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Table 2.
Known nonpower PN functions in GF(
p
n
)
function
conditions
proved in
x
10
− x
6
− x
2
p
=3
,n
odd
[5]
x
10
+
x
6
− x
2
p
=3
,n
odd
[2]
n−
1
a
i,j
x
p
i
+
p
j
,a
i,j
∈
GF(
p
n
)
.
i,j
=0
Functions of this type are referred to as
Dembowski-Ostrom
polynomials.
The addition of any PN function and any ane function is clearly still a PN
function. This yields the following question: which functions should be considered
to be equivalent? In order to distinguish functions, we present the concept of
ane equivalence. Two functions
f, g
:GF(
p
n
)
GF(
p
n
)are
ane equivalent
if there are two linearized permutation polynomials
L
and
M
andanane
polynomial
G
such that
→
M
+
G,
which defines an equivalence relation. Note that all the Coulter-Matthews
monomials are ane inequivalent, and they are never ane equivalent to a
Dembowski-Ostrom polynomial for
k
g
=
L
◦
f
◦
=1.
In [3], it is shown that the only mappings
f
:GF(
p
n
)
GF(
p
n
) with linear
→
f
(
x
+
a
)
f
(
a
) are given by a sum of a planar Dembowski-Ostrom
polynomial and a linearized polynomial.
We now recall the trace of an element
ξ
in GF(
p
n
)overGF(
p
)
−
f
(
x
)
−
n−
1
ξ
p
i
,
Tr(
ξ
)=
(2)
i
=0
which is automatically a member of GF(
p
).
3M inR su s
Theorem 1.
Let
k
be a positive integer,
p
be an odd prime, and Tr denote the
trace mapping defined by (2). Consider a Dembowski-Ostrom binomial of the
form (1):
f
(
x
)=
x
2
+
x
p
k
+
p
2
k
over GF
(
p
2
k
+1
)
.IfTr
(
a
2
−p
k
−p
2
k
)
1)
k
+1
(1 + 2
2
k
)
for all
a
GF(
p
2
k
+1
)
,
=(
−
∈
then
f
(
x
)
is PN.
Proof.
Since
f
(
x
) is Dembowski-Ostrom polynomial then it is PN if the equation
f
(
x
+
a
)
−
f
(
x
)
−
f
(
a
)=0has
x
= 0 as its only solution for any nonzero
GF(
p
2
k
+1
). We have
a
∈
Δ
(
x
)=
f
(
x
+
a
)
− f
(
x
)
− f
(
a
)=2
ax
+
a
p
2
k
x
p
k
+
a
p
k
x
p
2
k
.
