Cryptography Reference
In-Depth Information
which is not close to an integer. However,
℘
(
ω
2
/
10 +
1
2
ω
1
)=
−
213
.
00000
....
This yields the point
(
x, y
)=(
−
213
,
2592)
on
E
. An easy check shows that this is a point of order 10. Since the order of
the torsion subgroup divides 10, we have dete
rm
ined that the torsion in
E
(
Q
)
consists of the multiples of (
−
213
,
2592).
Exercises
(a) Show that
d
3
d
5
9.1
≡
(mod 12) for all integers
d
.
(b) Show that
5
d
d
3
+7
d
d
5
≡
0
(mod 12)
|
n
|
n
for all positive integers
n
.
(c) Show that
g
2
=
(2
π
)
4
(1 + 240
X
)
,
12
where
X
=
n
=1
σ
3
(
n
)
q
n
.
(d) Show that
g
3
=
(2
π
)
6
(1
−
504
Y
)
,
216
where
Y
=
n
=1
σ
5
(
n
)
q
n
.
(e) Show that
1728(2
π
)
−
12
Δ = (1 + 240
X
)
3
−
(1
−
504
Y
)
2
.
(9.41)
(f) Show that the right side of (9.41) is congruent to 144(5
X
+7
Y
)
mod 1728.
(g) Conclude that (2
π
)
−
12
Δ=
n
=0
d
n
q
n
,with
d
n
∈
Z
.
(h) Compute enough coecients to obtain that
(2
π
)
−
12
Δ=
q
(1 +
∞
e
n
q
n
)
n
=1
with
e
n
∈
Z
.
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