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In-Depth Information
Lemma 8.
|
R
d,
1
|
=
|
R
d,−
1
|
.
∈
F
p
such that its inverse element satis-
Proof.
Let (
γ,δ
)
∈
R
d
and let
u
fies
η
(
u
−
1
)=
−
1. By (8) and the analysis in Case 2, one has
S
(
uγ, uδ,
0) =
ω
uΠ
γ,δ
(
x
)
=
ρ∈
F
p
ρ
=0
η
(
u
−
1
ρΔ
d
)
ω
ρ
=
p−
1
N
γ,δ,
0
(
u
−
1
ρ
)
w
ρ
=
p
n
+
d
−
1
−
S
(
γ,δ,
0)
.
2
x∈
F
p
n
The above equality shows that for
j
∈{
1
,
−
}
,if(
γ,δ
)
∈
R
d,j
,then(
uγ, uδ
)
∈
1
R
d,−j
. Thus, one has
|
R
d,
1
|
|
R
d,−
1
|
=
.
Applying Proposition 5 and Lemma 8, we have
=
p
m−d
(
p
n
=
p
m−d
(
p
n
Proposition 9.
|
R
d
|
−
1)
and
|
R
d,±
1
|
−
1)
/
2
.
Proof.
By (7), it is sucient to determine the number of (
γ,δ
)
∈
F
p
m
×
F
p
n
\
such that Equation (5) has exactly
p
d
solutions in
{
(0
,
0)
}
F
p
n
. By the proof of
Proposition 7, this case can occur only if
γδ
=0.
x
p
d
−
1
∈
F
p
n
}
be the set of nonzero (
p
d
Let
W
=
{
|
x
−
1)-th powers in
F
p
n
.
By Proposition 5(1),
g
δ,γ
(
y
) = 0 has either 0, 1, 2, or
p
d
+1rootsin
F
p
n
.
F
p
n
, by Proposition 5(2)
and Remark 6, if one of the solutions belongs to
W
, then all these solutions are
also in
W
and Equation (5) has at least 2(
p
d
When
g
δ,γ
(
y
) = 0 has at least two different roots in
1) + 1 = 2
p
d
1 solutions. If
none of these solutions belong to
W
, then Equation (5) has only zero solution.
Thus, in this case the number of solutions to Equation (5) can never be
p
d
.
If
g
δ,γ
(
y
) = 0 has exactly one solution in
−
−
F
p
n
, by Proposition 5(3) and Remark 6,
δ
γ
x
,
g
δ,γ
(
y
)=0has
(
p
d
1) + 1 =
p
d
solutions. Since
y
=
Equation (5) has 1
×
−
−
F
p
n
if and only if Equation (4) has exactly one solution in
F
p
n
. Furthermore, in this case the unique solution to Equation (4) belongs to
exactly one solution in
F
p
m
by the analysis in the proof of Proposition 5(3). When Equation (4) has exactly one
solution in
F
p
m
, it has exactly one solution in
F
p
n
since this equation has either 0,
1, 2, or
p
d
+1rootsin
F
p
n
and its solutions in
F
p
n
\
F
p
m
occur in pairs. Therefore,
F
p
n
if and only if Equation (4) has exactly
g
δ,γ
(
y
) = 0 has exactly one solution in
F
p
m
.
By Lemma 4, the number of
c
one solution in
∈
F
p
m
such that Equation (4) has exactly one
γ
p
m
−
k
+1
δ
p
m
−
k
(
p
m
+1)
runs through
F
p
m
is
p
m−d
. For any fixed
γ
∈
F
p
m
,
c
=
solution in
F
p
m
exactly (
p
m
+1) times when
δ
runs through
F
p
n
. Therefore, when
γ
and
δ
run
F
p
m
and
F
p
n
, respectively, there are exactly (
p
m
1)
p
m−d
(
p
m
+1) =
throughout
−
p
m−d
(
p
n
−
1) elements in
R
d
. This together with Lemma 8 finish the proof.
Proposition 10 describes the sums of
i
-th powers of
S
(
γ,δ,
0) for 1
≤
i
≤
3.
Proposition 10.
(i)
S
(
γ,δ,
0) =
p
n
+
m
.
γ∈
F
p
m
δ
∈
F
p
n
S
(
γ,δ,
0)
2
=
p
n
+
m
,
p
≡
3mod4
,
(ii)
(2
p
n
1)
p
n
+
m
,p
−
≡
1mod4
.
γ∈
F
p
m
δ∈
F
p
n
(iii)
γ∈
F
p
m
S
(
γ,δ,
0)
3
=
p
n
+
m
(
p
n
+
d
+
p
n
p
d
)
.
−
δ∈
F
p
n