Cryptography Reference
In-Depth Information
Then
y
2
+
a
1
xy
+
a
3
y
=
x
3
+
a
2
x
2
+
a
4
x
+
a
6
.
Theinverse tra n sform ation is
u
=
2
q
(
x
+
c
)
−
(
d
2
/
2
q
)
y
q
+
u
(
ux
−
d
)
2
q
,
v
=
−
.
The point
(
u, v
)=(0
,q
)
corresponds tothe point
(
x, y
)=
∞
and
(
u, v
)=
(0
,
−
q
)
corresponds to
(
x, y
)=(
−
a
2
,a
1
a
2
−
a
3
)
.
PROOF
Most of the proof is a “straightforward” calculation that we omit.
For the image of the point (0
, −q
), see [28].
Example 2.2
Consider the equation
v
2
=
u
4
+1
.
(2.7)
Then
a
=1,
b
=
c
=
d
=0,and
q
=1. If
x
=
2(
v
+1)
u
2
y
=
4(
v
+1)
u
3
,
,
then we obtain the elliptic curve
E
given by
y
2
=
x
3
−
4
x.
The inverse transformation is
1+(2
x
3
/y
2
)
.
u
=2
x/y,
v
=
−
The point (
u, v
)=(0
,
1) corresponds to
∞
on
E
,and(
u, v
)=(0
,
−
1) corre-
sponds to (0
,
0). We will show in Chapter 8 that
E
(
Q
)=
{∞
,
(0
,
0)
,
(2
,
0)
,
(
−
2
,
0)
}
.
These correspond to (
u, v
)=(0
,
1)
,
(0
,
1), and points at infinity. Therefore,
the only finite rational points on the quartic curve are (
u, v
)=(0
,
−
±
1). It is
easy to deduce from this that the only integer solutions to
a
4
+
b
4
=
c
2
satisfy
ab
= 0. This yields Fermat's Last Theorem for exponent 4. We will
discuss this in more detail in Chapter 8.
It is worth considering briefly the situation at infinity in
u, v
.Ifwemake
the equation (2.7) homogeneous, we obtain
F
(
u, v, w
)=
v
2
w
2
− u
4
− w
4
=0
.
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