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From: quasi on 9 Aug 2010 03:23
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
From: quasi on 9 Aug 2010 14:05
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".

For the case K = Q, here's a rather fuzzy plan of attack ...

(1) Show that the conjecture holds in the field K(x,y,z) iff it holds
in the ring Z[x,y,z].

(2) Show that Z[x,y,z][i] is a UFD (where i^2 = -1).

(3) Show that if f in Z[x,y,z] is a counterexample, then some
irreducible factor of f would also be a counterexample (using
Brahmagupta's identity??). If so, we can assume f is irreducible.

(4) Show that since f is irreducible, f admits no more than one
expression as a sum of 2 squares in Z[x,y,z], hence by hypothesis,
exactly one, say f = g^2 + h^2.

(5) But any of the 6 permutations of the variables x,y,z leads to a
potentially new representation. To avoid the contradiction, show that
g^2 and h^2 must be symmetric.

quasi
From: Timothy Murphy on 9 Aug 2010 14:16
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.



From: quasi on 9 Aug 2010 16:14
On Mon, 09 Aug 2010 20:16 +0200, Timothy Murphy <gayleard(a)eircom.net>
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.

If I have time later, I'll try to find an actual counterexample based
on (my interpretation of) the ideas you've suggested.

Thanks.

quasi
From: Timothy Murphy on 9 Aug 2010 17:32
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).


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