IWSDA2015-talk-16Sept2015

Introduction
Exceptional APN functions
Lower Bounds
Recent results
Large Classes of Gold degree families that are not exceptional APN
Main results
Open Problems and Future Directions
Heeralal Janwa (Joint work with Moises Delgado) Progress on Exceptional APN Functions and Absolute Irreducibilit

Introduction
Lower Bounds
Recent results
Main results
Progress on Exceptional APN Functions and
Absolute Irreducibility of Inﬁnite Classes of
Multivariate Polynomials
Heeralal Janwa
(Joint work with Moises Delgado)
IWSDA2015, Bengluru, India
18 September, 2015

Introduction
Lower Bounds
Recent results
Main results
Deﬁnition 1
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 (1)
has at most 2 solutions.

Introduction
Lower Bounds
Recent results
Main results
Equivalently, for p = 2, the cardinality of {f (x + a)−f (x) : x ∈ L}
is at least 2n−1 for each a ∈ L∗. APN functions are important in
applications to cryptography, coding theory, combinatorics, ﬁnite
geometries, and other related areas.

Introduction
Lower Bounds
Recent results
Main results
The best known examples of APN functions are
xd Exponent d Constraints
Gold 2r + 1 (r, n) = 1
Kasami-Welch 22r − 2r + 1 (r, n) = 1,n odd
Welch 2r + 3 n = 2r + 1
Niho 2r + 2r/2 − 1 n = 2r + 1, r even
2r + 2(3r+1)/2 − 1 n = 2r + 1, r odd
Inverse 22r − 1 n = 2r + 1
Dobbertin 24r + 23r + 22r + 2r − 1 n = 5r
Table: Monomial APN Functions

Introduction
Lower Bounds
Recent results
Main results
Until 2006, the list of known APN functions on L = GF(2n) was
rather short.
Y. Edel, G. Kyureghyan and A. Pott established (by an exhaustive
search) the ﬁrst example of an APN function not equivalent to any
monomial APN functions. Their example is
x3
+ ux36
∈ GF(210
)[x],
where u ∈ wGF(25)∗ ∪ w2GF(25)∗ and w has order 3, is APN on
GF(210). Since then, several new inﬁnite families of polynomial
APN functions have been discovered.

Introduction
Lower Bounds
Recent results
Main results
Deﬁnition 2
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.
Aubry, McGuire and Rodier made the following conjecture:
Up to equivalence, the Gold and Kasami-Welch functions are
the only exceptional APN functions.

Introduction
Lower Bounds
Recent results
Main 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
Lower Bounds
Recent results
Main results
Deﬁnition 3
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
Lower Bounds
Recent results
Main results
A First Bound
Second Bound
Theorem 2 (J and Wilson 93, Rodier 2009)
Let f be a polynomial from F2m to F2m , d its degree. Suppose that
the surface X with aﬃne equation
φ(x, y, z) = 0 (2)
is absolutely irreducible. Then, if d < 0.45q1/4 + 0.5 and d ≥ 9 , f
is not APN.

Introduction
Lower Bounds
Recent results
Main results
A First Bound
Second Bound
Proposition 2 (J and Wilson 93, Rodier 2009)
Let f be a polynomial of F2m to itself, d its degree. Let us suppose
that d is not a power of 2 and that the curve X∞ with equation
φd (x, y, z) = 0
is absolutely irreducible. Then the surface φ(x, y, z) = 0 of
equation (2) is absolutely irreducible.

Introduction
Lower Bounds
Recent results
Main results
A First Bound
Second Bound
Theorem 3 (J and Wilson 93, Rodier 2009)
Let f be a polynomial from F2m to F2m , d its degree. Let us
suppose that d is not a power of 2 and that the surface X
φ(x, y, z) = 0
is regular in codimension one. Then if d ≥ 10 and d < q1/4 + 4, f
is not APN.

Introduction
Lower Bounds
Recent results
Main results
A First Bound
Second Bound
PROOF: From an improvement of a theorem of Deligne on Weil’s
conjectures by Ghorpade-Lachaud,
|X(k)−q2
−q−1| ≤ (d −4)(d −5)q3/2
+(−82+57d −13d2
+d3
)q
If q > 183 − 230d + 94d2 − 16d3 + d4 and d ≥ 6, then
X(k) > 3((d − 3)q + 1) and so f is not APN. Q.E.D.

