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Proposition 2.
Let
p
be an odd prime and
n
be a positive integer. Any function
F
of the form
a
kj
x
p
k
+
p
j
F
(
x
)=
0
≤k<j<n
F
p
n
is CCZ-inequivalent to
x
2
.
over
Proof.
Since
x
2
is a planar DO polynomial then, by Corollary 2, CCZ-equivalence
of
F
to
x
2
implies the linear equivalence, that is, the existence of linear permu-
tations
L
1
and
L
2
such that
L
1
(
x
)
2
+
L
2
F
(
x
)
=0
.
(8)
Let
n−
1
u
i
x
p
i
,
L
1
(
x
)=
(9)
i
=0
n−
1
v
i
x
p
i
.
L
2
(
x
)=
(10)
i
=0
Then equality (8) implies
⎛
a
kj
x
p
k
+
p
j
⎞
n−
1
u
i
x
p
i
2
p
i
n−
1
⎝
⎠
0=
+
v
i
i
=0
i
=0
0
≤k<j<n
n−
1
u
i
x
2
p
i
+2
u
i
u
j
x
p
i
+
p
j
+
v
i
a
p
i
kj
x
p
i
(
p
k
+
p
j
)
.
=
i
=0
0
≤i<j<n
0
≤k<j<n,
0
≤i<n
∈
F
p
n
then obviously
u
i
Since the identity above takes place for any
x
=0for
all 0
i<n
,thatis,
L
1
(
x
) = 0. This contradicts the condition that
L
1
is a
permutation. Hence
F
is CCZ-inequivalent to
x
2
.
≤
Corollary 3.
The functions
(i
∗
)
and
(ii
∗
)
are CCZ-inequivalent to
x
2
.
We give below a sucient condition on DO polynomials to be CCZ-inequivalent
to the PN functions of the case (ii).
Proposition 3.
Let
p
be an odd prime number,
n
,
n
and
t
positive integers
such that
n
<n
and
n/
gcd(
n, t
)
is odd. Let a function
F
:
F
p
n
→
F
p
n
be such
that
n
A
i
(
x
p
s
i
+1
)
,
F
(
x
)=
i
=0
n
, and the functions
A
i
,
0
≤ i ≤ n
,
are linear. If
t
=
s
i
and
t
=
n − s
i
for all
0
≤ i ≤ n
then the PN
function
G
(
x
)=
x
p
t
+1
is CCZ-inequivalent to
F
.
where
0
<s
i
<n
and
s
i
=
s
j
,
for all
i
=
j
,
0
≤
i, j
≤
