Cryptography Reference
In-Depth Information
with
A, B
∈
Z
.Then
1
8
h
(
j
)
1
12
h
(Δ)
1
2
h
(
P
)
h
(
P
)
−
−
−
0
.
973
≤
−
12
h
(
j
)+
1
1
≤
12
h
(Δ) + 1
.
07
for all
P ∈ E
(
Q
)
.Here
Δ=
−
16(4
A
3
+27
B
2
)
and
j
=
−
1728(4
A
)
3
/
Δ
.
For the curve
y
2
=
x
3
25
x
,wehaveΔ=10
6
and
j
= 1728. Therefore,
−
1
2
h
(
P
)
<
2
.
843
−
3
.
057
< h
(
P
)
−
for all
P ∈ E
(
Q
). The points (0
,
0)
,
(5
,
0)
,
(
−
5
,
0)
,
(
−
4
,
6) generate the group
E
(
Q
)
/
2
E
(
Q
). The first three of these points have canonical height 0 since
they are torsion points. The point (
−
4
,
6) has canonical height 0
.
94974
...
(this can be calculated using the series (8.5)). The proof of Theorem 8.17
shows that the points with canonical height at most 0
.
94974
...
generate
E
(
Q
). Theorem 8.23 says that such points have noncanonical height
h
(
P
)
<
8
.
02. Since
e
8
.
02
3041, the nonlogarithmic height of the
x
-coordinate is at
most 3041. Therefore, we need to find all points (
x, y
)
≈
∈
E
(
Q
) such that
x
=
a
b
with
Max(
|a|, |b|
)
≤
3041
.
It is possible to find all such points using a computer. The fact that the
denominator of
x
must be a perfect square can be used to speed up the
search. We find the points
(0
,
0)
,
(
−
5
,
0)
,
(5
,
0)
,
(
−
4
,
6)
(45
, −
300) = (
−
5
,
0) + (
−
4
,
6)
(25
/
4
,
75
/
8) = (0
,
0) + (
−
4
,
6)
(
−
5
/
9
,
−
100
/
27) = (5
,
0) + (
−
4
,
6)
(1681
/
144
,
−
62279
/
1728) = 2(
−
4
,
6)
and the negatives of these points.
Since these points generate
E
(
Q
), we
conclude that (0
,
0)
,
(5
,
0)
,
(
−
5
,
0)
,
(
−
4
,
6) generate
E
(
Q
).
REMARK 8.24
In Chapter 1, we needed to find an
x
such that
x
,
x −
5,
and
x
+ 5 were all squares. We did this by starting with the point (
4
,
6) and
finding the other point of intersection of the tangent line with the curve. In
effect, we computed
−
2(
−
4
,
6) = (
41
2
12
2
,
−
62279
1728
)
and miraculously obtained
x
=41
2
/
12
2
with the desired property. We now
see that this can be explained by the fact that
φ
is a homomorphism. Since
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