Proof. We have

$$ \begin{align*}A(X^{m})\equiv \sum_{j\in A}X^{jm}\equiv \sum_{\substack{j\in A\\ jm\equiv g\ \bmod\ N}} X^{g}\ \bmod\ (X^{N}-1),\end{align*} $$

and the last one is precisely the definition of the mask polynomial of the multiset $m\cdot A$ . If $m\mid N$ , then

$$ \begin{align*}\sum_{\substack{j\in A\\ jm\equiv g\ \bmod\ N}} X^{g}\equiv\sum_{\substack{j\in A\\ j\equiv gm^{-1}\ \bmod\ N/m}}X^{g}\equiv \sum_{j=0}^{N/m-1}\lvert A_{j\ \bmod\ N/m} \rvert X^{jm}\ \bmod\ (X^{N}-1),\end{align*} $$

as desired.

Proof.

1. (i) This almost follows by definition. The space of functions $f:A\to \mathbb {C}$ has dimension $\lvert A \rvert $ ; therefore, a spectrum must have the same cardinality. Furthermore, by equation (3.1), we get that any $b,b^{\prime }\in B$ with $b\neq b^{\prime }$ must satisfy $b-b^{\prime }\in Z(A)$ .

2. (ii) Since every element of $\mathbb {Z}_{N}$ can be expressed uniquely as $a+t$ with $a\in A$ , $t\in T$ , we get a bijection between $\mathbb {Z}_{N}$ and $A\times T$ , proving $N=\lvert A \rvert \lvert T \rvert $ . Furthermore, suppose that $a,a^{\prime }\in A$ and $t,t^{\prime }\in T$ satisfy $a-a^{\prime }=t-t^{\prime }$ . Then $a+t^{\prime }=a^{\prime }+t$ , so by uniqueness of representation of $a+t^{\prime }$ as an element of A and an element of T, we must have $a=a^{\prime }$ and $t=t^{\prime }$ , proving the desired fact.

3. (iii) We will show that $m(b-b^{\prime })\in Z(A)$ for every $m\in \mathbb {Z}_{N}^{\star }$ and $b,b^{\prime }\in B$ with $b\neq b^{\prime }$ . Assume that $b-b^{\prime }\in d\mathbb {Z}_{N}^{\star }$ ; hence, by equation (3.1), we get $A(\zeta _{N}^{dg})=0$ , where $b-b^{\prime }\equiv dg\ \bmod\ N$ , with $g\in \mathbb {Z}_{N}^{\star }$ . Then $\Phi _{N/d}(X)\mid A(X)$ ; thus,

   $$ \begin{align*}0=A(\zeta_{N}^{dgm})=A(\zeta_{N}^{m(b-b^{\prime})}),\end{align*} $$
   completing the proof.

Proof. This follows from Theorem 3.3(i) as $Z(A)$ contains all divisor classes $d\mathbb {Z}_{N}^{\star }$ for which there are $b,b^{\prime }\in B$ satisfying $b-b^{\prime }\in d\mathbb {Z}_{N}^{\star }$ , so by equation (2.1) $\operatorname {\mathrm {ord}}(b-b^{\prime })=N/d$ , or equivalently, $\zeta _{N}^{b-b^{\prime }}$ is a primitive $N/d$ -th root of unity, yielding

$$ \begin{align*}A(\zeta_{N/d})=A(\zeta_{\operatorname{\mathrm{ord}}(b-b^{\prime})})=0,\end{align*} $$

as desired.

Proof. Suppose that $A\subseteq \mathbb {Z}_{N}$ satisfies (T1) and (T2). Denote by $N_{s}$ the maximal divisor of N that is prime to s. For example, if

$$ \begin{align*}N=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dotsm p_{k}^{\alpha_{k}},\end{align*} $$

then

$$ \begin{align*}N_{p_{1}^{m}}=p_{2}^{\alpha_{2}}\dotsm p_{k}^{\alpha_{k}}, \;\;\;\; \forall m \text{ with } 1\leq m\leq \alpha_{1}.\end{align*} $$

Consider the polynomial

$$ \begin{align*}T(X)=\prod_{s\in S^{N}\setminus S_{A}^{N}}\Phi_{s}(X^{N_{s}}).\end{align*} $$

By construction, $S_{T}^{N}=S^{N}\setminus S_{A}^{N}$ and $\lvert A \rvert \lvert T \rvert =N$ . Next, we will show that $A(X)T(X)$ vanishes on any Nth root of unity. Let $X=\zeta _{d}$ , $d\mid N$ . We write

$$ \begin{align*}d=p_{1}^{\beta_{1}}\dotsm p_{k}^{\beta_{k}},\end{align*} $$

with $0\leq \beta _{j}\leq \alpha _{j}$ , $1\leq j\leq k$ . If $s=p_{j}^{\beta _{j}}\in S_{T}^{N}$ for some j so that $d=s\cdot d_{s}$ , then $\Phi _{s}(X^{N_{s}})$ divides $T(X)$ and $d_{s}$ divides $N_{s}$ ; therefore,

$$ \begin{align*}\Phi_{s}(\zeta_{d}^{N_{s}})=\Phi_{s}(\zeta_{s}^{g})=0,\end{align*} $$

for $g=N_{s}/d_{s}$ , which is prime to s, yielding $T(\zeta _{d})=0$ . Otherwise, every $p_{j}^{\beta _{j}}\in S_{A}^{N}$ , and since A satisfies (T2), we get $\Phi _{d}(X)\mid A(X)$ or, equivalently $A(\zeta _{d})=0$ . Combined with $\lvert A \rvert \lvert T \rvert =N$ , this implies

$$ \begin{align*}A(X)T(X)\equiv1+X+\dotsb+X^{N-1}\ \bmod\ (X^{N}-1);\end{align*} $$

thus, $T(X)$ is the mask polynomial of a set $T\subseteq \mathbb {Z}_{N}$ , which is a tiling complement of A, proving (I).

For (II), we define the set $B\subseteq \mathbb {Z}_{N}$ consisting of all elements of the form

(3.2) $$ \begin{align} \sum_{s\in S_{A}^{N}}k_{s}\frac{N}{s}, \end{align} $$

where for each $s\in S_{A}^{N}$ , $k_{s}$ ranges through the set ${\left \{{0,1,\dotsc ,p-1}\right \}}$ , where $p=\Phi _{s}(1)$ is the prime dividing s. Different choices of $k_{s}$ , $s\in S_{A}^{N}$ give different elements, yielding $\lvert A \rvert =\lvert B \rvert $ since A satisfies (T1). Moreover, an element in $B-B$ has the form (3.2), with the difference that $\lvert k_{s} \rvert <\Phi _{s}(1)$ holds instead of $0\leq k_{s}\leq \Phi _{s}(1)-1$ . The order of $k_{s}\frac {N}{s}$ with $\lvert k_{s} \rvert <\Phi _{s}(1)$ is exactly s. Based on this simple fact, the order of a nonzero element of $B-B$ is $s_{1}s_{2}\dotsm s_{\ell }$ , where $s_{1},\dotsc ,s_{\ell }\in S_{A}^{N}$ powers of distinct primes. Since A satisfies (T2), we will then have $A(\zeta _{\operatorname {\mathrm {ord}}(b-b^{\prime })})=0$ for every $b,b^{\prime }\in B$ , $b\neq b^{\prime }$ , so B is the spectrum of A by Theorem 3.3(i) and Corollary 3.4.

