Cryptography Reference
In-Depth Information
so
q
=
−
1and
19(
x, y
)=
−
(
x, y
)=(
x,
−
y
) for all (
x, y
)
∈
E
[5]
.
We need to check whether
(
x
,y
)
de
=(
x
361
,y
361
)+(
x, −y
)
=
±
2(
x
19
,y
19
)
de
=
±
(
x
,y
)
?
for all (
x, y
)
∈ E
[5]. The recurrence of Section 3.2 shows that the fifth division
polynomial is
ψ
5
= 32(
x
3
+2
x
+1)
2
(
x
6
+10
x
4
+20
x
3
ψ
3
=5
x
12
+10
x
10
+17
x
8
+5
x
7
+
x
6
+9
x
5
+12
x
4
+2
x
3
+5
x
2
+8
x
+8
.
20
x
2
−
−
8
x
−
8
−
8)
−
The equation for the
x
-coordinates yields
x
=
y
361
+
y
x
361
2
3
x
38
+2
2
y
19
2
?
≡
x
361
2
x
19
=
x
−
−
x
−
(mod
ψ
5
)
.
− x
When
y
2
is changed to
x
3
+2
x
+ 1, this reduces to a polynomial relation in
x
, which is then verified. Therefore,
a ≡±
2(mod
.
To determine the sign, we look at the
y
-coordinates.
The
y
-coordinate of
(
x
,y
)=(
x
361
,y
361
)+(
x,
−
y
) is computed to be
y
(9
x
11
+13
x
10
+15
x
9
+15
x
7
+18
x
6
+17
x
5
+8
x
4
+12
x
3
+8
x
+6)
(mod
ψ
5
)
.
The
y
-coordinate of (
x
,y
)=2(
x, y
)is
y
(13
x
10
+15
x
9
+16
x
8
+13
x
7
+8
x
6
+6
x
5
+17
x
4
+18
x
3
+8
x
+ 18)
(mod
ψ
5
)
.
A computation yields
(
y
+
y
19
)
/y ≡
0(mod
ψ
5
)
.
This means that
(
x
19
,
y
19
)=
(
x
,y
)
2(
x
q
,y
q
)(mod
ψ
5
)
.
≡
−
−
It follows that
a
2(mod5).
As we showed above, the information from
=2
,
3
,
5 is sucient to yield
a
=
−
7. Therefore, #
E
(
F
19
) = 27.
≡−
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