Primitive TriplesDate: 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. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/