More constructions of semi-bent and plateaued functions in polynomial forms

Abstract

Plateaued functions and their subclass semi-bent functions have useful applications in cryptography and communications. In this paper we give new constructions of quadratic semi-bent functions in polynomial forms on the finite field $\mathbb {F}_{2^n}$ for both odd and even n. We also present some characterizations of e-plateaued functions with few trace terms when n is even.

1 Introduction

In the 1960s and 1990s, two families of m-sequences having low cross correlation were introduced by Gold [1] and Boztas et al. [2] respectively. Each of them has period $2^n-1$ and a plateaued cross-correlation spectra. That is, for two such m-sequences $u(t)={\mathrm{Tr}}_1^n(\alpha ^t)$ and $v(t)={\mathrm{Tr}}_1^n(\beta ^t)$, where $\alpha$ and $\beta$ have order $2^n-1$ in the finite field $\mathbb {F}_{2^n}$, we have

$$\begin{aligned} C_{u,v}(\tau )=\sum \limits _{t=0}^{2^n-2}(-1)^{u(t+\tau )+v(t)}\in \left\{ -1, -1\pm 2^{\frac{n+1}{2}}\right\} . \end{aligned}$$

These families of sequences have the trace representations

$$\begin{aligned} f(x)={\mathrm{Tr}}_1^n \Big(x^{1+2^i}\Big)\; ({\mathrm{gcd}}(i,n)=1) \;\;\text{ and }\;\; f(x)=\sum \limits _{i=1}^{\frac{n-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^i}\Big) \end{aligned}$$

respectively, where ${\mathrm{Tr}}_1^n(x)=\sum _{i=0}^{n-1}x^{2^i}$. Such families of maximum-length sequences, whose cross-correlation spectra attain exactly the values above, have a wide range of applications in cryptography and code-division multiple-access communication systems [3, 4]. Such sequences can be represented by Boolean functions which we call semi-bent functions, using the terminology of Khoo et al. [5].

In order to construct more sequences having the nice property as the above two sequences, Khoo et al. [6] investigated the problem of determining the function

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{n-1}{2}}c_i{\mathrm{Tr}}_1^n\Big(x^{1+2^i}\Big),\;\;c_i\in \mathbb {F}_2 \end{aligned}$$

defined on $\mathbb {F}_{2^n}$ with n odd is semi-bent, where this sum has more than one term. To such a function a cyclic code of length $2^n-1$ was associated, spanned by

$$\begin{aligned} c(x), xc(x), \ldots , x^{n-1}c(x),\;\text{ where }\;c(x)=\sum \limits _{i=1}^{\frac{n-1}{2}}c_i\Big(x^i+x^{n-i}\Big). \end{aligned}$$

Then it was proved that f is semi-bent if and only if $\mathrm{gcd}(c(x), x^n+1)=x+1$. This gives a very convenient tool for determining whether a function f having certain number of trace terms is semi-bent or not.

Following this work, Charpin et al. studied the following function [7]:

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\lfloor \frac{n-1}{2}\rfloor }c_i{\mathrm{Tr}}_1^n\Big(x^{1+2^i}\Big),\;\;c_i\in \mathbb {F}_2.\end{aligned}$$

(1)

When n is odd, they provide some semi-bent functions with three or four trace terms. When n is even, they proved that f(x) is semi-bent if and only if $\mathrm{gcd}(c(x), x^n+1)=x^2+1$, where $c(x)=\sum _{i=1}^{\frac{n}{2}-1}c_i(x^i+x^{n-i})$. Moreover, they found that the concatenation of two suitably chosen such semi-bent functions will yield a semi-bent function with higher algebraic degrees. After this work, a lot of research has been devoted to finding new families of quadratic semi-bent and bent functions in the form of Eq. (1) [8,9,10,11,12,13,14,15,16]. In 2013, Dong et al. present some new constructions of quadratic semi-bent functions. For odd $n=pq$ with $p(3\not \mid p), q$ odd, they proved that the function

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^q}\Big), \end{aligned}$$

is semi-bent. For even $n=2m$ with odd m, a necessary and sufficient condition for the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)=\sum \limits _{i-1}^{\frac{m-1}{2}}c_i{\mathrm{Tr}}_1^n\Big(\beta x^{1+2^{2i}}\Big), \end{aligned}$$

to be semi-bent is given. For some special cases of $c_i(1\le i\le \frac{m-1}{2})$, they proved that f is semi-bent.

Motivated by the paper [8], we present new constructions of quadratic semi-bent and e-plateaued functions in polynomial forms. We study the function defined by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{s}c_i{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+\sum \limits _{j=1}^{t}d_j{\mathrm{Tr}}_1^n\Big( x^{1+2^{qj}}\Big), \end{aligned}$$

(2)

where $c_i, d_j\in \mathbb {F}_2$, $1\le s\le \frac{q-1}{2}$, $1\le t\le \frac{p-1}{2}$, $n=pq$, p, q odd, $\mathrm{gcd}(p, q)=1$, and the function defined by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{m-1}{2}}c_i{\mathrm{Tr}}_1^n\Big(\beta x^{1+2^{ei}}\Big), \end{aligned}$$

(3)

where $n=em$, $e=2^l$, m is odd, $c_i\in \mathbb {F}_2$$(1\le i\le \frac{m-1}{2})$, $\beta \in \mathbb {F}_{2^e}^{*}$. For odd n, we find five new classes of semi-bent functions of the form Eq. (2) by choosing suitable vectors $(c_1,\ldots , c_{s})\in \mathbb {F}_2^{s}$ and $(d_1, \ldots , d_{t})\in \mathbb {F}_2^{t}$. For even n, we give a necessary and sufficient condition under which f(x) given by Eq. (3) is e-plateaued and provide some new e-plateaued and semi-bent functions with few trace terms.

To the best of our knowledge, we give a list of the quadratic semi-bent functions on $\mathbb {F}_{2^n}$ as follows:

When n is odd, the following functions are semi-bent.

• (1) $f(x)={\mathrm{Tr}}_1^n(x^{1+2^i})$, $\mathrm{gcd}(i, n)=1$ [1].

• (2) $f(x)=\sum \limits _{i=1}^{\frac{n-1}{2}}{\mathrm{Tr}}_1^n(x^{1+2^i})$ [2].

• (3) $f(x)=\sum \limits _{i=1}^{\frac{n-1}{2}}c_i{\mathrm{Tr}}_1^n(x^{1+2^i})$, $c_i\in \mathbb {F}_2$ for $1\le i\le \frac{n-1}{2}$, $\mathrm{gcd}(\sum \limits _{i=1}^{\frac{n-1}{2}}c_i\Big(x^i+x^{n-i}\Big), x^n+1)=x+1$ [6].

• (4) $f(x)=\sum \limits _{i=0}^r{\mathrm{Tr}}_1^n(x^{1+2^{a+id}})$, $\mathrm{gcd}(2a+rd, n)=\mathrm{gcd}((r+1)d, n)=1$ [5].

• (5) $f(x)={\mathrm{Tr}}_1^n(x^{1+2^i})+{\mathrm{Tr}}_1^n(x^{1+2^j})$, $\mathrm{gcd}(i+j, n)=\mathrm{gcd}(i-j, n)=1$ [5].

• (6) $f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^q}\Big)$, $n=pq$, $p(3\not \mid p), q$ are odd positive integers such that $\mathrm{gcd}(p, q)=1$ [8].

• (7) $f(x)={\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n(x^{1+2^{qj}})$, $n=pq$, p, q odd, $\mathrm{gcd}(p, q)=1$, and i, j are two positive integers such that $\mathrm{gcd}(i, q)=\mathrm{gcd}(j, p)=1$ (Theorem 3 of this paper).

• (8) $f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n(x^{1+2^{pi}})+{\mathrm{Tr}}_1^n(x^{1+2^{qj}})$, $n=pq$, $p(3\not \mid p), q$ are odd positive integers such that $\mathrm{gcd}(p, q)=1$, $\mathrm{gcd}(j, p)=1$ (Theorem 4 of this paper).

• (9) $f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n(x^{1+2^{pi}})+{\mathrm{Tr}}_1^n(x^{1+2^{q}})+{\mathrm{Tr}}_1^n(x^{1+2^{qj}})$, $n=pq$, p, q odd and $\mathrm{gcd}(p,q)=1$, $j=2^l$ and l is a positive integer such that $\mathrm{gcd}(l,n)=1$ (Theorem 5 of this paper).

