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

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