A proof for part (III) is in [Reference Malikiosis and Kolountzakis23].

Proof. Since T is primitive, for a given $t\in T$ it would hold $t-T\nsubseteq p\mathbb {Z}_{N}$ and $t-T\nsubseteq q\mathbb {Z}_{N}$ , yielding $t^{\prime },t^{\prime \prime }\in T$ such that $t-t^{\prime }\notin p\mathbb {Z}_{N}$ , $t-t^{\prime \prime }\notin q\mathbb {Z}_{N}$ . If either $t-t^{\prime }\notin q\mathbb {Z}_{N}$ or $t-t^{\prime \prime }\notin p\mathbb {Z}_{N}$ , then we get $t-t^{\prime }\in \mathbb {Z}_{N}^{\star }$ or $t-t^{\prime \prime }\in \mathbb {Z}_{N}^{\star }$ , respectively. Otherwise, $t-t^{\prime }\in q\mathbb {Z}_{N}$ and $t-t^{\prime \prime }\in p\mathbb {Z}_{N}$ ; therefore, $t^{\prime }-t^{\prime \prime }\notin p\mathbb {Z}_{N}\cup q\mathbb {Z}_{N}$ , giving $t^{\prime }-t^{\prime \prime }\in \mathbb {Z}_{N}^{\star }$ , thus $(T-T)\cap \mathbb {Z}_{N}^{\star }\neq \varnothing $ . The converse is trivial.

Proof. Let j be as in the hypothesis, and without loss of generality assume that $A\subseteq j+\frac {N}{pq}\mathbb {Z}_{N}$ or, equivalently, $A=A_{j\ \bmod\ \frac {N}{pq}}$ . Translating A does not affect the conclusion, so we may further assume that $j=0$ . This confines A in the subgroup of order $pq$ . Dividing every element by $\frac {N}{pq}$ , we can assume that $N=pq$ . The problem has thus been reduced to the following: Assume that $A\subseteq \mathbb {Z}_{pq}$ is not a subset of a p- or a q-cycle, then $(A-A)\cap \mathbb {Z}_{pq}^{\star }\neq \varnothing $ . Indeed, this is the case, since A is primitive, so the conlcusion of the reduced problem follows from Lemma 5.4 with $m=n=1$ .

Proof. By hypothesis and Proposition 2.2, we have

$$ \begin{align*}A(X^{qd})\equiv P(X)\Phi_{p}(X^{N/p})\ \bmod\ (X^{N}-1).\end{align*} $$

Consider the intersections of $dA$ with the cosets of the subgroup of index $pq$ , namely

$$ \begin{align*}dA\cap(j+\frac{N}{pq}\mathbb{Z}_{N})={\left({dA}\right)}_{j\ \bmod\ \frac{N}{pq}},\end{align*} $$

where $0\leq j\leq \frac {N}{pq}-1$ . If ${\left ({dA}\right )}_{j\ \bmod\ \frac {N}{pq}}\neq \varnothing $ , then it intersects every coset of the subgroup $\frac {N}{q}\mathbb {Z}_{N}$ contained in $j+\frac {N}{pq}\mathbb {Z}_{N}$ . Indeed, if $da\in {\left ({dA}\right )}_{j\ \bmod\ \frac {N}{pq}}$ , then

$$ \begin{align*}qda+i\frac{N}{p}\in qdA\end{align*} $$

for $0\leq i\leq p-1$ by hypothesis, so, for each i, there is $a^{\prime }\in A$ such that $qda^{\prime }\equiv qda+i\frac {N}{p}\ \bmod\ N$ ; therefore,

$$ \begin{align*}da^{\prime}\equiv da+i\frac{N}{pq}+k\frac{N}{q}\ \bmod\ N\end{align*} $$

for some k. Each element of the form $da+i\frac {N}{pq}+k\frac {N}{q}$ belongs to a different q-cycle of $j+\frac {N}{pq}\mathbb {Z}_{N}$ , and since there are exactly p of them, $dA$ intersects them all. If every pair of elements $da,da^{\prime }\in {\left ({dA}\right )}_{j\ \bmod\ \frac {N}{pq}}$ lying on two different q-cycles belongs to the same p-cycle, then it is evident that ${\left ({dA}\right )}_{j\ \bmod\ \frac {N}{pq}}$ is a single p-cycle. By hypothesis, the multiset $(qd)\cdot A$ contains the entire p-cycle $qj+\frac {N}{p}\mathbb {Z}_{N}$ with some multiplicity (i.e., every element in this p-cycle appears with the same positive multiplicity in $(qd)\cdot A$ ), since ${\left ({dA}\right )}_{j\ \bmod\ \frac {N}{pq}}\neq \varnothing $ . We have proven that the elements $d\cdot A$ restricted to $j+\frac {N}{pq}\mathbb {Z}_{N}$ (i.e. the inverse image of $qj+\frac {N}{p}\mathbb {Z}_{N}$ under the multiplication by q map) are supported on a single p-cycle; therefore, this p-cycle must appear with the same multiplicity in $d\cdot A$ as $qj+\frac {N}{p}\mathbb {Z}_{N}$ appears in $(qd)\cdot A$ . Thus, if every ${\left ({dA}\right )}_{j\ \bmod\ \frac {N}{pq}}$ is supported on a p-cycle, we conlcude that $d\cdot A$ is also a union of p-cycles as a multiset, giving $A(\zeta _{N}^{d})=0$ , a contradiction.

This proves the existence of $da,da^{\prime }\in {\left ({dA}\right )}_{j\ \bmod\ \frac {N}{pq}}$ , lying on two different q-cycles, and two different p-cycles as well, for some j. Then ${\left ({dA}\right )}_{j\ \bmod\ \frac {N}{pq}}$ is not supported on a p- or q-cycle, so by Proposition 5.5

$$ \begin{align*}(dA-dA)\cap\frac{N}{pq}\mathbb{Z}_{N}^{\star}\neq\varnothing,\end{align*} $$

or equivalently, there are $a,a^{\prime }\in A$ such that $d(a-a^{\prime })\in \frac {N}{pq}\mathbb {Z}_{N}^{\star }$ , which shows that $\operatorname {\mathrm {ord}}(a-a^{\prime })=pqd$ ; indeed, as $pqd(a-a^{\prime })\equiv 0\ \bmod\ N$ , while $pd(a-a^{\prime })\in \frac {N}{q}\mathbb {Z}_{N}^{\star }$ and $qd(a-a^{\prime })\in \frac {N}{p}\mathbb {Z}_{N}^{\star }$ . This implies that $(A-A)\cap \frac {N}{pqd}\mathbb {Z}_{N}^{\star }\neq \varnothing $ , hence $B(\zeta _{pqd})=0$ for any spectrum B of A by Corollary 3.4, completing the proof.

