WMC2016-230pm-presented

Introduction
Exceptional APN functions
Recent results
Moises Delgado (Joint work Heeralal Janwa) Some new Results on the conjecture on exceptional APN functio

Introduction
Recent results
Some new Results on the conjecture on
exceptional APN functions:
the Gold case
Moises Delgado
(Joint work Heeralal Janwa)
WMC2016, Santander Spain, In Memory of Oscar Moreno
08 July 2016

Introduction
Recent results
Consider the cyclic codes of length 2s − 1 generated by
C
(t)
s = mω(T)mωt (T) where mω(T) (resp. mωt (T) ) is the
minimal polynomial of ω (resp. mωt (T) ). ω is a primitive element
of F2s . These codes have arisen in the context of Preparata and
Goethals codes [Baker, van Lint and Wilson], and in sequence
design [Niho, Welch and others]. By van Lint and Wilson, these
cyclic codes have minimum distance at most 5 when s ≥ 4. J. and
Wilson 1993 proposed this question and gave partial results.
A word of weight 4 of C
(t)
s corresponds to ωi + ωj + ωk + ωl = 0
and ωit + ωjt + ωkt + ωlt = 0.

Introduction
Recent results
The equations ωi + ωj + ωk + ωl = 0 and ωit + ωjt + ωkt + ωlt = 0
lead to the hyperplane section of the following Fermat variety:
xt
+ yt
+ zt
+ wt
= 0
z + y + z + w = 0
and the study of its nontrivial solutions.

Introduction
Recent results
Deﬁnition 1
φ(x, y, z) :=
f (x) + f (y) + f (z) + f (x + y + z)
(x + y)(x + z)(y + z)
φj (x, y, z) :=
xj + yj + zj + (x + y + z)j
(x + y)(x + z)(y + z)
Theorem 1
If f = Σfj tj then
φ(x, y, z) = Σfj φj (x, y, z).

Introduction
Recent results
Theorem 2 (J. and Wilson, 93)
If f (x) = x2k +1 is a Gold function, then
φ(x, y, z) =
α∈F2k −F2
(x + αy + (α + 1)z). (1)
Theorem 3 (van Lint and Wilson 86)
If t = 2k + 1, k ≥ 1 then C
(t)
s has minimum distance 5.

Introduction
Recent results
Theorem 4 (J and Wilson, 93)
If f (x) = x22k −2k +1 is a Kasami–Welch function, then
φ(x, y, z) =
α∈F2k −F2
Pα(x, y, z), (2)
where Pα(x, y, z) is absolutely irreducible of degree 2k + 1 over
GF(2k).
Theorem 5 (van Lint and Wilson 86, J and Wilson, 93)
If t = 22k − 2k + 1, k ≥ 1 then C
(t)
s has minimum distance 5
whenever (s, i) = 1.

Introduction
Recent results
Theorem 6 (J. and Wilson 93)
If φt is absolutely irreducible then C
(t)
s has minimum distance 5 for
only ﬁnite values of s.
Conjecture 1
(J and Wilson 91, J 92, J, McGuire and Wilson 95] If t is not a
Gold or a Kasam-Welchi exponent, then φt is absolutely irreducible.
Conjecture 2 (J and Wilson 91, J 92, J, McGuire and Wilson 95)
If t is not a Gold or Kasami-Welch exponent, then C
(t)
s has
minimum distance 5 for only ﬁnite values of s.

Introduction
Recent results
Absolute irreducibility testing (J and Wilson 91, J 92, J,
McGuire and Wilson 95 )
The main technique for testing absolute irreducibility is the
following algorithm:
Assume f (x, y, z) factors as P(x, y, z)Q(x, y, z).
Compute and classify multiplicities of each singular point.
Find intersection multiplicities.
If the sum of intersection multiplicies exceeds that predicted
by Bezout’s theorem, then factorization can not occur.
[J and Wilson 91, J 92, J, McGuire and Wilson 95] proved parts of
these conjecttures. and after severa lother contributions, it was
ﬁnally settled by Hernando and McGuire 2011.

