\}$ takes on at most four values, then there must $l\geq0$, and $i\geq 0$ such that $d(2^l+1)\equiv 2^i~{\rm mod} (2^k-1)$, see the arguments above Remark 3 in [@Hell09] for more information.

It is an interesting and important problem to determine all the exponents $d$ such that $F(x)=x^{d(2^k+1)}$ has some bent components. Inspired by the above arguments, one may expect that $d=2^{i}$ are the only ones.

[**Open Problem 1.**]{} Let $n=2k$, $d$ be a positive integer such that ${\rm gcd}(d,2^k-1)=1$. To show that if the $(n,n)$-function $F(x)=x^{d(2^k+1)}$ has some bent components, then $d \equiv 2^i~{\rm mod}~(2^k-1)$ for some $i\geq 0$.

Nontrivial $(n, m)$-functions with maximal number of bent components and their properties
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It has been found that for the trivial $(n,m)$-functions with maximal number of bent components, their differential spectrum takes two values $0$ and $2^k$ only, and the Walsh spectrum takes the values $0, \pm2^k$ and $2^{2k}$ only. In this section, we are interested in the constructions of nontrivial $(n,m)$-functions with maximal number of bent components. Below we will present in bivariate form a large class of such functions, whose bent components belong to the Maiorana-McFarland class $\mathcal{M}$ of bent functions, and in univariate form a class of such 