Proof. For any \(a \in V_{n}^{(p)}\), \[\label{1} \begin{align} \sum_{x \in V_{n}^{(p)}}\zeta_{p}^{f(x+a)-f(x)}&=\frac{1}{p^n}\sum_{x, y \in V_{n}^{(p)}}\zeta_{p}^{f(y)-f(x)}\sum_{z \in V_{n}^{(p)}}\zeta_{p}^{\langle z, x-y+a\rangle_{n}}\\ &=\frac{1}{p^n}\sum_{z \in V_{n}^{(p)}}\zeta_{p}^{\langle z, a\rangle_{n}}\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{-f(x)+\langle z, x\rangle_{n}}\sum_{y \in V_{n}^{(p)}}\zeta_{p}^{f(y)-\langle z, y\rangle_{n}}\\ &=\frac{1}{p^n}\sum_{z \in V_{n}^{(p)}}|W_{f}(z)|^{2}\zeta_{p}^{\langle z, a\rangle_{n}}=\frac{1}{p^n}\sum_{z \in S_{f}}p^{n+s}\zeta_{p}^{\langle z, a\rangle_{n}}=p^{s}\zeta_{p}^{\langle v, a\rangle_{n}}\sum_{z \in E}\zeta_{p}^{\langle z, a\rangle_{n}}. \end{align}\tag{1}\] It is easy to see that \(a\) is a linear structure if and only if \(|\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{f(x+a)-f(x)}|=p^n\). Then by Eq. 1 , \(a\) is a linear structure if and only if \(|\sum_{z \in E}\zeta_{p}^{\langle z, a\rangle_{n}}|=p^{n-s}\). Note that \(|E|=p^{n-s}\) and \(0 \in E\). Thus, \(a\) is a linear structure if and only if \(\langle z, a\rangle_{n}=0\) for all \(z \in E\), if and only if \(a \in E^{\bot}\), where \(E^{\bot}=\{x \in V_{n}^{(p)}: \langle z, x\rangle_{n}=0 \text{ for all } z \in E\}\) is a linear subspace. As easily seen, \(E^{\bot}=<E>^{\bot}\), where \(<E>\) is the linear space spanned by \(E\). Then \(E^{\bot}=\{0\}\) if and only if \(E\) contains a basis of \(V_{n}^{(p)}\). ◻

Proof. If \(m=1\), then the result follows from the definition of \(s\)-plateaued partitions. In the following, we consider the case of \(m \geq 2\).

\((i) \Rightarrow (ii)\): Let \(F: V_{n}^{(p)} \rightarrow V_{m}^{(p)}\) for which every \(r \in V_{m}^{(p)}\) has exactly \(k\) of sets \(A_{i}\) in \(\Gamma\) in its preimage set. Denote \(D_{F, r}=\bigcup_{i=1}^{k}A_{b_{i}^{(r)}}, r \in V_{m}^{(p)}\). For any \(c \in {V_{m}^{(p)}}^{*}\), let \(l^{(c)}(x)=\langle c, x\rangle_{m}, x \in V_{m}^{(p)}\). Denote \(D_{l^{(c)}, j}=\{x_{1}^{(c, j)}, \dots, x_{p^{m-1}}^{(c, j)}\}, j \in \mathbb{F}_{p}\). Then for any \(c \in {V_{m}^{(p)}}^{*}\) and \(j \in \mathbb{F}_{p}\), \(D_{F_{c}, j}=\bigcup_{t=1}^{p^{m-1}}\bigcup_{i=1}^{k}A_{b_{i}^{(x_{t}^{(c, j)})}}\). By the definition of \(s\)-plateaued partitions, \(F_{c}\) is a \(p\)-ary \(s\)-plateaued function, and then \(F\) is a vectorial \(s\)-plateaued function.

\((ii) \Rightarrow (i)\): Let \(f: V_{n}^{(p)} \rightarrow \mathbb{F}_{p}\) for which every \(j \in \mathbb{F}_{p}\) has exactly \(\frac{K}{p}\) of sets \(A_{i}\) in \(\Gamma\) in its preimage set. Denote \(D_{f, j}=\bigcup_{i=1}^{\frac{K}{p}}A_{d_{i}^{(j)}}, j \in \mathbb{F}_{p}\). Let \(c \in {V_{m}^{(p)}}^{*}\) be fixed. Let \(F: V_{n}^{(p)} \rightarrow V_{m}^{(p)}\) be given by \(D_{F, x_{t}^{(c, j)}}=\bigcup_{i=1+(t-1)k}^{tk}A_{d_{i}^{(j)}}, 1 \leq t \leq p^{m-1}, j \in \mathbb{F}_{p}\), where \(\langle c, x_{t}^{(c, j)}\rangle_{m}=j, 1 \leq t \leq p^{m-1}, j \in \mathbb{F}_{p}\). Then every \(r \in V_{m}^{(p)}\) has exactly \(k\) of sets \(A_{i}\) in \(\Gamma\) in the preimage set of \(F\), and \(F\) is a vectorial \(s\)-plateaued function. Notice that \(f(x)=F_{c}(x)\). Thus, \(f\) is a \(p\)-ary \(s\)-plateaued function, and \(\Gamma\) is an \(s\)-plateaued partition.

\((i) \Rightarrow (iii)\): Let \(g: V_{n}^{(p)} \rightarrow \mathbb{Z}_{p^m}\) for which every \(r \in \mathbb{Z}_{p^m}\) has exactly \(k\) of sets \(A_{i}\) in \(\Gamma\) in its preimage set. Denote \(D_{g, r}=\bigcup_{i=1}^{k}A_{b_{i}^{(r)}}, r \in \mathbb{Z}_{p^m}\). By [10], \(g\) is a generalized \(s\)-plateaued function if and only if for any \(H: \mathbb{F}_{p}^{m-1} \rightarrow \mathbb{F}_{p}\), \(h(x)=g^{(0)}(x)+H(g^{(1)}(x), \dots, g^{(m-1)}(x))\) is \(s\)-plateaued, where \(g(x)=g^{(0)}(x)p^{m-1}+g^{(1)}(x)p^{m-2}+\dots+g^{(m-2)}(x)p+g^{(m-1)}(x)\) and \(g^{(t)}: V_{n}^{(p)} \rightarrow \mathbb{F}_{p}, t=0, 1, \dots, m-1\). For \(j \in \mathbb{F}_{p}\), \(D_{h, j}=\bigcup_{j_{1}, \dots, j_{m-1} \in \mathbb{F}_{p}}\{x \in V_{n}^{(p)}: g^{(0)}(x)=j-H(j_{1}, \dots, j_{m-1}), g^{(1)}(x)=j_{1}, \dots, g^{(m-1)}(x)=j_{m-1}\}=\bigcup_{r \in R_{j}}D_{g, r}=\bigcup_{r \in R_{j}}\bigcup_{i=1}^{k}A_{b_{i}^{(r)}}\), where \(R_{j}=\{(j-H(j_{1}, \dots, j_{m-1}))p^{m-1}+j_{1}p^{m-2}+\dots+j_{m-1}: j_{1}, \dots, j_{m-1} \in \mathbb{F}_{p}\}\). Note that \(|R_{j}|=p^{m-1}\). Hence, \(h\) is a \(p\)-ary \(s\)-plateaued function generated by \(\Gamma\) and \(g\) is a generalized \(s\)-plateaued function.

\((iii) \Rightarrow (i)\): Let \(f: V_{n}^{(p)} \rightarrow \mathbb{F}_{p}\) for which every \(j \in \mathbb{F}_{p}\) has exactly \(\frac{K}{p}\) of sets \(A_{i}\) in \(\Gamma\) in its preimage set. Denote \(D_{f, j}=\bigcup_{i=1}^{\frac{K}{p}}A_{d_{i}^{(j)}}, j \in \mathbb{F}_{p}\). Let \(c=(1, c_{1}, \dots, c_{m-1}) \in {\mathbb{F}_{p}^{m}}^{*}\) be fixed. Let \(F=(f^{(0)}, \dots, f^{(m-1)}): V_{n}^{(p)} \rightarrow \mathbb{F}_{p}^{m}\) be given by \(D_{F, x_{t}^{(c, j)}}=\bigcup_{i=1+(t-1)k}^{tk}A_{d_{i}^{(j)}}, 1 \leq t \leq p^{m-1}, j \in \mathbb{F}_{p}\), where \(c \cdot x_{t}^{(c, j)}=j, 1 \leq t \leq p^{m-1}, j \in \mathbb{F}_{p}\). Then \(f(x)=F_{c}(x)=f^{(0)}(x)+c_{1}f^{(1)}(x)+\dots+c_{m-1}f^{(m-1)}(x)\). Let \(h: V_{n}^{(p)} \rightarrow \mathbb{Z}_{p^m}\) be defined by \(h(x)=f^{(0)}(x)p^{m-1}+\dots+f^{(m-2)}(x)p+f^{(m-1)}(x)\). For any \(r=r_{0}p^{m-1}+\dots+r_{m-2}p+r_{m-1} \in \mathbb{Z}_{p^m}\), where \(r_{0}, \dots, r_{m-2}, r_{m-1} \in \mathbb{F}_{p}\), \(D_{h, r}=D_{F, (r_{0}, \dots, r_{m-1})}, r \in \mathbb{Z}_{p^m}\). Then every \(r \in \mathbb{Z}_{p^m}\) has exactly \(k\) of sets \(A_{i}\) in \(\Gamma\) in the preimage set of \(h\), and \(h\) is a generalized \(s\)-plateaued function. Therefore, \(f\) is a \(p\)-ary \(s\)-plateaued function by [10], and \(\Gamma\) is an \(s\)-plateaued partition. ◻

