Paper 2019/307 Original: eprint.iacr.org/2019/307 ↗
Solving $x^{2^k+1}+x+a=0$ in $\mathbb{F}_{2^n}$ with $\gcd(n,k)=1$
Abstract
Let $N_a$ be the number of solutions to the equation $x^{2^k+1}+x+a=0$ in $\mathbb F_{n}$ where $\gcd(k,n)=1$. In 2004, by Bluher it was known that possible values of $N_a$ are only 0, 1 and 3. In 2008, Helleseth and Kholosha have got criteria for $N_a=1$ and an explicit expression of the unique solution when $\gcd(k,n)=1$. In 2014, Bracken, Tan and Tan presented a criterion for $N_a=0$ when $n$ is even and $\gcd(k,n)=1$. This paper completely solves this equation $x^{2^k+1}+x+a=0$ with only condition $\gcd(n,k)=1$. We explicitly calculate all possible zeros in $\mathbb F_{n}$ of $P_a(x)$. New criterion for which $a$, $N_a$ is equal to $0$, $1$ or $3$ is a by-product of our result.
Metadata
- Available format(s)
PDF- Category
- Foundations
- Publication info
- Preprint.
- History
- 2019-03-20: last updated
- 2019-03-20: published
- Short URL
- https://ia.cr/2019/307
- License
- https://creativecommons.org/licenses/by/4.0/
BibTeX
@misc{cryptoeprint:2019/307,
author = {Kwang Ho Kim and Sihem Mesnager},
title = {Solving $x^{2^k+1}+x+a=0$ in $\mathbb{F}_{2^n}$ with $\gcd(n,k)=1$},
howpublished = {Cryptology {ePrint} Archive, Paper 2019/307},
year = {2019},
url = {https://eprint.iacr.org/2019/307}
}