Introduction
Recent results
Deﬁnition 2
Let L = Fq, with q = pn. f : L → L is almost perfect nonlinear
(APN) on L if for all a, b ∈ L, a = 0
f (x + a) − f (x) = b (3)
has at most 2 solutions.

Introduction
Recent results
Deﬁnition 3
Let L = Fq, q = pn. A function f : L → L is called exceptional
APN if f is APN on L and also on inﬁnitely many extensions of L.
Theorem 7
J-W1993, J-M-W-1985,...., Hernando-McGuire 2011
A monomial function xd is exceptional APN if and only if d is a
Gold or a Kasami–Welch exponent.
Conjecture 3 (Aubry, McGuire and Rodier 2010)
Conjecture: Up to equivalence, the Gold and Kasami-Welch
functions are the only exceptional APN functions.

Introduction
Recent results
Proposition 1 (J and Wilson 93, Rodier 2009)
Let L = Fq, with q = 2n. A function f : L → L is APN if and only
if the aﬃne surface X with equation
f (x) + f (y) + f (z) + f (x + y + z) = 0
has all its rational points contained in the surface
(x + y)(x + z)(y + z) = 0.

Introduction
Recent results
Theorem 8 (J and Wilson 93, Rodier 2009)
Let f : L → L, L = F2n , a polynomial function of degree d.
Suppose the surface X
φ(x, y, z) :=
f (x) + f (y) + f (z) + f (x + y + z)
(x + y)(x + z)(y + z)
= 0
is absolutely irreducible (or has an absolutely irreducible
component over L), then f is not an exceptional APN function.
Proof. Apply Weil estimate in the case of curves; Lang-Weil or
improved Lachaud-Ghorpae in the case of surfaces.

Introduction
Recent results
Theorem 9 (Aubry, McGuire and Rodier 2010)
If the degree of f is odd and not a Gold or a Kasami-Welch
number, then f is not exceptional APN.

Introduction
Recent results
Theorem 10 (Aubry, McGuire and Rodier 2010)
Suppose f (x) = x2k +1 + g(x), where deg(g) ≤ 2k−1 + 1. Let
g(x) = 2k−1+1
j=0 aj xj . Suppose that there exists a nonzero
coeﬃcient aj of g such that φj (x, y, z) is absolutely irreducible.
Then φ(x, y, z) is absolutely irreducible and f is not exceptional
APN.

Introduction
Recent results
Using this result, in [?], we generalized theorem 4 as follows:
Theorem 11
For k ≥ 2, let f (x) = x2k +1 + h(x) ∈ L[x] where deg(h) is an odd
integer that is a Gold number or that is Gold number of the form
2l + 1 with (l, k) = 1. Then φ(x, y, z) is absolutely irreducible, and
f (x) can not be exceptional APN.
One of our results is remove the ﬁnal condition. We will need:
The following result was conjectured in DJDESI, and then proved
in IWSDA2015.
Theorem 12
If d is an odd integer, then φ2k +1 and φd are relatively prime for
all k ≥ 1 except when d = 2l + 1 and (l, k) > 1.

Introduction
Recent results
In the case when deg(h) is even, they were able to obtain the
following partial result.
Theorem 13
Suppose f (x) = x2k +1 + h(x) ∈ L[x] and deg(h) = 2k −1 + 2. Let
k be odd and relatively prime to n. If h(x) does not have the form
ax2k −1+2 + a2x3 then φ is absolutely irreducible, while if h(x) does
have this form, then either φ is absolutely irreducible or φ splits
into two absolutely irreducible factors that are both deﬁned over L.
Our second result is that we can remove the condition of k being
odd.

Introduction
Recent results
Theorem 14
For k ≥ 2, let f (x) = x2k +1 + h(x) ∈ L[x] where
deg(h) = 2l + 1 < 2k + 1. Then: If (k, l) = 1 and h contains a
term of degree m such that (φ2k +1, φm) = 1, then φ(x, y, z) is
absolutely irreducible and f is not exceptional APN.

