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In-Depth Information
f
2
(
x
)=
x
2
+
x
90
over GF(3
5
)
First, 90 = 3
2
+3
4
,so
k
= 2. Due to the Theorem 1, we need to check for all
a
GF(3
5
)whetherTr(
a
2
−p
k
−p
2
k
)=Tr(
a
−
88
)isequalto(
1)
k
+1
(1+2
2
k
)=
∈
−
88
,
3
5
−
1) = 22. That is, for any nonzero
a
,
a
−
88
is a 11th root of unity over GF(3). Moreover,
x
11
17
≡
1mod3.Next,gcd(
−
−
−
1 factorizes in mod
3as
x
11
1=(
x
+2)(
x
5
+2
x
3
+
x
2
+2
x
+2)(
x
5
+
x
4
+2
x
3
+
x
2
+2)
−
This means that either
a
−
88
= 1 (and so has trace 2) or is a root of one of
the quintics and so has trace equal to the negative of one of the coecients
of
x
4
, i.e., 0 or 2. It cannot be 1. Hence,
f
2
(
x
)isPN.
Note that the function
x
2
+
x
90
is PN and it is inequivalent to the known
PN functions over GF(3
5
), see [8] for inequivalency proof.
5 Conclusions
In this paper, we proposed an approach for assurance of perfect nonlinearity
which involves simply checking a trace condition. This approach offers different
perspective for viewing nonlinear mappings.
References
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n
with collineation groups of order
n
2
.
Math. Z. 103, 239-258 (1968)
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(1997)
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