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.
Previous Page Next Page