• (10) $f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n(x^{1+2^{pi}})+{\mathrm{Tr}}_1^n(x^{1+2^{qj}})+{\mathrm{Tr}}_1^n(x^{1+2^{qk}})$, $n=pq$, p, q odd and $\mathrm{gcd}(p,q)=1$, $j=2^{u-1}-2^{v-1}$, $k=2^{u-1}+2^{v-1}$, $u>v\ge 1$ (Theorem 6 of this paper).

• (11) $f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n(x^{1+2^{pi}})+{\mathrm{Tr}}_1^n(x^{1+2^{qj}})+{\mathrm{Tr}}_1^n(x^{1+2^{qk}})$, $n=pq$, p, q odd and $\mathrm{gcd}(p,q)=1$, $j=2^u$, $k=2^v$, $u>v\ge 1$, $\mathrm{gcd}(u-v,n)=1$ (Theorem 7 of this paper).

When $n=2m$ is even, the following functions are semi-bent:

• (1) $f(x)=\sum \limits _{i=1}^{\frac{n}{2}-1}c_i{\mathrm{Tr}}_1^n(x^{1+2^i})$, $c_i\in \mathbb {F}_2$, $\mathrm{gcd}(\sum \limits _{i=1}^{\frac{n}{2}-1}c_i(x^i+x^{n-i}), x^n+1)=x^2+1$ [7].

• (2) $f(x)={\mathrm{Tr}}_1^n(\alpha x^{1+2^i})$, $\alpha \in \mathbb {F}_{2^n}^{*}$, i even, m odd [10].

• (3) $f(x)={\mathrm{Tr}}_1^n(\alpha x^{1+2^i})$, $\alpha \in \{x^3\,|\,x\in \mathbb {F}_{2^n}^{*}\}$, i odd, m even [10].

• (4) $f(x)={\mathrm{Tr}}_1^n(\alpha x^{1+2^i})$, $\alpha \in \{x^3\,|\,x\in \mathbb {F}_{2^n}^{*}\}$, i odd, m odd and $\mathrm{gcd}(i, m)=1$ [10].

• (5) $f(x)={\mathrm{Tr}}_1^n(x^{1+2^i})+{\mathrm{Tr}}_1^n(x^{1+2^j})$, m odd, $1\le i<j\le m$, $\mathrm{gcd}(i+j, n)=\mathrm{gcd}(j-i, n)=1$ or $\mathrm{gcd}(i+j, n)=\mathrm{gcd}(j-i, n)=2$ [10].

• (6) $f(x)=\sum \limits _{i=1}^{\frac{m-1}{2}}{\mathrm{Tr}}_1^n(\beta x^{1+4^i})$, m odd, $\beta \in \mathbb {F}_{4}^{*}$ [8].

• (7) $f(x)=\sum \limits _{i=1}^{\frac{m-1}{2}}c_i{\mathrm{Tr}}_1^n(\beta x^{1+4^i})$, $c_i\in \mathbb {F}_2$, $\beta \in \mathbb {F}_4^{*}$, m odd, $\mathrm{gcd}(\sum \limits _{i=1}^{\frac{m-1}{2}}c_i(x^i+x^{m-i}), x^m+1)=x+1$ [8].

• (8) $f(x)=\sum \limits _{i=1}^{k}{\mathrm{Tr}}_1^n(\beta x^{1+4^{di}})$, $\beta \in \mathbb {F}_4^{*}$, m odd, $d\ge 1$, $1\le k\le \frac{m-1}{2}$, $\mathrm{gcd}(k+1, m)=\mathrm{gcd}(k, m)=\mathrm{gcd}(d, m)=1$ [8].

• (9) $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+4^i}+\beta x^{1+4^j})$, $\beta \in \mathbb {F}_4^{*}$, m odd, $1\le i<j\le \lfloor \frac{n}{4}\rfloor$, $\mathrm{gcd}(i+j, m)=\mathrm{gcd}(j-i, m)=1$ [8].

• (10) $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+4^i}+\beta x^{1+4^j}+\beta x^{1+4^t})$, $\beta \in \mathbb {F}_{4}^{*}$, m odd, $1\le i<j<t\le \lfloor \frac{n}{4}\rfloor$, $i+j=t$, $\mathrm{gcd}(i, m)=\mathrm{gcd}(j, m)=\mathrm{gcd}(t, m)=1$ [8].

• (11) $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+4^i}+\beta x^{1+4^j}+\beta x^{1+4^t})$, $\beta \in \mathbb {F}_{4}^{*}$, m odd, $1\le i<j<t\le \lfloor \frac{n}{4}\rfloor$, $i+j=2t$, $j-i=3^hp$, $3\not \mid p$, $n=3^kq$, $3\not \mid q$, $\mathrm{gcd}(2t, m)=1$, $h\ge k$ [8].

• (12) $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+4^i}+\beta x^{1+4^j}+\beta x^{1+4^t})$, $\beta \in \mathbb {F}_{4}^{*}$, m odd, $1\le i<j<t\le \lfloor \frac{n}{4}\rfloor$, $j-i=2t$, $t\ne i$, $j+i=3^up$, $3\not \mid p$, $n=3^vq$, $3\not \mid q$, $\mathrm{gcd}(2t, m)=1$, $u\ge v$ [8].

• (13) $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+4^i}+\beta x^{1+4^j}+\beta x^{1+4^t}+\beta x^{1+4^s})$, $\beta \in \mathbb {F}_{4}^{*}$, m odd, $1\le i,j,t,s\le \lfloor \frac{n}{4}\rfloor$, $i<j$, $t<s$, $i+j=t+s=r$, $t\ne i$, $\mathrm{gcd}(r, m)=\mathrm{gcd}(s-i, m)=\mathrm{gcd}(s-j, m)=1$ [8].

• (14) $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{2i}})$, $1\le i\le m-1$, $\mathrm{gcd}(i, m)=1$ and m odd (Corollary 5 of this paper).

• (15) $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{2i}}+\beta x^{1+2^{2j}}+\beta x^{1+2^{2t}})$, $\beta \in \mathbb {F}_{4}^{*}$, m odd, $1\le i<j<t\le m-1$, $i+j=2t$, $\mathrm{gcd}(t, m)=1$ (Corollary 9 of this paper).

• (16) $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{2i}}+\beta x^{1+2^{2j}}+\beta x^{1+2^{2t}})$, $\beta \in \mathbb {F}_{4}^{*}$, m odd, $1\le i<j<t\le m-1$, $j-i=2t$, $\mathrm{gcd}(t, m)=1$ (Corollary 10 of this paper).

2 Preliminaries

Let $\mathbb {F}_{2^n}$ be the finite field with $2^n$ elements, and we use $\mathcal {B}_n$ to denote the set of Boolean functions from $\mathbb {F}_{2^n}$ to $\mathbb {F}_2$. In this paper, we mainly investigate the Boolean function of the form

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{s}c_i{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+\sum \limits _{j=1}^{t}d_j{\mathrm{Tr}}_1^n\Big( x^{1+2^{qj}}\Big), \end{aligned}$$

where $c_i, d_j\in \mathbb {F}_2$, $1\le s\le \frac{q-1}{2}$, $1\le t\le \frac{p-1}{2}$, $n=pq$, p, q odd, $\mathrm{gcd}(p, q)=1$, and the function defined by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{m-1}{2}}c_i{\mathrm{Tr}}_1^n\Big(\beta x^{1+2^{ei}}\Big), \end{aligned}$$

where $n=em$, $e=2^l$, m is odd, $c_i\in \mathbb {F}_2$$(1\le i\le \frac{m-1}{2})$, $\beta \in \mathbb {F}_{2^e}^{*}$. The Walsh transform of f at $\lambda \in \mathbb {F}_{2^n}$ is given by

$$\begin{aligned} W_f(\lambda )=\sum \limits _{x\in \mathbb {F}_{2^n}}(-1)^{f(x)+{\mathrm{Tr}}_1^n(\lambda x)}. \end{aligned}$$

Definition 1

([17]) Let $f(x)\in \mathcal {B}_n$. For any $\lambda \in \mathbb {F}_{2^n}$, if $W_f(\lambda )\in \{0,\pm 2^{\frac{n+r}{2}}\}$, for some fixed r, $r=0,1,\ldots ,n$, then f(x) is called r-plateaued. 0-plateaued (when n is even) functions are called bent. 1-plateaued (when n is odd) and 2-plateaued (when n is even) functions are called semi-bent.