Proof. \((i)\) We first show that when \(p \geq 5\), or \(p=3\) and \(n+s\) is even, then the generated \(p\)-ary \(s\)-plateaued functions by \(\Gamma\) are all balanced or all unbalanced. Define map \(F: V_{n}^{(p)}\rightarrow \{1, 2, \dots, K\}\) as \[F(x)=i, \text{ if } x \in A_{i}, 1\leq i \leq K.\] By the definition of \(s\)-plateaued partitions, for any balanced map \(B\) from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\) (that is, \(|D_{B, j}|=\frac{K}{p}, j \in \mathbb{F}_{p}\)), \(B(F(x))\) is a \(p\)-ary \(s\)-plateaued function generated by \(\Gamma\). Let \(f, f'\) be any \(p\)-ary \(s\)-plateaued functions generated by \(\Gamma\). Denote \(D_{f, j}=\bigcup_{i=1}^{\frac{K}{p}}A_{b_{i}^{(j)}}\), and \(D_{f', j}=\bigcup_{i=1}^{\frac{K}{p}}A_{{b^{'}_{i}}^{(j)}}\), where \(j \in \mathbb{F}_{p}\). Then \(f(x)=\widetilde{B}(F(x))\) and \(f'(x)=\widetilde{B}'(F(x))\), where \(\widetilde{B}\) and \(\widetilde{B}'\) are the balanced maps from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\) defined by \[\widetilde{B}(b_{i}^{(j)})=j, \widetilde{B}'({b^{'}_{i}}^{(j)})=j, 1\leq i \leq \frac{K}{p}, j \in \mathbb{F}_{p}.\] Define the permutation \(\widetilde{P}\) of \(\{1, 2, \dots, K\}\) as \[\widetilde{P}({b^{'}_{i}}^{(j)})=b_{i}^{(j)}, 1 \leq i \leq \frac{K}{p}, j \in \mathbb{F}_{p}.\] Then \(f'(x)=\widetilde{B}(\widetilde{P}(F(x)))\). We claim that if \(\delta_{S_{B(P(F(x)))}}(0)=\delta_{S_{B(F(x))}}(0)\) for any balanced map \(B\) from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\) and any transposition permutation \(P\) of \(\{1, 2, \dots, K\}\), then \(\delta_{S_{f}}(0)=\delta_{S_{f'}}(0)\). Since any permutation can be decomposed into transposition permutations, then there are transposition permutations \(P_{1}, P_{2}, \dots, P_{t}\) such that \(\widetilde{P}=P_{1} P_{2} \cdots P_{t}\). If \(t=1\), then the claim holds. We consider the case of \(t\geq 2\). Define \(B^{(i)}=\widetilde{B}(P_{1} \cdots P_{i}), 1\leq i \leq t-1\). Then \(B^{(t-1)}, \dots, B^{(1)}\) are balanced maps from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\), and we obtain \[\begin{align} \delta_{S_{f'}}(0)&=\delta_{S_{\widetilde{B}(P_{1} \cdots P_{t}(F(x)))}}(0)=\delta_{S_{B^{(t-1)}(P_{t}(F(x)))}}(0)=\delta_{S_{B^{(t-1)}(F(x))}}(0)\\ &=\delta_{S_{B^{(t-2)}(P_{t-1}(F(x)))}}(0)=\delta_{S_{B^{(t-2)}(F(x))}}(0)=\cdots \\ &=\delta_{S_{B^{(1)}(F(x))}}(0)=\delta_{S_{\widetilde{B}(P_{1}(F(x)))}}(0)=\delta_{S_{\widetilde{B}(F(x))}}(0)=\delta_{S_{f}}(0). \end{align}\] Therefore, the claim holds. Let \(B\) be any fixed balanced map from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\), and \(P\) be any fixed transposition permutation of \(\{1, 2, \dots, K\}\). Denote \(P=(i_{0} i_{1})\), where \(1 \leq i_{0}<i_{1} \leq K\). In the following, we only need to prove that \(\delta_{S_{B(P(F(x)))}}(0)=\delta_{S_{B(F(x))}}(0)\) if \(p \geq 5\), or \(p=3\) and \(n+s\) is even. We use the proof by contradiction. W.l.o.g, assume that \(0 \in S_{B(P(F(x)))}, 0 \notin S_{B(F(x))}\). We have \[\label{p1321} \begin{align} \epsilon_{B(P(F))}(0)p^{\frac{n+s}{2}}\zeta_{p}^{(B(P(F)))^{*}(0)}&=W_{B(P(F))}(0)\\ &=\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{B(P(F(x)))}=\sum_{i=1}^{K}\zeta_{p}^{B(P(i))}|A_{i}|\\ &=\sum_{i \in \{1, 2, \dots, K\} \backslash \{i_{0}, i_{1}\}} \zeta_{p}^{B(i)}|A_{i}|+\zeta_{p}^{B(i_{1})}|A_{i_{0}}|+\zeta_{p}^{B(i_{0})}|A_{i_{1}}|, \end{align}\tag{2}\] and \[\label{p1322} \begin{align} 0=W_{B(F)}(0)&=\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{B(F(x))}=\sum_{i=1}^{K}\zeta_{p}^{B(i)}|A_{i}|\\ &=\sum_{i \in \{1, 2, \dots, K\} \backslash \{i_{0}, i_{1}\}} \zeta_{p}^{B(i)}|A_{i}|+\zeta_{p}^{B(i_{0})}|A_{i_{0}}|+\zeta_{p}^{B(i_{1})}|A_{i_{1}}|. \end{align}\tag{3}\] Subtracting Eq. 2 from Eq. 3 , we get \[\label{p1323} \epsilon_{B(P(F))}(0)p^{\frac{n+s}{2}}\zeta_{p}^{(B(P(F)))^{*}(0)}=(\zeta_{p}^{B(i_{0})}-\zeta_{p}^{B(i_{1})})(|A_{i_{1}}|-|A_{i_{0}}|).\tag{4}\] Case I: \(B(i_{0})=B(i_{1})\) or \(|A_{i_{0}}|=|A_{i_{1}}|\). This case contradicts Eq. 4 .
Case II: \(B(i_{0}) \neq B(i_{1})\) and \(|A_{i_{0}}| \neq |A_{i_{1}}|\). Denote \(|A_{i_{1}}|-|A_{i_{0}}|=p^{t}r\), where \(t \geq 0\) is an integer and \(r\) is a positive integer with \(p \nmid r\). Let \(L=\mathbb{Q}(\zeta_{p})\) and \(\mathcal{O}_{L}=\mathbb{Z}[\zeta_{p}]\). Notice that \(\epsilon_{B(P(F))}(0)p^{\frac{n+s}{2}} \in \mathcal{O}_{L}\) and \((\epsilon_{B(P(F))}(0)p^{\frac{n+s}{2}})^{2}=\pm p^{n+s}\). Since \(p \mathcal{O}_{L}=((1-\zeta_{p})\mathcal{O}_{L})^{p-1}\), then \((\epsilon_{B(P(F))}(0)p^{\frac{n+s}{2}} \mathcal{O}_{L})^{2}=((1-\zeta_{p})\mathcal{O}_{L})^{(p-1)(n+s)}\). By the uniqueness of the decomposition of \(\epsilon_{B(P(F))}(0)p^{\frac{n+s}{2}} \mathcal{O}_{L}\), we have \(\epsilon_{B(P(F))}(0)p^{\frac{n+s}{2}}\mathcal{O}_{L}=((1-\zeta_{p})\mathcal{O}_{L})^{\frac{(p-1)(n+s)}{2}}\). Then by Eq. 4 and \((\zeta_{p}^{B(i_{0})}-\zeta_{p}^{B(i_{1})})\mathcal{O}_{L}=(1-\zeta_{p}^{B(i_{1})-B(i_{0})})\mathcal{O}_{L}=(1-\zeta_{p})\mathcal{O}_{L}\), we obtain \[\label{p1324} ((1-\zeta_{p})\mathcal{O}_{L})^{\frac{(p-1)(n+s)}{2}}=((1-\zeta_{p})\mathcal{O}_{L})^{1+(p-1)t} \cdot r\mathcal{O}_{L}.\tag{5}\] Since \(p \nmid r\), then \((1-\zeta_{p}) \mathcal{O}_{L} \nmid r\mathcal{O}_{L}\). Then by Eq. 5 , we get \(r=1\) and then \(\frac{(p-1)(n+s)}{2}=1+(p-1)t\), which is impossible when \(p \geq 5\), or \(p=3\) and \(n+s\) is even (since \(\frac{p-1}{2}(n+s-2t)=1\) implies that \(p=3\) and \(n+s=2t+1\)). Thus, when \(p \geq 5\), or \(p=3\) and \(n+s\) is even, the generated \(p\)-ary \(s\)-plateaued functions by \(\Gamma\) are all balanced or all unbalanced.

In the following, for the case of \(p=2\), or \(p=3, 2 \nmid (n+s)\), by using the value distribution of \(p\)-ary plateaued functions, we analyze the possible values of \(|A_{i}|\) if \(\Gamma\) generates both balanced and unbalanced \(p\)-ary \(s\)-plateaued functions simultaneously. Let \(A_{i}, A_{j}\) be arbitrary fixed sets in \(\Gamma\) with \(|A_{i}| \neq |A_{j}|\) (such \(A_{i}, A_{j}\) must exist otherwise the generated \(p\)-ary plateaued functions are all balanced). W.l.o.g, suppose \(|A_{i}|>|A_{j}|\).

When \(p=2\), let \(f\) be a Boolean \(s\)-plateaued function generated by \(\Gamma\) with \(A_{i} \subseteq D_{f, 0}, A_{j} \subseteq D_{f, 1}\). Let \(g\) be the Boolean \(s\)-plateaued function generated by \(\Gamma\) with \(D_{g, 0}=A_{j} \cup D_{f, 0} \backslash A_{i}\). Then by the value distribution of Boolean \(s\)-plateaued functions, \(|A_{i}|-|A_{j}|=|D_{f, 0}|-|D_{g, 0}|=2^{\frac{n+s}{2}-1}\) or \(2^{\frac{n+s}{2}}\). By the arbitrariness of \(A_{i}\) and \(A_{j}\), only the following cases may occur:

• \(|A_{i}|, 1 \leq i \leq K\), take two different values \(X\) and \(X+\varepsilon 2^{\frac{n+s}{2}}\), where the frequency of \(X\) is greater than or equal to that of \(X+\varepsilon 2^{\frac{n+s}{2}}\), and \(\varepsilon \in \{\pm 1\}\) is a constant. In fact, this case cannot occur. Note that the frequency of \(X+\varepsilon 2^{\frac{n+s}{2}}\) is at least \(2\) otherwise the generated Boolean plateaued functions are all unbalanced. Let \(A_{i_{1}}, A_{i_{2}}, A_{i_{3}}, A_{i_{4}}\) be with \(|A_{i_{1}}|=|A_{i_{2}}|=X, |A_{i_{3}}|=|A_{i_{4}}|=X+\varepsilon 2^{\frac{n+s}{2}}\). Let \(f'\) be a Boolean \(s\)-plateaued function generated by \(\Gamma\) with \(A_{i_{1}} \cup A_{i_{2}} \subseteq D_{f', 0}\), \(A_{i_{3}} \cup A_{i_{4}} \subseteq D_{f', 1}\) . Let \(g'\) be the Boolean \(s\)-plateaued function generated by \(\Gamma\) with \(D_{g', 0}=A_{i_{3}} \cup A_{i_{4}} \cup D_{f', 0} \backslash (A_{i_{1}} \cup A_{i_{2}})\). Then \(|D_{f', 0}|-|D_{g', 0}|=\pm 2^{\frac{n+s}{2}+1}\), which contradicts the value distribution of Boolean plateaued functions.

• \(|A_{i}|, 1 \leq i \leq K\), take two different values \(X\) and \(X+\varepsilon 2^{\frac{n+s}{2}-1}\), where the frequency of \(X\) is greater than or equal to that of \(X+\varepsilon 2^{\frac{n+s}{2}-1}\), and \(\varepsilon \in \{\pm 1\}\) is a constant. The frequency of \(X+\varepsilon 2^{\frac{n+s}{2}-1}\) is at least \(2\) otherwise the generated Boolean plateaued functions are all unbalanced. We show the frequency of \(X+\varepsilon 2^{\frac{n+s}{2}-1}\) is \(2\). To the contrary, assume its frequency is at least \(3\). Then \(K \geq 6\) and \(\frac{K}{2} \geq 3\). Let \(A_{i_{t}}, 1 \leq t \leq 6\), be with \(|A_{i_{1}}|=|A_{i_{2}}|=|A_{i_{3}}|=X, |A_{i_{4}}|=|A_{i_{5}}|=|A_{i_{6}}|=X+\varepsilon 2^{\frac{n+s}{2}-1}\). Let \(f''\) be a Boolean \(s\)-plateaued function generated by \(\Gamma\) with \(A_{i_{1}} \cup A_{i_{2}} \cup A_{i_{3}} \subseteq D_{f'', 0}, A_{i_{4}} \cup A_{i_{5}} \cup A_{i_{6}} \subseteq D_{f'', 1}\). Let \(g''\) be the Boolean \(s\)-plateaued function generated by \(\Gamma\) with \(D_{g'', 0}=A_{i_{4}} \cup A_{i_{5}} \cup A_{i_{6}} \cup D_{f'', 0} \backslash (A_{i_{1}} \cup A_{i_{2}} \cup A_{i_{3}})\). Then \(|D_{f'', 0}|-|D_{g'', 0}|=\pm 3 \cdot 2^{\frac{n+s}{2}-1}\), which contradicts the value distribution of Boolean plateaued functions. Since the frequency of \(X+\varepsilon 2^{\frac{n+s}{2}-1}\) is \(2\), then by the value distribution of Boolean \(s\)-plateaued functions, we get \(X=\frac{2^{n}-\varepsilon 2^{\frac{n+s}{2}}}{K}\), and the corresponding values \(|A_{i}|, 1 \leq i \leq K\), can be obtained.

