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In-Depth Information
5. Ness, G.J., Helleseth, T.: A new family of four-valued cross correlation between
m
-sequences of different lengths. IEEE Trans. Inf. Theory 53(11), 4308-4313 (2007)
6. Bluher, A.W.: On
x
q
+1
+
ax
+
b
. Finite Fields and Their Applications 10(3), 285-305
(2004)
7. Helleseth, T., Kholosha, A.: On the equation
x
2
l
+1
+
x
+
a
=0overGF(2
k
). Finite
Fields and Their Applications 14(1), 159-176 (2008)
A
Remaining Proof of Proposition 3
By Corollary 1,
a
has the form of (6) and, by Proposition 1, Tr
ne
e
=0.
Using these facts and assuming that
n
is odd (note that the latter assumption
is involved only at this stage), we compute
(
v
0
)
n
−
1
2
n
−
1
2
B
2
(2
i
+1)
l
B
2
(2
i
+1)
l
n−
1
(
a
)
B
2
(2
i
+1)
l
+2
2
il
−
1
n−
1
(
a
)
B
2
(2
i
+1)
l
+2
2
il
=
ca
1
B
2
2
l
A
a
c
(
a
)
n
(
a
)
(
a
)
n
n
i
=0
i
=1
1
2
n
−
B
2
2
l
B
2
2
il
n−
1
(
a
)
B
2
2
l
+2
l
n−
1
(
a
)
B
2
2
il
+2
(2
i
−
1)
l
+
ca
1
B
2
2
l
+
cB
2
l
(
a
)
n
(
a
)
n
(
a
)
(
a
)
n
n
i
=1
1
2
n
−
n
(
a
)
B
2
l
B
2
(2
i
+1)
l
n−
1
(
a
)
B
2
l
+1
n−
1
(
a
)
B
2
(2
i
+1)
l
+2
2
il
+
cB
2
l
+
ca
0
B
n
(
a
)
+
c
(
a
)
(
a
)
n
n
i
=0
n
−
1
2
n
−
1
2
B
2
(2
i
+1)
l
B
2
2
il
(
11
=
cB
2
l
n−
1
(
a
)
B
2
(2
i
+1)
l
+2
2
il
n−
1
(
a
)
B
2
2
il
+2
(2
i
−
1)
l
+
cB
2
l
n
(
a
)
n
(
a
)
(
a
)
(
a
)
n
n
i
=0
i
=1
+
ca
0
B
n
(
a
)
B
2
l
B
2
2
l
n−
1
(
a
)
B
2
l
+1
n−
1
(
a
)
B
2
2
l
+2
l
+
ca
1
B
2
2
l
(
a
)
+
c
n
(
a
)
(
a
)
n
n
B
2
l
n−
+
c
a
0
B
2
l
1
(
a
)+(
a
0
B
2
l
1
(
a
))
2
l
1
(
a
)
B
2
l
+1
=
cB
2
l
n
(
a
)Tr
nl
n−
n−
+
c
l
B
2
n
(
a
)
(
a
)
n
=
c
(
a
n−
1
B
n−
1
(
a
))
2
l
+
B
2
l
n
+1
(
a
)
+
c
(
3
)
=0
,
(
∗
)
B
2
n
(
a
)
where (
) holds by Corollary 2 (since the value under the trace function is
an element of GF(2
ne
)) and since
B
n
+1
(
a
)=
a
0
B
2
l
∗
n−
1
(
a
) resulting from (5) if
Z
n
(
a
)=0.
Finally, to prove the trace identity for
v
µ
first note that, by (10), Tr
ne
(
B
n
(
a
)+
e
Z
n
(
a
)) = Tr
ne
(
B
n
(
a
)) = 0 if
Z
n
(
a
) = 0. Further,
e
1
2
n
−
n
−
1
2
n−
1
B
2
(2
i
+1)
l
B
2
jl
n
(
a
)
B
2
(2
i
+
j
+1)
l
n−
1
n−
1
(
a
)
B
2
(2
i
+1)
l
+2
2
il
−
1
(
a
)
Tr
ne
e
=
B
2
(2
i
+
j
+1)
l
+2
(2
i
+
j
)
l
(
a
)
(
a
)
n
n
i
=0
j
=0
i
=0