Introduction
Lower Bounds
Recent results
Main results
A First Bound
Second Bound
Absolute irreducibility testing (J and Wilson 91, J 92, J,
McGuire and Wilson 95 )
The main ingredient in all the proofs on APN functions (by
perhaps everyone) is the following absolute irreducibility testing.
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.

Introduction
Lower Bounds
Recent results
Main results
Theorem 4
If the degree of f is odd and not a Gold or a Kasami-Welch
number, then f is not exceptional APN.
Theorem 5
If the degree of f is 2e with e odd, and if f contains an odd degree
term, then f is not exceptional APN.
Theorem 6
If the degree of f is 4e with e ≥ 7 and e ≡ 3 (mod 4), then f is
not exceptional APN.

Introduction
Lower Bounds
Recent results
Main results
Theorem 7
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
Lower Bounds
Recent results
Main results
Theorem 8 (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), (3)
where Pα(x, y, z) is absolutely irreducible of degree 2k + 1 over
GF(2k).
PROOF:
Hensel lifting, Bezout’s theorem, Gauss’ lemma and so on for three
pages.

Introduction
Lower Bounds
Recent results
Main results
Theorem 9 (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). (4)

Introduction
Lower Bounds
Recent results
Main results
Lemma 1
For an integer k > 1, let n = 2k + 1, m = 22k − 2k + 1 be Gold
and Kasami-Welch, respectively, and d an integer with d ≡ 3
(mod 4). Then
a) If n1 = 2k1 + 1 and n2 = 2k2 + 1 are diﬀerent Gold numbers
such that (k1, k2) = 1, then (φn1 , φn2 ) = 1.
b) (φm1 , φm2 ) = 1 for diﬀerent Kasami-Welch numbers m1 and
m2.
c) (φn, φm) = 1, (φn, φd ) = 1, and (φm, φd ) = 1. In general, if
φd (x, y, z) is absolutely irreducible and is not one of the
factors of φm(x, y, z) (respectively of φn(x, y, z)), then
(φm, φd ) = 1 (respectively (φn, φd ) = 1).

Introduction
Lower Bounds
Recent results
Main results
Theorem 10
For k ≥ 2 and α = 0, let f (x) = x2k +1 + αx2k−1+3 + h(x) ∈ L[x],
where h(x) = 2k−1+1
j=0 aj xj and either a5 = 0 or there is a non
zero aj φj for some j = 5.
Then φ(x, y, z) is absolutely irreducible.
PROOF:
Becomes four times as complicated if the technique of [AMR] is
used. We need new techniques.

Introduction
Lower Bounds
Recent results
Main results
Hyperplane sections
Hyperplane section y − z = 0.
Lemma 2
Let φj (x, y, z) be as in (3). Then
a) For n = 2k + 1 > 3, φj (x, y, y) = (x + y)2k −2;
b) For n > 3 and n ≡ 3 (mod 4), x + y does not divide φj (x, y, y).
c) For n > 5 and n ≡ 1 (mod 4), φj (x, y, y) = (x + y)2l −2S(x, y),
such that x + y does not divide S(x, y), where j = 1 + 2l m, l ≥ 2,
and m is an odd number greater than 1.

Introduction
Lower Bounds
Recent results
Main results
The Kasami-Welch Degree Case
The case deg(h) ≡ 3 (mod 4).
The aﬃne transformation φ(x + 1, y + 1)
Main results: Gold case
Theorem 11
For k ≥ 2, let f (x) = x2k +1 + h(x) ∈ L[x], where deg(h) ≡ 3
(mod 4) and deg(h) < 2k + 1. Then, φ(x, y, z) is absolutely
irreducible.
Theorem 12
For k ≥ 2, let f (x) = x2k +1 + h(x) ∈ L[x]. Let d = deg(h). If
d≡ 1 (mod 4) and d < 2k + 1. If (φ2k +1, φd ) = 1, then φ(x, y, z)
is absolutely irreducible.

