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 $$f_{a,b}(x) = \mathrm{Tr}<em>1^{n}(a x^{{2^m-1} }) +\mathrm{Tr}_1^{{2}(bx^{{\frac{2^n-1}{3}}} }),$$ where $m$ is an even positive integer, $a\in \mathbb{F}</em>{2<sup>n}<sup>*$ and $b\in \mathbb{F}<em>4<sup>*$. We show that $f</em>{a,b}$ is a bent function if the Kloosterman sum $$K_{m}\left(a^{{2^{m+1}\right)=1+}} \sum_{x\in \mathbb{F}_{2^{m}^*}} (-1)^{{\mathrm{Tr}_1^{{m}(a^{{2^m+1}}}} x+ \frac{1}{x})}$$ 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.