Cryptography Reference
In-Depth Information
Successively letting (
S, T
)=(
P
+
Q, R
), (
P, Q
−
R
), (
P
+
R, Q
), and (
Q, R
)
yields the following equations:
h
(
P
+
Q
+
R
)+
h
(
P
+
Q
R
)=2
h
(
P
+
Q
)+2
h
(
R
)
−
2
h
(
P
)+2
h
(
Q
R
)=
h
(
P
+
Q
R
)+
h
(
P
−
−
−
Q
+
R
)
h
(
P
+
R
+
Q
)+
h
(
P
+
R − Q
)=2
h
(
P
+
R
)+2
h
(
Q
)
4
h
(
Q
)+4
h
(
R
)=2
h
(
Q
+
R
)+2
h
(
Q − R
)
.
Adding together all of these equations yields
2
h
(
P
+
Q
+
R
)
−
h
(
R
)
h
(
P
+
Q
)
−
=2
h
(
P
+
R
)
−
h
(
R
)
.
h
(
P
)
−
h
(
R
)+
h
(
Q
+
R
)
−
h
(
Q
)
−
Dividing by 2 and using the definition of the pairing yields the result.
Example 8.11
Let
E
be given by
y
2
=
x
3
+ 73. Let
P
=(2
,
9) and
Q
=(3
,
10). Then
P, P
=
.
9239
...
P, Q
=
−
0
.
9770
...
Q, Q
=
.
9927
....
Since
det
0
.
9239
−
0
.
9770
−
0
.
9770
=0
.
8865
···
=0
,
1
.
9927
the points
P
and
Q
are independent on
E
.
8.6 Fermat's Infinite Descent
The methods in this chapter have their origins in Fermat's
method of
infinite descent
. In the present section, we'll give an example of Fermat's
method and show how it relates to the calculations we have been doing.
Consider the equation
a
4
+
b
4
=
c
2
.
(8.12)
The goal is to show that it has no solutions in nonzero integers
a, b, c
. Recall
the parameterization of
Pythagorean triples
:
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