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