An Open Problem on the Bentness of Mesnager's Functions

TL;DR

This paper investigates the bentness of a specific class of binomial Boolean functions, providing a condition involving Kloosterman sums that determines when these functions are bent, thereby resolving an open problem posed by Mesnager.

Contribution

It establishes a precise condition involving Kloosterman sums under which Mesnager's functions are bent, advancing understanding of their cryptographic properties.

Findings

f_{a,b} is bent if Kloosterman sum equals 4

Provides a new criterion for bentness of Mesnager's functions

Employs advanced tools like Walsh coefficients, Gauss sums, and graph theory

Abstract

Let $n=2m$. In the present paper, we study the binomial Boolean functions of the form $$f_{a,b}(x) = \mathrm{Tr}_1^{n}(a x^{2^m-1 }) +\mathrm{Tr}_1^{2}(bx^{\frac{2^n-1}{3} }), $$ 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 $$K_{m}\left(a^{2^m+1}\right)=1+ \sum_{x\in \mathbb{F}_{2^m}^*} (-1)^{\mathrm{Tr}_1^{m}(a^{2^m+1} x+ \frac{1}{x})}$$ 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.