Re: solving x^3-ay^3=1 over integers (N)

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

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 elliptic curves:
First you need this gadget:

cubic(u)=[ellinit([0,-u^2/108]),(P)->my([X,Y]=P);[u/X/6+Y/X,u/X/6-Y/X]]

This maps the equation C: x^3+y^3=u to the Mordell equation E: y^2 = x^3-u^2/108

(you can check this formula easily).

You should do this
? [E,f] = cubic(41);

Now if P is a point on E, the f(P) is a point on x^3+y^3=41
Now, you can use ellrank and elltors to find the points on E;
? ellrank(E)
%9 = [0,0,2,[]]
? elltors(E)
%10 = [1,[],[]]

there is no solutions, so we find that there is no rational solutions to x^3+y^3=41,
and so no rational solution to x^3-41*y^3=1 either.

? [E,f]=cubic(37);
? ellrank(E)
%6 = [2,2,0,[[37/3,259/6],[37/9,407/54]]]
so there is an infinite number of rational solution.
? f([37/3,259/6])
%7 = [4,-3]
and indeed 4^3+(-3)^3 = 37
(4/3)^3 - (1/3)^3*37 = 1
which is a rational solution to your equation.

Cheers,
Bill