• \(|A_{i}|, 1 \leq i \leq K\), take three different values \(X\), \(X+2^{\frac{n+s}{2}-1}\) and \(X+2^{\frac{n+s}{2}}\). We claim that the frequency of \(X\) and \(X+2^{\frac{n+s}{2}}\) are both \(1\). To the contrary, assume that the frequency of \(X\) or \(X+2^{\frac{n+s}{2}}\) is at least \(2\). We only consider the case of the frequency of \(X\) is at least \(2\) since the other case is similar. Let \(A_{i_{1}}, A_{i_{2}}, A_{i_{3}}, A_{i_{4}}\) be with \(|A_{i_{1}}|=|A_{i_{2}}|=X, |A_{i_{3}}|=X+2^{\frac{n+s}{2}-1},|A_{i_{4}}|=X+2^{\frac{n+s}{2}}\). Let \(f'''\) be a Boolean \(s\)-plateaued function generated by \(\Gamma\) with \(A_{i_{1}} \cup A_{i_{2}} \subseteq D_{f''', 0}\), \(A_{i_{3}} \cup A_{i_{4}} \subseteq D_{f''', 1}\). Let \(g'''\) be the Boolean \(s\)-plateaued function generated by \(\Gamma\) with \(D_{g''', 0}=A_{i_{3}} \cup A_{i_{4}} \cup D_{f''', 0} \backslash (A_{i_{1}} \cup A_{i_{2}})\). Then \(|D_{g''', 0}|-|D_{f''', 0}|=3 \cdot 2^{\frac{n+s}{2}-1}\), which contradicts the value distribution of Boolean plateaued functions. Thus, the claim holds. Since the frequency of \(X\) and \(X+2^{\frac{n+s}{2}}\) are both \(1\), then by the value distribution of Boolean \(s\)-plateaued functions, we get \(X=\frac{2^n}{K}-2^{\frac{n+s}{2}-1}\). As \(X\) is an integer, \(K=2^m\) for some positive integer. Then the corresponding values \(|A_{i}|, 1 \leq i \leq K\), can be obtained.

When \(p=3\) and \(n+s\) is odd, let \(f\) be a ternary \(s\)-plateaued function generated by \(\Gamma\) with \(A_{i} \subseteq D_{f, 0}, A_{j} \subseteq D_{f, 1}\). Let \(g\) be a ternary \(s\)-plateaued function with \(D_{g, 0}=A_{j} \cup D_{f, 0} \backslash A_{i}\). Then by the value distribution of ternary \(s\)-plateaued functions, \(|A_{i}|-|A_{j}|=|D_{f, 0}|-|D_{g, 0}|=3^{\frac{n+s-1}{2}}\) or \(2 \cdot 3^{\frac{n+s-1}{2}}\). By the arbitrariness of \(A_{i}\) and \(A_{j}\), only the following cases may occur:

• \(|A_{i}|, 1 \leq i \leq K\), take two different values \(X\) and \(X+ \varepsilon 2 \cdot 3^{\frac{n+s-1}{2}}\), where the frequency of \(X\) is greater than or equal to that of \(X+\varepsilon 2 \cdot 3^{\frac{n+s-1}{2}}\), and \(\varepsilon \in \{\pm 1\}\) is a constant. With the similar arguments as for \(p=2\), it is easy to show that this case cannot occur.

• \(|A_{i}|, 1 \leq i \leq K\), take two different values \(X\) and \(X+ \varepsilon 3^{\frac{n+s-1}{2}}\), where the frequency of \(X\) is greater than or equal to that of \(X+\varepsilon 3^{\frac{n+s-1}{2}}\), and \(\varepsilon \in \{\pm 1\}\) is a constant. The frequency of \(X+\varepsilon 3^{\frac{n+s-1}{2}}\) is at least \(2\) otherwise the generated ternary plateaued functions are all unbalanced. When \(\frac{K}{3} \geq 3\), with the similar arguments as for \(p=2\), it is easy to show that the possible frequency of \(X+\varepsilon 3^{\frac{n+s-1}{2}}\) can only be \(2\). When \(\frac{K}{3} \geq 3\), or \(K=6\) and the frequency of \(X+\varepsilon 3^{\frac{n+s-1}{2}}\) is \(2\), let \(A_{i_{1}}, A_{i_{2}}\) be with \(|A_{i_{1}}|=|A_{i_{2}}|=X+\varepsilon 3^{\frac{n+s-1}{2}}\). Then \(|A_{i}|=X, i \neq i_{1}, i_{2}\). Let \(f''\) be a ternary \(s\)-plateaued function generated by \(\Gamma\) with \(A_{i_{1}} \subseteq D_{f'', 0}\), \(A_{i_{2}} \subseteq D_{f'', 1}\). Then \(|D_{f'', 0}|=|D_{f'', 1}| \neq |D_{f'', 2}|\), which contradicts the value distribution of ternary plateaued functions when \(n+s\) is odd. Thus, \(K=6\) and the frequency of \(X+\varepsilon 3^{\frac{n+s-1}{2}}\) is \(3\). By the value distribution of ternary \(s\)-plateaued functions, we get \(X=\frac{3^{n-1}-\varepsilon 3^{\frac{n+s-1}{2}}}{2}\), and the corresponding values \(|A_{i}|, 1 \leq i \leq 6\), can be obtained.

• \(|A_{i}|, 1 \leq i \leq K\), take three different values \(X\), \(X+3^{\frac{n+s-1}{2}}\) and \(X+2 \cdot 3^{\frac{n+s-1}{2}}\). With the similar arguments as for \(p=2\), the frequency of \(X\) and \(X+2 \cdot 3^{\frac{n+s-1}{2}}\) are both 1. Then by the value distribution of ternary \(s\)-plateaued functions, we get \(X=\frac{3^n}{K}-3^{\frac{n+s-1}{2}}\). As X is an integer, \(K=3^m\) for some positive integer. Then the corresponding values \(|A_{i}|, 1 \leq i \leq K\), can be obtained.

\((ii)\) If the generated \(p\)-ary \(s\)-plateaued functions are all balanced, then the sum of the cardinality of any \(\frac{K}{p}\) of sets \(A_{i}\) are \(p^{n-1}\). Hence, \(|A_{i}|, 1 \leq i \leq K\), are the same and \(|A_{i}|=\frac{p^n}{K}\). Since \(|A_{i}|\) is an integer, then \(K=p^m\) for some positive integer \(m\).

\((iii)\) Obviously, \(n+s\) is even if \(p=2\). Recall that \(K \neq 3\). In the following, we show that \(n+s\) is also even if \(p\) is odd. We first prove that when \(p \geq 5\) and \(n+s\) is odd, there is no unbalanced \(s\)-plateaued partition of \(V_{n}^{(p)}\). Let \(f\) be a \(p\)-ary \(s\)-plateaued function generated by \(\Gamma\). Let \(\alpha, \beta \in \mathbb{F}_{p} \backslash \{f^{*}(0)\}\) with \(\eta_{1}(f^{*}(0)-\alpha) \neq \eta_{1}(f^{*}(0)-\beta)\). Define the \(p\)-ary \(s\)-plateaued function \(g\) as \(D_{g, \alpha}=D_{f, \beta}, D_{g, \beta}=D_{f, \alpha}, D_{g, i}=D_{f, i}, i \in \mathbb{F}_{p} \backslash \{\alpha, \beta\}\). Since \(|D_{g, i}|=|D_{f, i}|, i \in \mathbb{F}_{p} \backslash \{\alpha, \beta\}\), then by the value distribution of \(p\)-ary plateaued functions, we have that \(f^{*}(0)=g^{*}(0), \epsilon_{f}(0)=\epsilon_{g}(0)\). As \(\eta_{1}(f^{*}(0)-\alpha) \neq \eta_{1}(f^{*}(0)-\beta)\), by the value distribution of \(p\)-ary plateaued functions, \(|D_{f, \alpha}| \neq |D_{g, \beta}|\), which is a contradiction. Now we prove that when \(p=3, 2 \nmid (n+s), K \geq 6\), there is no unbalanced \(s\)-plateaued partition of \(V_{n}^{(3)}\) of depth \(K\). Let \(A_{i_{1}}, A_{i_{2}}, A_{i_{3}}\) be arbitrary sets in \(\Gamma\). W.l.o.g., suppose \(|A_{i_{1}}| \geq |A_{i_{2}}| \geq |A_{i_{3}}|\). Let \(\{C_{1}, C_{2}, C_{3}\}\) be a partition of \(V_{n}^{(3)} \backslash (A_{i_{1}} \cup A_{i_{2}} \cup A_{i_{3}})\) for which \(C_{j} \;(j=1, 2, 3)\) is the union of \(\frac{K}{3}-1\) of sets \(A_{i}\) in \(\Gamma\). W.l.o.g, suppose \(|C_{1}| \geq |C_{2}| \geq |C_{3}|\). By \(|A_{i_{1}}|+|C_{1}| \geq |A_{i_{2}}|+|C_{2}| \geq |A_{i_{3}}|+|C_{3}|\) and the value distribution of ternary \(s\)-plateaued functions, we have that \(|A_{i_{1}}|+|C_{1}|=3^{n-1}+3^{\frac{n+s-1}{2}}\), \(|A_{i_{2}}|+|C_{2}|=3^{n-1}\), \(|A_{i_{3}}|+|C_{3}|=3^{n-1}-3^{\frac{n+s-1}{2}}\). Let \(f\) be the ternary \(s\)-plateaued function generated by \(\Gamma\) with \(D_{f, 0}=A_{i_{1}} \cup C_{3}\), \(D_{f, 1}=A_{i_{2}} \cup C_{2}\) and \(D_{f, 2}=A_{i_{3}} \cup C_{1}\). Then \(|D_{f, 1}|=3^{n-1}\) and thus \(|A_{i_{1}}|+|C_{3}|=|D_{f, 0}| \neq 3^{n-1}\) by the value distribution of ternary plateaued functions. It is easy to see that \(|A_{i_{1}}|+|C_{3}|=3^{n-1}-3^{\frac{n+s-1}{2}}\) implies \(|A_{i_{1}}|=|A_{i_{2}}|=|A_{i_{3}}|\), and \(|A_{i_{1}}|+|C_{3}|=3^{n-1}+3^{\frac{n+s-1}{2}}\) implies \(|C_{1}|=|C_{2}|=|C_{3}|\) and then \(|A_{i_{1}}|>|A_{i_{2}}|>|A_{i_{3}}|\). Further, by the arbitrariness of \(A_{i_{1}}, A_{i_{2}}, A_{i_{3}}\), we have that \(|A_{K}|>|A_{K-1}|>\cdots>|A_{1}|\), which is impossible by \(K \geq 6\) and the value distribution of ternary \(s\)-plateaued functions. Therefore, \(n+s\) is even.