Proof. Without loss of generality, we may assume that $q\nmid \lvert A \rvert $ . It holds $A(\zeta )\neq 0$ when $\zeta ^{q^{n}}=1$ ; otherwise, $A(X)$ would be divided by some cyclotomic polynomial $\Phi _{q^{k}}(X)$ for some $1\leq k\leq n$ , and as a consequence, $q=\Phi _{q^{k}}(1)$ would divide $A(1)=\lvert A \rvert $ , a contradiction. Next, consider the set

$$ \begin{align*}R={\left\{{r\in[1,m]:A(\zeta_{p^{r}}\zeta_{q^{k}})=0,\text{ for some }k, 0\leq k\leq n}\right\}}.\end{align*} $$

Define the following polynomial

(6.1) $$ \begin{align} F(X)=A(X)\prod_{\substack{1\leq r\leq m\\ r\notin R}}\Phi_{p^{r}}(X), \end{align} $$

which has nonnegative integer coefficients. We will show that $F(1)\leq p^{m}$ or, equivalently, $A(1)\leq p^{\lvert R \rvert }$ . To this end, consider the spectrum B of A, which satisfies $\lvert A \rvert =\lvert B \rvert $ and $B-B\subseteq {\left \{{0}\right \}}\cup Z(A)$ by Theorem 3.3(i). Since $A(\zeta )\neq 0$ when $\zeta ^{q^{n}}=1$ , we obtain $(B-B)\cap p^{m}\mathbb {Z}_{N}={\left \{{0}\right \}}$ ; therefore, any two distinct elements of B are not congruent modulo $p^{m}$ (giving a first estimate of $\lvert A \rvert \leq p^{m}$ ). So, we may write for every $b\in B$

$$ \begin{align*}b\equiv b_{0}+b_{1}p+\dotsb+b_{m-1}p^{m-1}\ \bmod\ p^{m},\ 0\leq b_{j}\leq p-1,\ 0\leq j\leq m-1\end{align*} $$

and observe that no two elements of B correspond to the same m-tuple $(b_{0},\dotsc ,b_{m-1})$ . Moreover, no two elements of B can agree on all p-adic digits $b_{r}$ for $m-r\in R$ . If such elements existed, say $b\neq b^{\prime }$ , then $p^{r}\mid \!\mid b-b^{\prime }$ for some $m-r\notin R$ , and since $(A,B)$ is a spectral pair, we should have $A(\zeta _{p^{m-r}}\zeta _{q^{k}})=0$ for some k by Corollary 3.4, contradicting the fact $m-r\notin R$ . This clearly shows that $\lvert A \rvert \leq p^{\lvert R \rvert }$ , and thus $F(1)\leq p^{m}$ .

Next, we will show that $\overline {\Phi }_{p^{r}}(X)\mid \overline {F}(X)$ in $\mathbb {F}_{q}[X]$ for $1\leq r\leq m$ . If $r\notin R$ , this follows immediately from equation (6.1). If $r\in R$ , then by definition $\Phi _{p^{r}q^{k}}(X)\mid A(X)$ for some k, hence $\Phi _{p^{r}q^{k}}(X)\mid F(X)$ , where $p^{r}q^{k}\mid N$ . Using standard properties of cyclotomic polynomials, we get

$$ \begin{align*}\Phi_{p^{r}q^{k}}(X)=\Phi_{pq}(X^{p^{r-1}q^{k-1}})=\frac{\Phi_{p}(X^{p^{r-1}q^{k}})}{\Phi_{p}(X^{p^{r-1}q^{k-1}})}.\end{align*} $$

Reducing coefficients $\ \bmod\ q$ , we obtain

$$ \begin{align*}\frac{\overline{\Phi}_{p}(X^{p^{r-1}})^{q^{k}}}{\overline{\Phi}_{p}(X^{p^{r-1}})^{q^{k-1}}}=\overline{\Phi}_{p}(X^{p^{r-1}})^{q^{k-1}(q-1)}=\overline{\Phi}_{p^{r}}(X)^{q^{k-1}(q-1)}.\end{align*} $$

Therefore,

$$ \begin{align*}\prod_{r=1}^{m}\overline{\Phi}_{p^{r}}(X)=X^{p^{m}-1}+\dotsb+1\mid \overline{F}(X)\end{align*} $$

in $\mathbb {F}_{q}[X]$ , say

$$ \begin{align*}\overline{F}(X)=(X^{p^{m}-1}+\dotsb+1)G(X).\end{align*} $$

Now, we reduce $\ \bmod\ (X^{p^{m}}-1)$ . Since

$$ \begin{align*}X^{\ell}(X^{p^{m}-1}+\dotsb+1)\equiv X^{p^{m}-1}+\dotsb+1\ \bmod\ (X^{p^{m}}-1)\end{align*} $$

for every $\ell $ , we get

(6.2) $$ \begin{align} \overline{F}(X)\equiv G(1)(X^{p^{m}-1}+\dotsb+1)\ \bmod\ (X^{p^{m}}-1), \end{align} $$

where $G(1)\neq 0$ (in $\mathbb {F}_{q}$ ) by assumption. Consider the remainder of the Euclidean division $F(X):X^{p^{m}}-1$ in $\mathbb {Z}[X]$ , say $R(X)$ . It is not hard to show that R has also nonnegative coefficients; indeed, if $F(X)=\sum _{k=0}^{\deg F}f_{k}X^{k}\in \mathbb {Z}_{\geq 0}[X]$ , then

$$ \begin{align*}R(X)=\sum_{k=0}^{p^{m}-1}{\left({\sum_{\ell\equiv k\ \bmod\ (p^{m}-1)}f_{\ell}}\right)}X^{k}\in\mathbb{Z}_{\geq0}[X].\end{align*} $$

All coefficients of $R(X)$ must be congruent to $c\equiv G(1)\ \bmod\ q$ due to equation (6.2). Without loss of generality, we may assume $0<c<q$ ; hence, positivity of the coefficients of R implies

$$ \begin{align*}R(X)=c(X^{p^{m}-1}+\dotsb+1)+qQ(X),\end{align*} $$

where $Q(X)\in \mathbb {Z}_{\geq 0}[X]$ . Therefore,

$$ \begin{align*}F(1)=R(1)=cp^{m}+qQ(1)\geq p^{m},\end{align*} $$

implying $F(1)=p^{m}$ and $Q(X)\equiv 0$ . From this we obtain

$$ \begin{align*}F(X)\equiv X^{p^{m}-1}+\dotsb+1\ \bmod\ (X^{p^{m}}-1)\end{align*} $$

in $\mathbb {Z}[X]$ . Clearly,

$$ \begin{align*}F(X)\sum_{s=0}^{q^{n}-1}X^{p^{m}s}\equiv X^{N-1}+\dotsb+1\ \bmod\ (X^{N}-1),\end{align*} $$

so, for the polynomial $C(X)\in \mathbb {Z}[X]$ with nonnegative coefficients defined by

$$ \begin{align*}C(X)={\left({\sum_{s=0}^{q^{n}-1}X^{p^{m}s}}\right)}\prod_{\substack{1\leq r\leq m\\ r\notin R}}\Phi_{p^{r}}(X),\end{align*} $$

we have

$$ \begin{align*}A(X)C(X)\equiv X^{N-1}+\dotsb+1\ \bmod\ (X^{N}-1),\end{align*} $$

implying that $C(X)$ is the mask polynomial of a set C which is the tiling complement of A. This completes the proof.

