Bill Allombert on Tue, 02 Aug 2022 15:26:22 +0200 |
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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 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