|Bill Allombert on Thu, 02 Mar 2017 11:26:46 +0100|
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|Re: Mathematica "Reduce" function|
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.