The r-plateaued functions exist only when $n-r$ is even, or equivalently, if n and r have the same parity [18]. It is well-known that all the quadratic functions are plateaued [19].

The quadratic Boolean functions on $\mathbb {F}_{2^n}$ are as follows:

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\lfloor \frac{n}{2}\rfloor }c_i{\mathrm{Tr}}_1^n\Big(x^{1+2^i}\Big), \;c_i\in \mathbb {F}_2. \end{aligned}$$

Any such Boolean function with n variables has rank 2t with $0\le t\le \lfloor \frac{n}{2}\rfloor$ [3], and the rank can be found as follows. Let

$$\begin{aligned} \Omega _f(x;y)=f(0)+f(x)+f(y)+f(x+y). \end{aligned}$$

(4)

Then the rank of f(x) is 2t if and only if the equation

$$\begin{aligned} \Omega _f(x;y)=0,\,\text{ for } \text{ any }\,y\in \mathbb {F}_{2^n} \end{aligned}$$

in x just has $2^{n-2t}$ solutions. The rank of quadratic Boolean functions is connected with the distribution of its Walsh transform values. Furthermore, the following theorem holds.

Lemma 1

([3]) Let $f(x)\in \mathcal {B}_n$ is a quadratic function, and the rank of f(x) is 2t, $0\le t\le \lfloor \frac{n}{2}\rfloor$, then the distribution of its Walsh transform values is given by

$$\begin{aligned} W_f(\lambda )=\left\{ \begin{array}{ll} 2^{n-t},&{}\;\;2^{2t-1}+2^{t-1}\;\mathrm{times,}\\ 0, &{} \;\;2^{n-2t}\;\mathrm{times,}\\ -2^{n-t}, &{}\;\;2^{2t-1}-2^{t-1}\; \mathrm{times}. \end{array}\right. \end{aligned}$$

From the above theorem, it is easy to see that a quadratic Boolean function is semi-bent if and only if the rank of f(x) is $n-2$ when n is even, or the rank of f(x) is $n-1$ when n is odd.

Definition 2

([20]) The polynomials $l(x)=\sum \limits _{i=0}^na_ix^i$ and $L(x)=\sum \limits _{i=0}^na_ix^{q^i}$ over $\mathbb {F}_{q^m}$ (q is a prime integer) are called q-associates of each other. More specifically, l(x) is the conventional q-associate of L(x) and L(x) is the linearized q-associate of l(x).

Lemma 2

([20]) Let $L_1(x)$ and L(x) be q-polynomials over $\mathbb {F}_{q}$ with conventional q-associates $l_1(x)$ and l(x). Then $L_1(x)$ divides L(x) holds if and only if $l_1(x)$ divides l(x).

Lemma 3

Let $\phi (x)=x+\frac{1}{x}$ be a function defined over $\mathbb {F}_{2^n}^{*}$. Then the following two statements hold.

• (i) If $\phi (x)=\phi (y)$ for two elements x and y in $\mathbb {F}_{2^n}^*$, then $x=y$ or $xy=1$.

• (ii) $\phi (x)=0$ if and only if $x=1$.

Proof

• (i) If $\phi (x)=\phi (y)$ for $x,y\in \mathbb {F}_{2^n}^{*}$, i.e.,

  $$\begin{aligned} x+\frac{1}{x}=y+\frac{1}{y}, \end{aligned}$$

  then

  $$\begin{aligned} x^2y+y=xy^2+x, \end{aligned}$$

  which implies

  $$\begin{aligned} (x+y)(xy+1)=0. \end{aligned}$$

  Therefore $x=y$ or $xy=1$.

• (ii) $\phi (x)=0$ if and only if $x+\frac{1}{x}=0$, which is equivalent to $x^2=1$. Since $\mathrm{gcd}(2,2^n-1)=1$, the equation $x^2=1$ has only one solution $x=1$ in $\mathbb {F}_{2^n}^{*}$.

Lemma 4

Let p, q and i, j be positive integers satisfying that p, q are odd and $\mathrm{gcd}(p,q)=1$ and $\mathrm{gcd}(i, q)=\mathrm{gcd}(j, p)=1$. Then $\mathrm{gcd}(pq, pi\pm qj)=1$.

Proof

If $\mathrm{gcd}(pq, pi\pm qj)\ne 1$, then there exists a prime integer t such that $t\,|\,\mathrm{gcd}(pq, pi\pm qj)$, i.e.,

$$\begin{aligned} t\,|\,pq \;\;\text{ or }\;\;t\,|\,pi\pm qj. \end{aligned}$$

(5)

Since t is a prime, by Eq. (5) we have $t\,|\,p$ or $t\,|\,q$.

If $t\,|\,p$, then by Eq. (5), we have $t\,|\,qj$. Therefore, $t\,|\,q$ or $t\,|\,j$. If $t\,|\,q$, then $t\,|\,\mathrm{gcd}(p, q)=1$, which is impossible. If $t\,|\,j$, then $t\,|\,\mathrm{gcd}(j, p)=1$, which is also a contradiction with the assumption that t is a prime.

If $t\,|\,q$, we can similarly deduce that $t=1$, a contradiction. This completes the proof.

3 New constructions of semi-bent functions on $\mathbb {F}_{2^n}$ with n odd

In this section, several classes of semi-bent functions are constructed on $\mathbb {F}_{2^n}$, where $n=pq$ and p, q are odd integers such that $\mathrm{gcd}(p,q)=1$.

Theorem 3

Let $n=pq$ with p, q odd and $\mathrm{gcd}(p,q)=1$. Then the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)={\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^{qj}}\Big) \end{aligned}$$

(6)

is semi-bent, where i, j are two positive integers such that $\mathrm{gcd}(i,q)=\mathrm{gcd}(j,p)=1$.

Proof

By Lemma 1, in order to prove that f(x) is a semi-bent function, we just need to prove that the rank of f(x) is $n-1$. By Eq. (4), we have

$$\begin{aligned} \Omega _f(x;y)={\mathrm{Tr}}_1^n\Big(y \Big(x^{2^{pi}}+x^{2^{pq-pi}}+x^{2^{qj}}+x^{2^{pq-qj}}\Big)\Big). \end{aligned}$$

Let $L(x)=x^{2^{pi}}+x^{2^{pq-pi}}+x^{2^{qj}}+x^{2^{pq-qj}}$, and it is easy to see that $x^2+x\,|\,L(x)$. To prove that the rank of f(x) is $n-1$, we need to show that L(x) has two solutions in $\mathbb {F}_{2^n}$ or equivalently to prove $\mathrm{gcd}(L(x),x^{2^{pq}}+x)=x^2+x$. By Lemma 2, we need to show

$$\begin{aligned} \mathrm{gcd}(l(x), x^{pq}+1)=x+1, \end{aligned}$$

where $l(x)=x^{pi}+x^{pq-pi}+x^{qj}+x^{pq-qj}$. To do this, we divide the remaining proof into three cases.

If $\beta \ne 1$ is root of $x^{pq}+1$ and $\beta ^p=1$, then $l(\beta )=\beta ^{qj}+\beta ^{-qj}\ne 0$. Otherwise, if $l(\beta )=0$, then $\beta ^{2qj}=1$. Since $\mathrm{gcd}(2, 2^n-1)=1$, we have $\beta ^{qj}=1$. Recall that $\beta ^{p}=1$, $\mathrm{gcd}(p,q)=1$ and $\mathrm{gcd}(p,j)=1$, we have $\beta =1$, which is a contradiction with the assumption $\beta \ne 1$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^q=1$, we can similarly deduce that $l(\beta )=\beta ^{pi}+\beta ^{-pi}\ne 0$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^p\ne 1$, $\beta ^{q}\ne 1$, then $l(\beta )=\beta ^{pi}+\beta ^{-pi}+\beta ^{qj}+\beta ^{-qj}$. If $l(\beta )=0$, i.e., $\phi (\beta ^{pi})=\phi (\beta ^{qj})$, where $\phi (x)$ is the function defined in Lemma 2. By Lemma 2, we have

$$\begin{aligned} \beta ^{pi}=\beta ^{qj}\; \text{ or }\;\beta ^{pi+qj}=1. \end{aligned}$$

(7)

Since $\mathrm{gcd}(p,q)=1$ and $\mathrm{gcd}(i,q)=\mathrm{gcd}(j,p)=1$, by Lemma 4 we have $\mathrm{gcd}(pq, pi\pm qj)=1$. Recall that $\beta ^{pq}=1$, then by Eq. (7), we have $\beta =\beta ^{\mathrm{gcd}(pq,pi+qj)}=1$ or $\beta =\beta ^{\mathrm{gcd}(pq,pi-qj)}=1$, both of which contradict with the assumption $\beta \ne 1$.

