From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: pythagorean triples
Date: 22 Feb 1996 17:20:21 GMT

One person writes: 
>Is there a listing of the first 100 pythagorean triples on the internet, 

Another writes:
>>gET READY- HERE THEY ARE.
>Sorted according to x
>
>u"	v"	x"	y"	z"	x^2	y^2	z^2
>	   (v^2-u^2) 2uv  (u^2+v^2)			
>					
>1	2	3	4	5	9	16	25
>2	3	5	12	13	25	144	169
>3	4	7	24	25	49	576	625
etc. (Way more than 100 posted).

1. You missed some.
2. The Bandwidth Police will be contacting you shortly. It would be more
compact and more informative to say how you computed these.

For the record: if (x,y,z) is a positive Pythagorean triple, _and_ x, y, and z
are relatively prime, _and_  y is even (note: x and y can't both be odd)
then there exist (relatively prime) integers  u  and  v  (of different parity,
with  v > u )  such that
	x=(v^2-u^2),  y=2uv,  z=(u^2+v^2)
Conversely, all such triples satisfy x^2+y^2=z^2, x,y,z > 0, y even, and
x, y, z  relatively prime.

In view of the conditions of the theorem, it appears you missed
triples such as  (4,3,5)  (obtained by inversion), and (6,8,10)  (obtained
by scaling). We may certainly agree that these are 'trivial', but
my guess is that the original poster is not at a level to know
about such modifications.

So perhaps it is better to point out that every triple (x,y,z) will be
of one of these forms:
	( abd, (a+b)(a-b)d/2, (a^2+b^2)d/2 )
	( 2uvd, (v+u)(v-u)d, (v^2+u^2)d )
for some integers  u,v,d,  or  a,b,d with  a  and  b  both odd.
(Here I have set  a=(v+u), b=(v-u). ) Thus you can sort the triples by  x
simply by considering for each  x  all its possible factorizations.
There will be many repeats unless you are careful to take  gcd(u,v)=1
and so on.

dave

==============================================================================

Date: Mon, 26 Feb 1996 12:10:42 -0500 (EST)
From: [Permission pending]
To: rusin@vesuvius.math.niu.edu
Subject: Re: pythagorean triples
Newsgroups: sci.math

Hi Dave.  I thought you might like to know another *easy* way to generate
all pythagorean triples, with no repeats in the list.

Theorem:  x^2 + y^2 = z^2  iff  k^2 = 2mn where m=z-y, n=z-x, k=x+y-z.
Proof: Note x = k + m,  y = k + n, and z = k + m + n.  Substituting these in
for x, y, and z gives (k+m)^2 + (k+n)^2 = (k+m+n)^2 which can be simplified
to k^2 = 2mn.  Every step is reversible.

How to generate all pythagorean triples: Take k to be even, then find 
every way to choose m and n so that k^2 = 2mn.  (Note that if k is odd, then
k^2 will be odd, and the left hand side is clearly even.)
Let x = k+m, y = k+n, z = k+m+n.  Here's a chart:

k	m	n 	x	y	z
2	1	2	3	4	5
4	1	8	5	12	13		
4	2	4	6	8	10
6	1	18	7	24	25

==============================================================================

Date: Mon, 26 Feb 96 12:02:23 CST
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: pythagorean triples

This is nice. I was trying to match the previous poster's intent to
sort the triples by  x, which is part of why what I suggested was more
cumbersome. Also note that if you consider  (x,y,z,)  and (y,x,z)  to
be distinct, then the examples you sent are incomplete -- it's _not_
sufficient to take  m <=n.
Thanks for responding.
dave