Proof. Without loss of generality, $p^{m}\mid \lvert A \rvert $ , and let B be a spectrum of A. Suppose that

$$ \begin{align*}S_{B}^{N}={\left\{{p^{m_{1}},\dotsc,p^{m_{\ell}},q^{n_{1}},\dotsc,q^{n_{k}}}\right\}},\end{align*} $$

where $1\leq m_{1}<\dotsb <m_{\ell }\leq m$ and $1\leq n_{1}<\dotsb <n_{k}\leq n$ . Since

$$ \begin{align*}\Phi_{q^{n_{1}}}(X)\dotsm\Phi_{q^{n_{k}}}(X)\mid B(X),\end{align*} $$

we obtain

$$ \begin{align*}p^{m} q^{k}\mid \lvert B \rvert=\lvert A \rvert\end{align*} $$

by putting $X=1$ , and using the hypothesis. For every j, consider $A_{j\ \bmod\ p^{m}}$ . Since $A_{j\ \bmod\ p^{m}}-A_{j\ \bmod\ p^{m}}\subseteq p^{m}\mathbb {Z}_{N}$ , we must have

(6.3) $$ \begin{align} A_{j\ \bmod\ p^{m}}-A_{j\ \bmod\ p^{m}}\subseteq{\left\{{0}\right\}}\cup p^{m}q^{n-n_{1}}\mathbb{Z}_{N}^{\star}\cup\dotsb\cup p^{m}q^{n-n_{k}}\mathbb{Z}_{N}^{\star} \end{align} $$

by Theorem 3.3(i). As usual, we consider the (finite) q-adic expansion of each $a\in \mathbb {Z}_{N}$ as

$$ \begin{align*}a\equiv a_{0}+a_{1}q+a_{2}q^{2}+\dotsb+a_{n-1}q^{n-1}\ \bmod\ q^{n}, \;\;\;\; 0\leq a_{i}\leq q-1, 0\leq i\leq n-1.\end{align*} $$

Suppose that $\lvert A_{j\ \bmod\ p^{m}} \rvert>q^{k}$ for some j. This would imply the existence of $a,a^{\prime }\in A_{j\ \bmod\ p^{m}}$ that would have the same q-adic digits at positions $n-n_{1},\dotsc ,n-n_{k}$ ; hence, $a-a^{\prime }$ would not belong to the above union, contradiction. Thus, $\lvert A_{j\ \bmod\ p^{m}} \rvert \leq q^{k}$ for all j, and in light of $p^{m}q^{k}\mid \lvert A \rvert $ , we must have

$$ \begin{align*}\lvert A_{j\ \bmod\ p^{m}} \rvert=q^{k},\ \forall j.\end{align*} $$

This further implies $A(\zeta _{p^{\ell }})=0$ , $1\leq \ell \leq m$ . Using the same argument as above, where we reverse the roles of A and B, we also obtain $B(\zeta _{p^{\ell }})=0$ , for $1\leq \ell \leq m$ , whence

$$ \begin{align*}\lvert B_{j\ \bmod\ p^{m}} \rvert=q^{k},\ \forall j.\end{align*} $$

Next, consider

$$ \begin{align*}S_{A}^{N}={\left\{{p,p^{2},\dotsc,p^{m},q^{s_{1}},\dotsc,q^{s_{t}}}\right\}},\end{align*} $$

where $1\leq s_{1}<\dotsb <s_{t}\leq n$ so that

$$ \begin{align*}\Phi_{q^{s_{1}}}(X)\dotsm\Phi_{q^{s_{t}}}(X)\mid A(X),\end{align*} $$

and putting $X=1$ , we get $q^{t}\mid \lvert A \rvert $ , which yields $t\leq k$ . We actually have $t=k$ since

(6.4) $$ \begin{align} B_{j\ \bmod\ p^{m}}-B_{j\ \bmod\ p^{m}}={\left\{{0}\right\}}\cup p^{m}q^{n-s_{1}}\mathbb{Z}_{N}^{\star}\cup\dotsb\cup p^{m}q^{n-s_{t}}\mathbb{Z}_{N}^{\star}, \end{align} $$

for every j, and $\lvert B_{j\ \bmod\ p^{m}} \rvert =q^{k}$ . If $t<k$ , then there would exist $b,b^{\prime }\in B_{j\ \bmod\ p^{m}}$ , $b\neq b^{\prime }$ whose q-adic expansions $\ \bmod\ q^{n}$ agree on all q-adic digits corresponding to $q^{n-s_{i}}$ , $1\leq i\leq t$ . But then

$$ \begin{align*}b-b^{\prime}\notin {\left\{{0}\right\}}\cup p^{m}q^{n-s_{1}}\mathbb{Z}_{N}^{\star}\cup\dotsb\cup p^{m}q^{n-s_{t}}\mathbb{Z}_{N}^{\star},\end{align*} $$

contradiction. Thus, $t=k$ and in particular, A satisfies (T1).

Next, consider the polynomial

$$ \begin{align*}\Lambda(X)=\prod_{i=1}^{k}\Phi_{q^{n_{i}}}(X^{p^{m}}).\end{align*} $$

As the coefficients of $\Lambda (X)$ are $0$ or $1$ , it is the mask polynomial of a set $\Lambda \subseteq \mathbb {Z}_{N}$ with cardinality $\lvert \Lambda \rvert =\Lambda (1)=q^{k}$ . By equation (6.3), $\Lambda $ is then the common spectrum of the sets $A_{j\ \bmod\ p^{m}}$ . It holds

$$ \begin{align*}\bigcup_{i=1}^{k} p^{m}q^{n_{i}-1}\mathbb{Z}_{N}^{\star}\subseteq \Lambda-\Lambda \subseteq{\left\{{0}\right\}}\cup Z(A_{j\ \bmod\ p^{m}})\end{align*} $$

for every j by Theorem 3.3(i); therefore,

$$ \begin{align*}A_{j\ \bmod\ p^{m}}(\zeta_{q^{n-n_{i}+1}})=0,\ 1\leq i\leq k,\ \forall j.\end{align*} $$

Thus, $A(\zeta _{q^{n-n_{i}+1}})=0$ , $1\leq i\leq k$ , proving eventually that $n_{i}+s_{i}=n+1$ , and

$$ \begin{align*}S_{A}^{N}={\left\{{p,p^{2},\dotsc,p^{m},q^{n-n_{1}+1},\dotsc,q^{n-n_{k}+1}}\right\}}.\end{align*} $$

Next, we fix j and consider the polynomial $F(X)$ satisfying

$$ \begin{align*}A_{j\ \bmod\ p^{m}}(X)\equiv X^{j} F(X^{p^{m}})\ \bmod\ (X^{N}-1).\end{align*} $$

Since $\Phi _{q^{n-n_{i}+1}}(X)\mid F(X^{p^{m}})$ for $1\leq i\leq k$ and $p^{m}$ is prime to $q^{n-n_{i}+1}$ , we also have $\Phi _{q^{n-n_{i}+1}}(X)\mid F(X)$ . Therefore, for $1\leq \ell \leq m$ , we have

$$ \begin{align*}A_{j\ \bmod\ p^{m}}(\zeta_{p^{\ell}q^{n-n_{i}+1}})=\zeta_{p^{\ell}q^{n-n_{i}+1}}^{j}F(\zeta_{p^{\ell}q^{n-n_{i}+1}}^{p^{m}})=\zeta_{p^{\ell}q^{n-n_{i}+1}}^{j}F(\zeta_{q^{n-n_{i}+1}}^{p^{m-\ell}})=0,\end{align*} $$

proving

$$ \begin{align*}A(\zeta_{p^{\ell}q^{n-n_{i}+1}})=0,\ 1\leq i\leq k,\ 1\leq \ell\leq m;\end{align*} $$

thus, A satisfies (T2) as well, hence tiles $\mathbb {Z}_{N}$ by virtue of Theorem 3.5(I), as desired.

