Information Technology Reference
In-Depth Information
∈
F
p
m
⊆
F
p
n
and gcd(
m
Since
c
k, n
)=
d
by Equation (2), by Lemma 4,
Equation (4) has either 0, 1, 2, or
p
d
+1 roots in
−
F
p
n
. Thus, so does
g
δ,γ
(
y
)=0.
(2) Let
y
1
,
y
2
be two different solutions of
g
δ,γ
(
y
)=0in
F
p
n
.Then
y
1
y
2
(
y
1
−
y
2
)
p
m
−
k
δ
p
n
−
k
,
i.e.,
y
1
y
2
=
δ
1
−p
n
−
k
(
y
1
−
y
2
)
1
−p
m
−
k
.
This together with Equation (2) imply (
y
1
y
2
)
(
γy
1
+
δ
)
y
2
δ
p
n
−
k
(
γy
2
+
δ
)
y
1
δ
p
n
−
k
)=
δ
(
y
1
−y
2
)
=(
−
)
−
(
−
p
n
−
1
1
=1
.
(3) Suppose that
y
0
is the unique solution of
g
δ,γ
(
y
)=0.Thus
x
0
=
p
d
−
γy
0
δ
−
∈
F
p
m
,
x
p
m
is the unique solution of Equation (4) in
F
p
n
.Since
c
is also a
0
solution of Equation (4). Then
x
0
=
x
p
m
0
, i.e.,
x
0
∈
F
p
m
. By Lemma 4, one has
p
m
p
m
p
m
−
1
−
1
−
1
=(
δ
p
n
−
k
−
1
y
p
m
−
k
+1
0
γy
0
δ
1=(
x
0
−
1)
=(
−
−
1)
)
1
.
Similarly, if
y
p
d
p
d
p
d
−
1
−
1
−
is a solution of
g
δ,γ
(
y
) = 0, one can verify that
y
p
m
δ
1
−p
m
is also a solution of
g
δ,γ
(
y
)=0.Thus,
y
0
=
y
p
m
0
δ
1
−p
m
, i.e.,
y
p
m
−
1
=
δ
p
m
−
1
.
By Equation (2), we
0
(
p
n
−
k
−
1)(
p
m
(
p
m
−
k
+1)(
p
m
p
m
−
k
p
d
p
n
−
1)
−
1)
−
1
−
1
(
p
n
−
1)
p
d
−
1
p
d
p
d
have 1 =
δ
y
−
1
=
y
. This leads to
y
−
1
=
0
0
0
p
m
−
k
p
d
−
1
(
p
n
−
1)
)
p
m
+
k
=1.
(
y
0
Remark 6.
(1) For given
γ
and
δ
with
γδ
= 0, by Proposition 5(3), if
g
δ,γ
(
y
)=0
F
p
n
, then the unique solution is (
p
d
has exactly one solution in
−
1)-th power
in
F
p
n
.
(2) By Proposition 5(2), if
g
δ,γ
(
y
) = 0 has at least two different solutions in
F
p
n
, then either all these solutions or none of them are (
p
d
1)-th powers in
F
p
n
. Over the finite field of characteristic 2, an analogy of Proposition 5(2) was
obtained in Proposition 2 of [8]. But the analogy of Proposition 5(3) does not
exist in [8].
−
With Proposition 5, rank(
Π
γ,δ
) can be determined as follows.
Proposition 7.
For
γ
=0
or
δ
=0
, the rank of
Π
γ,δ
(
x
)
is
n
,
n
−
d
,or
n
−
2
d
.
Proof.
The integer
p
n−
rank(
Π
γ,δ
)
is equal to the number of
z
∈
F
p
n
such that
Π
γ,δ
(
x
+
z
)=
Π
γ,δ
(
x
)holdsforall
x
∈
F
p
n
. This equation holds if and only if
γz
p
m
+
δz
p
k
+(
δz
)
p
n
−
k
=0
(5)
since Equation (5) implies
Tr
1
(
γz
p
m
+1
)+
Tr
1
(
δz
p
k
+1
)=0.
When
δ
= 0, one has
γ
= 0 and then Equation (5) has only zero solution. In
the sequel, we only consider the case
δ
=0.If
γ
= 0, Equation (5) is equivalent to
z
(
z
p
2
k
−
1
+
δ
1
−p
k
) = 0. The number of all solutions to this equation is
p
gcd(2
k,n
)
=
p
2
d
or 1 depending on whether
δ
1
−p
k
is a (
p
2
d
−
−
1)-th power in
F
p
n
or not.
=0,
γz
p
m
+
δz
p
k
+(
δz
)
p
n
−
k
=
z
p
k
(
δ
+
γz
p
m
−p
k
+
δ
p
n
−
k
z
p
n
−
k
−p
k
)=0.
Thus, we only need consider the number of nonzero solutions to the equation
For
γ
δ
+
γz
p
m
−p
k
+
δ
p
n
−
k
z
p
n
−
k
−p
k
=0
,
(6)
which becomes
g
δ,γ
(
y
)=
δ
p
n
−
k
y
p
m
−
k
+1
+
γy
+
δ
= 0 if let
y
=
z
p
k
(
p
m
−
k
−
1)
.By
Proposition 5(1),
g
δ,γ
(
y
) = 0 has either 0, 1, 2, or
p
d
+1 roots in
F
p
n
.Remark6
