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Proof.
The proof of (i) is trivial, and we only give the proof of (ii) and (iii).
(ii) It is true that
S
(
γ,δ,
0)
2
γ
∈
F
p
m
,δ
∈
F
p
n
ω
Tr
1
δ
(
x
p
k
+1
+
y
p
k
+1
)
ω
Tr
1
(
γ
(
x
p
m
+1
+
y
p
m
+1
)
)
=
x,y∈
F
p
n
γ∈
F
p
m
δ∈
F
p
n
=
p
n
+
m
T
2
|
where
T
2
consists of all solutions (
x, y
)
|
×
F
p
n
to the system of equations
x
p
m
+1
+
y
p
m
+1
=0
,
x
p
k
+1
+
y
p
k
+1
=0
.
∈
F
p
n
(10)
If
xy
=0,then
x
=
y
= 0 by (10). If
xy
= 0, again by (10), one has
=
(
y
)
p
k
(
p
m
−
k
−
1)
p
n
−
k
(
y
)
p
k
(
p
m
−
k
−
1)
= 1 which implies (
y
)
p
m
−
k
−
1
=1,then
x
y
∈
F
p
n
∩
F
p
m
−
k
=
F
p
d
since gcd(
m
−
k, n
)=
d
by (2). Let
x
=
ty
for some
∈
F
p
d
, then Equation (10) becomes
t
2
+1=0 since gcd(
m, k
)=
d
and
xy
t
=0.
For
p
≡
3mod4,
−
1 is a non-square element in
F
p
d
since
d
is odd. Thus,
t
2
+ 1 = 0 has no solutions and
|
T
2
|
=1.For
p
≡
1mod4,
−
1 is a square element
F
p
d
and
t
2
+1=0has2solutionsin
F
p
d
. Then,
=1+2(
p
n
1) = 2
p
n
in
|
T
2
|
−
−
1.
S
(
γ,δ,
0)
3
=
p
n
+
m
|
T
3
|
(iii) Similar analysis as in (ii) shows
,where
γ
∈
F
p
m
,δ
∈
F
p
n
T
3
consists of all solutions (
x, y, z
)
×
F
p
n
to the system of equations
x
p
m
+1
+
y
p
m
+1
+
z
p
m
+1
=0
,
x
p
k
+1
+
y
p
k
+1
+
z
p
k
+1
=0
.
∈
F
p
n
×
F
p
n
(11)
For
xyz
=0,then
x
=
y
=
z
= 0, or there are exactly two nonzero elements
in
. Thus, by (10), in this case the number of solutions to Equation (11)
is equal to 3(
{
x, y, z
}
|
T
2
|−
1) + 1 = 3
|
T
2
|−
2. For
xyz
= 0, the number of solutions to
Equation (11) is (
p
n
−
1) multiples of that to
x
p
m
+1
+
y
p
m
+1
+1=0
,
x
p
k
+1
+
y
p
k
+1
+1=0
(12)
∈
F
p
n
. Equation (12) implies
y
(
p
k
+1)
p
m
+
y
p
k
+1
y
(
p
m
+1)
p
k
y
p
m
+1
=
where
x, y
−
−
(
y
p
m
+
k
y
)(
y
p
m
y
p
k
)=0
.
By (2), one has gcd(
m
+
k, n
)=
d
.Then
y
−
−
∈
F
p
m
+
k
∩
F
p
n
=
F
p
d
,or
y
∈
F
p
m
∩
F
p
k
=
F
p
d
, i.e.,
y
∈
F
p
d
. Similarly, one has
x
∈
F
p
d
.Bythefactgcd(
m, k
)=
d
, Equation (12) is equivalent to
x
2
+
y
2
+1=0
,x,y
∈
F
p
d
.
(13)
∈
F
p
d
×
F
p
d
to
x
2
+
y
2
+1=0 is
By Lemma 1, the number of solutions (
x, y
)
p
d
+
v
(
1). Notice that the number of solutions to
xy
=0and
x
2
+
y
2
+1 = 0
−
1)
η
(
−
in
F
p
d
is 0 for
p
≡
3 mod 4, and 4 for
p
≡
1 mod 4. Then, the number of solutions
to Equation (13) is equal to
p
d
+1 for
p
3 mod 4, and (
p
d
≡
−
1)
−
4for
p
≡
1mod4.
=
p
n
+
d
+
p
n
p
d
.
Therefore, one has
|
T
3
|
−
