Re: Question on ternary quadratic form

[Markus Grassl <markus.grassl@ug.edu.pl>]

It took me a bit longer to find the corresponding functions in Pari - 
slightly different from Bill's solution:

? G=[1, 0,  0; 0,  5,  28; 0, 28, 157]
%1 =
[1  0   0]

[0  5  28]

[0 28 157]

? qflllgram(G)
%2 =
[1  0   0]

[0 -6 -11]

[0  1   2]


I'm not sure whether this works for all cases.

More generally, given a quadratic form, one can look for short vectors 
in the lattice with the give Gram matrix whose length is a square.
Those vectors can be used in a Gram-Schmidt like procedure, and the 
normalisation of the vectors does not introduce a square root if the the 
length of the vectors is a square.


Markus


Am 15/11/2023 um 22:28 schrieb hermann@stamm-wilbrandt.de:
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> 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.
> In order to determine the transformation I define:
>
> ? X
>
> [1 0 0]
>
> [0 a b]
>
> [0 c d]
>
> ?
>
> And get this:
>
> ? X~*Qt*X
>
> [1                                   0
> 0]
>
> [0            5*a^2 + 56*c*a + 157*c^2 (5*b + 28*d)*a + (28*c*b +
> 157*d*c)]
>
> [0 (5*b + 28*d)*a + (28*c*b + 157*d*c)            5*b^2 + 56*d*b +
> 157*d^2]
>
> ?
>
> Now I switch to wolframscript and determine a solution for the three
> equations:
>
> In[4]:= FindInstance[5*a^2 + 56*c*a + 157*c^2==1&&5*b^2 + 56*d*b +
> 157*d^2==1&&(
> 5*b + 28*d)*a + (28*c*b + 157*d*c)==0,{a,b,c,d},Integers]
>
> Out[4]= {{a -> -6, b -> -11, c -> 1, d -> 2}}
>
> In[5]:=
>
> With that I define matrix Y:
>
> ? Y
>
> [1  0   0]
>
> [0 -6 -11]
>
> [0  1   2]
>
> ?
>
> And really Y is transformation matrix to ternary quadratic form:
>
> ? Y~*Qt*Y
>
> [1 0 0]
>
> [0 1 0]
>
> [0 0 1]
>
> ?
>
>
> How can I determine integer {a,b,c,d} solutions in PARI/GP without using
> wolframscript?
>
>
> Regards,
>
> Hermann.