Re: Question on ternary quadratic form

[hermann@stamm-wilbrandt.de]

Thanks Bill, Denis and Markus for the answers.

I worked on determining sum of three squares for integer n
using ternary quadratic form as described in chapter 4 of
E. Grosswald. Representations of Integers as Sums of Squares. 
SpringerVerlag, New York, 1985, in my tqf.gp gist:
https://gist.github.com/Hermann-SW/f13b8adf7d7e3094f0b6db0bce29a7b8

qflllgram() is my favorite solution, better qflllgram(_, flag=1) to 
ensure integer solution.

The two transformations that I did are not needed when using
qflllgram(), it does the complete transformation to identity form!

In the gist initial ternary quadratic for Q for n=46 is
created as, with det=1 and representing n:

? Q

[ 6  37  1]

[37 229  0]

[ 1   0 46]

? matdet(Q)
1
? qfeval(Q,[0,0,1]~)
46
?

Now one call to qflllgram to determine G:

? G=qflllgram(Q,flag=1)

[-6 -43  93]

[ 1   7 -15]

[ 0   1  -2]

?

And a representation of n=46 as sum of 3 squares is ...

? G^-1*[0,0,1]~
[-6, 3, 1]~
?

... wow!

Now I need to eliminate all the stuff that did not work and
make tqf.gp gist compute sum of three squares for all n.

Regards,

Hermann.

On 2023-11-15 23:15, Denis Simon wrote:
> Hi Hermann,
>
> More generally, if you have a quadratic form that you suspect to be
> highly reducible,
> you shoud try to reduce it with qflllgram:
>
> ? M=[1,0,0;0,5,28;0,28,157];
> ? G=qflllgram(M)
> %2 =
> [1  0   0]
>
> [0 -6 -11]
>
> [0  1   2]
>
> ? G~*M*G
> %3 =
> [1 0 0]
>
> [0 1 0]
>
> [0 0 1]
>
> In your situation, you can do the same thing with the 2x2 block 
> [5,28;28,157].
>
> Cheers,
> Denis SIMON.
>
> ----- Mail original -----
> > De: "Bill Allombert" <Bill.Allombert@math.u-bordeaux.fr>
> > À: "pari-users" <pari-users@pari.math.u-bordeaux.fr>
> > Envoyé: Mercredi 15 Novembre 2023 23:00:39
> > Objet: Re: Question on ternary quadratic form
>
> > On Wed, Nov 15, 2023 at 10:28:25PM +0100, hermann@stamm-wilbrandt.de 
> > wrote:
> > > After two modifications I got this ternary quadratic form:
> > >
> > > ? Qt
> > >
> > > [1  0   0]
> > >
> > > [0  5  28]
> > >
> > > [0 28 157]
> > >
> > > ?
> > >
> > > I know it can be transformed to ternary identity form.
> >
> > You can use qfisom (if the form is positive definite):
> >
> > ? M=[1,0,0;0,5,28;0,28,157];
> > ? qfisom(M,matid(3))
> > %17 =
> > [0  1   6]
> >
> > [0 -2 -11]
> >
> > [1  0   0]
> >
> > Of course, if you look for solution of the form
> > 1 0 0
> > 0 a b
> > 0 c d
> >
> > Then you should do
> > ? qfbredsl2(Qfb(5,28+28,157))
> > %20 = [Qfb(1, 0, 1), [-6, 11; 1, -2]]
> >
> > which will be much faster.
> >
> > Cheers,
> > Bill.