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An Open Problem on the Bentness of Mesnager's Functions (2109.13421v1)

Published 28 Sep 2021 in cs.IT and math.IT

Abstract: Let $n=2m$. In the present paper, we study the binomial Boolean functions of the form $$f_{a,b}(x) = \mathrm{Tr}1{n}(a x{2m-1 }) +\mathrm{Tr}_1{2}(bx{\frac{2n-1}{3} }), $$ where $m$ is an even positive integer, $a\in \mathbb{F}{2n}*$ and $b\in \mathbb{F}4*$. We show that $ f{a,b}$ is a bent function if the Kloosterman sum $$K_{m}\left(a{2m+1}\right)=1+ \sum_{x\in \mathbb{F}_{2m}*} (-1){\mathrm{Tr}_1{m}(a{2m+1} x+ \frac{1}{x})}$$ 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.

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