DIFFERENTIALLY 4-UNIFORM FUNCTIONS 7
are respectively the intersections S1,∞ and S2,∞ of the cylinders S1 and S2 with
that hyperplane. The homogeneous equations of S1,∞ and S2,∞ are
Pxd (x, y, z) =
xd + yd + zd + (x + y + z)d
(x + y)(x + z)(y + z)
and
Pxd (x, y, t) =
xd + yd + td + (x + y + t)d
(x + y)(x + t)(y + t)
·
By Proposition 8, the intersection of the curve X∞ with the plane x + z + t = 0
(inside the hyperplane at infinity) is an absolutely irreducible component C of the
curve X∞ of multiplicity 1, defined over
F2.
So the only absolutely irreducible
component of X, say X1, which contains C is defined over
Fq.
Proposition 11. Let X be an absolutely irreducible projective surface of degree
1. Then the maximum number of rational points on X which are contained in
the hypersurface V H∞ is
(X (V H∞)) 8(deg(X )q + 1).
Proof. As deg(X ) 1, the surface X is not contained in any hyperplane.
Thus, a hyperplane section of X is a curve of degree deg(X ). Using the bound on
the maximum number of rational points on a general hypersurface of given degree
proved by Serre in [19], we get the result.
Theorem 12. Let q =
2m.
Consider a function f :
Fq
−→
Fq
of degree
d =
2r
1 with r 3. If 31 d
q1/8
+ 2, then δ(f) 4. For d 31, we get
δ(f) 4 for d = 7 and m 22 and also if d = 15 and m 30.
Proof. From an improvement of a result of S. Lang and A. Weil [15] proved
by S. Ghorpade and G. Lachaud [11, section 11], we deduce
|#X1(Fq)

q2
q 1| ((d
3)2
1)((d
3)2

2)q3/2
+ 36(2d
3)5q
(d
3)4q3/2
+ 36(2d
3)5q.
Hence
#X1(Fq)

q2
+ q + 1 (d
3)4q3/2
36(2d
3)5q.
Therefore, if
q2
+ q + 1 (d
3)4q3/2
36(2d
3)5q
8((d 3)q + 1),
then
#X(Fq)

#X1(Fq)
8((d 3)q + 1), and hence
X1(Fq)
V H∞ by
Proposition 11. As X is the set of affine points of the projective surface X, we
deduce that
X(Fq)
V and so the differential uniformity of f is at least 6 from
Theorem 3. This condition can be written
q (d
3)4q1/2
36(2d
3)5
8(d 3) 0.
This condition is satisfied when
q1/2 d4

12d3
+
54d2
+ 1044d + 5265 + 25920/d
if d 2, or d q1/8 + 2 if d 31.
7
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