Introduction
Lower Bounds
Recent results
Main results
Theorem 13
Let f (x) = xd + h(x) ∈ L[x], where d = 22k − 2k + 1,
deg(h) ≤ 22k−1 − 2k−1 + 1. Suppose that there exist aj = 0 such
that (φj , φn) = 1. Then φ(x, y, z) is absolutely irreducible.

Introduction
Lower Bounds
Recent results
Main results
The aﬃne transformation φ(x + 1, y + 1) We consider the aﬃne
transformation x ← x + 1, y ← y + 1 on the surface φ(x, y, z) = 0
(see (2)). Let
φj (x, y) = φj (x + 1, y + 1, 1).
Fact: For a Gold number n = 2k + 1 we have:
φn(x, y) =
α∈F2k −F2
(x + αy) (5)
The fact follows directly from equation
Next we will extend our results with two more theorems.

Introduction
Lower Bounds
Recent results
Main results
Theorem 14
Given a Gold number n, and a number m ≡ 5 (mod 8) and m > 5,
then (φn, φm) = 1.
As an application of Theorem 14:
Theorem 15
For k ≥ 2, let f (x) = x2k +1 + h(x) ∈ L[x] where d = deg(h) ≡ 5
(mod 8). Then φ(x, y, z) is absolutely irreducible.

Introduction
Lower Bounds
Recent results
Main results
Kasami exponents
Lemma 3
Let φj (x, y, z) be as before.
For j = 2k + 1 > 3, φj (x, 1, 1) = (x + 1)2k −2.
For j ≡ 3 (mod 4) > 3,
φj (x, 1, 1) = (x2m + x2m−1 + ... + x + 1)2, where j = 3 + 4m
for m ≥ 1.
For j ≡ 1 (mod 4) > 5,
φj (x, 1, 1) = (x + 1)2l −2(xm−1 + xm−2 + ... + x + 1)2l
, where
j = 1 + 2l m, l ≥ 2 and m > 1 is an odd number.

Introduction
Lower Bounds
Recent results
Main results
Theorem 16
For k ≥ 2, let f (x) = x22k −2k +1 + h(x) ∈ L[x] where
d = deg(h) ≡ 3 (mod 4) < 22k − 2k+1 + 3. Then:
If d ≤ 22k−1 − 2k−1 + 1 or
if d > 22k−1 − 2k−1 + 1 and (2k − 1, d−1
2 ) = 1,
then φ(x, y, z) is absolutely irreducible.

Introduction
Lower Bounds
Recent results
Main results
PROOF: Suppose that φ factors as
φ(x, y, z) = P(x, y, z)Q(x, y, z), where P and Q are polynomials
deﬁned on the algebraic closure of L. Then
22k −2k +1
j=3
aj φj (x, y, z) = (Ps +Ps−1 +...+P0)(Qt +Qt−1 +...+Q0)
(6)
where Pi , Qi the homogeneous components of P and Q. Assume
that s ≥ t, then
22k
− 2k
+ 1 > s ≥ 22k−1
− 2k−1
− 1 ≥ t > 0.
Let e = 22k − 2k + 1 − d > 2k − 2.

Introduction
Lower Bounds
Recent results
Main results
PsQt = Pα(x, y, z) (7)
where α ∈ F2k , α = 0, 1. Since Pα are diﬀerent absolutely
irreducible factors, Ps and Qt are relatively prime.
By the hypothesis on h(x), the homogeneous terms of degree r, for
d − 3 < r < s + t, equal zero. Then the terms of degree s + t − 1
are PsQt−1 + Ps−1Qt = 0. Hence, Ps divides Ps−1Qt and this
implies that Ps divides Ps−1. We conclude that Ps−1 = 0 as the
degree of Ps−1 is less than the degree of Ps. In addition Qt−1 = 0
as Ps = 0. Similarly,
Ps−2 = Qt−2 = 0, Ps−3 = Qt−3 = 0, ..., Ps−(e−1) = Qt−(e−1) = 0.