Proof. If A tiles, then it is spectral by Theorem 3.6. Suppose that A is spectral. If $q\nmid \lvert A \rvert $ , then A tiles by Theorem 6.1; if $q\mid \lvert A \rvert $ , then A tiles by Proposition 6.2.

Proof. Suppose that A has a spectrum B that is not primitive. Then a translate of B must be contained in a maximal subgroup of $\mathbb {Z}_{N}$ , i.e., either $p\mathbb {Z}_{N}$ or $q\mathbb {Z}_{N}$ . This translate is also a spectrum of A, so without loss of generality, we may assume $B\subseteq p\mathbb {Z}_{N}$ . This shows the existence of $B^{\prime }\subseteq \mathbb {Z}_{N}$ with $\lvert B \rvert =\lvert B^{\prime } \rvert $ and $B=pB^{\prime }$ . The matrix

$$ \begin{align*}\tfrac{1}{\sqrt{\lvert A \rvert}}{\left({\zeta_{N}^{ab}}\right)}_{a\in A, b\in B}=\tfrac{1}{\sqrt{\lvert A \rvert}}{\left({\zeta_{N/p}^{ab^{\prime}}}\right)}_{a\in A, b^{\prime}\in B^{\prime}}\end{align*} $$

is unitary by definition; hence, no two elements of A leave the same residue $\ \bmod\ N/p$ (as well as of $B^{\prime }$ ). Hence, $(A,B^{\prime })$ can be considered as a spectral in $\mathbb {Z}_{N/p}$ , so $A(X)\ \bmod\ (X^{N/p}-1)$ satisfies (T1) and (T2) by definition of $\mathscr {S}^{\prime }_{p,q}$ . This in particular shows that $A(\zeta _{p^{m}})\neq 0$ . Otherwise, (T1) would fail $\ \bmod\ N/p$ . Therefore, (T1) and (T2) are satisfied by $A(X)\ \bmod\ (X^{N}-1)$ as well.

Next, suppose that A is not primitive so that $A\subseteq p\mathbb {Z}_{N}$ without loss of generality. As before, $(A^{\prime },B)$ would be a spectral pair in $\mathbb {Z}_{N/p}$ , where $A=pA^{\prime }$ , so $A^{\prime }(X)$ satisfies (T1) and (T2) by definition of $\mathscr {S}^{\prime }_{p,q}$ . We have $A(X)\equiv A^{\prime }(X^{p})\ \bmod\ (X^{N}-1)$ , hence $A(\zeta _{p})\neq 0$ . The following show that A also satisfies (T1) and (T2):

$$ \begin{align*}A^{\prime}(\zeta_{p^{k}q^{\ell}})=0\Longrightarrow A(\zeta_{p^{k+1}q^{\ell}})=0,\end{align*} $$

and

$$ \begin{align*}A^{\prime}(\zeta_{q^{k}})=0\Longrightarrow A(\zeta_{pq^{k}})=A(\zeta_{q^{k}})=0.\end{align*} $$

For the latter, consider t such that $pt\equiv 1\ \bmod\ q^{k}$ . Then

$$ \begin{align*}A(\zeta_{q^{k}}^{t})=A^{\prime}(\zeta_{q^{k}}^{pt})=A^{\prime}(\zeta_{q^{k}})=0,\end{align*} $$

and by a Galois action, we get $A(\zeta _{q^{k}})=0$ as well. This completes the proof.

Proof. By Proposition 8.2, A and every spectrum B are primitive. By Lemma 5.4, we get $(A-A)\cap \mathbb {Z}_{N}^{\star }\neq \varnothing \neq (B-B)\cap \mathbb {Z}_{N}^{\star }$ , so by Theorem 3.3(i), we obtain $A(\zeta _{N})=B(\zeta _{N})=0$ .

Proof. Let B be a spectrum of A, and suppose that A is a union of p- or q-cycles exclusively. Without loss of generality, we may assume $Q\equiv 0$ so that

(8.3) $$ \begin{align} A(X)\equiv P(X)\Phi_{p}(X^{N/p})\ \bmod\ (X^{N}-1). \end{align} $$

For every $a\in A$ , we have $a+\frac {N}{p}\mathbb {Z}_{N}\subseteq A$ , hence $(A-A)\cap \frac {N}{p}\mathbb {Z}_{N}^{\star }\neq \varnothing $ , implying $B(\zeta _{p})=0$ and

$$ \begin{align*}\lvert B_{j\ \bmod\ p} \rvert=\frac{1}{p}\lvert B \rvert,\ \forall j.\end{align*} $$

Since $P(X)\in \mathbb {Z}_{\geq 0}[X]$ , the coefficients must be either $0$ or $1$ ; therefore, $P(X)$ is the mask polynomial of a set $P\subseteq \mathbb {Z}_{N}$ . Furthermore, equation (8.3) shows that any element of P is unique $\ \bmod\ N/p$ , so we may consider this as a subset in $\mathbb {Z}_{N/p}$ .

Next, we will show that each $B_{j\ \bmod\ p}$ is a spectrum for P. Since

$$ \begin{align*}B_{j\ \bmod\ p}-B_{j\ \bmod\ p}\subseteq p\mathbb{Z}_{N}\cap (Z(A)\cup{\left\{{0}\right\}}),\end{align*} $$

by Theorem 3.3(i), it suffices to show that $A(\zeta _{N}^{d})=0$ implies $P(\zeta _{N}^{d})=0$ whenever $p\mid d$ and $d\mid N$ , i.e., $A(X)$ and $P(X)$ have the same roots when we restrict to the $\frac {N}{p}$ -roots of unity. But this is true since (8.3) implies

$$ \begin{align*}A(X)\equiv pP(X)\ \bmod\ (X^{N/p}-1),\end{align*} $$

therefore, $A(\zeta _{N/p}^{d})=pP(\zeta _{N/p}^{d})$ for all d. We observe that each spectrum $-j+B_{j\ \bmod\ p}$ is a subset of the subgroup of $\mathbb {Z}_{N}$ of order $N/p$ , which shows that $P\subseteq \mathbb {Z}_{N/p}$ is spectral since $\lvert P \rvert =\lvert B_{j\ \bmod\ p} \rvert $ for all j.

By definition of $\mathscr {S}_{p,q}^{\prime }$ , P satisfies (T1) and (T2). We remind that $A(X)$ and $P(X)$ have precisely the same roots when we restrict to the $\frac {N}{p}$ -roots of unity. In addition, $A(X)$ vanishes for all other Nth roots of unity as $\Phi _{p}(X^{N/p})\mid A(X)$ , clearly showing that $A(X)$ satisfies (T1) and (T2) as well.

