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
.