From the analysis of above, we can see that $\mathrm{gcd}(l(x),x^{pq}+1)=x+1$. Thus the rank of f(x) is $n-1$, and this completes the proof.

Corollary 1

Let $n=pq$ with p, q distinct odd prime integers. Then the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)={\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^{qj}}\Big) \end{aligned}$$

(8)

is semi-bent for any integers i, j.

Theorem 4

Let $n=pq$ with p, q odd, $\mathrm{gcd}(p,q)=1$ and $3\not \mid p$. Then the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^{qj}}\Big) \end{aligned}$$

(9)

is semi-bent, where j is a positive integer such that $\mathrm{gcd}(j,p)=1$.

Proof

From Lemma 1, in order to prove that f(x) is a semi-bent function, we just need to prove that the rank of f(x) is $n-1$. By Eq. (9), we have

$$\begin{aligned} \begin{array}{rcl} f(x+y)&{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x+y)^{1+2^{pi}}+{\mathrm{Tr}}_1^n(x+y)^{1+2^{qj}}\Big)\\ {} &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}+y^{1+2^{pi}}\Big)+\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+y^{2^{pi}}x\Big)\\ {} &&+\,{\mathrm{Tr}}_1^n\Big(x^{1+2^{qj}}+y^{1+2^{qj}}+x^{2^{qj}}y+y^{2^{qj}}x\Big). \end{array} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{array}{rcl} \Omega _f(x;y)&{}=&{}f(0)+f(x)+f(y)+f(x+y)\\ {} &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+y^{2^{pi}}x\Big)+{\mathrm{Tr}}_1^n\Big(x^{2^{qj}}y+y^{2^{qj}}x\Big)\\ {} &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+yx^{2^{pq-pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{2^{qj}}y+yx^{2^{pq-qj}}\Big)\\ &{}=&{}{\mathrm{Tr}}_1^n\left(y\left(\sum \limits _{i=1}^{\frac{q-1}{2}}\Big(x^{2^{pi}}+x^{2^{pq-pi}}\Big)\right)\right)+{\mathrm{Tr}}{\mathrm{Tr}}_1^n\Big(y\Big(x^{2^{qj}}+x^{2^{pq-qj}}\Big)\Big)\\ &{}=&{} {\mathrm{Tr}}_1^n\left(y\left(\sum \limits _{i=1}^{q-1}x^{2^{pi}}+x^{2^{qj}}+x^{2^{pq-qj}}\right )\right). \end{array} \end{aligned}$$

Let $L(x)=\sum \limits _{i=1}^{q-1}x^{2^{pi}}+x^{2^{qj}}+x^{2^{pq-qj}}$, and it is easy to see that $x^2+x\,|\,L(x)$. To prove that the rank of f(x) is $n-1$, we need to show that $\mathrm{gcd}(L(x),x^{2^{pq}}+x)=x^2+x$, which is equivalent to show that

$$\begin{aligned} \mathrm{gcd}(l(x), x^{pq}+1)=x+1 \end{aligned}$$

from Lemma 2, where $l(x)=\sum \limits _{i=1}^{q-1}x^{pi}+x^{qj}+x^{pq-qj}$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^p=1$, then $l(\beta )=\beta ^{jq}+\beta ^{-jq}\ne 0$. Otherwise, we have $\beta ^{2jq}=1$. Since $\mathrm{gcd}(2,2^n-1)=1$, $\beta ^{jq}=1$. From the conditions $\beta ^p=1$ and $\mathrm{gcd}(j,p)=1$, we have $\beta =1$, which is a contradiction with the assumption.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^q=1$, then $l(\beta )=\frac{\beta ^{pq}+\beta ^p}{\beta ^p+1}+\beta ^{q(p-j)}+\beta ^{qj}=1\ne 0$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^p\ne 1$, $\beta ^q\ne 1$, thus

$$\begin{aligned} l(\beta )=\frac{\beta ^{pq}+\beta ^p}{\beta ^p+1}+\beta ^{q(p-j)}+\beta ^{qj}=1+\beta ^{q(p-j)}+\beta ^{qj}. \end{aligned}$$

If $l(\beta )=0$, then $\beta ^{3jq}=1$. Since $\beta ^{pq}=1$, $3\not \mid p$ and $\mathrm{gcd}(j,p)=1$, $\beta ^{\mathrm{gcd}(pq,3jq)}=\beta ^{\mathrm{gcd}(p,3j)q}=\beta ^q=1$, which contradicts with the assumption $\beta ^q\ne 1$. Hence we also have $l(\beta )\ne 0$ in this case.

From the analysis of above, we can see that $\mathrm{gcd}(l(x),x^{pq}+1)=x+1$. Thus the rank of f(x) is $n-1$, and this completes the proof.

For $j=1$ in Theorem 4, we have the following corollary.

Corollary 2

([8]) Let $n=pq$ with p, q odd, $\mathrm{gcd}(p,q)=1$ and $3\not \mid p$. Then the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^{q}}\Big) \end{aligned}$$

(10)

is semi-bent.

Theorem 5

Let $n=pq$ with p, q odd and $\mathrm{gcd}(p,q)=1$. Then the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^{q}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^{qj}}\Big) \end{aligned}$$

(11)

is semi-bent, where $j=2^l$ and l is a positive integer such that $\mathrm{gcd}(l,n)=1$.

Proof

From Lemma 1, in order to prove that f(x) is a semi-bent function, we just need to prove that the rank of f(x) is $n-1$. By Eq. (11), we have

$$\begin{aligned} \begin{array}{rcl} f(x+y)&{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big((x+y)^{1+2^{pi}}+{\mathrm{Tr}}_1^n((x+y)^{1+2^{q}})+{\mathrm{Tr}}_1^n((x+y)^{1+2^{qj}}\Big)\\ &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n(x^{1+2^{pi}}+y^{1+2^{pi}})+\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n(x^{2^{pi}}y+y^{2^{pi}}x)\\ &&+{}{\mathrm{Tr}}_1^n(x^{1+2^{q}}+y^{1+2^{q}}+x^{2^{q}}y+y^{2^{q}}x)\\ &&+{}{\mathrm{Tr}}_1^n(x^{1+2^{qj}}+y^{1+2^{qj}}+x^{2^{qj}}y+y^{2^{qj}}x). \end{array} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{array}{rcl} \Omega _f(x;y)&{}=&{}f(0)+f(x)+f(y)+f(x+y)\\ &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+y^{2^{pi}}x\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(x^{2^{q}}y+y^{2^{q}}x\Big)+{\mathrm{Tr}}_1^n\Big(x^{2^{qj}}y+y^{2^{qj}}x\Big)\\ &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+yx^{2^{pq-pi}}\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(x^{2^{q}}y+yx^{2^{pq-q}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{2^{qj}}y+yx^{2^{pq-qj}}\Big)\\ &{}=&{}{\mathrm{Tr}}_1^n \left(y\left(\sum \limits _{i=1}^{\frac{q-1}{2}}\Big(x^{2^{pi}}+x^{2^{pq-pi}}\Big)\right)\right)+{\mathrm{Tr}}_1^n\Big(y\Big(x^{2^{q}}+x^{2^{pq-q}}\Big)\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(y\Big(x^{2^{qj}}+x^{2^{pq-qj}}\Big)\Big)\\ &{}=&{} {\mathrm{Tr}}_1^n\left(y\left(\sum \limits _{i=1}^{q-1}x^{2^{pi}}+x^{2^{q}}+x^{2^{pq-q}}+x^{2^{qj}}+x^{2^{pq-qj}}\right)\right). \end{array} \end{aligned}$$

