2.9. Descent and the weak Mordell-Weil theorem 55
where all the congruences are modulo f(x) = x3 + Ax + B. The
congruences in the previous equation say that a polynomial of degree
1, call it g(x) = x1 x, is congruent to a polynomial of degree
2, call the last line h(x), modulo a polynomial of degree 3, namely
f(x). Then h(x) g(x) is a polynomial of degree 2, divisible by a
polynomial of degree 3. This implies that h(x) g(x) must be zero
and h(x) = g(x), i.e.,
x1 x = (2ac +
b2

Ac2)x2
+ (2ab
Bc2
2Abc)x +
(a2
2bcB).
If we match coefficients, we obtain the following equalities:
2ac +
b2

Ac2
= 0, (2.7)
2ab
Bc2
2Abc = −1, (2.8)
a2
2bcB = x1. (2.9)
If c = 0, then b = 0 by Eq. (2.7); therefore, p(x) = a +bx+cx2 = a is
a constant function, and so t1 = t2 = t3. By Eq. (2.6), it follows that
e1 = e2 = e3, which is a contradiction with our assumptions. Hence,
c must be non-zero. We multiply Eq. (2.8) by
1
c2
and Eq. (2.7) by
b
c3
to obtain
2ab
c2
B
2Ab
c
=
1
c2
, (2.10)
2ab
c2
+
b3
c3

Ab
c
= 0. (2.11)
We subtract Eq. (2.10) from Eq. (2.11) to get:
b
c
3
+ A
b
c
+ B =
1
c
2
.
Hence, the point P = (x0,y0) = (
b
c
,
1
c
) is a rational point on E(Q).
It remains to show that x(2P ) = x(Q). From Eq. (2.11) we deduce
that
a =
Ab
c

b3
c3
2b
c2
=
A
(
b
c
)2
2 ·
1
c
=
A x0
2
2y0
,
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