In the following, we compute the possible values of \(|A_{r}|, 1 \leq r \leq K\). Let \(A_{i}, A_{j}\) be arbitrary fixed sets in \(\Gamma\) with \(|A_{i}|\neq |A_{j}|\). W.o.l.g., suppose \(|A_{i}|>|A_{j}|\). Let \(f\) be a \(p\)-ary \(s\)-plateaued function generated by \(\Gamma\) with \(A_{i} \subseteq D_{f, 0}\), \(A_{j} \subseteq D_{f, 1}\). Let \(g\) be the \(p\)-ary \(s\)-plateaued function generated by \(\Gamma\) for which \(D_{g, 0}=A_{j} \cup D_{f, 0} \backslash A_{i}\), \(D_{g, 1}=A_{i} \cup D_{f, 1} \backslash A_{j}\), and when \(p \geq 3\), for \(k \neq i, j\), \(D_{g, k}=D_{f, k}\). By the value distribution of unbalanced \(p\)-ary \(s\)-plateaued functions, we obtain \(|A_{i}|-|A_{j}|=|D_{f, 0}|-|D_{g, 0}|=p^{\frac{n+s}{2}}\). By the arbitrariness of \(A_{i}, A_{j}\), we have that the values of \(|A_{r}|, 1 \leq r \leq K\), take two different values \(X\) and \(X+\varepsilon p^{\frac{n+s}{2}}\), where the frequency of \(X\) is greater than or equal to that of \(X+\varepsilon p^{\frac{n+s}{2}}\), and \(\varepsilon \in \{\pm 1\}\) is a constant. Using the value distribution of unbalanced \(p\)-ary \(s\)-plateaued functions, we get that the frequency of \(X+\varepsilon p^{\frac{n+s}{2}}\) is \(1\), and \(X=\frac{p^n-\varepsilon p^{\frac{n+s}{2}}}{K}\). Then the corresponding values \(|A_{r}|, 1 \leq r \leq K\), can be obtained. ◻

Proof. We first show that if \((i)\) holds, then \(\{D_{F, i}, i \in V_{m}^{(p)}\}\) is an \(s\)-plateaued partition. For any \(a \in V_{n}^{(p)}\) and \(i \in V_{m}^{(p)}\), by [30], \[\chi_{-a}(D_{F, i})=p^{n-m}\delta_{0}(a)+p^{-m}\sum_{c \in {V_{m}^{(p)}}^{*}}W_{F_{c}}(a)\zeta_{p}^{-\langle c, i\rangle_{m}}.\] Let \(B: V_{m}^{(p)} \rightarrow \mathbb{F}_{p}\) be any fixed balanced function. If \(a \notin S\), then \(\chi_{-a}(D_{F, i})=p^{n-m}\delta_{0}(a)\). In this case, we have \[W_{B(F)}(a)=\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{B(F(x))-\langle a, x\rangle _{n}}=\sum_{i \in V_{m}^{(p)}}\zeta_{p}^{B(i)}\chi_{-a}(D_{F, i})= p^{n-m}\delta_{0}(a)\sum_{i \in V_{m}^{(p)}}\zeta_{p}^{B(i)}=0.\]

If \(a \in S\), then \[\begin{align} \chi_{-a}(D_{F, i})&=p^{n-m}\delta_{0}(a)+\epsilon(a) p^{\frac{n+s}{2}-m}\sum_{c \in {V_{m}^{(p)}}^{*}}\zeta_{p}^{(F_{c})^{*}(a)-\langle c, i\rangle_{m}}\\ &=p^{n-m}\delta_{0}(a)+\epsilon(a) p^{\frac{n+s}{2}-m}\zeta_{p}^{g(a)}\sum_{c \in {V_{m}^{(p)}}^{*}}\zeta_{p}^{\langle c, G(a)-i\rangle_{m}}\\ &=p^{n-m}\delta_{0}(a)+\epsilon(a) p^{\frac{n+s}{2}-m}\zeta_{p}^{g(a)}(p^m \delta_{i}(G(a))-1). \end{align}\] In this case, we have \[\begin{align} W_{B(F)}(a)&=\sum_{i \in V_{m}^{(p)}}\zeta_{p}^{B(i)}\chi_{-a}(D_{F, i})\\ &=(p^{n-m}\delta_{0}(a)-\epsilon(a) p^{\frac{n+s}{2}-m}\zeta_{p}^{g(a)})\sum_{i \in V_{m}^{(p)}}\zeta_{p}^{B(i)}+\epsilon(a) p^{\frac{n+s}{2}}\zeta_{p}^{B(G(a))+g(a)}\\ &=\epsilon(a) p^{\frac{n+s}{2}}\zeta_{p}^{B(G(a))+g(a)}. \end{align}\] Therefore, \(B(F(x))\) is an \(s\)-plateaued function, and by the definition of \(s\)-plateaued partitions, \(\{D_{F, i}, i \in V_{m}^{(p)}\}\) is an \(s\)-plateaued partition.

In the following, we show that \((i)\) and \((ii)\) are equivalent. Obviously, \((i)\) implies \((ii)\). Suppose that \((ii)\) holds. Then there is \(g: S \rightarrow \mathbb{F}_{p}\) such that \[\label{Pr2321} (F_{d})^{*}(x)-(F_{e})^{*}(x)-(F_{d-e})^{*}(x)=-g(x), x \in S, d \neq e \in {V_{m}^{(p)}}^{*}.\tag{6}\] Let \(\{\beta_{1}, \dots, \beta_{m}\}\) be a basis of \(V_{m}^{(p)}\) over \(\mathbb{F}_{p}\). For any \(x \in S\), let \(G(x) \in V_{m}^{(p)}\) be defined by the equations: \(\langle \beta_{i}, G(x)\rangle_{m}=(F_{\beta_{i}})^{*}(x)-g(x), i=1, 2, \dots, m\). For any \(c \in {V_{m}^{(p)}}^{*}\), denote \(c=c_{i_{1}}\beta_{i_{1}}+\dots+c_{i_{t}}\beta_{i_{t}}\), where \(c_{i_{j}} \in \mathbb{F}_{p}^{*}, 1\leq j \leq t, 1\leq i_{1} < \dots <i_{t}\leq m, 1\leq t\leq m\). Then \[\label{Pr2322} G_{c}(x)=\sum_{j=1}^{t}c_{i_{j}}\langle \beta_{i_{j}}, G(x)\rangle_{m}=\sum_{j=1}^{t}c_{i_{j}}(F_{\beta_{i_{j}}})^{*}(x)-g(x)\sum_{j=1}^{t}c_{i_{j}}, x \in S.\tag{7}\] For any \(a \in \mathbb{F}_{p}^{*}\) and \(c \in {V_{m}^{(p)}}^{*}\), we first show \[\label{Pr2323} (F_{ac})^{*}(x)=a(F_{c})^{*}(x)-(a-1)g(x), x \in S.\tag{8}\] Obviously, when \(a=1\), Eq. 8 holds. Suppose Eq. 8 holds for \(a=k\), where \(1\leq k \leq p-2\). When \(a=k+1\), by Eq. 6 , \[\begin{align} (F_{(k+1)c})^{*}(x)&=(F_{kc})^{*}(x)+(F_{c})^{*}(x)-g(x)=k(F_{c})^{*}(x)-(k-1)g(x)+(F_{c})^{*}(x)-g(x)\\ &=(k+1)(F_{c})^{*}(x)-kg(x). \end{align}\] Then Eq. 8 also holds for \(a=k+1\). By induction, Eq. 8 holds. When \(t=1\), by Eqs. 7 and 8 , Eq. ?? holds. Suppose Eq. ?? holds for \(t=r\). When \(t=r+1\), denote \(c'=\sum_{j=1}^{r}c_{i_{j}}\beta_{i_{j}}\), by Eq. 6 , \[\begin{align} (F_{c})^{*}(x)&=(F_{c'+c_{i_{r+1}}\beta_{i_{r+1}}})^{*}(x)=(F_{c'})^{*}(x)+(F_{c_{i_{r+1}}\beta_{i_{r+1}}})^{*}(x)-g(x)\\ &=G_{c'}(x)+g(x)+G_{c_{i_{r+1}}\beta_{i_{r+1}}}(x)+g(x)-g(x)=G_{c}(x)+g(x). \end{align}\] Then Eq. ?? also holds for \(t=r+1\). By induction, Eq. ?? holds. This completes the proof. ◻

Proof. For any fixed \(c \in \mathbb{F}_{p^r}^{*}\) and \(y \in \mathbb{F}_{p^t}\), the map \(x\mapsto \mathrm{Tr}_{1}^{r}(c\phi(x, y))\) is linear from \(\mathbb{F}_{p^m}\) to \(\mathbb{F}_{p}\). Since each linear map from \(\mathbb{F}_{p^m}\) to \(\mathbb{F}_{p}\) is of the form \(\mathrm{Tr}_{1}^{m}(ax), x \in \mathbb{F}_{p^m}\), where \(a \in \mathbb{F}_{p^m}\), then for any \(c\in \mathbb{F}_{p^r}^{*}\) and \(y \in \mathbb{F}_{p^t}\), there exists \(\pi^{[c]}(y)\in \mathbb{F}_{p^m}\) such that \(\mathrm{Tr}_{1}^{r}(c\phi(x, y))=\mathrm{Tr}_{1}^{m}(\pi^{[c]}(y)x)\) for all \(x \in \mathbb{F}_{p^m}\). For any \(c \in \mathbb{F}_{p^r}^{*}\), \(F_{c}(x, y)=\mathrm{Tr}_{1}^{r}(c\phi(x, y))+H_{c}(y)=\mathrm{Tr}_{1}^{m}(\pi^{[c]}(y)x)+H_{c}(y)\). For \(a=(a_{1}, a_{2}) \in \mathbb{F}_{p^m} \times \mathbb{F}_{p^t}\), we have \[\begin{align} W_{F_{c}}(a)&=\sum_{x \in \mathbb{F}_{p^m}, y \in \mathbb{F}_{p^t}}\zeta_{p}^{\mathrm{Tr}_{1}^{m}(\pi^{[c]}(y)x)+H_{c}(y)-\mathrm{Tr}_{1}^{m}(a_{1}x)-\mathrm{Tr}_{1}^{t}(a_{2}y)}\\ &=\sum_{y \in \mathbb{F}_{p^t}}\zeta_{p}^{H_{c}(y)-\mathrm{Tr}_{1}^{t}(a_{2}y)}\sum_{x \in \mathbb{F}_{p^m}}\zeta_{p}^{\mathrm{Tr}_{1}^{m}((\pi^{[c]}(y)-a_{1})x)}\\ &=\begin{cases} 0, & \text{ if } a_{1} \notin S_{1},\\ p^{m}\zeta_{p}^{-\mathrm{Tr}_{1}^{t}(a_{2}(\pi^{[c]})^{-1}(a_{1}))+H_{c}((\pi^{[c]})^{-1}(a_{1}))}, & \text{ if } a_{1} \in S_{1}. \end{cases} \end{align}\] Then \(F\) is a vectorial \((m-t)\)-plateaued function for which \(S_{F_{c}}=S_{1} \times \mathbb{F}_{p^t}\), \(\epsilon_{F_{c}}(x, y)=1\), \((F_{c})^{*}(x, y)=-\mathrm{Tr}_{1}^{t}(y(\pi^{[c]})^{-1}(x))+\mathrm{Tr}_{1}^{r}(cH((\pi^{[c]})^{-1}(x))), x \in S_{1}, y \in \mathbb{F}_{p^t}\). Since \((\pi^{[d]})^{-1}(x)+(\pi^{[e]})^{-1}(x)=(\pi^{[d+e]})^{-1}(x)\) for any \(d \neq -e \in \mathbb{F}_{p^r}^{*}\) and \(x \in S_{1}\), and \(\mathrm{Tr}_{1}^{r}(cH((\pi^{[c]})^{-1}(x))), c \in \mathbb{F}_{p^r}^{*}\), are the same function from \(S_{1}\) to \(\mathbb{F}_{p}\), then \((F_{d})^{*}(x, y)+(F_{e})^{*}(x, y)-(F_{d+e})^{*}(x, y), d \neq -e \in \mathbb{F}_{p^r}^{*}\), are the same function from \(S_{1} \times \mathbb{F}_{p^t}\) to \(\mathbb{F}_{p}\). By Proposition 3, \(\{D_{F, i}, i \in \mathbb{F}_{p^r}\}\) is an \((m-t)\)-plateaued partition. ◻

