Re: Mathematica "Reduce" function

[Pedro Fortuny Ayuso <fortunypedro@uniovi.es>]

Now that is a really good solution.

I have forgotten all I (once possibly) knew about quadratic
forms.

Thanks a lot for your help!

Pedro.


On Thu, Mar 02, 2017 at 11:26:35AM +0100, Bill Allombert wrote:
> On Thu, Mar 02, 2017 at 10:00:48AM +0100, Pedro Fortuny Ayuso wrote:
> > Thanks to all.
> > 
> > My specific problem is trying to solve equations like
> > 
> > 6x^2 + 12y^2 +20z^2 = 0
> > 
> > over Z/(2^k)Z. That is, finding the points of that surface
> > over that ring.
> 
> Solutions of homogenous degree-2 equation in three variables can be
> parametrized as soon as one solution is known using qfparam:
> For example [0,1,1] is a (primitive) solution mod 2^5 so set
> 
> ? M=qfparam(matdiagonal([6,12,20]),[0,1,1])
> %9 = [0,-20,0;3,0,10;3,0,-10]
> ? v = y^2 * M*[1,x/y,(x/y)^2]~
> %10 = [-20*y*x,10*x^2+3*y^2,-10*x^2+3*y^2]~
> 
> so for all x,y, (-20*y*x,10*x^2+3*y^2,-10*x^2+3*y^2) is a solution mod
> 2^5:
> 
> ? content(6*(-20*y*x)^2+12*(10*x^2+3*y^2)^2+20*(-10*x^2+3*y^2)^2)
> %12 = 32
> 
> > Bill's reply of counting
> > 
> > length([[x,y,z]|x<-[0..2^k-1];y<-[0..2^k-1];z<-[0..2^k-1],6*x^2+12*y^2+20*z^2==0])
> 
> You are missing a reduction mod 2^k at the end.
> 
> > is the fastest but it ***looks like*** a lot slower than
> > Mathematica (but please notice I am working on a system
> > with pari/gp and my colleague on a different one with Mathematica,
> > so that it may have nothing to do with pari/Mathematica).
> 
> This is quite possible, I do not know what Mathematica is doing.
> what you can do is to check whether (6*x^2+12*y^2)/-20 is a square
> instead of iterating over z:
> 
> [[x,y,z]|x<-2*[0..2^(k-1)-1];y<-[0..2^k-1],issquare((6*x^2+12*y^2)/-20*Mod(1,2^k),&z)]
> 
> Cheers,
> Bill.
> 

-- 
Pedro Fortuny Ayuso
http://pfortuny.net


EPIG, Campus de Viesques, Gijon
Dpto. de Matematicas
Universidad de Oviedo