Cryptography Reference
In-Depth Information
x
−
y
=
x
+
y
,so
y
= 0. Similarly,
y
1
=
−
36 yields
x
= 0. The point with
(
x
1
,y
1
)=
y
, which means that
z
= 0. Therefore,
there are no solutions to
x
3
+
y
3
+
z
3
=0when
xyz
∞
corresponds to
x
=
−
=0.
2.5.3
Quartic Equations
Occasionally, we will meet curves defined by equations of the form
v
2
=
au
4
+
bu
3
+
cu
2
+
du
+
e,
(2.6)
with
a
K
,then
the equation (when it is nonsingular) can be transformed into a Weierstrass
equation by an invertible change of variables that uses rational functions with
coe
cients in the field
K
. Note that an elliptic curve
E
defined over a field
K
always has a point in
E
(
K
), namely
=0. Ifwehaveapoint(
p, q
) lying on the curve with
p, q
∈
(whose projective coordinates (0 : 1 : 0)
certainly lie in
K
). Therefore, if we are going to transform a curve
C
into
Weierstrass form in such a way that all coecients of the rational functions
describing the transformation lie in
K
, then we need to start with a point on
C
that has coordinates in
K
.
There are curves of the form (2.6) that do not have points with coordinates
in
K
. This phenomenon will be discussed in more detail in Chapter 8.
Suppose we have a curve defined by an equation (2.6) and suppose we have
apoint(
p, q
) lying on the curve. By changing
u
to
u
+
p
, we may assume
p
= 0, so the point has the form (0
,q
).
First, suppose
q
=0. If
d
= 0, then the curve has a singularity at (
u, v
)=
(0
,
0). Therefore, assume
d
∞
=0. Then
(
v
u
2
)
2
=
d
(
1
u
)
3
+
c
(
1
u
)
2
+
b
(
1
u
)+
a.
This can be easily transformed into a Weierstrass equation in
d/u
and
dv/u
2
.
The harder case is when
q
= 0. We have the following result.
THEOREM 2.17
Let
K
be a fi eld of characteristic not 2. C onsider the equation
v
2
=
au
4
+
bu
3
+
cu
2
+
du
+
q
2
with
a, b, c, d, q ∈ K
.Let
y
=
4
q
2
(
v
+
q
)+2
q
(
du
+
cu
2
)
(
d
2
u
2
/
2
q
)
x
=
2
q
(
v
+
q
)+
du
u
2
−
,
.
u
3
D efine
a
2
=
c −
(
d
2
/
4
q
2
)
,
a
4
=
−
4
q
2
a,
a
1
=
d/q,
3
=2
qb,
a
6
=
a
2
a
4
.
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