Proof. Let \(\phi: \mathbb{F}_{p^m} \times \mathbb{F}_{p^t} \rightarrow \mathbb{F}_{p^r}\) be defined by \(\phi(x, y)=\mathrm{Tr}_{r}^{m}(\psi(x, y)), x \in \mathbb{F}_{p^m}, y \in \mathbb{F}_{p^t}\). When \(y \in \mathbb{F}_{p^t}^{*}\), for any \(x_{1}, x_{2} \in \mathbb{F}_{p^m}\), from \((\psi(x_{1}, y)+\psi(x_{2}, y))\circ \pi(y)=\psi(x_{1}, y) \circ \pi(y)+ \psi(x_{2}, y) \circ \pi(y)=x_{1}+x_{2}=\psi(x_{1}+x_{2}, y) \circ \pi(y)\) and \(\pi(y) \neq 0\), we have \(\psi(x_{1}+x_{2}, y)=\psi(x_{1}, y)+\psi(x_{2}, y)\). When \(y=0\), then \(\psi(x_{1}+x_{2}, 0)=\psi(x_{1}, 0)+\psi(x_{2}, 0)=0\). Thus, for any \(y \in \mathbb{F}_{p^t}\), \(x \mapsto \phi(x, y)\) is linear. In the following, we determine \(\pi^{[c]}: \mathbb{F}_{p^t} \rightarrow \mathbb{F}_{p^m}\) with \(\mathrm{Tr}_{1}^{r}(c\phi(x, y))=\mathrm{Tr}_{1}^{m}(\pi^{[c]}(y)x), x \in \mathbb{F}_{p^m}, y \in \mathbb{F}_{p^t}\). When \(x=0\), let \(\lambda_{x}=0\); when \(x \in \mathbb{F}_{p^m}^{*}\), let \(\lambda_{x} \in \mathbb{F}_{p^m}^{*}\) be given by \(x \star \lambda_{x}^{-1}=1\). Define \(\rho: \mathbb{F}_{p^m} \rightarrow \mathbb{F}_{p^m}\) as \(\rho(x)=\pi^{-1}(\lambda_{x}^{-1})\). Then \(\rho\) is a permutation of \(\mathbb{F}_{p^m}\) with \(\rho(0)=0\). For any fixed \(c \in \mathbb{F}_{p^r}^{*}\) and \(x \in \mathbb{F}_{p^m}^{*}\), let \(z_{c, x}=\rho^{-1}(c^{-1}x)\). Then \(\pi^{-1}(\lambda_{z_{c, x}}^{-1})=\rho(z_{c, x})=c^{-1}x\) and \(\lambda_{z_{c, x}}^{-1}=\pi(c^{-1}x)\). For \(y=d\beta_{i} \in \mathbb{F}_{p^t}^{*}\), where \(d \in \mathbb{F}_{p^r}^{*}\) and \(1 \leq i \leq \frac{p^t-1}{p^r-1}\), we have \(\pi(c^{-1}y)=\pi(c^{-1}d\beta_{i})=c^{-u}d^{u}\alpha_{\tau(i)}=c^{-u}\pi(d \beta_{i})=c^{-u}\pi(y)\), and thus \(\lambda_{z_{c, y}}^{-1}=c^{-u}\pi(y)\). Let \(v=p^{r-j_{0}}\). For any \(c \in \mathbb{F}_{p^r}^{*}\) and \(y \in \mathbb{F}_{p^t}^{*}\), since \(P^{T}\) is right \(\mathbb{F}_{p^r}\)-linear, we have \(1=z_{c, y} \star \lambda_{z_{c, y}}^{-1}=z_{c, y} \star c^{-u}\pi(y)=c^{-u}(z_{c, y} \star \pi(y))\), and then \(c^{u}=z_{c, y} \star \pi(y)=\rho^{-1}(c^{-1}y) \star \pi(y)\), which implies that \(c=\rho^{-1}(c^{-v}y) \star \pi(y)\) for any \(c \in \mathbb{F}_{p^r}^{*}\). Therefore, for any \(x \in \mathbb{F}_{p^m}\) and \(y \in \mathbb{F}_{p^t}^{*}\), we have \[\label{Th0321} \begin{align} \mathrm{Tr}_{1}^{r}(c\phi(x,y))&=\mathrm{Tr}_{1}^{m}(c \psi(x, y))=\mathrm{Tr}_{1}^{m}(\psi(x, y)(\rho^{-1}(c^{-v}y) \star \pi(y)))\\ &=\mathrm{Tr}_{1}^{m}(\rho^{-1}(c^{-v}y)(\psi(x, y) \circ \pi(y))) =\mathrm{Tr}_{1}^{m}(\rho^{-1}(c^{-v}y)x). \end{align}\tag{9}\] By \(\psi(x, 0)=0, \rho(0)=0\), Eq. 9 also holds for \(y=0\). Thus, for any \(c \in \mathbb{F}_{p^r}^{*}\) and \(y \in \mathbb{F}_{p^t}\), we have \(\pi^{[c]}(y)=\rho^{-1}(c^{-v}y)\). Since \(\rho\) is a permutation of \(\mathbb{F}_{p^m}\), then \(\pi^{[c]}\) is an injection from \(\mathbb{F}_{p^t}\) to \(\mathbb{F}_{p^m}\). For any \(c \in \mathbb{F}_{p^r}^{*}\), \(c^{-v}\mathbb{F}_{p^t}=\mathbb{F}_{p^t}\), and \(\pi^{[c]}(\mathbb{F}_{p^t})=\rho^{-1}(\mathbb{F}_{p^t})\) for all \(c \in \mathbb{F}_{p^r}^{*}\). Denote \(S_{1}=\rho^{-1}(\mathbb{F}_{p^t})\), which contains \(0\) as \(\rho(0)=0\). For any \(x \in S_{1}\), it is easy to see that \((\pi^{[c]})^{-1}(x)=c^{v}\rho(x)\). Note that as \((\pi^{[c]})^{-1}(S_{1})= \mathbb{F}_{p^t}\), we have \(\rho(x) \in \mathbb{F}_{p^t}\) for any \(x \in S_{1}\). Since \(v =p^{r-j_{0}}\), then \((\pi^{[d]})^{-1}(x)+(\pi^{[e]})^{-1}(x)=(\pi^{[d+e]})^{-1}(x)\) for any \(d \neq -e \in \mathbb{F}_{p^r}^{*}\). For any \(y \in \mathbb{F}_{p^t}^{*}\), denote \(y=\gamma_{y}\beta_{i_{y}}\), where \(\gamma_{y} \in \mathbb{F}_{p^r}^{*}\) and \(1 \leq i_{y} \leq \frac{p^r-1}{p^t-1}\). Then for \(x \in S_{1}^{*}\), \(\mathrm{Tr}_{1}^{r}(cH((\pi^{[c]})^{-1}(x)))=\mathrm{Tr}_{1}^{r}(cH(c^{v}\rho(x)))=\mathrm{Tr}_{1}^{r}(cH(c^{v}\gamma_{\rho(x)}\beta_{i_{\rho(x)}}))=\mathrm{Tr}_{1}^{r}(\gamma_{\rho(x)}^{-u}H(\beta_{i_{\rho(x)}}))\), which is independent of \(c\). For \(x=0\), \(\mathrm{Tr}_{1}^{r}(cH((\pi^{[c]})^{-1}(x)))=\mathrm{Tr}_{1}^{r}(cH(c^{v}\rho(x)))=0\). By Lemma 2, \(\{D_{F, i}, i \in \mathbb{F}_{p^r}\}\) is an \((m-t)\)-plateaued partition. ◻

Proof. With the same notations as in Lemma 2, \(\pi^{[c]}(y)=c\pi(y), y \in \mathbb{F}_{p^t}\). By the definition of \(\pi\), for every \(c \in \mathbb{F}_{p^r}^{*}\), \(c\pi\) is injective and \(\{c\pi(y), y \in \mathbb{F}_{p^t}\}=\{0, \gamma\alpha_{i}, \gamma \in \mathbb{F}_{p^r}^{*}, 1 \leq i \leq \frac{p^t-1}{p^r-1}\}\). We denote this set by \(S_{1}\). For any \(x \in S_{1}^{*}\), denote \(x=\gamma_{x} \alpha_{i_{x}}\), where \(\gamma_{x} \in \mathbb{F}_{p^r}^{*}, 1 \leq i_{x} \leq \frac{p^t-1}{p^r-1}\). From the definition of \(\pi\) and the properties of \(\theta\) and \(H\), for any \(x \in S_{1}^{*}\) and \(d \neq -e \in \mathbb{F}_{p^r}^{*}\), \[\begin{align} &(\pi^{[d]})^{-1}(x)+(\pi^{[e]})^{-1}(x)-(\pi^{[d+e]})^{-1}(x)=\pi^{-1}(d^{-1}x)+\pi^{-1}(e^{-1}x)-\pi^{-1}((d+e)^{-1}x)\\ &=\pi^{-1}(d^{-1}\gamma_{x}\alpha_{i_{x}})+\pi^{-1}(e^{-1}\gamma_{x}\alpha_{i_{x}})-\pi^{-1}((d+e)^{-1}\gamma_{x}\alpha_{i_{x}})\\ &=\theta^{-1}(d^{-1}\gamma_{x})\beta_{\tau^{-1}(i_{x})}+\theta^{-1}(e^{-1}\gamma_{x})\beta_{\tau^{-1}(i_{x})}-\theta^{-1}((d+e)^{-1}\gamma_{x})\beta_{\tau^{-1}(i_{x})}=0, \end{align}\] and \[\begin{align} \mathrm{Tr}_{1}^{r}(cH((\pi^{[c]})^{-1}(x)))&=\mathrm{Tr}_{1}^{r}(cH((\pi^{-1}(c^{-1}x)))=\mathrm{Tr}_{1}^{r}(cH(\pi^{-1}(c^{-1}\gamma_{x}\alpha_{i_{x}})))\\ &=\mathrm{Tr}_{1}^{r}(cH(\theta^{-1}(c^{-1}\gamma_{x})\beta_{\tau^{-1}(i_{x})}))=\mathrm{Tr}_{1}^{r}(\gamma_{x}H(\beta_{\tau^{-1}(i_{x})})). \end{align}\] Then by Lemma 2, \(\{D_{F, i}, i \in \mathbb{F}_{p^r}\}\) is an \((m-t)\)-plateaued partition. Since \(\{\alpha_{1}, \dots, \alpha_{\frac{m}{r}}\}\) is a basis of \(\mathbb{F}_{p^m}\) over \(\mathbb{F}_{p^r}\), then \(S_{1} \times \mathbb{F}_{p^t}\) contains a basis of \(\mathbb{F}_{p^m} \times \mathbb{F}_{p^t}\) over \(\mathbb{F}_{p}\). Therefore any generated \(p\)-ary \((m-t)\)-plateaued function has no nonzero linear structure by the proof of Lemma 2, Proposition 1 and Remark 3. This completes the proof. ◻

