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(mod 3) can be proved similarly (essentially, the roles of P ( z 2 k )and Q ( z 2 k )are
exchanged). If k
In what follows, we treat the case k
0 (mod 3), the case ( m
k )
0(mod3),wehave2 k
1
2 (mod 3) and from Lemma 5
P ( z 2 k )=2 kn ω c δ ψ ( z 2 k )+ a
(3)
for some c
Z 8 and a
V n .Now k
0 (mod 3) implies m
k
m (mod 3),
so 2( m
0 (mod 3) or 1 (mod 3). Hence by Lemma 5
Q ( z 2 k )= 2 ( m−k ) n ω d (
k )
1) b·φ ( z 2 k )
if m
0(mod3)
(4)
2 ( m−k ) n ω d (
1) b·φ ( z 2 k ) i wt( ψ ( z 2 k ))
if m
2(mod3)
for some d ∈ Z 8 and b ∈ V n . Combining (2), (3), and (4), we arrive at
N
( f )( u 1 ,...,u 2 m )
ω c + d
1) h ( z 2 k )+ z 2 k ·u 2 k + b·φ ( z 2 k ) i wt( z 2 k ) δ ψ ( z 2 k )+ a
(
if m
0(mod3)
z 2 k ∈V n
=
ω c + d
1) h ( z 2 k )+ z 2 k ·u 2 k + b·φ ( z 2 k ) δ ψ ( z 2 k )+ a
if m
2(mod3 .
(
z 2 k
V n
In either case the term inside the sum is zero unless z 2 k = ψ 1 ( a ). Therefore,
|N
( f )( u 1 ,...,u 2 m )
|
= 1, as was claimed.
Example 6. Take m =2and k =1inTheorem4.Then f reads
f ( x 1 ,x 2 ,y 1 ,y 2 )= y 1 ·
ψ ( x 1 )+ φ ( x 1 )
·
y 2 + y 2 ·
x 2 + h ( x 1 ) .
In this way we can construct bent-negabent functions in 4 n variables of degree
ranging from 2 to n .
In general, whenever m
1 (mod3),wecanuseTheorem 4 to construct bent-
negabent functions in 2 mn variables of degree ranging from 2 to n . This yields
bent-negabent functions in 2 t variables for every t
2and t
1(mod6);if
t
1(mod3),wecantake n =1and m = t ,andif t
1(mod3)and t
1
(mod 6), we can take n =2and m = t/ 2.
In the remainder of this section we apply Theorem 3 to construct further
bent-negabent functions by taking a partial dual of f givenin(1).Wetherefore
have to prove that the partial dual of f exists with respect to certain subspaces
of V and to find an explicit expression for this function.
Write V = U
W , where dim W = k and k
mn . Suppose that we have a
function τ : V
V .Wecanseparate τ on U and W by defining
|
W
|
functions
τ z : U
U and
|
U
|
functions τ x : W
W such that
τ ( x, z )=( τ z ( x ) x ( z )) ,
x
U, z
W.
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