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GF(2
l
). Since GF(2
ne
)
GF(2
l
)=GF(2
e
), we have
Z
n
(
v
)
and thus,
Z
n
(
v
)
∈
∈
GF(2
e
). Therefore,
c
Z
n
(
a
)
a
1
B
2
2
l
(
a
)+
B
2
l
A
a
(
V
a
)=
n
(
a
)+
a
0
B
n
(
a
)+
Z
n
(
a
)
(9)
n
(
3
=
c
Z
n
(
a
)
a
1
B
2
2
l
n−
1
(
a
)+
a
1
a
0
B
2
2
l
n−
2
(
a
)+
B
2
l
n
(
a
)+
a
0
B
n
(
a
)+
Z
n
(
a
)
(
4
=
c
Z
n
(
a
)
B
n
+1
(
a
)+
a
0
B
2
l
n−
1
(
a
)+
Z
n
(
a
)
=0
.
Showing that in our case
V
a
is the only zero of
A
a
(
x
)isdoneexactlytheway
presented in [7, Proposition 2].
Using Corollary 1, we can obtain the number of
a
GF(2
ne
)
∗
∈
such that
Z
n
(
a
)
=0(notethat
Z
n
(0) = 1). Observe that this number is identical to
N
0
from Theorem 1 that is equal to the number of
a
GF(2
ne
)
∗
such that
l
a
(
x
)=0
has exactly one root in GF(2
ne
) (see explanations following Theorem 1). There-
fore, if
A
a
(
x
) has exactly one zero in GF(2
ne
) then its homogeneous part
l
a
(
x
)
has the same number of zeros and so
a
is necessarily such that
Z
n
(
a
)
∈
=0.
GF(2
ne
)
Finally, to prove the trace identity for
V
a
first note that for any
v
∈
(
B
n
(
v
)+
Z
n
(
v
))
(
5
=Tr
ne
B
n
(
v
)+
B
n
+1
(
v
)+
v
0
B
2
l
Tr
ne
e
n−
1
(
v
)
(10)
e
(
3
=Tr
ne
e
B
n
(
v
)+
B
n
(
v
)+
v
n−
1
B
n−
1
(
v
)+
v
0
B
2
l
n−
1
(
v
)
v
n−
1
B
n−
1
(
v
)+(
v
n−
1
B
n−
1
(
v
))
2
l
=Tr
ne
e
=0
.
Therefore, since
c
and
Z
n
(
v
) are both in GF(2
e
), then
c
+
c
B
n
(
v
)+
Z
n
(
v
)
Z
n
(
v
)
Tr
ne
(
V
v
)=Tr
ne
c
Z
n
(
v
)
Tr
ne
=Tr
e
(
nc
)+Tr
e
(
B
n
(
v
)+
Z
n
(
v
))
e
=Tr
e
(
nc
)
.
This completes the proof.
GF(2
ne
)
∗
. Then polynomial
Proposition 3.
Let
n
be odd and take any
a
∈
A
a
(
x
)
has exactly
2
e
zeros in
GF(2
ne
)
if and only if
Z
n
(
a
)=0
and
B
n
(
a
)
=0
.
Moreover, these zeros are the fol lowing
n
−
1
2
B
2
(2
i
+1)
l
n−
1
(
a
)
B
2
(2
i
+1)
l
+2
2
il
−
1
v
µ
=
c
+
μB
n
(
a
)
(
a
)
n
i
=0
GF(2
e
)
and for each zero of this type
Tr
ne
(
v
µ
)=0
.Also
with
μ
∈
|
M
2
e
|
=
2
k−e
−
1
.
