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and the fact gcd(
p
k
(
p
m−k
1)
,p
n
1) =
p
d
−
−
−
1 show that Equation (6) has 0,
p
d
1, 2(
p
d
1), or (
p
d
1)(
p
d
+1)=
p
2
d
−
−
−
−
1 nonzero solutions in
F
p
n
. Then,
Equation (5) has 1,
p
d
,2
p
d
1, or
p
2
d
solutions. Since 2
p
d
−
−
1isnotapower
of
p
and hence is not the number of solutions of an
F
p
-linearization polynomial,
the number of solutions to Equation (5) is equal to 1,
p
d
,or
p
2
d
.
Therefore, the rank of
Π
γ,δ
(
x
)iseither
n
,
n
−
d
,or
n
−
2
d
.
In order to further determine the rank distribution of
Π
γ,δ
(
x
), we define
R
i
=
(
γ,δ
)
|
rank(
Π
γ,δ
)=
n
−
i,
(
γ,δ
)
∈
F
p
m
×
F
p
n
\{
(0
,
0)
}
(7)
for
i
=0,
d
,and2
d
. To achieve this goal, we need to evaluate the sum
S
(
γ,δ,
)=
x
ω
Π
γ,δ
(
x
)+
Tr
1
(
x
)
,
(8)
∈
F
p
n
where
γ
∈
F
p
,let
N
γ,δ,
(
ρ
) denote the number of
solutions to
Π
γ,δ
(
x
)+
Tr
1
(
x
)=
ρ
. Then, (8) can be expressed as
∈
F
p
m
,δ,
∈
F
p
n
.For
ρ
p−
1
N
γ,δ,
(
ρ
)
ω
ρ
.
S
(
γ,δ,
)=
(9)
ρ
=0
Let
f
(
x
)=
Π
γ,δ
(
x
) be as in (1). For convenience, for
i
∈{
0
,d,
2
d
}
, we define
j
=1
a
j
,
where
n−i
n
−
i
2
1)
n
−
i
Δ
i
=(
−
denotes the largest integer not exceeding
2
n−i
2
,andthecoecients
a
j
are defined by (3).
In what follows, we will study the values of
S
(
γ,δ,
0) according to the rank of
Π
γ,δ
(
x
), and then use them to determine the rank distribution.
Case 1.
(
γ,δ
)
R
0
: in this case, rank(
Π
γ,δ
)=
n
and every
a
i
in (3) is
nonzero. Since det(
Π
γ,δ
)(det(
B
))
2
∈
i
=1
n
i
=1
n
=
a
i
, one has
η
(det(
Π
γ,δ
)) =
η
(
a
i
).
Then by Lemma 1,
N
γ,δ,
0
(
ρ
)=
p
n−
1
+
v
(
ρ
)
p
n
−
2
η
(
Δ
0
)
and then by (9), we have
S
(
γ,δ,
0) =
η
(
Δ
0
)
p
2
since
ρ
=0
v
(
ρ
)
ω
ρ
=
p
.
p−
1
Case 2.
(
γ,δ
)
∈
R
d
:sincerank(
Π
γ,δ
)=
n
−
d
, without loss of generality, we
i
=1
a
i
i
=1
a
i
y
i
,
and
n−d
n−d
=0and
a
i
=0for
n
−
d<i
≤
n
. Then,
Π
γ,δ
(
x
)=
assume
by Lemma 1, for odd
d
, one has
N
γ,δ,
0
(
ρ
)=
p
d
p
n−d−
1
+
p
n
−
d
−
1
η
(
ρΔ
d
)
=
2
η
(
ρΔ
d
)
.
By (9) and Lemma 2,
S
(
γ,δ,
0) =
η
(
Δ
d
)
(
p
n−
1
+
p
n
+
d
−
1
1)
p
−
2
p
n
+
2
.
−
2
i
=1
n−
2
d
Case 3.
(
γ,δ
)
∈
R
2
d
: similarly, let
a
i
=0and
a
i
=0for
n
−
2
d<i
≤
n
.
η
(
Δ
2
d
)
and
S
(
γ,δ,
0) =
η
(
Δ
2
d
)
p
2
+
d
.
Then,
N
γ,δ,
0
(
ρ
)=
p
n−
1
+
v
(
ρ
)
p
n
+2
d
−
2
2
For each
i
∈{
0
,d,
2
d
}
, we define
R
i,j
=
{
(
γ,δ
)
∈
R
i
|
η
(
Δ
i
)=
j
}
for
j
=
±
1.
Since
d
is odd, we have