Proof. By Theorem 5.2, the multiset $\frac {N}{pqd}\cdot B$ is a disjoint union of p- and q-cycles. Each such cycle is a subset of a unique $pq$ -cycle (i.e., a coset of the subgroup $\frac {N}{pq}\mathbb {Z}_{N}$ ) as $j+\frac {N}{p}\mathbb {Z}_{N}\subseteq j+\frac {N}{pq}\mathbb {Z}_{N}$ , for example. Therefore, the same conclusion holds for the elements of $\frac {N}{pqd}\cdot B$ (counting multiplicities) in an arbitrary $pq$ -cycle. If $\frac {N}{pqd}\cdot B\cap (j+\frac {N}{pq}\mathbb {Z}_{N})$ is not supported on a single p- or q-cycle, then there are $b,b^{\prime }\in B$ such that $\frac {N}{pqd}(b-b^{\prime })\in \frac {N}{pq}\mathbb {Z}_{N}^{\star }$ by virtue of Proposition 5.5. This yields $b-b^{\prime }\in d\mathbb {Z}_{N}^{\star }$ , contradicting the fact that $A(\zeta _{N}^{d})\neq 0$ , using Theorem 3.3(i). Now, every element of $\frac {N}{pqd}\cdot B_{i\ \bmod\ d}$ belongs to the $pq$ -cycle $\frac {iN}{pqd}+\frac {N}{pq}\mathbb {Z}_{N}$ ; therefore, it must also be a disjoint union of p- and q-cycles. In our case, it must be a single p- or q-cycle with a certain multiplicity. If $\frac {N}{pqd}\cdot B_{i\ \bmod\ d}$ is a p-cycle with a multiplicity, then this p-cycle must be

$$ \begin{align*}\frac{iN}{pqd}+k\frac{N}{q}+\frac{N}{p}\mathbb{Z}_{N}=\frac{N}{pqd} B_{i\ \bmod\ d}\end{align*} $$

for some k, which clearly shows that $\lvert B_{i+kpd+jqd\ \bmod\ pqd} \rvert $ is constant in j, and

$$ \begin{align*}B_{i+k^{\prime}pd+jqd\ \bmod\ pqd}=\varnothing,\end{align*} $$

for $k^{\prime }\not \equiv k\ \bmod\ q$ . This yields the absorption

$$ \begin{align*}B_{i+jqd\ \bmod\ pd}=B_{i+kpd+jqd\ \bmod\ pqd}\end{align*} $$

for all j, which shows that $B_{i\ \bmod\ d}$ is equidistributed $\ \bmod\ pd$ . If on the other hand, $\frac {N}{pqd}\cdot B_{i\ \bmod\ d}$ is a q-cycle with a multiplicity, then this q-cycle must be

$$ \begin{align*}\frac{iN}{pqd}+j\frac{N}{p}+\frac{N}{q}\mathbb{Z}_{N}=\frac{N}{pqd} B_{i\ \bmod\ d}\end{align*} $$

for some j, which shows that $\lvert B_{i+jqd+kpd\ \bmod\ pqd} \rvert $ is constant in k, and

$$ \begin{align*}B_{i+j^{\prime}qd+kpd\ \bmod\ pqd}=\varnothing,\end{align*} $$

for $j^{\prime }\not \equiv j\ \bmod\ p$ . This yields

$$ \begin{align*}\lvert B_{i\ \bmod\ d} \rvert=\lvert B_{i+jqd\ \bmod\ pd} \rvert=p\lvert B_{i+jqd+kpd\ \bmod\ pqd} \rvert\end{align*} $$

for this value of j and every k, which shows that $B_{i\ \bmod\ d}$ is absorbed $\ \bmod\ pd$ . Absorption $\ \bmod\ pd$ must occur for at least one i. Otherwise, we would have $B(\zeta _{pd})=0$ , contradiction.

Proof. Denote by T the set whose mask polynomial is $\Phi _{p^{m-k}}(X^{q^{n}})$ so that

$$ \begin{align*}T={\left\{{0,p^{m-k-1}q^{n},2p^{m-k-1}q^{n},\dotsc,(p-1)p^{m-k-1}q^{n}}\right\}}.\end{align*} $$

Since $B_{i\ \bmod\ p^{k}}$ is absorbed $\ \bmod\ p^{k+1}$ for every i, we have $B(\zeta _{p^{k+1}})\neq 0$ . Therefore, by Theorem 3.3(i), we get

$$ \begin{align*}(A-A)\cap\frac{N}{p^{k+1}}\mathbb{Z}_{N}^{\star}=\varnothing.\end{align*} $$

Moreover,

$$ \begin{align*}T-T={\left\{{\ell p^{m-k-1}q^{n}:\lvert \ell \rvert<p}\right\}}\subseteq {\left\{{0}\right\}}\cup\frac{N}{p^{k+1}}\mathbb{Z}_{N}^{\star};\end{align*} $$

therefore, $(A-A)\cap (T-T)={\left \{{0}\right \}}$ . Hence, every element in $A+T$ has a unique representation as $a+t$ with $a\in A$ , $t\in T$ (proof is similar to that of Theorem 3.3(ii)). This shows that $\overline {A}(X)$ is a mask polynomial of the set $\overline {A}=A\oplus T$ . Furthermore, since each $B_{i\ \bmod\ p^{k}}$ is absorbed $\ \bmod\ p^{k+1}$ , if $p^{k}\mid b-b^{\prime }$ for some $b,b^{\prime }\in B$ , then $p^{k+1}\mid b-b^{\prime }$ (definition of absorption). In other words,

(9.1) $$ \begin{align} (B-B)\cap (p^{k}\mathbb{Z}_{N}^{\star}\cup p^{k}q\mathbb{Z}_{N}^{\star}\cup\dotsb\cup p^{k}q^{n}\mathbb{Z}_{N}^{\star})=\varnothing. \end{align} $$

The set

(9.2) $$ \begin{align} S={\left\{{0,p^{k}q^{n},2p^{k}q^{n},\dotsc,(p-1)p^{k}q^{n}}\right\}} \end{align} $$

has mask polynomial $\Phi _{p^{k+1}}(X^{q^{n}})$ and satisfies

(9.3) $$ \begin{align} S-S={\left\{{\ell p^{k}q^{n}:\lvert \ell \rvert<p}\right\}}\subseteq {\left\{{0}\right\}}\cup p^{k}q^{n}\mathbb{Z}_{N}^{\star}; \end{align} $$

therefore, $(B-B)\cap (S-S)={\left \{{0}\right \}}$ and, similarly as before, we deduce that $\overline {B}(X)$ is the mask polynomial of the set $\overline {B}=B\oplus S$ . We observe that $\lvert \overline {A} \rvert =\lvert \overline {B} \rvert $ .

