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From: master1729 on 9 Aug 2010 13:51
quasi wrote :

> Prove or disprove:
>
> If f in K(x,y,z) is a symmetric rational function
> such that
>
> f = g^2 + h^2
>
> for some g,h in K(x,y,z), then g^2 and h^2 are
> symmetric.
>
> quasi

i modify that to f,g and h are irreducible in the ring of their ufd domain coefficients.

my 50 cent.
From: achille on 9 Aug 2010 19:05
On Aug 10, 5:32 am, Timothy Murphy <gayle...(a)eircom.net> wrote:
> quasi wrote:
> >>quasi wrote:
>
> >>> Prove or disprove:
>
> >>> If f in K(x,y,z) is a symmetric rational function such that
>
> >>>    f = g^2 + h^2
>
> >>> for some g,h in K(x,y,z), then g^2 and h^2 are symmetric.
>
> >>> quasi
>
> >>I would have thought not.
> >>Suppose f = g^2 + y^2 with g,h symmetric.
> >>Take a function F(x) of one variable
> >>which can be expressed in the form F(x) = G(x)^2 + H(x)^2
> >>in several different ways.
> >>Now consider F(x)F(y)F(z)f(x,y,z).
> >>I would have thought you could take different
> >>expressions for F(x),F(y),F(z) as a sum of 2 squares,
> >>and get f(x) as a sum of non-symmetric squares.
>
> > I think I see the essence of your idea, but I don't see why you need
> > the original f in the product. Why not just let f be defined by
>
> >    f(x,y,z) = F(x)F(y)F(z)
>
> > where F has the non-uniqueness property you specified. It would then
> > seem that f would have at least 8 different representations as a sum
> > of 2 squares, and it does seem intuitive that they won't all be
> > symmetric.
>
> You are quite right.
> I was trying to modify the given function f(x,y,z),
> but there is no need to do this, as you say.
> One can just consider F(x)F(y)F(z).

The simplest example (or counterexample) using F(x) = x^2 + 1:

(x^2+1)*(y^2+1)*(z^2+1) = (x*y*z-z+y+x)^2 + (y*z+x*z-x*y+1)^2.

From: achille on 9 Aug 2010 22:56
On Aug 10, 7:05 am, achille <achille_...(a)yahoo.com.hk> wrote:
>
> The simplest example (or counterexample) using F(x) = x^2 + 1:
>
> (x^2+1)*(y^2+1)*(z^2+1) = (x*y*z-z+y+x)^2 + (y*z+x*z-x*y+1)^2.

A better counterexample:

f(x,y,z) = 1 + ((x-y)*(y-z)*(z-x))^2

and f(x,y,z) is irreducible over Q[x,y,z].
From: quasi on 10 Aug 2010 00:45
On Mon, 9 Aug 2010 19:56:21 -0700 (PDT), achille
<achille_hui(a)yahoo.com.hk> wrote:

>On Aug 10, 7:05 am, achille <achille_...(a)yahoo.com.hk> wrote:
>>
>> The simplest example (or counterexample) using F(x) = x^2 + 1:
>>
>> (x^2+1)*(y^2+1)*(z^2+1) = (x*y*z-z+y+x)^2 + (y*z+x*z-x*y+1)^2.

A nice counterexample.

>A better counterexample:
>
> f(x,y,z) = 1 + ((x-y)*(y-z)*(z-x))^2
>
>and f(x,y,z) is irreducible over Q[x,y,z].

However this last example is not a counterexample.

The claim was that g^2 and h^2 must be symmetric, and in your last
example, they are.

quasi
From: quasi on 10 Aug 2010 02:41
On Mon, 09 Aug 2010 13:05:17 -0500, quasi <quasi(a)null.set> wrote:

>On Mon, 09 Aug 2010 02:23:50 -0500, quasi <quasi(a)null.set> wrote:
>
>>Prove or disprove:
>>
>>If f in K(x,y,z) is a symmetric rational function such that
>>
>> f = g^2 + h^2
>>
>>for some g,h in K(x,y,z), then g^2 and h^2 are symmetric.
>
>I'll conjecture that the answer is "yes".

But Tim Murphy pointed the way to a counterexample, and achille
followed up and actually produced one, so the conjecture, as stated
above, is dead.

But what if we switch from rational functions to polynomials and then
also require that f is irreducible?

So here's a revised version ...

Prove or disprove:

Let K be a field. If f in K[x,y,z] is an irreducible symmetric
polynomial such that

f = g^2 + h^2

for some g,h in K[x,y,z], then g^2 and h^2 are symmetric.

Remark:

I think my previously posted "fuzzy plan of attack" might be work for
this version.

quasi
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