Bill Allombert on Tue, 02 Aug 2022 16:06:45 +0200 |
[Date Prev] [Date Next] [Thread Prev] [Thread Next] [Date Index] [Thread Index]
Re: solving x^3-ay^3=1 over integers (N) |
On Tue, Aug 02, 2022 at 02:48:40PM +0200, Ralph Beckmann wrote: > Now you have me thrilled.. 😊 > > yes, I am very curious and happy to learn. > > Thank you so much for your time! > however, note that I have not heard about GRH (beyond what I just read in > Wikipedia) and the "elliptic curves" I have come across only once when I was > trying to factorize a C200 (to no avail..) > but then I am happy to "use" things even before I have reached the full understanding (which is why I have now started to study math) For norm equation: Let b a solution of b^3=u and K = Q(b) the elements of K can be written as x+b*y+b^2*z and their norm is N(x+b*y+b^2*z) = x^3+u*y^3+u^2*z^3-3*u*x*y*z (a nice formula by itself) so you want to solve N(x+b*y+b^2*z)=1 and z=0 It happens that the set of solution of N(x+b*y+b^2*z)=1 is a group. If you restrict x,y,z so that x+b*y+b^2*z is an algebraic integer, then the group has a single generator, that can be computed this way ? bnf=bnfinit(x^3-37);g=bnf.fu[1] %6 = Mod(-3*x+10,x^3-37) (up to sign, if the norm is -1, multiply by -1) Here we have z=0 already, so we find 10^3-37*3^3 = 1 More generally, the general solution will be g^n with n in Z, and the component z_n of g^n will obey a third-term linear recurrence, and if you are lucky you can find for which n z_n=0, but in general it is very hard. Cheers, Bill