From: jr@redmink.demon.co.uk (John R Ramsden)
Subject: Re: ax^2+by^2 = cz^2
Date: Sun, 11 Feb 2001 08:07:11 GMT
Newsgroups: sci.math
Summary: Conditions under which a homogeneous polynomial has integer solutions

ahmedfares@my-deja.com wrote:
>
> If a,b,c are integers with gcd(a,b,c) = 1 what condition should a,b,c
> satisfie in order that there will be an integer solution to the
> equation ax^2+by^2 = cz^2  with xyz =/= 0 ? i.e non trivial solution.
>
> For example when (a,b,c)=(1,1,3) the equation is
> x^2+y^2 = 3z^2 and it has only the trivial solution.
>
> Thanks,
> Ahmed

The nec and suff conditions for a.x^2 + b.y^2 + c.z^2 = 0
(a, b, c integers) to have integer solutions with x, y, z
not all zero are:

 1)  a, b, c not all of the same sign

 2)  -bc, -ca, -ab quadratic residues mod a, b, c respectively

So in your example, with a, b, c = 1, 1, -3, the first condition
is satisfied, but not the second (since -ab = -1 is not a quadratic
residue mod 3).

If one of the coefficients is zero then condition 2 must be interpreted
to mean that the negated (and hence positive) product of the other two
must actually be a square.

ObExercise: Show that these conditions are consistent with the equation
m.x^2 + n.y^2 = (m + n).z^2 having non-trivial integer solutions for any
pair of integers m, n.

There are similar N&S conditions for quadratic forms in four and five
variables. But I'm not sure if anyone has extended this to any number
of variables.

A good reference to this is "History of the Theory of Numbers" and
"Introduction to the T of N" by L E Dickson, both published by Dover
and indispensable for anyone who is interested in this kind of thing.


Cheers

John Ramsden
==============================================================================

From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: ax^2+by^2 = cz^2
Date: 12 Feb 2001 02:01:24 GMT
Newsgroups: sci.math

ahmedfares@my-deja.com wrote:
> If a,b,c are integers with gcd(a,b,c) = 1 what condition should a,b,c
> satisfie in order that there will be an integer solution to the
> equation ax^2+by^2 = cz^2  with xyz =/= 0 ? i.e non trivial solution.

In article <3a86419b.67979041@news.demon.co.uk>,
John R Ramsden <jr@redmink.demon.co.uk> wrote:
>The nec and suff conditions for a.x^2 + b.y^2 + c.z^2 = 0
>(a, b, c integers) to have integer solutions with x, y, z
>not all zero are:
>
> 1)  a, b, c not all of the same sign
>
> 2)  -bc, -ca, -ab quadratic residues mod a, b, c respectively
[...]
>There are similar N&S conditions for quadratic forms in four and five
>variables. But I'm not sure if anyone has extended this to any number
>of variables.

Sure they have: a single quadratic form has a nontrivial rational solution
iff it has a nontrivial p-adic solution for each prime  p  of the
rationals (including the infinite prime, which is more or less your
condition (1).) The same is true over number fields. The quadratics can
have any number of variables. The coefficients do not have to be
square-free, although that makes it easier to state the conclusion
without reference to p-adic completions. (Note: there is a subtlety at
the prime  2  with can be ignored since the number of 'bad' primes
is even, and so any single prime can be ignored in the analysis.)

This is the 'Hasse principle' or 'local-to-global principle', which
applies in some other contexts too but famously not in all Diophantine
problems, e.g. there are cubics which have p-adic points for all  p
but no rational points.

As it turns out, quadratic forms in 5 or more variables always have
p-adic zeros for every finite prime  p, so as long as the form is
not positive-definite or negative-definite it also has a rational zero.

A source I like for this material is Borevich and Shafarevich's 
"Number Theory", where this material closes chapter 1. The closing 
nugget there is also interesting: there is a paper by B. J. Birch 
(Mathematika 4 (1957) 102-105) in which he showed that homogeneous forms 
of _odd_ degree also always have a rational zero, as long as the 
number of variables is sufficiently large (depending on the degree). 
No such result can be true for even degrees greater than 2, however.

dave