isomorphic to $\mathbb{F}_{2^{k}}\subseteq \mathbb{F}_{2^{n}}$ as the vector spaces over $\mathbb{F}_{2}$, one can easily see that $F(x)$ is EA-equivalent to a function $G(x)$ with $S(G)=\mathbb{F}_{2^{k}}$. It indicates that one only needs to consider those functions $G$ with $S(G)=\mathbb{F}_{2^{k}}$ in the sense of EA-equivalent.

Let $n=2k$, $s$ be positive integers, and $F(x)=x^s$ an $(n,n)$-function. If ${\rm Tr}^{n}_{1}(\alpha F(x))$ is a Boolean bent function for any $\alpha \in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$ ($S(F)=\mathbb{F}_{2^{k}}$), then $2^k+1~|~s$, and $d=\frac{s}{2^k+1}$ satisfying that ${\rm gcd}(d,2^{k}-1)=1$. In particular, $F(x)$ admits the maximal number of bent components if and only if it is a vectorial bent $(n,k)$-function.

Assume to the contrary, $2^k+1\not|s$, then there exists an element $x_{0}^s\not\in \mathbb{F}_{2^{k}}, $ where $x_{0}\in \mathbb{F}^{\ast}_{2^{n}}$. We have $$\begin{aligned}
W_{F}(\lambda,x_{0}^s)
             &\!=\!& \sum\limits_{x\in \mathbb{F}_{2^{n}}}(-1)^{{\rm Tr}^{n}_{1}((x_{0}x)^s+\lambda x)}\\
             &\!=\!&  \sum\limits_{x\in \mathbb{F}_{2^{n}}}(-1)^{{\rm Tr}^{n}_{1}(x^s)+{\rm Tr}^{n}_{1}(\lambda x^{-1}_{0}x)}\\
             &\!=\!& W_{F}(\lambda x_{0}^{-1},1).
             \end{aligned}$$ However, by the assumption and Theorem \[number\], one has ${\rm Tr}^{n}_{1}(\alpha F(x))$ is not bent for any $\alpha\in \mathbb{F}_{2^{k}}$, and then the equation that $W_{F}(\lambda,x_{0}^s)=W_{F}(\lambda x_{0}^{-1},1)$ must induce a contradiction because of $x_{0}^{s}\in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}} $, and $1\in \mathbb{F}_{2^{k}}$.

Denote $\frac{s}{2^k+1}$ by $d$, and $x^{2^{k}}$ by $\overline{x}$, we shall show ${\rm gcd}(d,2^{k}-1)=1$. Now $F(x)=x^s=x^{d(2^{k}+1)}=(x\overline{x})^{d}.$ Let $\gamma \in  \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$, then the elments $x  \in  \mathbb{F}_{2^{n}}$ can be written uniquely 