Inverse Elliptic Curves Problem

[Max Alekseyev]

On Nov 27, 2007 3:30 PM, Artur <grafix at csl.pl> wrote:

> Yes I give x-coordinates of the integer points.
> Elliptic equation is more general that reduced case y^2=x^3+n but I
> don't know exact form (if I will be know I will be don't ask)

For a general form of elliptic curves see
http://mathworld.wolfram.com/EllipticCurve.html

For the special case of y^2=x^3+n, your problem has a simple (or no) solution.
Namely, substitute the x-coordinates that you have in mind into this
equation to get a system of equations:
y1^2 = x1^3 + n
y2^2 = x2^3 + n
...
yk^2 = xk^3 + n
where y1,y2,...,yk,n are unknowns.
Taking the difference of any two equations i, j, you'll get:
yi^2 - yj^2 = xi^3 - xj^3
or
(yi-yj)(yi+yj) = xi^3 - xj^3.
All solutions to this equation come from the prime factorization of
xi^3 - xj^3. For any (positive or negative) divisor d of xi^3 - xj^3
of the same oddness with its co-divisor d'=(xi^3 - xj^3)/d, we have a
solution:
yi = (d+d')/2
yj = (d'-d)/2.

For each pair of original equations (i,j), find all solutions (yi,yj)
and represent then as edges in a k-partite graph G that connect a
vertex yi from i-th path with a vertex yj from j-th part.
The system of equations has a solution if and only if you have a
clique of size k in G. Each such clique gives values (y1,y2,...,yk)
that is a solution to the system (the value of n is derived easily),
and vice versa.

Regards,
Max