Introduction
Lower Bounds
Recent results
Main results
The term of degree d − 3 is:
PsQt−e + Ps−eQt = ad φd (x, y, z) (8)
We show that Qt−e = 0. Suppose d ≤ 22k−1 − 2k−1 + 1. Then
e = 22k − 2k + 1 − d ≥ 22k−1 − 2k−1. Since t ≤ 22k−1 − 2k−1 − 1,
then t − e < 0. Thus Qt−e = 0.

Introduction
Lower Bounds
Recent results
Main results
We assume that d > 22k−1 − 2k−1 + 1 and (2k − 1, d−1
2 ) = 1.
Now consider y = z = 1.
PsQt = (x + 1)2k −2
(x2k −2
+ x2k −3
+ ... + 1)2k
(9)
PsQt−e + Ps−eQt = ad (x2m
+ x2m−1
+ ... + 1)2
(10)
where Ps, Qt are functions of x of degree s, t respectively and
d = 3 + 4m.

Introduction
Lower Bounds
Recent results
Main results
x + 1 x2m + x2m−1 + ... + 1. The roots of x2k −2 + x2k −3 + ... + 1,
x2m + x2m−1 + ... + 1 are also lth root of unity for l = 2k − 1 or
l = d−1
2 . Then, by hypothesis, the left hand side of (9) and (10)
are relatively prime, so (Ps, Qt) = 1 which implies that
s = 2k(2k − 2) and t = 2k − 2.
Thus, t − e < 0 implying that Qt−e = 0. Q.E.D.

Introduction
Lower Bounds
Recent results
Main results
Consider the particular subcase where deg(h) is a Gold number.
Theorem 17
For k ≥ 2, let f (x) = x22k −2k +1 + h(x) ∈ L[x] where
d = deg(h) = 2m + 1 < 22k − 2k + 1. Then φ(x, y, z) is absolutely
irreducible.

Introduction
Lower Bounds
Recent results
Main results
A new conjecture on φi
Based on overwhelming evidence, we believe that (φ2k +1, φd ) = 1
is always the case, and therefore, theorem 12 is unconditionally
true. From the evidence, we propose the following conjecture:
Conjecture 1
If d≡ 1 (mod 4) and d is not a Gold or Kasami exponent, then,
φ2k +1, φd are relatively prime for all k ≥ 1.

Introduction
Lower Bounds
Recent results
Main results
Theorem 18
Given a Gold number n and any odd number m (not a Gold
number), then (φn, φm) = 1.
Proof:
Since (φn, φm) = 1 ⇔ (φn, φm) = 1, we will work with the
functions φ.
Let n = 2k + 1, m = 2i l + 1, where l > 1 is an odd integer.
Let a ∈ F2k , a = 0, a = 1. By (6), we will prove the theorem by
showing that no term (x + ay) divides φm. By deﬁnition of the
functions φj , φj , in (3) and (6), (x + ay) divides φm(x, y) if and
only if (x + ay) divides f (x, y) = φm(x, y)(x)(y)(x + y).
Writing f (x, y) = (x + 1)m + (y + 1)m + 1 + (x + y + 1)m as a
sum of homogeneous terms:

Introduction
Lower Bounds
Recent results
Main results
f (x, y) = Fm(x, y) + Fm−1(x, y) + ... + F2i +1(x, y) (11)
Then (x + ay)|f (x, y) if and only if (x + ay) divides each
homogeneous term Fr in (7).
From (3), by direct computations, we have:
Fm−1(x, y) = xm−1
+ ym−1
+ (x + y)m−1
(12)
Fm(x, y) = xm
+ ym
+ (x + y)m
(13)
Then, for (x + ay) to divides (8) and (9), it should happen that
Fm−1(ay, y) = (ay)m−1 + ym−1 + (ay + y)m−1 = 0 and
Fm(ay, y) = (ay)m + ym + (ay + y)m = 0; which respectively
implies that

Introduction
Lower Bounds
Recent results
Main results
(a + 1)l
+ al
+ 1 = 0 (14)
(a + 1)l+1
+ al+1
+ 1 = 0 (15)
Substituting (10) in (11)
(al
+ 1)(a + 1) = al+1
+ 1
al+1
+ al
+ a + 1 = al+1
+ 1
al−1
= 1 (16)
ie, a is a (l − 1)-th root of unity. Furthermore, using this in (10)
(a + 1)l−1
= 1 (17)
ie, a + 1 is also a (l − 1)-th root of unity.
Thus, if a does not satisfy either (12) or (13), the theorem is
proved.

