|Bill Allombert on Wed, 23 Aug 2017 00:22:31 +0200|
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|Re: Using bnfisintnorm to check if a number is of the form P|
On Tue, Aug 22, 2017 at 05:41:41PM -0400, Charles Greathouse wrote: > I have a homogeneous polynomial P(x,y) and a number n. I would like to find > the integers (a, b) such that P(a, b) = n. Why not use thue ? ? thue(z^3+2,25) %2 = [[3,-1]] > The basic idea is to represent P in univariate form P(x,1) and > call bnfisintnorm(bnfinit(P), n). But I don't know how to interpret the > output of bnfisintnorm, and the help entry doesn't really say anything. > > For example, > > bnfcertify(K = bnfinit(z^3+2,1)) > bnfisintnorm(K, 25) > > gives > > [-z^2 - 2*z + 1, z + 3] > > but it's not clear to me how to determine from this that 3^3 + 2*(-1)^3 == > 25. You need to compute the formula for the norm form: ? norm(Mod(a+b*x+c*x^2,x^3+2)) %3 = a^3+6*c*b*a+(-2*b^3+4*c^3) So for a=3,b=1,c=0, one get (3)^3-2*(1)^3 = 25 Cheers, Bill.