Thanks, Fred, for the quick response on -Pythag. My solution was: every primitive solution (x,y,z have no common factor) is of form (ac,bc,ab) where (a,b,c) is a Pythagorean Triple. At 05:02 AM 9/8/03 -0400, you wrote:
At 04:23 AM 9/8/03, David Gale wrote:
The solutions of x^2 + y^2 = z^2 the well known "Pythagorean triples" goes back to Euclid. I'd like to know if anyone in the history of Diophantine equations has looked at the same problem with 2 replaced by -2. That is, find all solutions in integers of 1/x^2 + 1/y^2 = 1/z^-2. The solution is easy and rather nice.
Oh yes, I've certainly seen this one before. A little web sleuthing shows that the question has been posed quite a few times. Here are two places that state and answer the general question (many others just seek the least solution, (15,20,12)):
Brainteasers, University of Adelaide, April 1999 (posed October 1998) http://www.eng.adelaide.edu.au/newsletter/apr99/pg14.html
Problem Posing Variations on Fermat's Last Theorem, Michael de Villiers, University of Durban-Westville, originally published December 1995 http://mzone.mweb.co.za/residents/profmd/probpose.pdf (pp.3-4)
Curiously, these inverted Pythagorean triples seem to be popular in the southern hemisphere.
-- Fred W. Helenius <fredh@ix.netcom.com>
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