Pythagorean Triple

Date: 8/28/96 at 17:6:55
From: Patrick Nethercot
Subject: Pythagorean Triple

Does anybody know the formula for finding the three lengths in a 
Pythagorean triple?

Patrick 

Date: 8/30/96 at 10:58:38
From: Doctor Jerry
Subject: Re: Pythagorean Triple

I can give a "constructive" method of deriving the formulas

a = p^2+q^2
b = p^2-q^2
c = 2pq

for generating the Pythagorean triple (a,b,c) - so that 
a^2 = b^2+c^2 -- from integers p and q.

Seek to express a and b in terms of integers x and y, so that

a = x+y
b = x-y

To satisfy a^2=b^2+c^2, 

c^2=(x+y)^2-(x-y)^2=4xy.

If x and y are integers, a and b will be integers.  The number c will 
be an integer only if sqrt(pq) is an integer. So, take x = p^2 and 
y = q^2.  Then  c = 2pq.

I note that c is even but it's not necessarily the shortest side.  
Take p = 4 and q = 3, for example.  Then 

a = p^2+q^2 = 25
b = p^2-q^2 = 7
c = 2pq = 24.

As an side note, the ancient Babylonians had a table of Pythagorean 
triples, generated, so it argued, by the above formulas.  See Otto 
Neugebauer's The Exact Sciences in Antiquity, where I saw the above 
material for the first time.

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/