Let $n=2k$, $F(x)=x^{d(2^k+1)}$, ${\rm gcd}(d,2^k-1)=1$. It is easy to see the following two facts: 1) for any $\alpha \in \mathbb{F}_{2^{k}}$, ${\rm Tr}^{n}_{1}(\alpha F(x))$ is not bent; 2) the Walsh spectrums are the same for ${\rm Tr}^{n}_{1}(\alpha F(x))$, ${\rm Tr}^{n}_{1}(\beta F(x))$, where $\alpha,\beta \in \mathbb{F}_{2^{n}} \backslash \mathbb{F}_{2^{k}}$. Therefore, it suffices to consider only one component function ${\rm Tr}^{n}_{1}(\alpha F(x))$ for $\alpha \in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$ when one wants to compute the number of the bent components of $F(x)$. As it has been shown in [@Hell09] by Helleseth et al., if $d(2^l+1)\equiv 2^i~{\rm mod} (2^k-1)$ for some integers $l\geq 1$, $i\geq 0$ (in this case ${\rm gcd}(d,2^k-1)=1$), for any $\alpha \in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$, $\{W_{F}(\lambda,\alpha)~|~ \lambda \in \mathbb{F}^{\ast}_{2^n} \}$ takes on either exactly three or four values depending on whether $l$ and $n$ are coprime or not. It means that for those $d$, $F(x)$ does not have any bent components. Note that if $l=0$, and $d(2^l+1)\equiv 2^i~{\rm mod} (2^k-1)$, then $d \equiv 2^j~{\rm mod} (2^k-1)$ for some $j$. In this case, $F(x)=x^{d(2^k+1)}=x^{2^{i}(2^k+1)}$ has $2^n-2^k$ bent components (Gold function), and $\{W_{F}(\lambda,\alpha)~|~ \lambda \in \mathbb{F}^{\ast}_{2^n} \}$ takes on exactly two values. Based on those observations above, for $n>8$, they conjectured that if $\{W_{F}(\lambda,\alpha)~|~ \lambda \in \mathbb{F}^{\ast}_{2^n} 