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.

References in corpus (1)

• A conjecture about Gauss sums and bentness of binomial Boolean functions