Finding normal bases over finite fields with prescribed trace self-orthogonal relations

Abstract: Normal bases and self-dual normal bases over finite fields have been found to be very useful in many fast arithmetic computations. It is well-known that there exists a self-dual normal basis of $\mathbb{F}{2^n}$ over $\mathbb{F}_2$ if and only if $4\nmid n$. In this paper, we prove there exists a normal element $\alpha$ of $\mathbb{F}{2^n}$ over $\mathbb{F}{2}$ corresponding to a prescribed vector $a=(a_0,a_1,...,a{n-1})\in \mathbb{F}2^n$ such that $a_i={Tr}{2^n|2}(\alpha^{1+2^i})$ for $0\leq i\leq n-1$, where $n$ is a 2-power or odd, if and only if the given vector $a$ is symmetric ($a_i=a_{n-i}$ for all $i, 1\leq i\leq n-1$), and one of the following is true. 1) $n=2^s\geq 4$, $a_0=1$, $a_{n/2}=0$, $\sum\limits_{1\leq i\leq n/2-1, (i,2)=1}a_i=1$; 2) $n$ is odd, $(\sum\limits_{0\leq i\leq n-1}a_ix^i,x^n-1)=1$. Furthermore we give an algorithm to obtain normal elements corresponding to prescribed vectors in the above two cases. For a general positive integer $n$ with $4|n$, some necessary conditions for a vector to be the corresponding vector of a normal element of $\mathbb{F}{2^n}$ over $\mathbb{F}{2}$ are given. And for all $n$ with $4|n$, we prove that there exists a normal element of $\mathbb{F}_{2^n}$ over $\mathbb{F}_2$ such that the Hamming weight of its corresponding vector is 3, which is the lowest possible Hamming weight.