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3 Another Family of PN Multinomials
In this section we show that one of the ways to construct PN mappings is to ex-
tend a known family of APN functions over
F
2
n
to a family of PN functions over
F
p
n
for odd primes
p
. Below we construct a class of PN functions by following
the pattern of APN multinomials over
F
2
2
k
presented in [4].
Theorem 2.
Let
p
be an odd prime,
s
and
k
positive integers,
n
=2
k
,and
gcd(
k
+
s, n
)=gcd(
k
+
s, k
)
.If
b
∈
F
p
n
is not a square,
c
∈
F
p
n
\
F
p
k
,and
r
i
∈
F
p
k
,
0
≤
i<k
, then the function
k−
1
(
bx
p
s
+1
)+
cx
p
k
+1
+
r
i
x
p
k
+
i
+
p
i
F
(
x
)=Tr
2
k
i
=1
is PN over
F
p
n
.
∈
F
p
n
the equation
Δ
(
x
) = 0 has only 0
Proof.
We have to show that for any
a
as a solution when
Δ
(
x
)=
F
(
x
+
a
)
−
F
(
x
)
−
F
(
a
)
b
(
x
p
s
a
+
xa
p
s
)
+
c
(
x
p
k
a
+
xa
p
k
)+
k−
1
r
i
(
x
p
k
+
i
a
p
i
+
x
p
i
a
p
k
+
i
)
.
=Tr
2
k
i
=1
After replacing
x
by
ax
we get
ba
p
s
+1
(
x
p
s
+
x
)
+
ca
p
k
+1
(
x
p
k
+
x
)
Δ
1
(
x
)=
Δ
(
ax
)=Tr
2
k
k−
1
r
i
a
p
k
+
i
+
p
i
(
x
p
k
+
i
+
x
p
i
)
.
+
i
=1
Δ
1
(
x
)
p
k
Since
Δ
1
(
x
)=0then
Δ
1
(
x
)
−
=0,thatis,
(
ca
p
k
+1
c
p
k
a
p
k
+1
)(
x
p
k
+
x
)=0
.
−
Thus,
c
p
k
)
a
p
k
+1
(
x
p
k
+
x
)=0
(
c
−
and, therefore,
x
p
k
=
−
x
since
c
\
F
p
k
.
Substituting
x
p
k
=
∈
F
p
2
k
−
x
in
Δ
1
(
x
)=0weobtain
Δ
1
(
x
)=
ba
p
s
+1
(
x
p
s
+
x
)+
b
p
k
a
p
s
+
k
+
p
k
(
x
p
s
+
k
+
x
p
k
)
=(
ba
p
s
+1
b
p
k
a
p
s
+
k
+
p
k
)(
x
p
s
+
x
)
.
−