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Bill Allombert on Wed, 15 Nov 2023 23:00:46 +0100
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Re: Question on ternary quadratic form
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- To: pari-users@pari.math.u-bordeaux.fr
- Subject: Re: Question on ternary quadratic form
- From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
- Date: Wed, 15 Nov 2023 23:00:39 +0100
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- In-reply-to: <9bbdc86703960aaf2b5421cbf58e6b65@stamm-wilbrandt.de>
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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.