides of any triple.

Date: 05/22/99 at 08:10:07
From: Doctor Jerry
Subject: Re: The general formula for pythagorean triples

Hi Charlie,

The following result may be what you're looking for.  G. L. Hardy was 
also British. I don't know about Wright.

Pythagorean triples can be generated using the following theorem, 
taken from Hardy and Wright's _The Theory of Numbers_.

The most general solution of the equation 

   x^2+y^2 = z^2

satisfying the conditions

   x > 0, y > 0, z > 0
   x and y have no common factors
   x is divisible by 2


   x = 2a*b   y = a^2-b^2   z = a^2+b^2

where a and b are integers, one even and the other odd, with no common 
factors, and  a > b > 0. There is a 1-1 correspondence between 
different values of a and b and different values of x,y,z.

- Doctor Jerry, The Math Forum   

Date: 05/24/99 at 08:22:16
From: guy joseph
Subject: Pythagorean triples

Please could you send me the general formula for all Pythagorean 
triples, and if possible an explanation, some examples and a proof?

Thank you in advance.

Date: 05/24/99 at 14:32:00
From: Doctor Anthony
Subject: Re: Pythagorean triples

Pythagorean Triplets
We wish to satisfy the Diophantine equation

    x^2 + y^2 = z^2

If we let  x = a^2 - b^2

           y = 2ab

           z = a^2 + b^2

Then x^2 = a^4 - 2a^2.b^2 + b^4      y^2 = 4a^2.b^2

    x^2 + y^2 =  a^4 + 2a^2.b^2 + b^4

              = (a^2 + b^2)^2

              =  z^2

Thus choosing any pair (a,b) we generate a Pythagorean triplet.

However, always choose these such that a > b.

Example (2,1) gives  x = 4 - 1 = 3
                     y = 2x2x1 = 4
                     z = 4 + 1 = 5

Example (3,2) gives  x = 9 - 4 = 5
                     y = 2x3x2 = 12
                     z = 9 + 4 = 13

Example (3,1) gives  x = 9 - 1 = 8
                     y = 2x3x1 = 6
                     z = 9 + 1 = 10

- Doctor Anthony, The Math Forum   
Associated Topics:
High School Number Theory

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