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Case 2.
(
γ,δ
)
=(0
,
0):
Case 2.1.
(
γ,δ
)
b
i
2
a
i
∈
R
0
: a substitution
y
i
=
z
i
−
≤
i
≤
n
leads to
for 1
i
=1
(
a
i
y
i
+
b
i
y
i
)=
ρ ⇐⇒
i
=1
a
i
z
i
=
λ
γ,δ,
+
ρ,
where
λ
γ,δ,
=
i
=1
n
n
n
b
i
4
a
i
. Then, for
any
ρ
∈
F
p
and given (
γ,δ
)
R
0
, by Lemma 1, one has
N
γ,δ,
(
ρ
)=
p
n−
1
+
v
(
λ
γ,δ,
+
ρ
)
p
n
−
2
η
(
Δ
0
)
.
∈
(15)
p
since
CB
is non-
singular. Notice that
λ
γ,δ,
is a quadratic form with
n
variables
b
i
for 1
When
runs through
F
p
n
,(
b
1
,b
2
,
···
,b
n
) runs through
F
≤
i
≤
n
.
Then, for any given (
γ,δ
)
∈
R
0
, by Lemma 1, when
runs through
F
p
n
, one has
i
=1
n
b
i
4
a
i
=
ρ
occurring
p
n−
1
+
v
(
ρ
)
p
n
−
2
η
(
Δ
0
)times
λ
γ,δ,
=
(16)
i
=1
n
for each
ρ
1
∈
F
p
since
η
(
)=
η
(
a
i
). Thus, when
runs through
i
=1
a
i
n
4
n
1)
p
n
−
2
η
(
Δ
0
)
occurs
p
n−
1
+(
p −
F
p
n
, by (15) and (16),
N
γ,δ,
(0) =
p
n−
1
+(
p
−
1)
p
n
−
2
η
(
Δ
0
)times,and
N
γ,δ,
(0) =
p
n−
1
p
n
−
2
η
(
Δ
0
) occurs (
p
1)(
p
n−
1
−
−
−
p
n
−
2
η
(
Δ
0
)) times.
By (9) and (15),
S
(
γ,δ,
)=
η
(
Δ
0
)
p
2
ω
−λ
γ,δ,
since
ρ∈
F
p
v
(
λ
γ,δ,
+
ρ
)
ω
ρ
+
λ
γ,δ,
=
p
.Noticethat
v
(
−
λ
γ,δ,
)=
v
(
λ
γ,δ,
). By (16), for given (
γ,δ
)
∈
R
0
,when
runs
F
p
n
,
S
(
γ,δ,
)=
η
(
Δ
0
)
p
2
ω
ρ
occurs
p
n−
1
+
v
(
ρ
)
p
n
−
2
η
(
Δ
0
)times for
through
each
ρ
∈
F
p
.
n−d
i
=1
i
=1
n
R
d
:inthiscase,
Π
γ,δ
(
x
)+
Tr
1
(
x
)=
a
i
y
i
+
Case 2.2.
(
γ,δ
)
∈
b
i
y
i
.If
∈
F
p
,
N
γ,δ,
(
ρ
)=
p
n−
1
there exists some
b
i
=0for
n
−
d<i
≤
n
, then for any
ρ
and by (9)
,S
(
γ,δ,
) = 0. Further, for given (
γ,δ
)
∈
R
d
,when
runs through
F
p
n
,thereareexactly
p
n
p
n−d
−
choices for
such that there is at least one
b
i
=0with
n
−
d<i
≤
n
since
CB
is nonsingular.
If
b
i
=0forall
n
−
d<i
≤
n
, a similar analysis as in Case 2.1 shows that
i
=1
i
=1
i
=1
n
−
d
n
−
d
n
−
d
b
i
4
a
i
(
a
i
y
i
+
b
i
y
i
)=
ρ
a
i
z
i
=
λ
γ,δ,
+
ρ,
where
λ
γ,δ,
=
⇐⇒
and
b
i
2
a
i
z
i
=
y
i
+
for 1
≤
i
≤
n
−
d
. Then, for any
ρ
∈
F
p
and given (
γ,δ
)
∈
R
d
,by
d
,
N
γ,δ,
(
ρ
)=
p
d
(
p
n−d−
1
+
p
n
−
d
−
1
Lemma 1, for odd
n
−
η
((
λ
γ,δ,
+
ρ
)
Δ
d
)), i.e.,
2
N
γ,δ,
(
ρ
)=
p
n−
1
+
p
n
+
d
−
1
η
((
λ
γ,δ,
+
ρ
)
Δ
d
)
.
(17)
2
n−d
p
For given (
γ,δ
)
∈
R
d
, by Lemma 1, when (
b
1
,b
2
,
···
,b
n−d
) runs through
F
,
i
=1
n−d
b
i
4
a
i
=
ρ
occurring
p
n−d−
1
+
η
(
ρ
)
p
n
−
d
−
1
(18)
λ
γ,δ,
=
η
(
Δ
d
)times
2
for each
ρ
∈
F
p
.Thus,
η
(
λ
γ,δ,
) = 0 occurs
p
n−d−
1
times, and
±
1 occur
p−
1
2
p
n
−
d
−
1
(
p
n−d−
1
±
η
(
Δ
d
)) times, respectively. Therefore, when (
b
1
,b
2
,
···
,b
n−d
)
2
