Re: another simple question

[Kurt Foster <drsardonicus@earthlink.net>]

On Apr 16, 2024, at 10:45 PM, American Citizen wrote:

> Recently, we saw a point with thousands of digits posted with the  
> question, what elliptic curve might have this point?
>
> The answer was to use lindep to determine the elliptic curve, and so  
> everything went fine.
>
> However, I am dubious that this general approach using lindep on  
> [y^2, xy, y, x^3, x^2, x, 1] will work when we have points of lesser  
> heights.
>
> Suppose we have the point [1/2, 5/8] ??
>
> I strongly suspect that thousands perhaps millions, maybe more?  
> elliptic curves can be found containing this point.
>
> How can we effectively pare down the possibilities to something we  
> can work with?
>
> Randall

As Denis indicates, the general solution is a vector space of  
dimension 6.  Here is a computation giving a basis for x = 1/2 and y =  
5/8 with "small" (LLL reduced) integer entries.  Alas, a solution for  
which all coefficients are integers, and the coefficients of y^2 and - 
x^3 are both 1, seems not to be in the cards:

> ? x=1/2;y=5/8;v=[y^2,x*y,y,-x^3,-x^2,-x,-1];
> ? w=v*denominator(v);
> ? M=matrix(7,7,i,j,if(i==1,w[j],0));
> ? K=matkerint(M)
> %4 =
> [0 0 0 0 0 -4]
>
> [0 0 0 0 -2 1]
>
> [1 1 1 -1 1 0]
>
> [1 -1 1 1 0 0]
>
> [0 1 0 1 0 -1]
>
> [1 1 -1 0 0 0]
>
> [0 0 1 -1 0 -1]
>
> ?

The dependencies are the columns of K.

If you have one "nice" solution where the coefficients of y^2 and -x^3  
are both 1, a slight variation of the above calculation gives a 4- 
dimensional LLL reduced lattice of integer "translates" giving other  
curves, with the coefficients of y^2 and -x^3 also equal to 1.  Take

v = [x*y, y, -x^2, - x, -1];
w = v*denominator(v);
M=matrix(5,5,i,j,if(i==1,w[j],0));
K=matkerint(M);

This worked quickly for the "game" point.  However, the entries of K  
were nowhere near as small as the original coefficients.