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that has two solutions
cδ
and
cδ
−
1
which do not belong to GF(2
e
). Thus,
Tr
e
(
c
2
)=Tr
e
(
c
)=1.
Define function
χ
(
x
)=(
1)
Tr
2
k
(
ax
2
l
+1
)+Tr
k
(
x
2
k
+1
)
on GF(2
2
k
) and linearized
polynomial
L
a
(
x
)=
a
2
l
x
2
2
l
+
x
2
k
+
l
+
ax
. Note that for any
u, v
−
GF(2
2
k
)with
∈
L
a
(
u
)=
L
a
(
v
)=0wehave
a
(
uv
2
l
+
u
2
l
v
)
+Tr
k
(
uv
2
k
+
u
2
k
v
)
1)
Tr
2
k
χ
(
u
+
v
)=
χ
(
u
)
χ
(
v
)(
−
u
2
l
L
a
(
v
)
1)
Tr
2
k
=
χ
(
u
)
χ
(
v
)(
−
=
χ
(
u
)
χ
(
v
)
.
Therefore,
χ
(
x
) defines homomorphism on the set of zeros of
L
a
(
x
) and, thus,
since
L
a
(
x
) is a linearized polynomial,
χ
(
x
) is either identically 1 or is balanced
on this set. Now we can compute
1)
Tr
2
k
(
a
(
x
2
l
+1
+
y
2
l
+1
))+Tr
k
(
x
2
k
+1
+
y
2
k
+1
)
S
0
(
a
)
2
=
(
−
x,y∈
GF(2
2
k
)
a
((
v
+
y
)
2
l
+1
+
y
2
l
+1
)
(
v
+
y
)
2
k
+1
+
y
2
k
+1
1)
Tr
2
k
+Tr
k
=
(
−
y,v∈
GF(2
2
k
)
a
(
vy
2
l
+
v
2
l
y
+
v
2
l
+1
)+
yv
2
k
+Tr
k
(
v
2
k
+1
)
1)
Tr
2
k
−
=
(
y,v∈
GF(2
2
k
)
y
2
l
L
a
(
v
)
1)
Tr
2
k
(
av
2
l
+1
)+Tr
k
(
v
2
k
+1
)
1)
Tr
2
k
=
(
−
(
−
v∈
GF(2
2
k
)
y∈
GF(2
2
k
)
=2
2
k
χ
(
v
)
v∈
GF(2
2
k
)
,L
a
(
v
)=0
=2
2
k
#
GF(2
2
k
)
{
v
∈
|
L
a
(
v
)=0
}
or
0
.
GF(2
k
) and, thus,
L
a
(
x
)=
a
2
k
+
l
x
2
2(
k
+
l
)
+
x
2
k
+
l
+
ax
is similar
to
l
a
(
x
) from (8). By Note 1,
L
a
(
x
) has either 1, 2
e
or 2
2
e
zeros in GF(2
2
k
)if
l/e
is even because
Now recall that
a
∈
gcd(
k
+
l,
2
k
)=
e
gcd(
k/e
+
l/e,
2
k/e
)=
e
gcd(
k/e
+
l/e, k/e
)=
e
since
k/e
is odd.
Nowweshowthat
L
a
(
x
)cannothave2
e
zeros in GF(2
2
k
)if
l/e
is even. This
obviously holds for
a
=0.Take
a
= 0 and assume the opposite. Then there
GF(2
2
k
)
∗
exists some
V∈
with
L
a
(
V
) = 0 and all zeros of
L
a
(
x
)areexactly
2
k
)=
L
a
(
)
2
k
GF(2
e
)
GF(2
k
) and,
{
μ
V|
μ
∈
}
.Notethat
L
a
(
V
V
=0since
a
∈
2
k
−
1
GF(2
e
). Take
ξ
being a primitive element of GF(2
2
k
) and assume
thus,
V
∈
2
k
=
ξ
i
.Then
−
1
GF(2
e
) if and only if 2
2
k
1 divides
i
(2
k
1)(2
e
V
V
∈
−
−
−
1)
which is equivalent to 2
k
+ 1 divide
i
(2
e
1) and, further, to 2
k
+ 1 divide
i
since
−
gcd(2
k
+1
,
2
e
GF(2
k
)
∗
.Takingany
δ
=0
it can be checked directly that
L
a
(
a
−
1
δ
) = 0 if and only if
G
(
δ
)=0,where
−
1) = 1 if
k/e
is odd. Therefore,
V∈
