Abstract. We are concerned about a conjecture proposed in the middle of the seventies by Hellesseth in the framework of maximal sequences and theirs cross-correlations. The conjecture claims the existence of a zero outphase Fourier coefficient. We give some divisibility properties in this direction

Philippe Langevin

Yves Aubry

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On a Conjecture of Helleseth
Yves Aubry1,2 and Philippe Langevin1,
1 Institut de Mathématiques de Toulon, Université du Sud Toulon-Var, France
2 Insitut de Mathématiques de Luminy, Université d’Aix-Marseille, France
Abstract. We are concerned about a conjecture proposed in the middle
of the seventies by Hellesseth in the framework of maximal sequences
and theirs cross-correlations. The conjecture claims the existence of a
zero outphase Fourier coefficient. We give some divisibility properties in
this direction.
1 Two Conjectures of Helleseth
Let L be a finite field of order q > 2 and characteristic p. Let μ be the canonical
additive character of L i.e.
μ(x) = exp(2iπTr (x)/p)
where Tr is the trace function with respect to the finite field extension L/Fp.
The Fourier coefficient of a mapping f : L → L is defined at a ∈ L by
̂f(a) =
∑
x∈L
μ(ax+ f(x)). (1)
The distribution of these values is called the Fourier spectrum of f . Note that
when f is a permutation the phase Fourier coefficient ̂f(0) is equal to 0.
The mapping f(x) = xs is called the power function of exponent s, and it is
a permutation if and only if (s, q − 1) = 1. Moreover, if s ≡ 1 mod (p − 1) the
Fourier coefficients of f are rational integers. Helleseth made in [3] the following
conjecture on the quantity (related to Dedekind determinant, see [9])
D(f) =
∏
a∈L×
̂f(a). (2)
Conjecture 1 (Helleseth). Let L be a field of cardinal q > 2. If f is a power
permutation of L of exponent s ≡ 1 mod (p− 1) then D(f) = 0.
For p = 2, it generalizes Dillon’s conjecture (see [2]) which corresponds to the
case s = q − 2 ≡ −1 (mod q − 1), and known to be true because it is related to
the vanishing of Kloosterman sums and the class number hq of the imaginary
quadratic number field Q(
√
1− 4q) (see [5,8]). Note also that in odd character-
istic the Kloosterman sums do not vanish (see [7]) except if p = 3 (see [5]).
In the same paper [3], Helleseth proposed a second conjecture:
 The authors would like to thank the anonymous reviewers for their valuable com-
ments and suggestions to improve this manuscript.
T. Muntean, D. Poulakis, and R. Rolland (Eds.): CAI 2013, LNCS 8080, pp. 113–118, 2013.
c© Springer-Verlag Berlin Heidelberg 2013
114 Y. Aubry and P. Langevin
Conjecture 2. If [L : Fp] is a power of 2 then the spectrum of a power permuta-
tion of exponent not a power of p modulo q − 1 takes at least four values.
In this note, we prove some results concerning the divisibility properties of the
Fourier coefficients of a power permutation in connection with Conjecture 1. Our
results can be seen as a proof “modulo ” of Conjecture 1 for certain primes .
2 Boolean Function Case
In this section, we assume p = 2. In [10], the second author has computed the
Fourier spectra of power permutations for all the fields of characteristic 2 with
degree less or equal to 25 without finding any counter-example to the above con-
jectures. More curiously, if we denote by nbz (s) the number of vanishing Fourier
coefficients of the power function of exponent s then the numerical experience
suggests that:
nbz (s) ≥ nbz (−1) = hq.
At this point, it is interesting to notice that Helleseth’s conjecture can not be
extended to the set of all permutations. Indeed, let m be a positive integer and
let g : Fm2 → F2 be a Boolean function in m variables. One defines the Walsh
coefficient of g at a ∈ Fm2 by :
gW(a) =
∑
x∈Fm2
(−1)a.x+g(x).