Let $L(x)=\sum \limits _{i=1}^{q-1}x^{2^{pi}}+x^{2^{q}}+x^{2^{pq-q}}+x^{2^{qj}}+x^{2^{pq-qj}}$, and it is easy to see that $x^2+x\,|\,L(x)$. To prove that the rank of f(x) is $n-1$, we need to show that $\mathrm{gcd}(L(x),x^{2^{pq}}+x)=x^2+x$, which equals

$$\begin{aligned} \mathrm{gcd}(l(x), x^{pq}+1)=x+1 \end{aligned}$$

from Lemma 2, where $l(x)=\sum \limits _{i=1}^{q-1}x^{pi}+x^{{q}}+x^{pq-q}+x^{qj}+x^{pq-qj}$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^p=1$, then $l(\beta )=\beta ^q+\beta ^{-q}+\beta ^{jq}+\beta ^{-jq}$. Let $w=\beta ^q+\beta ^{-q}$, then $l(\beta )=w+w^{j}$, where $j=2^l$ and $w\ne 0, 1$. We claim that $l(\beta )\ne 0$. Otherwise, we have $w=w^{2^l}$. Since $\mathrm{gcd}(2^l-1,2^n-1)=1$, $w=0$ or 1, which is a contradiction.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^q=1$, then $l(\beta )=1\ne 0$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^p\ne 1$, $\beta ^q\ne 1$, thus

$$\begin{aligned} l(\beta )=1+\beta ^q+\beta ^{-q}+\beta ^{jq}+\beta ^{-jq}. \end{aligned}$$

If $l(\beta )=0$, then $w+w^{2^l}=1$. Since n is odd, ${\mathrm{Tr}}_1^n(1)=1$. But ${\mathrm{Tr}}_1^n(w+w^{2^l})=0$, leading to a contradiction. Hence we also have $l(\beta )\ne 0$ in this case.

From the analysis of above, we can see that $\mathrm{gcd}(l(x),x^{pq}+1)=x+1$. Thus the rank of f(x) is $n-1$, and this completes the proof.

Theorem 6

Let $n=pq$, with p, q odd and $\mathrm{gcd}(p,q)=1$. Let $j=2^{u-1}-2^{v-1}$ and $k=2^{u-1}+2^{v-1}$, where u, v are positive integers such that $u>v$. Then

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^{qj}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{1+2^{qk}}\Big) \end{aligned}$$

(12)

is semi-bent on $\mathbb {F}_{2^n}$.

Proof

From Lemma 1, in order to prove that f(x) is a semi-bent function, we just need to prove that the rank of f(x) is $n-1$. By Eq. (12), we have

$$\begin{aligned} \begin{array}{rcl} f(x+y)&{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big((x+y)^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n\Big((x+y)^{1+2^{qj}}\Big)+{\mathrm{Tr}}_1^n\Big((x+y)^{1+2^{qk}}\Big)\\ &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}+y^{1+2^{pi}}\Big)+\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+y^{2^{pi}}x\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(x^{1+2^{qj}}+y^{1+2^{qj}}+x^{2^{qj}}y+y^{2^{qj}}x\Big)\\ &&+\,{}{\mathrm{Tr}}_1^n\Big(x^{1+2^{qk}}+y^{1+2^{qk}}+x^{2^{qk}}y+y^{2^{qk}}x\Big). \end{array} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{array}{rcl} \Omega _f(x;y)&{}=&{}f(0)+f(x)+f(y)+f(x+y)\\ &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+y^{2^{pi}}x\Big)+{\mathrm{Tr}}_1^n\Big(x^{2^{qj}}y+y^{2^{qj}}x\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(x^{2^{qk}}y+y^{2^{qk}}x\Big)\\ &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+yx^{2^{pq-pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{2^{qj}}y+yx^{2^{pq-qj}}\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(x^{2^{qk}}y+yx^{2^{pq-qk}}\Big)\\ &{}=&{}{\mathrm{Tr}}_1^n\left(y\left(\sum \limits _{i=1}^{\frac{q-1}{2}}\Big(x^{2^{pi}}+x^{2^{pq-pi}}\Big)\right)\right)+{\mathrm{Tr}}_1^n\Big(y\Big(x^{2^{qj}}+x^{2^{pq-qj}}\Big)\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(y\Big(x^{2^{qk}}+x^{2^{pq-qk}}\Big)\Big)\\ &{}=&{} {\mathrm{Tr}}_1^n\left(y\left(\sum \limits _{i=1}^{q-1}x^{2^{pi}}+x^{2^{qj}}+x^{2^{pq-qj}}+x^{2^{qk}}+x^{2^{pq-qk}}\right)\right). \end{array} \end{aligned}$$

Let $L(x)=\sum \limits _{i=1}^{q-1}x^{2^{pi}}+x^{2^{qj}}+x^{2^{pq-qj}}+x^{2^{qk}}+x^{2^{pq-qk}}$, and it is easy to see that $x^2+x\,|\,L(x)$. To prove that the rank of f(x) is $n-1$, we need to show that $\mathrm{gcd}(L(x),x^{2^{pq}}+x)=x^2+x$, which equals

$$\begin{aligned} \mathrm{gcd}(l(x), x^{pq}+1)=x+1 \end{aligned}$$

from Lemma 2, where $l(x)=\sum \limits _{i=1}^{q-1}x^{pi}+x^{{qj}}+x^{pq-qj}+x^{qk}+x^{pq-qk}$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^p=1$, then $l(\beta )=\beta ^{qj}+\beta ^{-qj}+\beta ^{qk}+\beta ^{-qk}$. If $l(\beta )=0$, then $\phi (\beta ^{qj})=\phi (\beta ^{qk})$, where $\phi$ is the function defined in Lemma 2. By Lemma 2, we have $\beta ^{qj}=\beta ^{qk}$ or $\beta ^{q(j+k)}=1$. Since $\mathrm{gcd}(j\pm k,p)=1$, $\beta ^p=1$ and $\mathrm{gcd}(p,q)=1$, we have $\beta =1$, which is a contradiction with the assumption $\beta \ne 1$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^q=1$, then $l(\beta )=1\ne 0$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^p\ne 1$, $\beta ^q\ne 1$, thus

$$\begin{aligned} l(\beta )=1+\beta ^{qj}+\beta ^{-qj}+\beta ^{qk}+\beta ^{-qk}. \end{aligned}$$

If $l(\beta )=0$, then

$$\begin{aligned} \beta ^{qj}+\beta ^{-qj}+\beta ^{qk}+\beta ^{-qk}=1 \end{aligned}$$

(13)

By Eq. (13), we have

$$\begin{aligned} \beta ^{q(j+k)}+\beta ^{q(k-j)}+\beta ^{2qk}+\beta ^{qk}=1. \end{aligned}$$

(14)

Since $k+j=2^u$ and $k-j=2^v$, Eq. (14) can be rewritten as

$$\begin{aligned} (\beta ^{q})^{2^u}+(\beta ^{q})^{2^v}+(\beta ^{qk})^2+\beta ^{qk}=1. \end{aligned}$$

(15)

It is clear that ${\mathrm{Tr}}_1^n((\beta ^{q})^{2^u}+(\beta ^{q})^{2^v}+(\beta ^{qk})^2+\beta ^{qk})=0$. But ${\mathrm{Tr}}_1^n(1)=1$, leading to a contradiction. Hence we also have $l(\beta )\ne 0$ in this case.

From the analysis of above, we can see that $\mathrm{gcd}(l(x),x^{pq}+1)=x+1$. Thus the rank of f(x) is $n-1$, and this completes the proof.

Theorem 7

Let $n=pq$, with p, q odd and $\mathrm{gcd}(p,q)=1$. Let $j=2^u$ and $k=2^v$, where u, v are positive integers such that $u>v$ and $\mathrm{gcd}(u-v,n)=1$. Then

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+{\mathrm{Tr}}_1^n(x^{1+2^{qj}})+{\mathrm{Tr}}_1^n\Big(x^{1+2^{qk}}\Big) \end{aligned}$$

(16)

is semi-bent on $\mathbb {F}_{2^n}$.

