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 form $$f_{a,b}(x)={\mathrm{Tr}}_1^n(a{x}^{2^m-1})+{\mathrm{Tr}}_1^2(b{x}^{\frac{2^n-1}{3}}),$$ where $m$ is an even positive integer, $a\in {\mathbb{F}}_{2^n}^{\ast }$ and $b\in {\mathbb{F}}_4^{\ast }$. We show that $f_{a,b}$ is a bent function if the Kloosterman sum $$K_m\left(a^{2^m+1}\right)=1+\sum _{x\in {\mathbb{F}}_{2^m}^{\ast }}(-1{)}^{{\mathrm{Tr}}_1^m(a^{2^m+1}x+\frac{1}{x})}$$ 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.