Proof. For any \(c \in \mathbb{F}_{p^r}^{*}\), \(T_{c}(x, y, z)=P(F(x, y))_{c}+R_{c}(z), x \in \mathbb{F}_{p^m}, y \in \mathbb{F}_{p^t}, z \in V_{k}^{(p)}\), and then for any \(a=(a_{1}, a_{2}, a_{3}) \in \mathbb{F}_{p^m} \times \mathbb{F}_{p^t} \times V_{k}^{(p)}\), \[W_{T_{c}}(a)=W_{P(F)_{c}}(a_{1}, a_{2})W_{R_{c}}(a_{3}).\] Since \(R\) is a vectorial dual-bent function satisfying Condition A, then \((R_{c})^{*}(a_{3})=(R^{*})_{c}(a_{3}), \epsilon_{R_{c}}=\epsilon\), where \(\epsilon \in \{\pm 1\}\) is a constant. Combining Theorem 3, Lemma 2, Proposition 3 (together with their proofs) and Remark 3, we have that \(P(F(x, y))_{c}\) is \((m-t)\)-plateaued for which \(\epsilon_{P(F)_{c}}(a_{1}, a_{2})=1, a_{1}\in S_{1}, a_{2} \in \mathbb{F}_{p^t}\), and there exist functions \(G: S_{1} \times \mathbb{F}_{p^t} \rightarrow \mathbb{F}_{p^r}\) and \(g: S_{1} \times \mathbb{F}_{p^t} \rightarrow \mathbb{F}_{p}\) such that \[(P(F(x, y))_{c})^{*}(a_{1}, a_{2})=\mathrm{Tr}_{1}^{r}(cP(G(a_{1}, a_{2})))+g(a_{1}, a_{2}), a_{1} \in S_{1}, a_{2} \in \mathbb{F}_{p^t},\] where \(S_{1}=\{0, \gamma \alpha_{i}, \gamma \in \mathbb{F}_{p^r}^{*}, 1 \leq i \leq \frac{p^t-1}{p^r-1}\}\) contains a basis of \(\mathbb{F}_{p^m}\) over \(\mathbb{F}_{p}\). Hence, for any \(c \in \mathbb{F}_{p^r}^{*}\), \(T_{c}\) is an \((m-t)\)-plateaued function with \(S_{T_{c}}=S_{1} \times \mathbb{F}_{p^t} \times V_{k}^{(p)}\) (denoted by \(S\)) and for any \(a=(a_{1}, a_{2}, a_{3}) \in S\), \(\epsilon_{T_{c}}(a)=\epsilon\), and \[(T_{c})^{*}(a)=\mathrm{Tr}_{1}^{r}(cP(G(a_{1}, a_{2})))+g(a_{1}, a_{2})+\mathrm{Tr}_{1}^{r}(cR^{*}(a_{3})).\] Denote \(G'(a)=P(G(a_{1}, a_{2}))+R^{*}(a_{3})\) and \(g'(a)=g(a_{1}, a_{2})\). Then \((T_{c})^{*}(a)=G'_{c}(a)+g'(a)\) for any \(a \in S\). By Proposition 3, \(\{D_{T, i}, i \in \mathbb{F}_{p^r}\}\) is an \((m-t)\)-plateaued partition. Since \(S\) contains a basis of \(\mathbb{F}_{p^m} \times \mathbb{F}_{p^t} \times V_{k}^{(p)}\) over \(\mathbb{F}_{p}\), then any generated \(p\)-ary \((m-t)\)-plateaued function has no nonzero linear structure by Proposition 1 and Remark 3. ◻

Proof. For any \(c \in \mathbb{F}_{p^r}^{*}\) and \(a=(a_{1}, a_{2}, a_{3}, a_{4}) \in \mathbb{F}_{p^m} \times \mathbb{F}_{p^t} \times \mathbb{F}_{p^k} \times \mathbb{F}_{p^k}\), by the proof of [22], \[W_{T_{c}}(a)=p^{k}\zeta_{p}^{\mathrm{Tr}_{1}^{k}(cv Q(-a_{3}^{-1}a_{4}))}W_{(F(\mathrm{Tr}_{r}^{k}(u Q(-a_{3}^{-1}a_{4})); x, y))_{c}}(a_{1}, a_{2}).\] Combining Theorem 3, Lemma 2, Proposition 3 (together with their proofs) and Remark 3, \(F(j; x, y) \;(j \in \mathbb{F}_{p^r})\) is a vectorial \((m-t)\)-plateaued function for which for any \(c \in \mathbb{F}_{p^r}^{*}\), \(S_{F(j; x, y)_{c}}=S_{1} \times \mathbb{F}_{p^t}\) and there exist functions \(G^{(j)}: S_{1} \times \mathbb{F}_{p^t} \rightarrow \mathbb{F}_{p^r}\) and \(g^{(j)}: S_{1} \times \mathbb{F}_{p^t} \rightarrow \mathbb{F}_{p}\) such that \[(F(j; x, y)_{c})^{*}(a_{1}, a_{2})=\mathrm{Tr}_{1}^{r}(cP^{(j)}(G^{(j)}(a_{1}, a_{2})))+g^{(j)}(a_{1}, a_{2}), a_{1} \in S_{1}, a_{2} \in \mathbb{F}_{p^t},\] where \(S_{1}=\{0, \gamma \alpha_{i}, \gamma \in \mathbb{F}_{p^r}^{*}, 1 \leq i \leq \frac{p^t-1}{p^r-1}\}\) contains a basis of \(\mathbb{F}_{p^m}\) over \(\mathbb{F}_{p}\). Therefore, for any \(c \in \mathbb{F}_{p^r}^{*}\), \(T_{c}\) is an \((m-t)\)-plateaued function with \(S_{T_{c}}=S_{1} \times \mathbb{F}_{p^t} \times \mathbb{F}_{p^k} \times \mathbb{F}_{p^k}\) (denoted by \(S\)) and for any \(a=(a_{1}, a_{2}, a_{3}, a_{4}) \in S\), \[\begin{align} (T_{c})^{*}(a)&=\mathrm{Tr}_{1}^{r}\Big(cP^{(\mathrm{Tr}_{r}^{k}(u Q(-a_{3}^{-1}a_{4})))}\big(G^{(\mathrm{Tr}_{r}^{k}(u Q(-a_{3}^{-1}a_{4})))}(a_{1}, a_{2})\big)\Big)+g^{(\mathrm{Tr}_{r}^{k}(u Q(-a_{3}^{-1}a_{4})))}(a_{1}, a_{2})\\ & \;\;\;+\mathrm{Tr}_{1}^{k}(cv Q(-a_{3}^{-1}a_{4})). \end{align}\] Denote \[G(a)=P^{(\mathrm{Tr}_{r}^{k}(u Q(-a_{3}^{-1}a_{4})))}\big(G^{(\mathrm{Tr}_{r}^{k}(u Q(-a_{3}^{-1}a_{4})))}(a_{1}, a_{2})\big)+\mathrm{Tr}_{r}^{k}(vQ(-a_{3}^{-1}a_{4}))\] and \[g(a)=g^{(\mathrm{Tr}_{r}^{k}(u Q(-a_{3}^{-1}a_{4})))}(a_{1}, a_{2}).\] Then \((T_{c})^{*}(a)=G_{c}(a)+g(a)\) for any \(c \in \mathbb{F}_{p^r}^{*}\) and \(a=(a_{1}, a_{2}, a_{3}, a_{4}) \in S\). By Proposition 3, \(\{D_{T, i}, i \in \mathbb{F}_{p^r}\}\) is an \((m-t)\)-plateaued partition. Since \(S\) contains a basis of \(\mathbb{F}_{p^m} \times \mathbb{F}_{p^t} \times \mathbb{F}_{p^k} \times \mathbb{F}_{p^k}\) over \(\mathbb{F}_{p}\), then any generated \(p\)-ary \((m-t)\)-plateaued function has no nonzero linear structure by Proposition 1 and Remark 3. ◻

Proof. Let \(\sigma_{-1}\) be the automorphism of \(L\) defined by \(\sigma_{-1}(\zeta_{p})=\zeta_{p}^{-1}\). Then for any \(\alpha \in L\), \(\sigma_{-1}(\alpha)=\overline{\alpha}\), and \(\alpha\) is a real number if and only if \(\sigma_{-1}(\alpha)=\alpha\). From \(a\mathcal{O}_{L}=b(1-\zeta_{p})^{k}\mathcal{O}_{L}\), we have that \(a=b(1-\zeta_{p})^{k}u\), where \(u\) is a unit of \(\mathcal{O}_{L}\). It is a well-known result that every unit of \(\mathcal{O}_{L}\) can be written as the product of a real unit and a root of unity [26], [27]. Namely, \(u=r\zeta_{p}^{i}\) for some real unit \(r\) and \(0 \leq i \leq p-1\). Then \((1-\zeta_{p})^{k}\zeta_{p}^{i}=\frac{a}{br}\) is a real number. Thus, \((1-\zeta_{p})^{k}\zeta_{p}^{i}=\sigma_{-1}((1-\zeta_{p})^{k}\zeta_{p}^{i})=(1-\zeta_{p}^{-1})^{k}\zeta_{p}^{-i}=(-1)^{k}(1-\zeta_{p})^{k}\zeta_{p}^{-k-i}\), which implies that \(k\) is even (otherwise, \(\zeta_{p}^{2i+k}=(-1)^{k}=-1\) which is impossible). ◻

Proof. By Proposition 3, \((ii) \Rightarrow (i)\) holds. In the following, we prove \((i) \Rightarrow (ii)\). As \(-A_{i}=A_{i}, 1 \leq i \leq K\), all \(p\)-ary \(s\)-plateaued functions \(f\) generated by \(\Gamma\) satisfy \(f(x)=f(-x)\), and then \(|D_{f, f(0)}|\) is odd and \(|D_{f, j}| \;(j \neq f(0))\) is even. Thus \(\Gamma\) is unbalanced. By Proposition 2, \(n+s\) is even. Define \(\widetilde{F}: V_{n}^{(p)} \rightarrow \{1, 2, \dots, K\}\) as \(\widetilde{F}(x)=i, x \in A_{i}, 1 \leq i \leq K\). By the definition of \(s\)-plateaued partitions, \(B(\widetilde{F}(x))\) is a \(p\)-ary \(s\)-plateaued function for any balanced map from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\). We first show that the Walsh supports of all \(p\)-ary \(s\)-plateaued functions generated by \(\Gamma\) are the same. Let \(a\) be an arbitrary fixed element in \(V_{n}^{(p)}\). With the same arguments as in the proof of Proposition 2 \((i)\), we only need to prove that \(\delta_{S_{B(P(\widetilde{F}(x)))}}(a)=\delta_{S_{B(\widetilde{F}(x))}}(a)\) for any balanced map \(B\) from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\) and any transposition permutation \(P\) of \(\{1, 2, \dots, K\}\). Let \(B\) be an arbitrary fixed balanced map from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\) and \(P=(i_{0}i_{1})\) be an arbitrary fixed transposition permutation. We use the proof by contradiction. W.l.o.g., assume that \(a \in S_{B(P(\widetilde{F}))}, a \notin S_{B(\widetilde{F})}\). We have \[\label{Th4321} \begin{align} \epsilon_{B(P(\widetilde{F}))}(a)p^{\frac{n+s}{2}}\zeta_{p}^{(B(P(\widetilde{F})))^{*}(a)} &=W_{B(P(\widetilde{F}))}(a)=\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{B(P(\widetilde{F}(x)))-\langle a, x\rangle_{n}}=\sum_{i=1}^{K}\zeta_{p}^{B(P(i))}\chi_{-a}(A_{i})\\ &=\sum_{i \in \{1, \dots, K\} \backslash \{i_{0}, i_{1}\}}\zeta_{p}^{B(i)}\chi_{-a}(A_{i}) +\zeta_{p}^{B(i_{1})}\chi_{-a}(A_{i_{0}})+\zeta_{p}^{B(i_{0})}\chi_{-a}(A_{i_{1}}), \end{align}\tag{10}\] and \[\label{Th4322} \begin{align} 0=W_{B(\widetilde{F})}(a) &=\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{B(\widetilde{F}(x))-\langle a, x\rangle_{n}}=\sum_{i=1}^{K}\zeta_{p}^{B(i)}\chi_{-a}(A_{i})\\ &=\sum_{i \in \{1, \dots, K\} \backslash \{i_{0}, i_{1}\}}\zeta_{p}^{B(i)}\chi_{-a}(A_{i}) +\zeta_{p}^{B(i_{0})}\chi_{-a}(A_{i_{0}})+\zeta_{p}^{B(i_{1})}\chi_{-a}(A_{i_{1}}). \end{align}\tag{11}\] Subtracting Eq. 10 from Eq. 11 , we have \[\label{Th4323} \epsilon_{B(P(\widetilde{F}))}(a)p^{\frac{n+s}{2}}\zeta_{p}^{(B(P(\widetilde{F})))^{*}(a)}=(\zeta_{p}^{B(i_{0})}-\zeta_{p}^{B(i_{1})})(\chi_{-a}(A_{i_{1}})-\chi_{-a}(A_{i_{0}})).\tag{12}\] Case I: \(B(i_{0})=B(i_{1})\) or \(\chi_{-a}(A_{i_{0}})=\chi_{-a}(A_{i_{1}})\). This case contradicts Eq. 12 .
Case II: \(B(i_{0}) \neq B(i_{1})\) and \(\chi_{-a}(A_{i_{0}}) \neq \chi_{-a}(A_{i_{1}})\). Let \(L=\mathbb{Q}(\zeta_{p})\) and \(\mathcal{O}_{L}=\mathbb{Z}[\zeta_{p}]\). Since \(-A_{i}=A_{i}\), then \(\sigma_{-1}(\chi_{-a}(A_{i}))=\sigma_{-1}(\sum_{x \in A_{i}}\zeta_{p}^{\langle -a, x\rangle_{n}})=\sum_{x \in A_{i}}\zeta_{p}^{\langle a, x\rangle_{n}}=\sum_{x \in A_{i}}\zeta_{p}^{\langle a, -x\rangle_{n}}=\chi_{-a}(A_{i})\), where \(\sigma_{-1}\) is the automorphism of \(L\) defined by \(\sigma_{-1}(\zeta_{p})=\zeta_{p}^{-1}\). Thus, \(\chi_{-a}(A_{i}) \in \mathcal{O}_{L}\) is a real number. Since \(p\mathcal{O}_{L}=((1-\zeta_{p})\mathcal{O}_{L})^{p-1}\) and \((\zeta_{p}^{B(i_{0})}-\zeta_{p}^{B(i_{1})})\mathcal{O}_{L}=(1-\zeta_{p}^{B(i_{1})-B(i_{0})})\mathcal{O}_{L}=(1-\zeta_{p})\mathcal{O}_{L}\), by Eq. 12 , we have \[\label{Th4324} ((1-\zeta_{p})\mathcal{O}_{L})^{\frac{(p-1)(n+s)}{2}}=(1-\zeta_{p})\mathcal{O}_{L} \cdot (\chi_{-a}(A_{i_{1}})-\chi_{-a}(A_{i_{0}}))\mathcal{O}_{L}.\tag{13}\] Then \((\chi_{-a}(A_{i_{1}})-\chi_{-a}(A_{i_{0}}))\mathcal{O}_{L}=(1-\zeta_{p})^{\frac{(p-1)(n+s)}{2}-1}\mathcal{O}_{L}\), which is impossible by Lemma 3 since \(\chi_{-a}(A_{i_{1}})-\chi_{-a}(A_{i_{0}})\) is a real number and \(\frac{(p-1)(n+s)}{2}-1\) is odd. Therefore, the Walsh supports of all \(p\)-ary \(s\)-plateaued functions generated by \(\Gamma\) are the same (denoted by \(S\)). Note that \(0 \in S\) as \(\Gamma\) is unbalanced.

