International Association for Cryptologic Research (IACR)
Publication date
27/09/2021
Field of study
Let n=2m. In the present paper, we study the binomial Boolean functions of the form fa,bβ(x)=Tr1nβ(ax2mβ1)+Tr12β(bx32nβ1β),
where m is an even positive integer, aβF2nββ and bβF4ββ.
We show that fa,bβ is a bent function if the Kloosterman sum
Kmβ(a2m+1)=1+xβF2mββββ(β1)Tr1mβ(a2m+1x+x1β)
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