Bill Allombert on Tue, 02 Aug 2022 16:06:45 +0200

 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
%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

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