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 $\GF {n}$ 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 $\GF{n}$ 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.

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