Let \(a\) be an arbitrary fixed element in \(S\). Now we show that for all \(p\)-ary \(s\)-plateaued functions \(f\) generated by \(\Gamma\), \(\epsilon_{f}(a)\) are the same. With the similar arguments as in the proof of Proposition 2 \((i)\), we only need to prove that \(\epsilon_{B(P(\widetilde{F}(x)))}(a)=\epsilon_{B(\widetilde{F}(x))}(a)\) for any balanced map \(B\) from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\) and any transposition permutation \(P\) of \(\{1, 2, \dots, K\}\). Let \(B\) be an arbitrary fixed balanced map from \(\{1, 2, \dots, K\}\) to \(\mathbb{F}_{p}\) and \(P=(i_{0}i_{1})\) be an arbitrary fixed transposition permutation. We use the proof by contradiction. W.l.o.g, assume that \(\epsilon_{B(P(\widetilde{F}(x)))}(a)=1, \epsilon_{B(\widetilde{F}(x))}(a)=-1\). By Eq. 10 , we can obtain \[\label{Th4325} p^{\frac{n+s}{2}}(\zeta_{p}^{(B(P(\widetilde{F})))^{*}(a)}+\zeta_{p}^{(B(\widetilde{F}))^{*}(a)}) =(\zeta_{p}^{B(i_{0})}-\zeta_{p}^{B(i_{1})})(\chi_{-a}(A_{i_{1}})-\chi_{-a}(A_{i_{0}})).\tag{14}\] Case I: \(B(i_{0})=B(i_{1})\) or \(\chi_{-a}(A_{i_{0}})=\chi_{-a}(A_{i_{1}})\). This case contradicts Eq. 14 .
Case II: \(B(i_{0}) \neq B(i_{1})\), \(\chi_{-a}(A_{i_{0}}) \neq \chi_{-a}(A_{i_{1}})\) and \((B(P(\widetilde{F})))^{*}(a)=(B(\widetilde{F}))^{*}(a)\). Since \(p \mathcal{O}_{L}=((1-\zeta_{p})\mathcal{O}_{L})^{p-1}\) and \((\zeta_{p}^{B(i_{0})}-\zeta_{p}^{B(i_{1})}) \mathcal{O}_{L}=(1-\zeta_{p})\mathcal{O}_{L}\), then we have \((\chi_{-a}(A_{i_{1}})-\chi_{-a}(A_{i_{0}}))\mathcal{O}_{L}=2(1-\zeta_{p})^{\frac{(p-1)(n+s)}{2}-1}\mathcal{O}_{L}\), which is impossible by Lemma 3.
Case III: \(B(i_{0}) \neq B(i_{1})\), \(\chi_{-a}(A_{i_{0}}) \neq \chi_{-a}(A_{i_{1}})\) and \((B(P(\widetilde{F})))^{*}(a) \neq (B(\widetilde{F}))^{*}(a)\). Since \(p \mathcal{O}_{L}=((1-\zeta_{p})\mathcal{O}_{L})^{p-1}\), \((\zeta_{p}^{B(i_{0})}-\zeta_{p}^{B(i_{1})}) \mathcal{O}_{L}=(1-\zeta_{p})\mathcal{O}_{L}\) and \((\zeta_{p}^{(B(P(\widetilde{F})))^{*}(a)}+\zeta_{p}^{(B(\widetilde{F}))^{*}(a)})\mathcal{O}_{L}=\mathcal{O}_{L}\), then we have \((\chi_{-a}(A_{i_{1}})-\chi_{-a}(A_{i_{0}}))\mathcal{O}_{L}=(1-\zeta_{p})^{\frac{(p-1)(n+s)}{2}-1}\mathcal{O}_{L}\), which is impossible by Lemma 3. Therefore, for all \(p\)-ary \(s\)-plateaued functions \(f\) generated by \(\Gamma\), \(\epsilon_{f}(a)\) are the same (denoted by \(\epsilon(a)\)).

In the following, we compute the values \(\chi_{-a}(A_{i}), 1 \leq i \leq K\), where \(a\) is an arbitrary fixed element in \(S\). Let \(f\) be an arbitrary fixed \(p\)-ary \(s\)-plateaued function generated by \(\Gamma\). Note that \(cf \;(c \in \mathbb{F}_{p}^{*})\) is also a \(p\)-ary \(s\)-plateaued function generated by \(\Gamma\). Then \(S_{cf}=S_{f}=S\). By [31], for any \(c \in \mathbb{F}_{p}^{*}\), we have \(S_{cf}=cS_{f}\) (which yields that \(\mathbb{F}_{p}^{*}S=S\)) and \((cf)^{*}(a)=cf^{*}(c^{-1}a)\). Since \(f(x)=f(-x)\), then \(W_{f}(a)=W_{f}(-a)\), and thus \(f^{*}(a)=f^{*}(-a)\). When \(p=3\), then \((cf)^{*}(a)=cf^{*}(a), c \in \mathbb{F}_{p}^{*}\). When \(p \geq 5\), we claim that \((cf)^{*}(a)=cf^{*}(a), c \in \mathbb{F}_{p}^{*}\) also holds. For any permutation \(P\) of \(\mathbb{F}_{p}\), \(P(f)\) is also a \(p\)-ary \(s\)-plateaued function generated by \(\Gamma\). Then \(\epsilon_{P(f)}(a)=\epsilon(a)\). Define \(M_{j}^{(a)}\triangleq \frac{1-\zeta_{p}^{(P_{0,j}(f))^{*}(a)-f^{*}(a)}}{1-\zeta_{p}^{j}}\), where \(j \in \mathbb{F}_{p}^{*}\) and \(P_{0,j}=(0j)\) is the transposition permutation of \(\mathbb{F}_{p}\). From \(\epsilon(a)p^{\frac{n+s}{2}}\zeta_{p}^{(P_{0,j}(f))^{*}(a)}=W_{P_{0, j}(f)}(a)=\sum_{i \in \mathbb{F}_{p} \backslash \{0, j\}}\chi_{-a}(D_{f, i})\zeta_{p}^{i}+\chi_{-a}(D_{f, 0})\zeta_{p}^{j}+\chi_{-a}(D_{f, j})\) and \(\epsilon(a)p^{\frac{n+s}{2}}\zeta_{p}^{f^{*}(a)}=W_{f}(a)=\sum_{i \in \mathbb{F}_{p} \backslash \{0, j\}}\chi_{-a}(D_{f, i})\zeta_{p}^{i}+\chi_{-a}(D_{f, 0})+\chi_{-a}(D_{f, j})\zeta_{p}^{j}\), we get \(\chi_{-a}(D_{f, j})=\chi_{-a}(D_{f, 0})-\epsilon(a)p^{\frac{n+s}{2}}\zeta_{p}^{f^{*}(a)}M_{j}^{(a)}\) for any \(j \in \mathbb{F}_{p}^{*}\). Then for any \(c \in \mathbb{F}_{p}^{*}\), \[\epsilon(a)p^{\frac{n+s}{2}}\zeta_{p}^{(cf)^{*}(a)}=W_{cf}(a)=\sum_{j \in \mathbb{F}_{p}}\chi_{-a}(D_{f, j})\zeta_{p}^{cj}=-\epsilon(a)p^{\frac{n+s}{2}}\zeta_{p}^{f^{*}(a)}\sum_{j \in \mathbb{F}_{p}^{*}}M_{j}^{(a)}\zeta_{p}^{cj}.\] By Lemma 4, \(M_{j}^{(a)}=1, j \in \mathbb{F}_{p}^{*}\), or there is a unique \(j_{a} \in \mathbb{F}_{p}^{*}\) such that \(M_{j_{a}}^{(a)}=-\zeta_{p}^{-j_{a}}, M_{j}^{(a)}=0, j \in \mathbb{F}_{p}^{*} \backslash \{j_{a}\}\). Denote \(E=\{a \in S: M_{j}^{(a)}=1, j \in \mathbb{F}_{p}^{*}\}\). Then when \(a \in E\), \(\sum_{j \in \mathbb{F}_{p}^{*}}M_{j}^{(a)}\zeta_{p}^{cj}=-1\), and then \((cf)^{*}(a)=f^{*}(a)\); when \(a \in S \backslash E\), \(\sum_{j \in \mathbb{F}_{p}^{*}}M_{j}^{(a)}\zeta_{p}^{cj}=-\zeta_{p}^{(c-1)j_{a}}\), and then \((cf)^{*}(a)=f^{*}(a)+(c-1)j_{a}\). Thus, \((cf)^{*}(a)=cg(a)+h(a)\), where \(g: S \rightarrow \mathbb{F}_{p}\) is given as \(g(a)=0, a \in E\), and \(g(a)=j_{a}, a \in S \backslash E\), and \(h: S \rightarrow \mathbb{F}_{p}\) is given as \(h(a)=f^{*}(a), a \in E\), and \(h(a)=f^{*}(a)-j_{a}, a \in S \backslash E\). Then for any \(a \in S\), \(-g(a)+h(a)=(-f)^{*}(a)=-f^{*}(-a)=-f^{*}(a)=-g(a)-h(a)\), which implies that \(h(a)=0\) and \(g(a)=f^{*}(a)\). Thus, \((cf)^{*}(a)=cf^{*}(a), c \in \mathbb{F}_{p}^{*}\). Then by [30], for any \(j \in \mathbb{F}_{p}\), \[\begin{align} \chi_{-a}(D_{f, j})&=p^{n-1}\delta_{0}(a)+p^{-1}\sum_{c \in \mathbb{F}_{p}^{*}}W_{cf}(a)\zeta_{p}^{-cj}=p^{n-1}\delta_{0}(a)+\epsilon(a)p^{\frac{n+s}{2}-1}\sum_{c \in \mathbb{F}_{p}^{*}}\zeta_{p}^{c(f^{*}(a)-j)}\\ &=p^{n-1}\delta_{0}(a)+\epsilon(a)p^{\frac{n+s}{2}-1}(p\delta_{j}(f^{*}(a))-1). \end{align}\] For any fixed \(a \in S\), since \(\chi_{-a}(D_{f, j}), j \in \mathbb{F}_{p}\), take exactly two different values for any \(s\)-plateaued function \(f\) generated by \(\Gamma\), where the value \(p^{n-1}\delta_{0}(a)+\epsilon(a)p^{\frac{n+s}{2}-1}(p-1)\) occurs one time and the value \(p^{n-1}\delta_{0}(a)-\epsilon(a)p^{\frac{n+s}{2}-1}\) occurs \((p-1)\) times, we can obtain \[\chi_{-a}(A_{i})=\left\{ \begin{align} \frac{p^n}{K}\delta_{0}(a)-\epsilon(a)\frac{p^{\frac{n+s}{2}}}{K}+\epsilon(a)p^{\frac{n+s}{2}}, & \text{ if } i=\widetilde{G}(a),\\ \frac{p^n}{K}\delta_{0}(a)-\epsilon(a)\frac{p^{\frac{n+s}{2}}}{K}, & \text{ if } i \in \{1, 2, \dots, K\} \backslash \{\widetilde{G}(a)\}, \end{align} \right.\] where \(\widetilde{G}\) is some function from \(S\) to \(\{1, 2, \dots, K\}\). For \(a \in S^{*}\), since \(\chi_{-a}(A_{i}) \in \mathbb{Z}[\zeta_{p}] \cap \mathbb{Q}=\mathbb{Z}\), then \(K\) is a divisor of \(p^{\frac{n+s}{2}}\) and \(K\) is a power of \(p\) (denoted by \(K=p^m\)). Thus, we can view \(\Gamma\) as \(\Gamma=\{A_{i}, i \in V_{m}^{(p)}\}\) and (1) and (2) hold as \(F_{c}\) is a \(p\)-ary \(s\)-plateaued function generated by \(\Gamma\), and \(\chi_{-a}(A_{i})=p^{n-m}\delta_{0}(a)-\epsilon(a)p^{\frac{n+s}{2}-m}+\delta_{i}(G(a))\epsilon(a)p^{\frac{n+s}{2}}, a \in S, i \in V_{m}^{(p)}\), where \(G\) is some function from \(S\) to \(V_{m}^{(p)}\). Then for any \(a \in S\), \(W_{F_{c}}(a)=\sum_{i \in V_{m}^{(p)}}\chi_{-a}(A_{i})\zeta_{p}^{\langle c, i\rangle_{m}}=\epsilon(a)p^{\frac{n+s}{2}}\zeta_{p}^{G_{c}(a)}\), which shows that (3) holds. This completes the proof. ◻