Proof

From Lemma 1, in order to prove that f(x) is a semi-bent function, we just need to prove that the rank of f(x) is $n-1$. Through similar calculations as Theorem 5, we have

$$\begin{aligned} \begin{array}{rcl} \Omega _f(x;y)&{}=&{}f(0)+f(x)+f(y)+f(x+y)\\ &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+y^{2^{pi}}x\Big)+{\mathrm{Tr}}_1^n\Big(x^{2^{qj}}y+y^{2^{qj}}x\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(x^{2^{qk}}y+y^{2^{qk}}x\Big)\\ &{}=&{}\sum \limits _{i=1}^{\frac{q-1}{2}}{\mathrm{Tr}}_1^n\Big(x^{2^{pi}}y+yx^{2^{pq-pi}}\Big)+{\mathrm{Tr}}_1^n\Big(x^{2^{qj}}y+yx^{2^{pq-qj}}\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(x^{2^{qk}}y+yx^{2^{pq-qk}}\Big)\\ &{}=&{}{\mathrm{Tr}}_1^n\left(y\left(\sum \limits _{i=1}^{\frac{q-1}{2}}\Big(x^{2^{pi}}+x^{2^{pq-pi}}\Big)\right)\right)+{\mathrm{Tr}}_1^n\Big(y(x^{2^{qj}}+x^{2^{pq-qj}})\Big)\\ &&+\,{\mathrm{Tr}}_1^n\Big(y\Big(x^{2^{qk}}+x^{2^{pq-qk}}\Big)\Big)\\ &{}=&{} {\mathrm{Tr}}_1^n\left(y\left(\sum \limits _{i=1}^{q-1}x^{2^{pi}}+x^{2^{qj}}+x^{2^{pq-qj}}+x^{2^{qk}}+x^{2^{pq-qk}}\right)\right). \end{array} \end{aligned}$$

Let $L(x)=\sum \limits _{i=1}^{q-1}x^{2^{pi}}+x^{2^{qj}}+x^{2^{pq-qj}}+x^{2^{qk}}+x^{2^{pq-qk}}$, and it is easy to see that $x^2+x\,|\,L(x)$. To prove that the rank of f(x) is $n-1$, we need to show that $\mathrm{gcd}(L(x),x^{2^{pq}}+x)=x^2+x$, which equals

$$\begin{aligned} \mathrm{gcd}(l(x), x^{pq}+1)=x+1 \end{aligned}$$

from Lemma 2, where $l(x)=\sum \limits _{i=1}^{q-1}x^{pi}+x^{{qj}}+x^{pq-qj}+x^{qk}+x^{pq-qk}$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^p=1$, then $l(\beta )=\beta ^{qj}+\beta ^{-qj}+\beta ^{qk}+\beta ^{-qk}$. Let $w=\beta ^q+\beta ^{-q}$, then $l(\beta )=w^{2^u}+w^{2^v}$. If $l(\beta )=0$, then $w^{2^u}=w^{2^v}$. Note that $w\ne 0$, then we have $w^{2^u-2^v}=w^{2^v(2^{u-v}-1)}=1$. Since $\mathrm{gcd}(2^v(2^{u-v}-1),2^n-1)=\mathrm{gcd}(2^{u-v}-1,2^n-1)=2^{\mathrm{gcd}(u-v,n)}-1=1$, we have $w=1$. This is impossible, because the equality $w=1$ will lead to $\beta =1$, which is a contradiction with the assumption $\beta \ne 1$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^q=1$, then $l(\beta )=1\ne 0$.

If $\beta \ne 1$ is a root of $x^{pq}+1$ and $\beta ^p\ne 1$, $\beta ^q\ne 1$, thus

$$\begin{aligned} l(\beta )=1+\beta ^{qj}+\beta ^{-qj}+\beta ^{qk}+\beta ^{-qk}=w^{2^u}+w^{2^v}. \end{aligned}$$

If $l(\beta )=0$, then $w^{2^u}+w^{2^v}=1$. Note that ${\mathrm{Tr}}_1^n(w^{2^u}+w^{2^v})=0$. But ${\mathrm{Tr}}_1^n(1)=1$, leading to a contradiction. Hence we also have $l(\beta )\ne 0$ in this case.

From the analysis of above, we can see that $\mathrm{gcd}(l(x),x^{pq}+1)=x+1$. Thus the rank of f(x) is $n-1$, and this completes the proof.

4 New constructions of e-plateaued and semi-bent functions on $\mathbb {F}_{2^n}$ with n even

In this section, we give some new constructions of quadratic e-plateaued and semi-bent functions in polynomial forms with even n. We suppose $n=em$, where e and m are even and odd positive integers respectively in this section.

Theorem 8

For any $\beta \in \mathbb {F}_{2^e}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{m-1}{2}}{\mathrm{Tr}}_1^{n}\Big(\beta x^{1+2^{ei}}\Big) \end{aligned}$$

(17)

is e-plateaued.

Proof

By Lemma 1, in order to prove that f(x) is an e-plateaued function, we just need to prove that the rank of f(x) is $n-e$. By Eq. (17), we have

$$\begin{aligned} \begin{array}{rcl} \Omega _f(x;y)&{}=&{}f(0)+f(x)+f(y)+f(x+y)\\ {} &{}=&{}\sum \limits _{i=1}^{\frac{m-1}{2}}{\mathrm{Tr}}_1^n\Big(\beta xy^{2^{ei}}+\beta x^{2^{ei}}y\Big)\\ {} &{}=&{}\sum \limits _{i=1}^{\frac{m-1}{2}}{\mathrm{Tr}}_1^n\Big(\beta x^{2^{em-ei}}y+\beta x^{2^{ei}}y\Big)\\ {} &{}=&{}\sum \limits _{i=1}^{\frac{m-1}{2}}{\mathrm{Tr}}_1^n\Big(\beta x^{2^{em-ei}}y+\beta x^{2^{ei}}y\Big)\\ {} &{}=&{}\sum \limits _{i=1}^{m-1}{\mathrm{Tr}}_1^{n}\Big(\beta yx^{2^{ei}}\Big)\\ {} &{}=&{}{\mathrm{Tr}}_1^n\Big(\beta y\Big({\mathrm{Tr}}_e^n(x)+x\Big)\Big).\end{array} \end{aligned}$$

It follows that

$$\begin{aligned} \Omega _f(x;y)=0, \;\text{ for } \text{ any }\;y\in \mathbb {F}_{2^n} \end{aligned}$$

(18)

holds if and only if

$$\begin{aligned} {\mathrm{Tr}}_e^n(x)+x=0. \end{aligned}$$

So $x={\mathrm{Tr}}_e^n(x)\in \mathbb {F}_{2^e}$, and for any $x\in \mathbb {F}_{2^e}$, ${\mathrm{Tr}}_e^n(x)=x{\mathrm{Tr}}_e^n(1)=x$. This implies that ${\mathrm{Tr}}_e^n(x)+x=0$ holds if and only if $x\in \mathbb {F}_{2^e}$. Hence Eq. (18) has only $2^e$ solutions, so the rank of f(x) is $n-e$. By Lemma 1 and Definition 1, f(x) is an e-plateaued function.

For $e=2$, we have the following corollary.

Corollary 3

([8]) For any $\beta \in \mathbb {F}_{2^2}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{m-1}{2}}{\mathrm{Tr}}_1^n\Big(\beta x^{1+2^{2i}}\Big) \end{aligned}$$

is a semi-bent function.

Now we consider the general case. We study the function defined by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{m-1}{2}}c_i{\mathrm{Tr}}_1^n\Big(\beta x^{1+2^{ei}}\Big), \end{aligned}$$

(19)

where $c_i\in \mathbb {F}_2, \;(1\le i\le \frac{m-1}{2}),\,\beta \in \mathbb {F}_{2^e}^{*}$.

