The solution to an open problem on the bentness of Mesnager's functions

Abstract

Let n=2m. In the present paper, we study the binomial Boolean functions of the formfa,b(x)=Tr1n(ax2m−1)+Tr12(bx2n−13), where m is an even positive integer, a∈F2n⁎ and b∈F4⁎. We show that fa,b is a bent function if the Kloosterman sumKm(a2m+1)=1+∑x∈F2m⁎(−1)Tr1m(a2m+1x+1x) equals 4, thus settling an open problem of Mesnager. The proof employs tools including computing Walsh coefficients of Boolean functions via multiplicative characters, divisibility properties of Gauss sums, and graph theory.