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G
(
x
)=
b
2
k
+
l
x
2
2(
k
+
l
)
1
+
b
and
b
=
a
−
2
k
+
l
.Notethat
G
(
x
)hasthe
form of polynomial
g
(
x
) from Theorem 1. Thus,
G
(
x
)has2
e
zeros in GF(2
2
k
)
with
a
1
+
b
2
x
2
k
+
l
−
−
V
being one of them and, by the trace condition from Theorem 1 (ii),
(
a
2
k
+
l
(
a
)
−
(2
k
+
l
+1)
)
Tr
2
k
e
=0whichiswrong.
Finally, we prove the value distribution of
S
0
(
a
)
2
.By(12),
S
0
(
a
)
2
=2
2
k
if and
only if
A
a
(
x
) has just one zero in GF(2
k
). Therefore, by Proposition 2,
S
0
(
a
)
2
is
equal to 2
2
k
for
|M
1
|
=
V
2
k
+2
e
2
k
+
e
2
k
+1
−
−
−
1
values of
a
.If
l/e
is even then suppose
S
0
(
a
)
2
takes on value zero
r
times and the value 2
2(
k
+
e
)
occurs
t
times. Taking
the sum-of-squares identity from Lemma 1 giving 2
2
k
2
2
e
+2
2(
k
+
e
)
t
=2
2
k
(2
k
|
M
1
|
−
1)
+
t
=2
k
and knowing also that
r
+
|
M
1
|
−
1, we are directly led to the claimed
distribution.
Corollary 3.
Under the conditions of Proposition 5, let
n
=
k/e
. Then the
distribution of
S
0
(
a
)
for
l/e
being even is as follows:
1)
Tr
k
(
aB
n
(
a
)
2
l
+1
/Z
n
(
a
)
2
)
2
k
(
−
−
if
Z
n
(
a
)
=0
0
if
Z
n
(
a
)=0
and
B
n
(
a
)
=0
2
k
+
e
±
if
B
n
(
a
)=0
.
= 0 then, by Proposition 2,
A
a
(
x
) has exactly one zero in GF(2
k
)
Proof.
If
Z
n
(
a
)
that is equal to
V
a
=
cB
n
(
a
)
/Z
n
(
a
)andTr
k
(
V
a
)=Tr
e
(
c
)=1since
n
is odd.
Further,
aB
n
(
a
)
2
l
+1
Z
n
(
a
)
2
2
l
+1
a
ac
−
2
Tr
k
V
=Tr
k
GF(2
e
) as proved in Proposition 2.
Further, assume
Z
n
(
a
)=0and
B
n
(
a
)
since
c, Z
n
(
a
)
∈
= 0. By Proposition 1 and Corol-
GF(2
e
)withTr
ne
e
GF(2
ne
)
lary 1, there exists some
v
∈
\
(
v
)
=0suchthat
a
=
v
2
2
l
+1
0
/
(
v
0
+
v
1
)
2
l
+1
. Using Proposition 1 we obtain
2
Tr
ne
e
(
v
0
)
2
v
1
v
0
v
0
+
v
1
aB
n
(
a
)
2
l
+1
=N
ne
.
e
Note that Tr
ne
e
(
v
)=Tr
nl
l
GF(2
ne
).
By Proposition 3,
A
a
(
x
) has exactly 2
e
zeros in GF(2
k
)thathaveparticular
form
v
µ
corresponding to every
μ
(
v
)andN
ne
e
(
v
)=N
nl
l
(
v
)if
v
∈
GF(2
e
)andTr
k
(
v
µ
) = 0. Checking the trace
calculations in the proof of Proposition 3 (see Appendix A), it is easy to see that
∈
B
2
l
n−
1
(
a
)
B
n
+1
(
a
)
B
2
l
+1
Tr
ne
e
B
n
(
a
)) = Tr
ne
(
V
e
(
a
)
n
(
v
1
+Tr
ne
(
v
0
))
3
v
0
v
0
+
v
1
(
7
=N
ne
e
Tr
ne
e
e
Tr
ne
e
(
v
0
)
2
v
1
v
0
v
0
+
v
1
=N
ne
e
1+Tr
ne
e
(
v
0
)Tr
ne
(
v
−
0
)
,
e
=
c
−
1
(
v
µ
+
μB
n
(
a
)). Now it is quite technical to show that
where
V
