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

Let n = 2m. In the present paper, we study the binomial Boolean functions of the form ( fa,b (x) = Trnax2m-1) ( + Tr2bx 2n -1 ) 3 1 1 , where mis an even positive integer, a is an element of F*2n and b is an element of F4*. We show that fa,b is a bent function if the Kloosterman sum (a2m+1) E Km = 1 + x is an element of F*2m ( ) (-1)Trm a2m +1 x+ 1 1 x 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.(c) 2023 Elsevier Inc. All rights reserved.