Introduction
Lower Bounds
Recent results
Main results
Suppose now that a satisfy (12) and (13). Then, let us consider
the term Fm−(2i +1) in (7).
Fm−(2i +1)(x, y)= m
2i +1 (xm−(2i +1) + ym−(2i +1) + (x + y)m−(2i +1))
First, using Lucas’s theorem, let us prove that m
2i +1 =1.

Introduction
Lower Bounds
Recent results
Main results
This theorem uses the convention that a
b = 0 if a < b.
Let l = ar 2r + ar−12r−1 + ... + a12 + 1 be the base 2 expansion of
l. Then, the expansion of m is
m = 21l + 1 = ar 2i+r + ar−12i+r−1 + ... + a12i+1 + 2i + 1. Using
the mentioned theorem we have that m
2i +1 =1.
Now, let us prove that x + ay does not divide Fm−(2i +1). For
x + ay to divide it, it should happen that Fm−(2i +1)(ay, y) = 0,
however,
Fm−(2i +1)(ay, y) = (ay)m−(2i +1) + ym−(2i +1) + (ay + y)m−(2i +1)
= ((a + 1)m−(2i +1) + am−(2i +1) + 1)ym−(2i +1)
= ((a + 1)2i (l−1) + a2i (l−1) + 1)ym−(2i +1)
= ym−(2i +1),
since a, a + 1 are (l − 1)-th roots of unity.

Introduction
Lower Bounds
Recent results
Main results
Using this results we can generalize theorems 7 and 8 in the
following theorem.
Theorem 19
For k ≥ 2, let f (x) = x2k +1 + h(x) ∈ L[x], where deg(h) is any
odd number (nor a Gold number). Then φ(x, y, z) is absolutely
irreducible.

Introduction
Lower Bounds
Recent results
Main results
Evidence for the new conjecture
We deﬁne φj (x, y) = φj (x + 1, y + 1, 1). Then:
For a Gold number j = 2k + 1 we have:
φj (x, y) =
α∈F2k −F2
(x + αy) (18)
For any number j ≡ 5 (mod 8):
φj (x, y) = x4
y + xy4
+
1+4l
r=6
Fr (x, y) (19)
where j = 1 + 4l, l > 1 is any odd number and Fr is zero or
homogeneous of degree r.

Introduction
Lower Bounds
Recent results
Main results
Theorem 20
Given a Gold number n and a number m ≡ 5 (mod 8), then
(φn, φm) = 1.

Introduction
Lower Bounds
Recent results
Main results
PROOF: Since (φn, φm) = 1 ⇔ (φn, φm) = 1, we will work with
the functions φ.
Let n = 2k + 1, α ∈ Fk, α = 0, 1 and m = 4l + 1, where l > 1 is
an odd integer.
By eq. (18), (φn, φm) = 1 ⇔ (x + αy) φm(x, y). We will prove
that no factor of the form (x + αy) divides φm.
Suppose that (x + αy) divides φm(x, y). Then (x + αy) divides
F(x, y) = φm(x, y)xy(x + y).
Writing F(x, y) as a sum of homogeneous terms:
F(x, y) = F5(x, y) + F6(x, y) + ... + F1+4l (x, y)

Introduction
Lower Bounds
Recent results
Main results
Then, (x + αy)|F(x, y) if and only if (x + αy)|Fr (x, y) for all r.
By eq. (19), (x − α)|x4y + xy4 implies α3 + 1 = 0, then α is a
3rd -root of unity.
Consider the following three cases to prove the theorem:
l ≡ 0 (mod 3)
l ≡ 1 (mod 3)
l ≡ 2 (mod 3)

Introduction
Lower Bounds
Recent results
Main results
Case l ≡ 0 (mod 3)
Let l ≡ 0 (mod 3), then l = 3q, m = 4(3q) + 1.
If (x + αy) divides φm, by the Remainder theorem φm(αy, y) = 0.
However,
φm(αy, y) = (αy + 1)m
+ (y + 1)m
+ 1 + (α2
y + 1)m
the term Fm−1 of degree m − 1 is:
Fm−1 = m
1 [αm−1 + 1 + (α2)m−1]ym−1 = ym−1, contradicting
that φm(αy, y) = 0.