Identifying L with the F2-vector space F
m
2 , the Boolean function g has a trace
representation i.e. there exists a mapping f : L → L such that g(x) = TrL(f(x))
for all x in L. Of course, the trace representation is not unique. Moreover, if g is
balanced then g can be represented by a permutation of L. In all the cases, the
Walsh spectrum of g and the Fourier spectrum of f are identical.
In [6], an example of a ten-variables Boolean function with a very atypical
Walsh spectrum (see Tab. 1) is given. This Boolean function is balanced and its
Walsh coefficients vanish only once. This numerical example, say g, implies the
existence of a permutation f of F1024 (not a power permutation) such that
g(x) = TrF1024f(x),
whence the Fourier spectrum of f is equal to the Walsh spectrum of g, and thus
∑
x∈F1024 μ(ax+ f(x)) = 0 for all a ∈ F×1024.
A possible generalization of the conjecture of Helleseth could be the following
one:
Conjecture 3. If f is a permutation of L then
∏
λ∈Ltimes D(λf) = 0.
Note that Conjecture 2 is know to be true in characteristic 2 since recent works
of Daniel Katz in [4] and Tao Feng in [12]. The next conjecture that appeared
in the paper by Pursley and Sarwate (see [11]) is still open
On a Conjecture of Helleseth 115
Table 1. An example of Walsh spectrum having only one Walsh coefficient equal to
zero (see [6])
Walsh -48 -44 -40 -36 -32 -28 -24 - 20 -16 -12
mult. 5 30 85 70 115 100 31 62 20 10
Walsh 0 8 16 20 24 28 32 36 40 44
mult. 1 5 25 20 85 90 90 80 50 50
Conjecture 4. If f is a power permutation of L where [L : F2] is even then
supa∈L ̂f(a) ≥ 2√q.
In the sequel, if λ ∈ L then we denote by ̂f(a) the Fourier coefficient of x →
λf(x). If f is a power permutation of exponent s, denoting by t the inverse of s
modulo q − 1, for all y ∈ L×, we have :
̂fλ(a) =
∑
x∈L
μ(λxs + ax) =
∑
x∈L
μ(λysxs + axy) = ̂f(aλ−t). (3)
Hence, one of the specifities of power permutations among the permutations of
L is that the spectrum of λf does not depend on λ ∈ L×.
We conclude this section by giving a divisibility result. Recall that a function
f defined over a field L of characteristic 2 is said to be almost perfect nonlinear
(APN) if for all u ∈ L× the derivative x → f(x + u) + f(x) is two-to-one. It
is for example the case of f(x) = x3 over any field L and of f(x) = x−1 when
[L : F2] is odd.
Theorem 1. Let f be a power permutation over a field L of even characteristic
of cardinal q ≡ 2, 4 mod 5. If f is almost perfect nonlinear then there exists
a ∈ L× such that ̂f(a) ≡ 0 mod 5 i.e.
D(f) ≡ 0 mod 5.
Proof. It is well-known (see [1]) that an APN function f satisfies
∑
λ∈L×
∑
a∈L
̂fλ(a)
4 = 2q3(q − 1). (4)
Since the spectrum of f does not depend on λ, it implies that:
∑
a∈L
̂fλ(a)
4 = 2q3. (5)
Assuming D(f) ≡ 0 mod 5, we get the congruence
q − 1 = 2q3 (mod 5)
implying q ≡ 2, 4 mod 5.
116 Y. Aubry and P. Langevin
3 Hyperplane Section
The key point of view of this note is to consider the number, say Nn(u, v), of
solutions in Ln of the system
{
x1 + x2 + . . . + xn = u
f(x1) + f(x2) + . . . + f(xn) = v.
(6)
By a counting principle using characters, we can state:
Lemma 1. Let f be a permutation of L. The number Nn(u, v) of solutions in
Ln of the system (6) verifies
q2Nn(u, v) = q
n +
∑
α∈L×
∑
β∈L×
̂fβ(α)
nμ̄(αu + βv).
Proof. For any function f : X −→ G where X is a set and G is a finite abelian
group, the number N of solutions in X of f(x) = y for y ∈ G is
N =
1
| G |
∑
x∈X
∑
χ∈ ̂G
χ(f(x)− y)
where ̂G denotes the group of characters of G.