Theorem 9

If $e=2^l$ for some positive integer l, then for any $\beta \in \mathbb {F}_{2^e}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by Eq. (19) is e-plateaued if and only if

$$\begin{aligned} \mathrm{gcd}\left( \sum \limits _{i=1}^{\frac{m-1}{2}}c_i(x^{{i}}+x^{{m-i}}), x^{m}+1\right) =x+1. \end{aligned}$$

Proof

By Lemma 1, we only need to prove that the rank of f(x) is $n-e$. Similar to the proof of Theorem 7, the rank of f(x) is $n-e$ if and only if the equation

$$\begin{aligned} \sum \limits _{i=1}^{\frac{m-1}{2}}c_i\Big(x^{2^{ei}}+x^{2^{em-ei}}\Big) \end{aligned}$$

(20)

has only $2^e$ solutions in $\mathbb {F}_{2^n}$. For any $x\in \mathbb {F}_{2^e}$, it is obvious that $x^{2^{ei}}+x^{2^{em-ei}}=0\,(1\le i\le \frac{m-1}{2})$ holds. So

$$\begin{aligned} x^{2^e}+x\,|\,\sum \limits _{i=1}^{\frac{m-1}{2}}c_i\Big(x^{2^{ei}}+x^{2^{em-ei}}\Big). \end{aligned}$$

In order to show that Eq. (20) has only $2^e$ solutions in $\mathbb {F}_{2^n}$, we just need to prove that

$$\begin{aligned} \mathrm{gcd}\left( \sum \limits _{i=1}^{\frac{m-1}{2}}c_i\Big(x^{2^{ei}}+x^{2^{em-ei}}\Big), x^{2^n}+x\right) =x^{2^e}+x\end{aligned}$$

(21)

holds. By Lemma 3, Eq. (21) holds if and only if

$$\begin{aligned} \mathrm{gcd}\left( \sum \limits _{i=1}^{\frac{m-1}{2}}c_i\Big(x^{{ei}}+x^{{em-ei}}\Big), x^{em}+1\right) =x^{e}+1=(x+1)^e. \end{aligned}$$

(22)

Eq. (22) holds if and only if

$$\begin{aligned} \mathrm{gcd}\left( \sum \limits _{i=1}^{\frac{m-1}{2}}c_i\Big(x^{{i}}+x^{{m-i}}\Big), x^{m}+1\right) =x+1. \end{aligned}$$

Theorem 10

If $e=2^l$ for some positive integer l, then for any $\beta \in \mathbb {F}_{2^e}^{*}$, $r\ge 1$, $1\le k\le \frac{m-1}{2}$, the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^k{\mathrm{Tr}}_1^n\Big(\beta x^{1+2^{eri}}\Big) \end{aligned}$$

is e-plateaued if and only if

$$\begin{aligned} \mathrm{gcd}(k+1,m)=\mathrm{gcd}(k,m)=\mathrm{gcd}(r,m)=1. \end{aligned}$$

Proof

Similar to the proof of Theorem 8, f(x) is e-plateaued if and only if

$$\begin{aligned} \mathrm{gcd}(L(x), x^m+1)=x+1, \;\;\text{ where }\;L(x)=\sum \limits _{i=1}^k\Big(x^{ri}+x^{m-ri}\Big). \end{aligned}$$

We have

$$\begin{aligned} \begin{array}{rcl} L(x)&{}=&{}\sum \limits _{i=1}^k\Big(x^{ri}+x^{-ri}\Big)\\ {} &{}=&{}\sum \limits _{i=0}^{2k}\frac{x^{ri}}{x^{rk}}+1\\ &{}=&{} \frac{x^{(2k+1)r}+1}{x^{rk}(x^r+1)}+1\\ {} &{}=&{}\frac{x^{(2k+1)r}+x^{rk+r}+x^{rk}+1}{x^{rk}(x^r+1)} \\ &{}=&{}\frac{(x^{rk+r}+1)(x^{rk}+1)}{x^{rk}(x^r+1)}. \end{array} \end{aligned}$$

Thus,

$$\begin{aligned} \mathrm{gcd}(L(x),x^m+1)=\mathrm{gcd}\left( \frac{(x^{rk+r}+1)(x^{rk}+1)}{x^{rk}(x^r+1)}, x^m+1\right) =x+1 \end{aligned}$$

holds if and only if

$$\begin{aligned} \mathrm{gcd}(r(k+1),m)=\mathrm{gcd}(rk,m)=\mathrm{gcd}(r,m)=1, \end{aligned}$$

which equals

$$\begin{aligned} \mathrm{gcd}(k+1,m)=\mathrm{gcd}(k,m)=\mathrm{gcd}(r,m)=1. \end{aligned}$$

Corollary 4

([8]) For any $\beta \in \mathbb {F}_{2^2}^{*}$, $r\ge 1$, $1\le k\le \frac{m-1}{2}$, the function defined on $\mathbb {F}_{2^n}$ by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^k{\mathrm{Tr}}_1^n\Big(\beta x^{1+2^{2ri}}\Big) \end{aligned}$$

is semi-bent if and only if

$$\begin{aligned} \mathrm{gcd}(k+1,m)=\mathrm{gcd}(k,m)=\mathrm{gcd}(r,m)=1. \end{aligned}$$

Theorem 11

If $e=2^l$ for some positive integer l, then for any $\beta \in \mathbb {F}_{2^e}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{ei}})\,(1\le i\le m-1)$ is e-plateaued if and only if $\mathrm{gcd}(i,m)=1$ and m is odd.

Proof

Let $l(x)=x^i+x^{m-i}$. By Theorem 8, the function is e-plateaued if and only if $\mathrm{gcd}(l(x),x^m+1)=x+1$. As $x^i+x^{m-i}=x^i(x^{m-2i}+1)$ and $\mathrm{gcd}(x^i,x^m+1)=1$. The equality $\mathrm{gcd}(l(x),x^m+1)=x+1$ holds if and only if $\mathrm{gcd}(i,m)=1$ and m is odd.

Corollary 5

For any $\beta \in \mathbb {F}_{2^2}^{*}$, the function defined by $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{2i}})$$(1\le i\le m-1)$, is semi-bent if and only if $\mathrm{gcd}(i, m)=1$ and m odd.

When $k=1$ and $r=1$ in Theorem 10, we have the following corollary.

Corollary 6

If $e=2^l$ for some positive integer l, then for any $\beta \in \mathbb {F}_{2^e}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{ei}}+\beta x^{1+2^{ej}})\,(1\le i<j\le \lfloor \frac{n}{4}\rfloor )$ is e-plateaued if and only if $\mathrm{gcd}(m,j+i)=1$, $\mathrm{gcd}(m,j-i)=1$ and m is odd.

Corollary 7

If $e=2^l$ for some positive integer l, then for any $\beta \in \mathbb {F}_{2^e}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{ei}}+\beta x^{1+2^{ej}})$ is e-plateaued for any i, j with $1\le i<j\le \lfloor \frac{n}{4}\rfloor$ if and only if m is an odd prime integer.

Corollary 8

([8]) For any $\beta \in \mathbb {F}_{2^2}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{2i}}+\beta x^{1+2^{2j}})\,(1\le i<j\le \lfloor \frac{n}{4}\rfloor )$ is semi-bent if and only if $\mathrm{gcd}(m,j+i)=1$, $\mathrm{gcd}(m,j-i)=1$ and m is odd.

Theorem 12

If $e=2^l$ for some positive integer l, then for any $\beta \in \mathbb {F}_{2^e}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{ei}}+\beta x^{1+2^{ej}}+\beta x^{1+2^{et}})\,(1\le i<j<t\le \lfloor \frac{n}{4}\rfloor , i+j=t)$ is e-plateaued if and only if $\mathrm{gcd}(m,i)=1$, $\mathrm{gcd}(m,j)=1$ and $\mathrm{gcd}(m,t)=1$.

Proof

Let $l(x)=x^{i}+x^{m-i}+x^{j}+x^{m-j}+x^{t}+x^{m-t}$. By Theorem 8, the function is e-plateaued if and only if $\mathrm{gcd}(l(x),x^m+1)=x+1$. Note that

$$\begin{aligned} \begin{array}{rcl} l(x)&{}=&{}(1+x^i)(1+x^j)+1+x^m+x^m(1+x^{-i})(1+x^{-j})\\ {} &{}=&{}(1+x^i)(1+x^j)(1+x^{m-i-j})+1+x^m. \end{array} \end{aligned}$$

Then the equality $\mathrm{gcd}(l(x),x^m+1)=x+1$ holds if and only if $\mathrm{gcd}(m,i)=1$, $\mathrm{gcd}(m,j)=1$ and $\mathrm{gcd}(m,t)=1$.

Theorem 13

If $e=2^l$ for some positive integer l, then for any $\beta \in \mathbb {F}_{2^e}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{ei}}+\beta x^{1+2^{ej}}+\beta x^{1+2^{et}})\,(1\le i<j\le m-1, i+j=2t)$ is e-plateaued if and only if $\mathrm{gcd}(t,m)=1$.

