From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math,sci.math.symbolic
Subject: Re: Math problem
Date: 26 Dec 1997 07:23:17 GMT

In article <883039210.1040449545@dejanews.com>,  <nat@netcom.com> wrote:
>Given A*x^2 + B*x*y + C*y^2 = z^2, and given values for A, B, and C,
>I would like to be able to find all integer solutions of x, y, and z.
>
>For example: -23*x^2 + 22*x*y + 17*y^2 = z^2
>
>Is there a general way to solve this?

See e.g.
	http://www.math.niu.edu/~rusin/known-math/95/quadratics
[URL updated 1999/01 -- djr]

Your particular example may be reduced to "polar form"
	u^2-2v^2=-23z^2
by completing the square (that is, by making some linear changes of
variables). It is sufficient to solve this p-adically for p=2 and p=23,
but one can proceed "by inspection", noting that u=3, v=4, z=1 is a 
solution. In terms of the original variables we have only a _rational_
solution x=y=1/4, z=1; but by scaling we obtain an integral solution
	x=1, y=1, z=4
to the original equation.

We can treat this as an inhomogeneous quadratic in two rational variables:
	-23*p^2 + 22*p*q + 17*q^2 = 1
where  p = x/z, q = y/z. Once we have one rational solution (p0, q0) to a
multivariate quadratic, we may obtain all others by finding the intersection
of the conic with the line  (p,q) = (p0 + t, q0 + mt), as a function of  m.
In our example, using (p0,q0)=(1/4, 1/4) we compute 
	t = (6-14m)/(-23+22m+17m^2),
so that the rational points on the curve are parameterized by the formulae
	p = (1-34m+17m^2)/(-92+88m+68m^2)
	q = (-23+46m-39m^2)/(-92+88m+68m^2)
for all rational  m.

You can now work backwards if you like and recover the integral solutions: They
are of the form
	x = (   n^2-34mn+17m^2)d
	y = (-23n^2+46mn-39m^2)d
	z = (-92n^2+88mn+68m^2)d
where  m  and  n  range over the integers, and  d  could be any rational 
number making  x, y, and z  integral. (Equivalently, d = d1/d2 where  d1 is 
any integer, and  d2  divides the GCD of   (n^2+34mn+17m^2), 
(-23n^2+46mn-39m^2), and  (-92n^2+88mn+68m^2); that  GCD  is 1 or 2 if  
n  and  m  are relatively prime.)

For related topics see e.g.
	http://www.math.niu.edu/~rusin/known-math/index/11DXX.html

dave