Introduction
Recent results
Next we will discuss the case when deg(h) is an even number (the
hardly discussed case, see the Remark at the end). The following
theorem is the version of theorem 2 when k is an even number.
Theorem 15
For k ≥ 2, let f (x) = x22k +1 + h(x) ∈ L[x] where
deg(h) = 22k−1 + 2. Let h(x) = 22k−1+2
j=0 aj xj . If there is a
nonzero coeﬃcient aj such that (φ22k +1, φj ) = 1. Then φ is
absolutely irreducible.

Introduction
Recent results
Suppose by way of contradiction that φ is not absolutely
irreducible. Then, φ(x, y, z) = P(x, y, z)Q(x, y, z). Writing P, Q
as a sum of homogeneous terms:
22k +1
j=0
aj φj (x, y, z) = (Ps +Ps−1 +...+P0)(Qt +Qt−1 +...+Q0) (4)
where Pi , Qi are zero or homogeneous of degree i and
s + t = 22k − 2. Assuming that s ≥ t, then
22k − 2 > s ≥ 22k −2
2 ≥ t > 0.
From the equation (3) we have that:
PsQt = φ22k +1. (5)

Introduction
Recent results
Since φ22k +1 is equal to the product of diﬀerent linear factors [?],
then Ps and Qt are relatively primes. Since h(x) is assumed to
have degree 22k−1 + 2, the homogeneous terms of degree r, for
22k−1 − 1 < r < 22k − 2, are equal to zero. Then equating the
terms of degree s + t − 1 gives PsQt−1 + Ps−1Qt = 0. Hence we
have that Ps divides Ps−1Qt and this implies that Ps divides Ps−1,
since Ps and Qt are relatively primes. We conclude that Ps−1 = 0
since the degree of Ps−1 is less than the degree of Ps. Then we
also have that Qt−1 = 0 as Ps = 0.
Similarly, equating the terms of degree s + t − 2, s + t − 3, ..., s + 1
we get:
Ps−2 = Qt−2 = 0, Ps−3 = Qt−3 = 0, ..., Ps−(t−2) = Q1 = 0
The equation of degree s is:
PsQ0 + Ps−tQt = as+3φs+3 (6)
Lets consider two cases to prove the absolute irreducibility of
φ(x, y, z).

Introduction
Recent results
Case s > t
Then s > 22k−1 − 1 and φs+3 = 0. Then the equation (5)
becomes:
PsQ0 + Ps−tQt = 0.
Then, using the previous argument, Ps−t = Q0 = 0. It means that
Q = Qt is homogeneous.
By the equation (4) and the factorization of φ22k +1 in linear factors
(see [7] for this factorization) we have that for all j, φj (x, y, z) is
divisible by x + αy + (α + 1)z for some α ∈ F2k − F2, which is a
contradiction by the hypothesis of the theorem.
Case s = t = 22k−1 − 1
For this case the equation (5) becomes:
PsQ0 + P0Qt = as+3φ22k−1+2. (7)
If P0 = 0 or Q0 = 0, then we have that Q = Qt or P = Ps. Then
by similar arguments of the ﬁrst case we have a contradiction. If

Introduction
Recent results
both P0, Q0 are different from zero, let us consider the intersection
of φ with the line z = 0, y = 1. Then the equation (4) and (5)
become:
PsQt =
α∈F22k −F2
(x + α) (8)
PsQ0 + P0Qt = as+3(x + 1)(x)
α∈F22k−2 −F2
(x + α)2
(9)
This last equation is derived from a result in [?] that states:
φ22k−1+2 = (x + y)(x + z)(y + z)φ2
22k−2+1
It is easy to show that F22k ∩ F22k−2 = F22 . Let x = α0 ∈ F4,
α0 = 0, 1. Then from (7) we have that Ps(α0) = 0 or Qt(α0) = 0.
If Ps(α0) = 0, then Qt(α0) = 0 (since PsQt is a product of
different linear factors) and from equation (8) we have
P0Qt(α0) = 0, a contradiction since both P0, Qt(α0) are different

Introduction
Recent results
from zero. The case Qt(α0) = 0 is analogous. Therefore φ(x, y, z)
is absolutely irreducible.

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