Proof. \((i) \Rightarrow (ii)\): According to Theorem 6 and its proof, \(f\) is an \(s\)-plateaued function, where \(n+s\) is even, and for any \(c \in \mathbb{F}_{p}^{*}\), \(S_{f}=S_{cf}=cS_{f}\) and \(\epsilon_{cf}(a)=\epsilon_{f}(a), (cf)^{*}(a)=cf^{*}(a)\) for all \(a \in S_{f}\). Then by [31], for any \(c \in \mathbb{F}_{p}^{*}\) and \(a \in S_{f}\), \(\epsilon_{f}(c^{-1}a)=\epsilon_{cf}(a)=\epsilon_{f}(a), cf^{*}(c^{-1}a)=(cf)^{*}(a)=cf^{*}(a)\), which implies that \(\epsilon_{f}(c^{-1}a)=\epsilon_{f}(a), f^{*}(c^{-1}a)=f^{*}(a)\). Thus, for any \(c \in \mathbb{F}_{p}^{*}\) and \(x \in V_{n}^{(p)}\), we have \[\begin{align} p^{n}\zeta_{p}^{f(cx)}&=\sum_{a \in S_{f}}W_{f}(a)\zeta_{p}^{\langle a, cx\rangle_{n}}=\sum_{a \in S_{f}}W_{f}(c^{-1}a)\zeta_{p}^{\langle a,x\rangle_{n}}=\sum_{a \in S_{f}}\epsilon_{f}(c^{-1}a)p^{\frac{n+s}{2}}\zeta_{p}^{f^{*}(c^{-1}a)+\langle a, x\rangle_{n}}\\ &=\sum_{a \in S_{f}}\epsilon_{f}(a)p^{\frac{n+s}{2}}\zeta_{p}^{f^{*}(a)+\langle a, x\rangle_{n}}=\sum_{a \in S_{f}}W_{f}(a)\zeta_{p}^{\langle a, x\rangle_{n}}=p^{n}\zeta_{p}^{f(x)}, \end{align}\] and hence \(f(cx)=f(x)\), that is, \(f\) is of \((p-1)\)-form.

\((ii) \Rightarrow (i)\): Since \(f\) is an \(s\)-plateaued function of \((p-1)\)-form, where \(n+s\) is even, then for any \(a \in S_{f}\) and \(c \in \mathbb{F}_{p}^{*}\), we have \[W_{f}(a)=\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{f(x)-\langle a, x\rangle_{n}}=\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{f(cx)-\langle a, cx\rangle_{n}}=\sum_{x \in V_{n}^{(p)}}\zeta_{p}^{f(x)-\langle ca, x\rangle_{n}}=W_{f}(ca),\] which implies that \(cS_{f}=S_{f}\) and \(\epsilon_{f}(ca)=\epsilon_{f}(a), f^{*}(ca)=f^{*}(a)\). Then by [31], for any \(c \in \mathbb{F}_{p}^{*}\), \(S_{cf}=cS_{f}=S_{f}\), and for any \(a \in S_{f}\), \(\epsilon_{cf}(a)=\epsilon_{f}(c^{-1}a)=\epsilon_{f}(a), (cf)^{*}(a)=cf^{*}(c^{-1}a)=cf^{*}(a)\). By Theorem 6, \(\{D_{f, i}, i \in \mathbb{F}_{p}\}\) is an \(s\)-plateaued partition. ◻

Proof. For a \(p\)-ary bent function \(f: V_{n}^{(p)} \rightarrow \mathbb{F}_{p}\), \(S_{f}=V_{n}^{(p)}\). Then the result follows from Theorem 6. ◻

Proof. Without loss of generality, we assume that \(0 \in A_{0}\). Denote \(U_{i}=A_{i} \cup \{0\}, i \in {V_{n/2}^{(p)}}^{*}\). Define \(F: V_{n}^{(p)} \rightarrow V_{n/2}^{(p)}\) as \(F(x)=i, x \in A_{i}, i \in V_{n/2}^{(p)}\). By Corollary 2 and the proof of Theorem 6, \((F_{c})^{*}(x)=G_{c}(x), \epsilon_{F_{c}}(x)=\epsilon(x), x \in V_{n}^{(p)}, c \in {V_{n/2}^{(p)}}^{*}\), and for any \(a \in V_{n}^{(p)}\) and \(i \in V_{n/2}^{(p)}\), \[\chi_{-a}(A_{i})=p^{\frac{n}{2}}\delta_{0}(a)-\epsilon(a)+\delta_{i}(G(a))\epsilon(a)p^{\frac{n}{2}},\] where \(G\) is some function from \(V_{n}^{(p)}\) to \(V_{n/2}^{(p)}\) and \(\epsilon(a) \in \{\pm 1\}\) is independent of \(c\). By Remark 7 and the condition that \(A_{i} \neq \{0\}\) for all \(i \in V_{n/2}^{(p)}\) when \(p=3\), we have \(\epsilon(0)=1\) and \(|A_{i}|=p^{\frac{n}{2}}-1, i \in {V_{n/2}^{(p)}}^{*}\). We claim that \(\epsilon(a)=1\) for all \(a \in V_{n}^{(p)}\). To the contrary, assume that there exists \(a \in {V_{n}^{(p)}}^{*}\) such that \(\epsilon(a)=-1\). Let \(i \in V_{n/2}^{(p)}\) be fixed with \(i \neq 0, G(a)\). Then \[1=\chi_{-a}(A_{i})=\sum_{x \in A_{i}}\zeta_{p}^{\langle -a, x\rangle_{n}}=\sum_{j \in \mathbb{F}_{p}}n_{j}\zeta_{p}^{j},\] where \(n_{j}=|\{x \in A_{i}: \langle -a, x\rangle_{n}=j\}|\). Since \(1+x+x^{2}+\dots+x^{p-1}\) is the minimal polynomial of \(\zeta_{p}\) over \(\mathbb{Q}\), then \(n_{0}-1=n_{1}=n_{2}=\dots=n_{p-1}\). Then by \(|A_{i}|=p^{\frac{n}{2}}-1\), \(n_{1}=\frac{p^{\frac{n}{2}}-2}{p}\) is an integer, which is impossible. Thus, \(\epsilon(a)=1\) for any \(a \in V_{n}^{(p)}\). Then for any \(a \in V_{n}^{(p)}\) and \(i \in {V_{n/2}^{(p)}}^{*}\), \(\chi_{-a}(A_{i}) \in \{p^{\frac{n}{2}}-1, -1\}\) and thus \(\chi_{-a}(U_{i}) \in \{p^{\frac{n}{2}}, 0\}\). Let \(i \in {V_{n/2}^{(p)}}^{*}\) be fixed. From \(|U_{i}|=p^{\frac{n}{2}}\) and \(0 \in U_{i}\), \(\chi_{-a}(U_{i})=\sum_{x \in U_{i}}\zeta_{p}^{-\langle a, x\rangle_{n}}=p^{\frac{n}{2}}\) if and only if \(a \in U_{i}^{\perp}\), where \(U_{i}^{\perp}=\{a \in V_{n}^{(p)}: \langle a, x\rangle_{n}=0 \text{ for all } x \in U_{i}\}\) is a linear subspace. Denote \(m=|\{a \in V_{n}^{(p)}: \chi_{-a}(U_{i})=p^{\frac{n}{2}}\}|\). Then \(p^{n}=\sum_{x \in U_{i}}\sum_{a \in V_{n}^{(p)}}\zeta_{p}^{-\langle a, x\rangle_{n}}=\sum_{a \in V_{n}^{(p)}}\chi_{-a}(U_{i})=p^{\frac{n}{2}}m\), which implies that \(m=p^{\frac{n}{2}}\). Thus, the dimension of \(U_{i}^{\perp}\) is \(\frac{n}{2}\). Note that \(<U_{i}>^{\perp}=U_{i}^{\perp}\), where \(<U_{i}>\) is the linear space spanned by \(U_{i}\). Then the dimension of \(<U_{i}>\) is \(\frac{n}{2}\), and by \(|U_{i}|=p^{\frac{n}{2}}\), we have that \(U_{i}\) is a linear subspace of dimension \(\frac{n}{2}\). Obviously, \(U_{i} \cap U_{j}=\{0\}\) for any \(i \neq j \in {V_{n/2}^{(p)}}^{*}\). Thus, \(\{U_{i}, i \in {V_{n/2}^{(p)}}^{*}\}\) is a partial spread. By [32], the partial spread \(\{U_{i}, i \in {V_{n/2}^{(p)}}^{*}\}\) can be extended to a spread \(\{U_{i}, i \in {V_{n/2}^{(p)}}^{*}, W, V\}\). Then \(A_{0}=W\cup V\). Therefore, \(\{A_{i}, i \in V_{n/2}^{(p)}\}=\{W \cup V, U_{i}^{*}, {i \in V_{n/2}^{(p)}}^{*}\}\), which is obtained from the spread \(\{U_{i}, i \in {V_{n/2}^{(p)}}^{*}, W, V\}\). This completes the proof. ◻