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GF(2
ne
)
having the form of (6) with
n>
2
and
Corollary 2.
For any
V
∈
Tr
ne
e
(
v
0
)
=0
we have
B
2
l
n−
B
2
l
n−
1
(
V
)
B
2
l
+1
1
(
V
)
B
n
+1
(
V
)
B
2
l
+1
Tr
ne
e
=0=Tr
ne
e
,
(
V
)
(
V
)
n
n
where the second identity holds if and only if
n
is odd.
Proof.
Using (7), it can be verified directly that
n
j
=2
v
j
B
2
l
v
1
n−
1
(
V
)
B
2
l
+1
1+
v
1
v
0
=N
ne
e
and
Tr
ne
e
(
v
0
)
2
(
V
)
n
B
2
l
n−
1
(
V
)
B
n
+1
(
V
)
B
2
l
+1
v
1
Tr
ne
e
=
(
v
0
)
2
+1
(
V
)
n
GF(2
ne
)havingtheformof(6)and
n>
2. Now note that
for any
V
∈
n
Tr
ne
e
=Tr
ne
e
v
1
Tr
ne
(
v
0
)+
v
1
=Tr
ne
e
(
v
0
)
2
+Tr
ne
(
v
0
)=0
.
v
1
v
j
e
e
j
=2
The second trace is obviously equal 0 if
n
is odd and is equal 1 if
n
is even.
Lemma 2.
Assume
A
a
(
x
)
hasazeroin
GF(2
k
)
.Take
v
0
being any zero of
A
a
(
x
)
in
GF(2
k
)
. Then for any zero
v
of
A
a
(
x
)
in
GF(2
k
)
holds
Tr
k
(
v
)=Tr
k
(
v
0
)
.
Proof.
The trace identity follows by observing that any zero of
A
a
(
v
) is obtained
as a sum of
v
0
and a zero of its homogeneous part
a
2
l
x
2
2
l
+
x
2
l
+
ax
.Toprovethe
identity it therefore su
ces to show that Tr
k
(
v
1
) = 0 for any
v
1
with
a
2
l
v
2
2
l
+
1
v
2
1
+
av
1
= 0. This follows from
)=Tr
k
(
v
2
l
+2
l
1
)=Tr
k
(
a
2
l
v
2
2
l
+2
l
1
+
av
2
l
+1
1
Tr
k
(
v
1
)=Tr
k
(
v
2
l
+1
)=0
,
1
as claimed.
We will need the following result that can be obtained combining Theorems 5.6
and 6.4 in [6].
Theorem 1 ([6]).
Take polynomials over
GF(2
k
)
f
(
x
)=
x
2
l
+1
+
b
2
x
+
b
2
g
(
x
)=
b
−
1
f
(
bx
2
l
1
)=
b
2
l
x
2
2
l
1
+
b
2
x
2
l
−
−
−
1
+
b
a
nd
with
b
=0
and
gcd(
l, k
)=
e
. Then exactly one of the following holds
(i)
f
(
x
)
has none or two zeros in
GF(2
k
)
and
g
(
x
)
has none zeros in
GF(2
k
)
;
(ii)
f
(
x
)
has one zero in
GF(2
k
)
,
g
(
x
)
has
2
e
1
zeros in
GF(2
k
)
and each
rational root
δ
of
g
(
x
)
satisfies
Tr
e
(
b
−
1
δ
−
(2
l
+1)
)
−
=0
;