Next, we will show that $\overline {A}(\zeta _{\operatorname {\mathrm {ord}}(\beta -\beta ^{\prime })})=0$ for every $\beta ,\beta ^{\prime }\in \overline {B}$ , $\beta \neq \beta ^{\prime }$ . Every $\beta \in \overline {B}$ has a unique representation as $b+s$ , where $b\in B$ and $s\in S$ ; we write also $\beta ^{\prime }=b^{\prime }+s^{\prime }$ , $b^{\prime }\in B$ , $s^{\prime }\in S$ . By equation (9.3), $s-s^{\prime }\in {\left \{{0}\right \}}\cup p^{k}q^{n}\mathbb {Z}_{N}^{\star }$ , while $v_{p}(b-b^{\prime })\neq k$ by (9.1). If $s=s^{\prime }$ , then $\beta -\beta ^{\prime }=b-b^{\prime }$ and

$$ \begin{align*}\overline{A}(\zeta_{\operatorname{\mathrm{ord}}(b-b^{\prime})})=A(\zeta_{\operatorname{\mathrm{ord}}(b-b^{\prime})})\Phi_{p^{m-k}}(\zeta_{\operatorname{\mathrm{ord}}(b-b^{\prime})}^{q^{n}})=0\end{align*} $$

by Corollary 3.4 since B is a spectrum of A. So, assume $s\neq s^{\prime }$ so that $s-s^{\prime }\in p^{k}q^{n}\mathbb {Z}_{N}^{\star }$ . Since $v_{p}(b-b^{\prime })\neq k$ , we will either have $p^{k+1}\mid b-b^{\prime }$ or $p^{k}\nmid b-b^{\prime }$ . In the former case, $v_{p}(\beta -\beta ^{\prime })=k$ , yielding $\operatorname {\mathrm {ord}}(\beta -\beta ^{\prime })=p^{m-k}q^{r}$ , where $0\leq r\leq n$ . Hence,

$$ \begin{align*}\overline{A}(\zeta_{\operatorname{\mathrm{ord}}(\beta-\beta^{\prime})})=A(\zeta_{\operatorname{\mathrm{ord}}(\beta-\beta^{\prime})})\Phi_{p^{m-k}}(\zeta_{p^{m-k}q^{r}}^{q^{n}})=A(\zeta_{\operatorname{\mathrm{ord}}(\beta-\beta^{\prime})})\Phi_{p^{m-k}}(\zeta_{p^{m-k}}^{q^{n-r}})=0.\end{align*} $$

In the latter case, $\operatorname {\mathrm {ord}}(s-s^{\prime })=p^{m-k}\mid \operatorname {\mathrm {ord}}(b-b^{\prime })$ ; therefore, $\operatorname {\mathrm {ord}}(b-b^{\prime })=\operatorname {\mathrm {ord}}(b-b^{\prime }+s-s^{\prime })=\operatorname {\mathrm {ord}}(\beta -\beta ^{\prime })$ . Hence, again, Corollary 3.4 gives $\overline {A}(\zeta _{\operatorname {\mathrm {ord}}(\beta -\beta ^{\prime })})=0$ . One more application of Corollary 3.4 gives the desired conclusion as $\overline {A}(\zeta _{\operatorname {\mathrm {ord}}(\beta -\beta ^{\prime })})=0$ for every $\beta ,\beta ^{\prime }\in \overline {B}$ , $\beta \neq \beta ^{\prime }$ . The second part of the lemma follows from the fact that $\overline {B}(\zeta _{p^{k+1}})=0$ .

Proof. Induction on $\lvert K_{p}(B) \rvert $ ; the case $\lvert K_{p}(B) \rvert =1$ is Lemma 9.3, and the proof for the induction step is almost identical. For the second part, we proceed by contradiction; suppose that $K_{p}(\overline {B})\neq \varnothing $ . By the second part of Lemma 9.3, we definitely have $K_{p}(B)\cap K_{p}(\overline {B})=\varnothing $ as, for every $k\in K_{p}(B)$ , it holds $\overline {B}(\zeta _{p^{k+1}})=0$ ; hence, each $\overline {B}_{i\ \bmod\ p^{k}}$ is equidistributed $\ \bmod\ p^{k+1}$ (and not absorbed).

Consider next some arbitrary $\ell \notin K_{p}(B)$ , $0\leq \ell \leq m$ . By definition, there is i such that $B_{i\ \bmod\ p^{\ell }}$ is not absorbed $\ \bmod\ p^{\ell +1}$ . Then, as $B\subseteq \overline {B}$ , it is obvious that $\overline {B}_{i\ \bmod\ p^{k}}$ is not absorbed $\ \bmod\ p^{k+1}$ .

Proof. If both A and B are absorption-free, there is nothing to prove as we can just take $S=T={\left \{{0}\right \}}$ . So, suppose first that B is not absorption-free. The process described in Lemma 9.3 and subsequently in Corollary 9.4 ‘kills’ all absorption in the following sense: If $B_{i\ \bmod\ p^{k}}$ is absorbed $\ \bmod\ p^{k+1}$ for all i, then $\overline {B}_{i\ \bmod\ p^{k}}$ is equidistributed $\ \bmod\ p^{k+1}$ for all i since $\overline {B}(\zeta _{p^{k+1}})=0$ , and thus, $\overline {B}$ is absorption-free at $p^{k}$ . Moreover, for every $k\in [0,m-1]$ , there is i such that $\overline {B}_{i\ \bmod\ p^{k}}$ is not absorbed $\ \bmod\ p^{k+1}$ . With the goal to eliminate all absorption from A and B, we successively multiply $B(X)$ by polynomials of the form $\Phi _{p^{k+1}}(X^{q^{n}})$ or $\Phi _{q^{\ell +1}}(X^{p^{m}})$ , whose product is a mask polynomial of a set S, and, respectively, $A(X)$ by polynomials of the form $\Phi _{p^{m-k}}(X^{q^{n}})$ or $\Phi _{q^{n-\ell }}(X^{p^{m}})$ , whose product is a mask polynomial of a set T. We thus obtain a pair of absorption-free sets $(A\oplus S,B\oplus T)$ , which is also spectral, by virtue of Lemma 9.3 or Corollary 9.4.

Proof. Let $(A,B)$ be a spectral pair in $\mathbb {Z}_{N}$ . If A is absorption-free, then it tiles $\mathbb {Z}_{N}$ by hypothesis. If not, there are $S,T\subseteq \mathbb {Z}_{N}$ by Lemma 9.6 such that $(A\oplus T, B\oplus S)$ is an absorption-free spectral pair. By hypothesis, this shows the existence of $R\subseteq \mathbb {Z}_{N}$ such that $(A\oplus T)\oplus R=\mathbb {Z}_{N}$ or, alternatively, $A\oplus (T\oplus R)=\mathbb {Z}_{N}$ , that is, A tiles $\mathbb {Z}_{N}$ .

Proof. Since $A\in \mathscr {F}(N)$ and $N\in \mathscr {S}_{p,q}^{\prime }$ , we obtain that A cannot be expressed as a union of p- or q-cycles exclusively by virtue of Theorem 5.2, Corollary 8.3 and Proposition 8.4. Thus, $m_{p}(A), m_{q}(A)\geq 0$ . Let $x< m_{p}(A)$ and $A(\zeta _{N}^{p^{x}})=0$ . Then $p^{x}\cdot A$ is a union of p- and q-cycles, not exclusively of q-cycles by Theorem 5.2 (second part) because $A(\zeta _{N}^{p^{m_{p}(A)}})\neq 0$ . So, there are $a,a^{\prime }\in A$ such that

$$ \begin{align*}p^{x}(a-a^{\prime})\equiv\frac{N}{p}\ \bmod\ N\end{align*} $$

or, equivalently,

$$ \begin{align*}a-a^{\prime}\equiv\frac{N}{p^{x+1}}\ \bmod\ \frac{N}{p^{x}},\end{align*} $$

which in turn implies $a-a^{\prime }\in \frac {N}{p^{x+1}}\mathbb {Z}_{N}^{\star }$ , whence $B(\zeta _{p^{x+1}})=0$ by Theorem 3.3. This proves one direction of equation (10.4).

