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yields a pair ( v, w )foreach x .Thereare q
1 choices for x , hence there are
1pairs( v, w ) satisfying w 2
v 2 =4. Let( v, w ) be such a pair. Consider
q
the congruences
u 2
2 p + pv 2
(mod q )and nu 2
2 p + pv 2
(mod q ) .
If 2 p + pv 2
0(mod q ), then exactly one of these has a solution, and it has
2 solutions. If 2 p + pv 2
0(mod q ), then both congruences have 1 solution.
Therefore, each of the q − 1pairs( v, w ) contributes 2 to the sum N + N ,so
N + N =2( q − 1).
The strategy now is the following. If N> 0, we're done. If N > 0,
then C can be transformed into an elliptic curve with approximately N
points. Hasse's theorem then gives a bound on N , which will show that
N =2( q − 1) − N > 0, so there must be points on C 1 ,p,p .
LEMMA 8.30
If q ≥ 11 ,then N> 0 .
PROOF If N =0then N =2( q− 1) > 0, by Lemma 8.29. In Section 2.5.4,
we showed how to start with the intersection of two quadratic surfaces and
a point and obtain an elliptic curve.
Therefore, we can transform C
t o
an elliptic curve E . By Hasse's theorem, E has less than q +1+2 q
points. We need to check that every point on C gives a point on E .Inthe
parameterization
w = 2+2 t 2
1 − t 2
4 t
1 − t 2 ,
v =
(8.16)
of w 2
v 2
2). All of
the other points ( v, w ) correspond to finite values of t . No (finite) pair ( v, w )
corresponds to t = ± 1 (the lines through (0 , 2) of slope t = ± 1 are parallel to
the asymptotes of the hyperbola). Substituting the parameterization (8.16)
into nu 2
= 4, the value t =
corresponds to ( v, w )=(0 ,
− pv 2 =2 p yields the curve
u 1 = 2 p
Q :
n ( t 4 +6 t 2 +1) ,
t 2 ) u .Apointon C
where u 1 =(1
2) yields a finite
point on the quartic curve Q .Since C has 2( q − 1) > 1pointsmod q ,there
is at least one finite point on Q . Section 2.5.3 describes how to change Q
to an elliptic curve E (thecasewhere Q is singular does not occur since Q
is easily shown to be nonsingular mod q when q =2 ,p ). Every point mod q
on Q (including those at infinity, if they are defined over F q ) yields a point
(possibly )on E
with ( v, w )
=(0 ,
(points at infinity on Q
yield points of order 2 on E ).
 
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