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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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