More on the sum-freedom of the multiplicative inverse function

July 19, 2024 Β· Declared Dead Β· πŸ› Designs, Codes and Cryptography

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Authors Claude Carlet, Xiang-dong Hou arXiv ID 2407.14660 Category math.NT Cross-listed cs.IT Citations 6 Venue Designs, Codes and Cryptography Last Checked 4 months ago
Abstract
In two papers entitled ``Two generalizations of almost perfect nonlinearity" and ``On the vector subspaces of $\mathbb F_{2^n}$ over which the multiplicative inverse function sums to zero", the first author has introduced and studied the notion of sum-freedom of vectorial functions, which expresses that a function sums to nonzero values over all affine subspaces of $\Bbb F_{2^n}$ of a given dimension $k\geq 2$, and he then focused on the $k$th order sum-freedom of the multiplicative inverse function $x\in \Bbb F_{2^n}\mapsto x^{2^n-2}$. Some general results were given for this function (in particular, the case of affine spaces that do not contain 0 was solved positively), and the cases of $k\in \{3,n-3\}$ and of $k$ not co-prime with $n$ were solved as well (negatively); but the cases of those linear subspaces of dimension $k\in [\![ 4;n-4]\!]$, co-prime with $n$, were left open. The present paper is a continuation of the previous work. After studying, from two different angles, the particular case of those linear subspaces that are stable under the Frobenius automorphism, we deduce from the second approach that, for $k$ small enough (approximately, $3\le k\leq n/10$), the multiplicative inverse function is not $k$th order sum-free. Finally, we extend a result previously obtained in the second paper mentioned above, and we deduce in particular that, for any even $n$ and every $2\leq k\leq n-2$, the multiplicative inverse function is not $k$th order sum-free.
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