Introduction
Lower Bounds
Recent results
Main results
Let l ≡ 1 (mod 3), then l = 1 + 3q,
m = 4(1 + 3q) + 1 = 5 + 4(3q).
If (x + αy) divides φm, φm(αy, y) = 0. However the term Fm−5 of
degree m − 5 in φm(αy, y) is:
Fm−5 = m
5 [αm−5 + 1 + (α2)m−5]ym−5 = ym−5. Contradiction.

Introduction
Lower Bounds
Recent results
Main results
If l ≡ 2 (mod 3), then l = 2 + 3q, m = 4(2 + 3q) + 1 = 9 + 4(3q).
If (x + αy) divides φm, φm(αy, y) = 0; but computing the term
Fm of degree m in φm(αy, y):
Fm = m
0 [αm + 1 + (α2)m]ym = ym. Q.E.D.

Introduction
Lower Bounds
Recent results
Main results
Theorem 21
For k ≥ 2, let f (x) = x2k +1 + h(x) ∈ L[x] where d = deg(h) ≡ 5
(mod 8) < 2k + 1. Then φ(x, y, z) is absolutely irreducible.
Theorem 22
All polynomials of the form f (x) = x65 + h(x) are not exceptional
APN for all odd degree polynomials h.

Introduction
Lower Bounds
Recent results
Main results
Some Remarks
In Theorem 12, (φ2k +1, φd ) = 1 is a necessary condition for f (x)
not to be exceptional APN.
This observation allows us to apply several key results. Janwa,
McGuire and Wilson proved the absolute irreducibility of φd , when
d ≡ 3 (mod 4).
They also established absolute irreducibility for several other
inﬁnite cases with d ≡ 1 (mod 4). Moreover for d ≡ 5
(mod 8) > 13, if the maximal cyclic code of odd length m, Bm
(where d = 2l m + 1, l > 0), has no codewords of weight 4 then,
φd (x, y, z) is absolutely irreducible. For many values of m it is
possible that Bl has no codewords of weight 4, for example, if m is
a prime congruent to ±3 (mod 8). For more details and inﬁnite
classes of examples, see [1].

Introduction
Lower Bounds
Recent results
Main results
Some Remarks
1 Until recently, it was thought that φd (x, y, z) was absolutely
irreducible for the values of d ≡ 5 (mod 8). F. Hernando and
G. McGuire [2], with the help of MAGMA, found that the
polynomial φ205(x, y, z) factors in F2[x, y, z].
2 Janwa and Wilson [3] proved, using diﬀerent methods
including Hensel’s lemma implemented on a computer, that
φd (x, y, z) is absolutely irreducible for 3 < d < 100, provided
that d is not a Gold or a Kasami-Welch number.
3 Subsequent results, (F´erard, Oyono, and Rodier [4] and
Delgado and Janwa [5]) supplement our results for the
Kasami degree polynomials.

Introduction
Lower Bounds
Recent results
Main results
Some Remarks
Thank you for your attention.

Introduction
Lower Bounds
Recent results
Main results
Some Remarks
Heeralal Janwa, Gary M. McGuire, and Richard M. Wilson.
Double-error-correcting cyclic codes and absolutely irreducible
polynomials over GF(2).
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Fernando Hernando and Gary McGuire.
Proof of a conjecture on the sequence of exceptional numbers,
classifying cyclic codes and APN functions.
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H. Janwa and R. M. Wilson.
Hyperplane sections of Fermat varieties in P3
in char. 2 and
some applications to cyclic codes.
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codes (San Juan, PR, 1993), volume 673 of Lecture Notes in
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Introduction
Lower Bounds
Recent results
Main results
Some Remarks
Eric F´erard, Roger Oyono, and Franccois Rodier.
Some more functions that are not APN inﬁnitely often. The
case of Gold and Kasami exponents.
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volume 574 of Contemp. Math., pages 27–36. Amer. Math.
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H. Janwa M. Delgado.
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