DIFFERENTIALLY 4-UNIFORM FUNCTIONS 5 satisfy: 0 = Pf (x, y, t) = xd + yd + td + (x + y + t)d (x + y)(x + t)(y + t) = xd + yd + (x + y + z)d + zd (x + y)(y + z)(x + z) = Pf (x, y, z), so they belong to S1. Proposition 6. The projective plane curve C7 is isomorphic to the projective plane curve C with equation Pf (x, y, z) = xd + yd + zd + (x + y + z)d (x + y)(x + z)(y + z) = 0. Proof. The projection from the vertex of the cone S1 defines an isomorphism of the projective plane H7 with equation x + y + z + t = 0 onto the plane with equation t = 0, and it maps C7 onto the curve C with equation Pf (x, y, z) = 0. Proposition 7. Let C be a plane curve of degree deg(C) and which is not contained in V . Then: (C V )(Fq) 7 deg(C). Proof. The variety V is the union of seven projective planes. Each plane cannot contain more than deg(C) points, therefore V contains at most 7 deg(C) rational points in C. In order to get a lower bound for the number of rational points over Fq on the curve C, hence on the curve X, we need to know if C is absolutely irreducible or not. This question has been discussed by H. Janwa, G. McGuire and R. M. Wilson in [14] and very recently by F. Hernando and G. McGuire in [10]. Proposition 8. If d = 2r 1 with r 3, then the projective curve X has an absolutely irreducible component C defined over F2 in the plane x + z + t = 0 and this component C is isomorphic to the curve C. Proof. One checks that the intersection of the cone S1 with the plane x + z + t = 0 is the same as the intersection of the cone S2 with that plane. Hence one can show, as in Proposition 6, that the intersection of the curve X with the plane x + z + t = 0 is isomorphic to the curve C. Furthermore, it is proved in [14] that the curve C is absolutely irreducible since, deg(C) = 2r 1 3 (mod 4). Hence we can state Theorem 9. Consider the function f : Fq −→ Fq defined by f(x) = xd with d = 2r 1 and r 3. If 5 d q1/4 + 4.6 , then f has differential uniformity strictly greater than 4. Proof. The curve C is an absolutely irreducible plane curve of arithmetic genus πC = (d 4)(d 5)/2. According to [1] (see also [2] for a more general statement), the number of rational points of the (possibly singular) absolutely ir- reducible curve C satisfies |#C (Fq) (q + 1)| 2πC q1/2. 5
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