50 2. Elliptic curves

The current implementation of the algorithm is more fully explained

in Cremona’s book [Cre97].

Let E/Q be a curve given by

y2

=

x3

+ Ax + B, with A, B ∈ Z.

The most general case of the method of descent is quite involved,

so we will concentrate on a particular case where the calculations

are much easier: we will assume that E(Q) has 4 distinct rational

points of 2-torsion (including O). As we saw before (Theorem 2.5.5,

or Exercise 2.12.6), a point P = (x, y) ∈ E(Q) is of 2-torsion if and

only if y = 0 and

x3

+ Ax + B = 0 (or P = O). Thus, if E(Q) has 4

distinct rational points of order 2, that means that

x3

+ Ax + B has

three (integral) roots and it factors completely over Z:

x3

+ Ax + B = (x − e1)(x − e2)(x − e3)

with ei ∈ Z. Since

x3

+Ax+B does not have an

x2

term, we conclude

that e1 + e2 + e3 = 0.

Suppose, then, that E :

y2

= (x − e1)(x − e2)(x − e3), where the

roots satisfy ei ∈ Z and e1 + e2 + e3 = 0. We would like to find a

solution (x0,y0) ∈ E with x0,y0 ∈ Q, i.e.,

y0

2

= (x0 − e1)(x0 − e2)(x0 − e3).

Thus, each term (x0 − ei) must be almost a square, and we can make

this precise by writing

(x0 − e1) =

au2,

(x0 − e2) =

bv2,

(x0 − e3) =

cw2,

y0

2

=

abc(uvw)2,

where a, b, c, u, v, w ∈ Q, the numbers a, b, c ∈ Q are square-free, and

abc is a square (in Q).

Example 2.9.1. Let

E :

y2

=

x3

− 556x + 3120 = (x − 6)(x − 20)(x + 26)

so that e1 = 6, e2 = 20 and e3 = −26. The point (x0,y0) =

(

164184

289

,

66469980

4913

) is rational and on E. We can write

x0 − e1 =

164184

289

− 6 = 2 ·

285

17

2

and, similarly, x0 − e2 = (

398

17

)2

and x0 − e3 = 2 · (

293

17

)2.

Thus,

following the notation of the preceeding paragraphs

a = 2, b = 1, c = 2, u =

285

17

, v =

398

17

, w =

293

17

.