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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 ) .
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