解一个椭圆曲线方程

nyy

谁能用pari/gp来解一下？？？？？？？

nyy

http://pari.math.u-bordeaux.fr/p ... 5.1/refcard-ell.pdf

这儿有帮助文件，谁能来解一下？？？？？？？

nyy

hujunhua 发表于 2023-5-8 19:44
令`Y=2y+x,X=x-2`, 方程化为\[
Y^2=X(4X^2-3X-16)\tag1
\]由于`\gcd(X,4X^2-3X-16)|16`, 故\[\begin{sp ...

Elliptic curve structures
An elliptic curve is given by a Weierstrass model

y^2 + a1 xy + a3 y = x^3 + a2 x^2 + a4 x + a6,

whose discriminant is nonzero. Affine points on E are represented as two-component vectors [x,y]; the point at infinity, i.e. the identity element of the group law, is represented by the one-component vector [0].

Given a vector of coefficients [a1,a2,a3,a4,a6], the function ellinit initializes and returns an ell structure. An additional optional argument allows to specify the base field in case it cannot be inferred from the curve coefficients. This structure contains data needed by elliptic curve related functions, and is generally passed as a first argument. Expensive data are skipped on initialization: they will be dynamically computed when (and if) needed, and then inserted in the structure. The precise layout of the ell structure is left undefined and should never be used directly. The following member functions are available, depending on the underlying domain.

All domains.

* a1, a2, a3, a4, a6: coefficients of the elliptic curve.

* b2, b4, b6, b8: b-invariants of the curve; in characteristic != 2, for Y = 2y + a_1x+a3, the curve equation becomes Y^2 = 4 x^3 + b2 x^2 + 2b4 x + b6 = : g(x).

* c4, c6: c-invariants of the curve; in characteristic != 2,3, for X = x + b2/12 and Y = 2y + a_1x+a3, the curve equation becomes Y^2 = 4 X^3 - (c4/12) X - (c6/216).

* disc: discriminant of the curve. This is only required to be nonzero, not necessarily a unit.

* j: j-invariant of the curve.

http://pari.math.u-bordeaux.fr/dochtml/html/Elliptic_curves.html

这个似乎就是你变换的依据吧！