Information Technology Reference
In-Depth Information
Clearly the number of solutions in
x
to the equation
Δ
(
x
)=0isthesameas
the number of solutions in
x
to the equation
Δ
(
ax
) = 0, so we replace
x
with
ax
and get
Δ
(
ax
)=2
a
2
x
+
a
p
k
+
p
2
k
(
x
p
k
+
x
p
2
k
)
.
Let
Δ
1
(
x
)=
x
p
k
+
x
p
2
k
+2
cx
=0
,
where
c
=
a
2
−p
k
−p
2
k
.Takingthe
p
-th power of the above equation consecutively
2
k
times, we obtain a set of equations which can be put into a matrix form as
follows
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
1
1
2
c
x
p
2
k
.
.
.
x
.
.
.
.
.
.
1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
=0
.
.
.
.
.
.
1
.
.
.
.
.
.
.
.
.
2
c
p
2
k
1
1
In order this system to have a unique (one) solution, the determinant of the
above matrix should be nonzero and this determinant is given by
1)
k
(1 + 2
2
k
)
.
det = 2Tr(
c
)+2(
−
Since
c
=
a
2
−p
k
p
2
k
, the conclusion follows.
−
4 Case Studies
In this section, we present two examples to illustrate our main result.
f
1
(
x
)=
x
2
+
x
56
over GF(7
3
)
First, 56 = 7
1
+7
2
,so
k
= 1. Due to the Theorem 1, we need to check for all
a
GF(7
3
)whetherTr(
a
2
−p
k
−p
2
k
)=Tr(
a
−
54
)isequalto(
1)
k
+1
(1+2
2
k
)=
∈
−
−
1) = 18. That is, for any nonzero
a
,
a
−
54
5. Next, gcd(
−
54
,
7
3
is a 19th
root of unity over GF(7). Moreover,
x
19
−
1 factorizes in mod 7 as
x
19
1=(
x
+6)(
x
3
+2
x
+6)(
x
3
+3
x
2
+3
x
+6)(
x
3
+4
x
2
+
x
+6)
·
−
(
x
3
+4
x
2
+4
x
+6)(
x
3
+5
x
2
+6)(
x
3
+6
x
2
+3
x
+6)
.
This means that either
a
−
54
= 1 (and so has trace 3) or is a root of one of
the cubics and so has trace equal to the negative of one of the coecients of
x
2
, i.e., 0, 1, 2, 3, or 4. It cannot be 5. Hence,
f
1
(
x
)isPN.
Note that the function
x
2
+
x
56
is PN and it can be shown to be equivalent
to the monomial
x
2
.
