Cryptography Reference
In-Depth Information
1.
−
a<b
≤
a
2.
a
≤
c
,
3. if
a
=
c
then
b
≥
0.
The first of these expresses the condition that
−
1
/
2
≤
(
τ
)
<
1
/
2, while the
second says that
|τ|≥
1. The case where
a
=
c
corresponds to
τ
lying on the
unit circle, and
b>
0 says that it lies on the left half. It can be shown (see
[16]) that there is a one-to-one correspondence between the ideals
I
that we
are considering (endomorphism ring exactly
R
) and those triples satisfying
a>
0, gcd(
a, b, c
)=1,
b
2
−
4
ac
=
−
171, and conditions (1), (2), and (3).
Let's count these triples. The strategy is to consider (
b
2
+ 171)
/
4andtryto
factor it as
ac
with
a, b, c
satisfying (1), (2), and (3):
b
(
b
2
+ 171)
/
4
ac
1
43
1
43
±
3
45
5
9
5
49
7
7
Thetriple(
a, b, c
)=(3
,
3
,
15), which arose in the above calculations, is not
listed since gcd(
a, b, c
)
O
K
,whichisan
ideal for the larger ring
O
K
, as mentioned above). There are no values for
a, c
when
b
=
±
7. When
|b|≥
9, the condition
|b|≤a ≤ c
can no longer be
satisfied. We have therefore found all triples. They correspond to values of
τ
,
call them
τ
1
,τ
2
,τ
3
,τ
4
:
= 1 (and it corresponds to the ideal 3
1+
√
−
(
a, b, c
)=(1
,
1
,
43)
←→ τ
1
=
−
171
2
3+
√
−
(
a, b, c
)=(5
,
3
,
9)
←→ τ
2
=
−
171
10
(
a, b, c
)=(5
, −
3
,
9)
←→ τ
3
=
3+
√
−
171
10
(
a, b, c
)=(7
,
5
,
7)
←→ τ
4
=
−
5+
√
−
171
14
.
Note that
j
(
τ
0
)=
j
(
τ
1
)since
τ
0
=
τ
1
+ 1. Compute the values
j
(
τ
2
)=
−
417
.
33569403605596400916623167906655644314607149466
...
+
i
3470
.
100833725097578092463768970644185234184993550
...
j
(
τ
3
)=
−
417
.
33569403605596400916623167906655644314607149466
...
i
3470
.
100833725097578092463768970644185234184993550
...
j
(
τ
4
) = 154
.
683676758820235444376830811774357548921993728906
....
−
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