Drexel dragonThe Math ForumDonate to the Math Forum


Math Forum
 Ask Dr. Math
 Discussions
 Internet Newsletter
 MathTools
 Problems of the Week
 Teacher2Teacher
 Teacher Exchange
 Workshops

Search All of the Math Forum:
Browse our Internet Mathematics Library

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.num-analysis.independent

Topic: a^2+b^2 = c^2+d^2
Replies: 17   Last Post: May 22, 2012 1:33 AM

Advanced Search
Reply to this Topic Reply to this Topic

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Titus Piezas III

Posts: 281
Registered: 12/13/04
a^2+b^2 = c^2+d^2
Posted: Jul 16, 2006 4:25 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Hello all,

It's been proven by Lehmer (1900) that, let P be the number of
primitive solutions to the Pythagorean triple a^2+b^2 = c^2 with
hypotenuse c less than a bound N. Then P/N = 1/(2pi) as N -> infinity.

The question is this: How about the equation,

a^2+b^2 = c^2+d^2 = z

with a,b,c,d all non-zero. Let p be the number of primitive solutions.
Does the ratio p/Sqrt[z] approach a real constant as z -> infinity? Can
it be expressed as a rational multiple of pi?

--Titus

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2013. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.