Proof

Let $L(x)=x^{i}+x^{m-i}+x^{j}+x^{m-j}+x^{t}+x^{m-t}$. By Theorem 8, the function is e-plateaued if and only if $\mathrm{gcd}(L(x),x^m+1)=x+1$. Note that

$$\begin{aligned} \begin{array}{rcl} L(x)&{}=&{}x^i+x^j+x^{\frac{i+j}{2}}+x^{m-i}+x^{m-j}+x^{m-\frac{i+j}{2}}\\ &{}=&{}x^i+x^j+x^{\frac{i+j}{2}}+x^{m-(i+j)}\Big(x^i+x^j+x^{\frac{i+j}{2}}\Big)\\ &{}=&{}x^i\Big(1+x^{j-i}+x^{\frac{j-i}{2}}\Big)\Big(1+x^{m-(i+j)}\Big), \end{array} \end{aligned}$$

(23)

and $\mathrm{gcd}(x^i, x^m+1)=1$, we have $\mathrm{gcd}(L(x), x^m+1)=\mathrm{gcd}((1+x^{j-i}+x^{\frac{j-i}{2}})(1+x^{m-(i+j)}), x^m+1)$. Since m is odd, ${\mathrm{Tr}}_1^m(1)=1$. Consequently, ${\mathrm{Tr}}_1^m(1+x^{j-i}+x^{\frac{j-i}{2}})=1$. That is, for any $a\in \mathbb {F}_{2^m}$, $1+a^{j-i}+a^{\frac{j-i}{2}}\ne 0$. Hence, $\mathrm{gcd}(x+x^{2^{j-i}}+x^{2^{\frac{j-i}{2}}}, x^{2^m}+x)=1$. By Lemma 2, we have $\mathrm{gcd}(1+x^{j-i}+x^{\frac{j-i}{2}}, x^m+1)=1$. Therefore, $\mathrm{gcd}(L(x), x^m+1)=\mathrm{gcd}(x^{m-(i+j)}+1, x^m+1)$. By Theorem 8, f(x) is e-plateaued if and only if $\mathrm{gcd}(x^{m-(i+j)}+1, x^m+1)=x+1$, which is equivalent to the condition $\mathrm{gcd}(i+j, m)=\mathrm{gcd}(2t, m)=\mathrm{gcd}(t, m)=1$.

Corollary 9

If $n=2m$ with $m(>1)$ odd, then for any $\beta \in \mathbb {F}_{2^2}^{*}$,

$$\begin{aligned} f(x)={\mathrm{Tr}}_1^n\Big(\beta x^{1+2^{2i}}+\beta x^{1+2^{2j}}+\beta x^{1+2^{2t}}\Big) \end{aligned}$$

$(i+j=2t, 1\le i<j\le m-1)$ is semi-bent if and only if $\mathrm{gcd}(t, m)=1$.

Theorem 14

If $e=2^l$ for some positive integer l, then for any $\beta \in \mathbb {F}_{2^e}^{*}$, the function defined on $\mathbb {F}_{2^n}$ by $f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{ei}}+\beta x^{1+2^{ej}}+\beta x^{1+2^{et}})\,(1\le i<j\le m-1, j-i=2t)$ is e-plateaued if and only if $\mathrm{gcd}(t, m)=1$.

Proof

Let $L(x)=x^{i}+x^{m-i}+x^{j}+x^{m-j}+x^{t}+x^{m-t}$. By Theorem 8, the function is e-plateaued if and only if $\mathrm{gcd}(L(x),x^m+1)=x+1$. Consider

$$\begin{aligned} \begin{array}{rcl} x^tL(x)&{}=&{}x^{t+i}+x^{t+j}+x^{2t}+x^{m-i+t}+x^{m-j+t}+x^{m} \\ &{}=&{}x^{t+j}+x^{m-i+t}+x^{2t}+x^{t+i}+x^{m-(j-t)}+x^m\\ &{}=&{}x^{t+j}+x^{m-i+t}+x^{2t}+x^{\frac{i+j}{2}}+x^{m-\frac{i+j}{2}}+x^m\\ &{}=&{}\Big(x^{2t}+1)(x^{\frac{i+j}{2}}+x^{m-\frac{i+j}{2}}+1\Big)+x^m+1. \end{array} \end{aligned}$$

(24)

Since $\mathrm{gcd}(x^t, x^m+1)=1$, we have

$$\begin{aligned} \begin{array}{rcl} \mathrm{gcd}(L(x), x^m+1)&{}=&{}\mathrm{gcd}(x^tL(x), x^m+1)\\ {} &{}=&{}\mathrm{gcd}\Big((x^{2t}+1)\Big(x^{\frac{i+j}{2}}+x^{m-\frac{i+j}{2}}+1\Big), x^m+1\Big). \end{array} \end{aligned}$$

Suppose that a is a root of $x^tL(x)$, $a\notin \mathbb {F}_2$ and $a^m=1$, then $a^{\frac{i+j}{2}}+a^{m-\frac{i+j}{2}}+1\ne 0$. Otherwise, we have $a^{\frac{i+j}{2}}(a^{\frac{i+j}{2}}+a^{m-\frac{i+j}{2}}+1)=0$, i.e.,

$$\begin{aligned} 1+a^{\frac{i+j}{2}}+a^{i+j}=0. \end{aligned}$$

(25)

Note that ${\mathrm{Tr}}_1^m(1)=1$, and ${\mathrm{Tr}}_1^m(a^{\frac{i+j}{2}}+a^{i+j})=0$. This induces a contradiction with Eq. (25).

By the analysis above, we have

$$\begin{aligned} \begin{array}{rcl} \mathrm{gcd}(L(x),x^m+1)&{}=&{}\mathrm{gcd}(x^tL(x), x^m+1)\\ {} &{}=&{}\mathrm{gcd}(x^{2t}+1, x^m+1). \end{array} \end{aligned}$$

Consequently, f(x) is e-plateaued if and only if $\mathrm{gcd}(x^{2t}+1, x^m+1)=x+1$, which is equivalent to the condition $\mathrm{gcd}(t, m)=1$.

Since m is odd, ${\mathrm{Tr}}_1^m(1)=1$. Consequently, ${\mathrm{Tr}}_1^m(1+x^{j-i}+x^{\frac{j-i}{2}})=1$. That is, for any $a\in \mathbb {F}_{2^m}$, $1+a^{j-i}+a^{\frac{j-i}{2}}\ne 0$. Hence, $\mathrm{gcd}(x+x^{2^{j-i}}+x^{2^{\frac{j-i}{2}}}, x^{2^m}+x)=1$. By Lemma 2, we have $\mathrm{gcd}(1+x^{j-i}+x^{\frac{j-i}{2}}, x^m+1)=1$. Therefore, $\mathrm{gcd}(L(x), x^m+1)=\mathrm{gcd}(x^{m-(i+j)}+1, x^m+1)$. By Theorem 8, f(x) is e-plateaued if and only if $\mathrm{gcd}(x^{m-(i+j)}+1, x^m+1)=x+1$, which is equivalent to the condition $\mathrm{gcd}(i+j, m)=\mathrm{gcd}(2t, m)=\mathrm{gcd}(t, m)=1$.

Corollary 10

If $n=2m$ with $m(>1)$ odd, then for any $\beta \in \mathbb {F}_{2^2}^{*}$,

$$\begin{aligned} f(x)={\mathrm{Tr}}_1^n(\beta x^{1+2^{2i}}+\beta x^{1+2^{2j}}+\beta x^{1+2^{2t}}) \end{aligned}$$

$(j-i=2t, 1\le i<j\le m-1)$ is semi-bent if and only if $\mathrm{gcd}(t, m)=1$.

5 Concluding remarks

In this paper, we study the function defined by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{q-1}{2}}c_i{\mathrm{Tr}}_1^n\Big(x^{1+2^{pi}}\Big)+\sum \limits _{i=1}^{\frac{p-1}{2}}d_i{\mathrm{Tr}}_1^n\Big( x^{1+2^{qi}}\Big), \end{aligned}$$

where $c_i, d_j\in \mathbb {F}_2$, $1\le i\le \frac{q-1}{2}$, $1\le j\le \frac{p-1}{2}$, $n=pq$, p, q odd, $\mathrm{gcd}(p, q)=1$, and the function defined by

$$\begin{aligned} f(x)=\sum \limits _{i=1}^{\frac{m-1}{2}}c_i{\mathrm{Tr}}_1^n\Big(\beta x^{1+2^{ei}}\Big), \end{aligned}$$

where $n=em$, $e=2^l$, m is odd, $c_i\in \mathbb {F}_2$$(1\le i\le \frac{m-1}{2})$, $\beta \in \mathbb {F}_{2^e}^{*}$. We prove that these two kinds of functions contain semi-bent ones in certain cases. Moreover, we present some characterizations of e-plateaued functions with few trace terms when n is even. Furthermore, their are still some problems that need to be studied such as how to obtain semi-bent functions with higher degree by the primary constructions.