For any α ∈ L, we denote by μα the additive character of L defined by
μα(x) = μ(αx), then we have:
q2Nn(u, v) =
∑
x1,x2,...,xn
∑
β∈L
∑
α∈L
μ̄β(v −
n
∑
i=1
f(xi))μ̄α(u−
n
∑
i=1
xi)
=
∑
β
∑
α
(
∑
y∈L
μ(βf(y) + αy)
)n
μ̄(αu + βv)
=
∑
β
∑
α
̂fβ(α)
nμ̄(αu + βv)
=
∑
α
̂f0(α)
nμ̄(αu) +
∑
β =0
∑
α
̂fβ(α)
nμ̄(αu + βv)
= qn +
∑
α=0
∑
β =0
̂fβ(α)
nμ̄(αu + βv).
Proposition 1. Assuming the Fourier coefficients of λf , λ ∈ L, are integers.
Let  = p be a prime such that ∏λ∈L× D(λf) ≡ 0 mod . Then
q2N−1(u, v) ≡ 1 + (qδ0(u)− 1)(qδ0(v)− 1) mod 
where δa(b) is equal to 1 if b = a and 0 otherwise.
On a Conjecture of Helleseth 117
Proof. By the Fermat’s little Theorem, we have the congruence
̂fλ(a)
−1 ≡ 1− δ0(a) mod .
Hence, by Lemma (1), we have:
q2N−1(u, v) = q−1 +
∑
α=0
∑
β =0
̂fβ(α)
−1μ̄(αu + βv)
≡ 1 +
∑
α=0
∑
β =0
μ̄(αu + βv) mod 
and we conclude remarking that
∑
α∈L× μ̄(αu) = qδ0(u)− 1.
4 Divisibility of Fourier Coefficients
In [3], it is proved that for the exponents s ≡ 1 (mod p− 1), the Fourier coeffi-
cients are multiple of p. In this section, we are interested in divisibility properties
modulo a prime  = p.
Assuming that the Fourier coefficients of any permutation f are rational in-
tegers, we can see that if 3 does not divide D(f) then we have necessarily q ≡ 2
mod 3. Indeed, using Parseval relation, we can write
1 ≡ q2 =
∑
a∈L
| ̂f(a)|2 ≡ q − 1 mod 3.
Theorem 2. Let f be a power permutation of Fpn (with p
n > 2) of exponent
s = 1 mod (p− 1). Then
D(f) ≡ 0 mod 3.
Moreover, if n is a power of a prime  and p ≡ 2 mod  then
D(f) ≡ 0 mod .
Proof. First point. Since p divides D(f), we may assume that p = 3. Suppose
that D(f) ≡ 0 mod 3. Applying Proposition 1 with  = 3, we get
∀u ∈ L×, ∀v ∈ L×, N2(u, v) ≡ 0 (mod ). (7)
In order to obtain a contradiction, we prove the existence of v ∈ L× such that
N2(1, v) = 0. The mapping x → (1 − x)s + xs sends x and 1 − x to the same
point. An element v in the image has at least 2 preimages except when x = 1−x,
which can only happen when p is odd and x = 1/2. So this means that if
p = 2, the cardinality of the image is less or equal to q/2 elements, while if
p is odd, the image of the map has at most (q + 1)/2 elements. If q > 3 the
complementary of the image contains at least two elements whence a nonzero v
such that N(1, v) = 0.
118 Y. Aubry and P. Langevin
Second point. Suppose now that n is a power of a prime  and p ≡ 2
mod . The Frobenius automorphism acts on the solutions of the system (6)
with u = 0, v = 1. Since s ≡ 1 mod (p − 1), the system has no Fp-solutions,
thus N−1(0, 1) ≡ 0 mod . On the other hand, by Proposition 1, if D(f) ≡ 0
mod  then
q2N−1(0, 1) ≡ 2− q ≡ 2− p mod .
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On a Conjecture of Helleseth

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