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=(1+
x
)
k
q
−
1
x
2
s
q
−
1
1+(
x
2
s
q
−
1
)
2
a
q
−
2
1+
x
2
s
q
−
1
=0
.
Hence,
x
d
is not APN if
(
gcd
(
l
1
−
1
,
2
m
=1or,
gcd
(2
a
0
1
,
2
m
−
1)
−
−
1)
=1)and
1
,
2
m
=1or,
gcd
(2
a
1
1
,
2
m
(
gcd
(
l
2
−
−
1)
−
−
1)
=1)and
···
and
(
gcd
(
l
q
−
1
,
2
m
=1or,
gcd
(2
a
q
−
1
1
,
2
m
−
1)
−
−
1)
=1)and
gcd
(2
a
q
−
2
,
2
m
=1.
That is,
x
d
is not APN if
(
gcd
(
l
1
−
−
1)
1
,
2
m
−
1)
=1or,
gcd
(
a
0
,m
)
=1)and
1
,
2
m
(
gcd
(
l
2
−
−
1)
=1or,
gcd
(
a
1
,m
)
=1)and
···
and
(
gcd
(
l
q
−
1
,
2
m
−
1)
=1or,
gcd
(
a
q−
1
,m
)
=1)and
gcd
(
a
q
−
1
,m
)
=1.
Table 1.
F
(
X
)=
X
d
,d
=2
p
−
1
,p≥
0
p d
Result:
m
such that
X
d
is not APN
1 1
X
can not be APN for any
m
2 3 no information for
X
3
3 7
X
7
can not be APN if
m
is even
4 15
X
15
can not be APN if 3
| m
5 31
X
31
can not be APN if
m
is even
6 63
X
63
can not be APN if 5
| m
7 127
X
127
can not be APN if 2
| m
or 3
| m
8 255
X
255
can not be APN if 7
| m
9 511
X
511
can not be APN if
m
is even
10 1023
X
1023
can not be APN if 3
| m
Table 2.
F
(
X
)=
X
d
,d
=2
p
−
1andodd
d
Result:
m
such that
X
d
is not APN
5
X
5
can not be APN if 3
|
(2
m
−
1) i.e.,
m
is even
9
X
9
can not be APN if 7
|
(2
m
−
1) i.e., 3
| m
11
X
11
can not be APN if 2
| m
or 3
| m
13 no information for
X
13
17
X
17
can not be APN if gcd(15
,
2
m
−
1)
= 1 i.e.,
m
is even
19
X
19
can not be APN if 2
| m
21
X
21
can not be APN if 19
|
(2
m
−
1) and 2
| m
23
X
23
can not be APN if 2
| m
or 3
| m
25 no information for
X
25
27 no information for
X
27
29
X
29
can not be APN if 2
| m
and gcd(27
,
2
m
−
1)
= 1 (i.e., 2
| m
is enough)
Following Theorem 5, we can specify some
m
such that
X
d
is not APN for given
d>
0. At first we present a simple case when
d
=2
p
1.
X
2
p
−
1
is not APN if
−
