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Proof.
Assume that
F
and
G
are CCZ-equivalent. Since
G
is a planar DO poly-
nomial then, by Corollary 2, CCZ-equivalence implies the existence of linear
permutations
L
1
and
L
2
, defined by (9)-(10), such that
G
L
1
(
x
)
+
L
2
F
(
x
)
=0
.
We get
⎛
⎝
⎞
⎠
0=
n−
1
u
i
x
p
i
p
t
+1
p
i
n
n−
1
A
i
(
x
p
s
i
+1
)
+
v
i
i
=0
i
=0
i
=0
n−
1
n
u
i
u
p
t
j
x
p
i
+
p
j
+
t
+
A
i
(
x
p
s
i
+1
)
,
=
i,j
=0
i
=0
where
A
i
,0
n
,
are some linear functions. Since the latter expression is
equal to 0 then the terms of the type
x
2
p
i
,0
≤ i<n
, should vanish and we get
≤
i
≤
u
i
u
p
t
i−t
=0
,
0
≤
i<n.
(11)
n
then canceling all terms of the type
Since
t
=
s
i
and
t
=
n
−
s
i
for all 0
≤
i
≤
x
p
i
(
p
t
+1)
,0
≤
i<n
,weget
u
i
u
p
t
u
i
+
t
u
p
t
=
−
i−t
,
0
≤
i<n.
(12)
i
Equalities (11) and (12) imply
L
1
= 0. Indeed, if
u
i
=0forsome
i
then from
(11) we get
u
i−t
= 0 while from (12) we get
u
i−t
=0.But
L
1
is a permutation
and cannot be constantly 0. This contradiction shows that the functions
F
and
x
p
t
+1
are CCZ-inequivalent.
Corollary 4.
The functions
(i
∗
)
and
(ii
∗
)
are CCZ-inequivalent to
x
p
t
+1
when
2
k/
gcd(2
k, s
)
is even.
To prove that functions (i
∗
)and(ii
∗
) are in general CCZ-inequivalent to the
functions corresponding to the Dickson semifields we need the following fact
which was checked with a computer.
Proposition 4.
The commutative semifields defined by the following planar DO
polynomials have the middle nuclei of order
p
2
and the left nuclei of order
p
:
(1)
the functions
(ii
∗
)
with
p
=3
and
n
=6
;
(2)
the functions
(i
∗
)
with
p
=3
and
n
=8
;
(3)
the functions
(i
∗
)
with
p
=5
and
n
=6
.
The work is in progress now to prove that all functions (i
∗
)and(ii
∗
) define
commutative semifields with the middle nuclei of order
p
2
.
Note that the PN functions of case (ii) are not defined for
n
=2
m
while the
PN functions (i
∗
)and(ii
∗
) are. This already shows that in general the semifields
