Primitive Triples

Date: 04/24/2003 at 16:21:30
From: Stan
Subject: Primitive Triples

In a primitive Pythagorean triple (x, y, z), why must x+y and x-y be
congruent to either 1 or 7 mod 8 ?

x+y and x-y provide odd answers since one is even and one is odd, so 
that eliminates some of the possible values mod 8. But I'm still left 
with 1,3,5,7 mod 8. I feel that I could also eliminate 3 mod 8 since 
one of x or y must either be three or a multiple of three.

Date: 04/25/2003 at 04:29:49
From: Doctor Jacques
Subject: Re: Primitive Triples

Hi Stan,

First a word of caution: when you write:

>I feel that I could also eliminate 3 mod 8 since one of x or y must 
>either be three or a multiple of three.

This is not true: you cannot mix congruences with different moduli 
like that. If, for example, x = 12, x is a multiple of 3, but we do 
not have x = 3 (mod 8).

Let (x, y, z) be a primitive Pythagorean triple. As you know, exactly 
one of (x, y) is even.

Assume first that y is even. We can write:

  z^2 = x^2 + y^2
      = x^2 - y^2 + 2y^2

The two key facts that we will use are:

* If a is odd, a^2 = 1 (mod 8)
* If a is even, a^2 = 0 (mod 4), and 2a^2 = 0 (mod 8)

In this case, this shows that:

* As z is odd, z^2 = 1 (mod 8)
* As y is even, 2y^2 = 0 (mod 8)

We can write:

  x^2 - y^2 = (x + y)(x - y) = 1  (mod 8)

As x + y and x - y are both odd, it is easy to see that we always 
have:

  x + y = x - y (mod 8)

If x is even instead, we can use the same argument to show that

  y + x = y - x  (mod 8)

or

  x + y = -(x - y) (mod 8)

which means that we have, in all cases:

(*)  x + y = (+/-) (x - y)  (mod 8)

Now, it is known that all the primitive Pythagorean triples can be 
expressed in the form:

  x = r^2 - s^2
  y = 2rs

where r and s are relatively prime, and exactly one of them is even. 
You can see the proof of this in the Dr. Math FAQ:

   Pythagorean Triples
   http://mathforum.org/dr.math/faq/faq.pythag.triples.html 

We have:

  x + y = r^2 - s^2 + 2rs
        = (r + s)^2 - 2s^2

As r + s is odd, we have always:

  (r + s)^2 = 1  (mod 8)

If s is even, 2s^2 = 0 (mod 8) and we have:

  x + y = 1   (mod 8)

If s is odd, 2s^2 = 2 (mod 8), and we have:

  x + y = -1 = 7 (mod 8)

and you can conclude the proof using the congruence (*) above.

Does this help?  Write back if you'd like to talk about this more, or 
if you have any other questions.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 04/25/2003 at 09:35:01
From: Stan
Subject: Thank you (Primitive Triples)

Thanks Dr. Jacques.  I see now what you are saying.