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This completes the proof of the second equality of Proposition 4.27.
We now show that the number of points on v 2 =( k/ 4) u 4 +1 can be ex-
pressed in terms of Jacobi sums. By separating out the terms with u =0and
the terms with v = 0, we obtain that the number of points is
# {v | v 2 =1 } +# {u | u 4 = 4 /k}
+
a + b =1
a,b =0
v 2 = a
u 4 =
#
{
v
|
}
#
{
u
|
4 b/k
}
1
3
1
3
χ 4 ( 4 /k ) +
a + b =1
a,b
χ 2 (1) j +
χ 2 ( a ) j
χ 4 ( 4 b/k )
=
j =0
=0
j =0
=0
=0
1
3
4 /k ) +
b
3
χ 2 (1) j +
4 b/k )
=
χ 4 (
χ 4 (
j =0
=0
=0 , 1
=0
+
a
1
χ 2 ( a ) j
( p
2)
j =0
=0 , 1
4 /k ) 2 J ( χ 2 4 )+ χ 4 (
4 /k ) 3 J ( χ 2 4 )
+ χ 4 (
4 /k ) J ( χ 2 4 )+ χ 4 (
(Separate out the terms with j =0and = 0. These yield the sums over
and over j , respectively. The terms with j = =0,whichsumto p − 2, are
counted twice, so subtract p − 2. The terms with j, = 0 contribute to the
Jacobi sums.)
1
3
χ 2 ( a ) j +
χ 4 ( 4 b/k )
=
( p − 2)
j =0
a
=0
=0
b
=0
−χ 2 ( 4 /k )+ χ 4 ( 4 /k ) J ( χ 2 4 )+ χ 4 ( 4 /k ) 3 J ( χ 2 4 )
=( p − 1) + ( p − 1) ( p − 2)
4 /k ) 3 J ( χ 2 4 )
χ 2 (
4 /k )+ χ 4 (
4 /k ) J ( χ 2 4 )+ χ 4 (
(by Lemma 4.26)
= p +1 − δ + χ 4 ( 4 /k ) J ( χ 2 4 )+ χ 4 ( 4 /k ) 3 J ( χ 2 4 ) .
For the last equality, we used the fact that
4 /k )=1+ χ 2 (1 /k )= 0if k is not a square
1+ χ 2 (
2if k is a square mod p ,
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