Solving $x^{2^k+1}+x+a=0$ in $\mathbb{F}_{2^n}$ with $\gcd(n,k)=1$

DOI10.1016/J.FFA.2019.101630zbMATH Open1507.11109arXiv1903.07481MaRDI QIDQ6315814

Kwang Ho Kim, Sihem Mesnager

Publication date: 18 March 2019

Abstract: Let $N_a$ be the number of solutions to the equation $x^{2^k+1}+x+a=0$ in $GFn$ where $gcd(k,n)=1$. In 2004, by Bluher cite{BLUHER2004} it was known that possible values of $N_a$ are only 0, 1 and 3. In 2008, Helleseth and Kholosha cite{HELLESETH2008} 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 cite{BRACKEN2014} 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 $GFn$ 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.

Mathematics Subject Classification ID

Polynomials over finite fields (11T06) Special polynomials in general fields (12E10)

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