Next, suppose that the reverse implication of equation (10.4) does not hold, and let y be the smallest exponent such that $A(\zeta _{N}^{p^{y}})\neq 0=B(\zeta _{p^{y+1}})$ . For every $x\leq y$ and every i, we have the obvious partition

$$ \begin{align*}B_{i\ \bmod\ p^{x}}=\bigcup_{k=1}^{p} B_{i+kp^{x}\ \bmod\ p^{x+1}}.\end{align*} $$

If $A(\zeta _{N}^{p^{x}})=0$ so that $B(\zeta _{p^{x+1}})=0$ , the above partition is an equidistribution $\ \bmod\ p^{x+1}$ , that is,

$$ \begin{align*}\lvert B_{i\ \bmod\ p^{x}} \rvert=p\lvert B_{i+kp^{x}\ \bmod\ p^{x+1}} \rvert\end{align*} $$

for every k. If $A(\zeta _{N}^{p^{x}})\neq 0\neq B(\zeta _{p^{x+1}})$ , we certainly do not have equidistribution for at least one choice of j; if we do not have absorption $\ \bmod\ p^{x+1}$ , i.e., $B_{i\ \bmod\ p^{x}}=B_{i+kp^{x}\ \bmod\ p^{x+1}}$ , then there are $k,k^{\prime }$ with $k\not \equiv k^{\prime }\ \bmod\ p$ such that $B_{i+kp^{x}\ \bmod\ p^{x+1}}$ and $B_{i+k^{\prime }p^{x}\ \bmod\ p^{x+1}}$ are nonempty. Let

$$ \begin{align*}b\in B_{i+kp^{x}\ \bmod\ p^{x+1}}, b^{\prime}\in B_{i+k^{\prime}p^{x}\ \bmod\ p^{x+1}}\end{align*} $$

be arbitrary. It holds $p^{x}\mid \!\mid b-b^{\prime }$ by assumption, and $b-b^{\prime }\notin p^{x}\mathbb {Z}_{N}^{\star }$ due to $A(\zeta _{N}^{p^{x}})\neq 0$ . This shows $q\mid b-b^{\prime }$ ; therefore, all elements of $B_{i+kp^{x}\ \bmod\ p^{x+1}}$ and $B_{i+k^{\prime }p^{x}\ \bmod\ p^{x+1}}$ have the same residue $\ \bmod\ q$ . The same holds for all other elements of $B_{i\ \bmod\ p^{x}}$ , establishing

$$ \begin{align*}B_{i\ \bmod\ p^{x}}\subseteq B_{j\ \bmod\ q}\end{align*} $$

for some j. Now, for each $i\in [0,p^{y}-1]$ , define $x(i)$ to be the smallest exponent x such that $B_{i\ \bmod\ p^{x}}$ is neither absorbed nor equidistributed $\ \bmod\ p^{x+1}$ , and then define $j(i)$ such that

(10.6) $$ \begin{align} B_{i\ \bmod\ p^{x(i)}}\subseteq B_{j(i)\ \bmod\ q}. \end{align} $$

If for every $x\leq y$ , $B_{i\ \bmod\ p^{x}}$ is either absorbed or equidistributed $\ \bmod\ p^{x+1}$ , define $x(i)=y$ ; since $B_{i\ \bmod\ p^{y}}$ is equidistributed $\ \bmod\ p^{y+1}$ and $A(\zeta _{N}^{p^{y}})\neq 0$ , we conclude that there is again some $j(i)$ such that equation (10.6) still holds. On $[0,p^{y}-1]$ , we define an equivalence relation: It holds $i\sim i^{\prime }$ if and only if $i\equiv i^{\prime }\ \bmod\ p^{x(i)}$ (it is not hard to show that this satisfies the properties of an equivalence relation as $i\equiv i^{\prime }\ \bmod\ p^{x(i)}$ implies $x(i)=x(i^{\prime })$ ). Let $I\subseteq [0,p^{y}-1]$ be a complete system of representatives; it holds

$$ \begin{align*}B=\bigsqcup_{i\in I}B_{i\ \bmod\ p^{x(i)}}.\end{align*} $$

The important fact about the above partition is that each $B_{i\ \bmod\ p^{x(i)}}$ has cardinality $\frac {\lvert B \rvert }{p^{k(i)}}$ for some positive integer $k(i)$ . Indeed, as for every $x<x(i)$ , we have either $\lvert B_{i\ \bmod\ p^{x}} \rvert =\lvert B_{i\ \bmod\ p^{x+1}} \rvert $ (absorption) or $\lvert B_{i\ \bmod\ p^{x}} \rvert =p\lvert B_{i\ \bmod\ p^{x+1}} \rvert $ (equidistribution). Then, by equation (10.6), we have

$$ \begin{align*}B_{0\ \bmod\ q}=\bigsqcup_{i\in I, j(i)=0}B_{i\ \bmod\ p^{x(i)}},\end{align*} $$

and taking cardinalities on both sides, we obtain

$$ \begin{align*}\frac{\lvert B \rvert}{q}=\sum_{i\in I, j(i)=0}\frac{\lvert B \rvert}{p^{k(i)}},\end{align*} $$

which then leads to an equation of the form $1/q=s/p^{t}$ , which has no solutions in integers, contradicting the fact that $A(\zeta _{N}^{p^{y}})\neq 0=B(\zeta _{p^{y+1}})$ . This completes the proof.

Proof. We will show first that $A(\zeta _{q^{n}})=A(\zeta _{p^{m}})=0$ . Suppose that it doesn’t hold, say $A(\zeta _{q^{n}})\neq 0$ , so that $m_{p}(A)=m$ . By equation (10.2), we get

$$ \begin{align*}S_{B}^{N}(p)-1 = R_{A}^{N}(p).\end{align*} $$

Using the same argument as in the proof of Lemma 10.3, replacing y by m, we get

$$ \begin{align*}B=\bigsqcup_{i\in I}B_{i\ \bmod\ p^{x(i)}},\end{align*} $$

where each $B_{i\ \bmod\ p^{x(i)}}$ has cardinality $\frac {\lvert B \rvert }{p^{k(i)}}$ and is contained in $B_{j(i)\ \bmod\ q}$ ; this is also true if $x(i)=m$ as $A(\zeta _{N}^{p^{m}})\neq 0$ . We establish a contradiction in the same way.

This shows $m_{p}(A)<m$ and $m_{q}(A)<n$ . Since $B(\zeta _{p^{m_{p}(A)+1}})\neq 0$ by (10.4), the polynomial

$$ \begin{align*}\overline{A}(X)\equiv A(X)\Phi_{p}(X^{p^{m-m_{p}(A)-1}q^{n}})\ \bmod\ (X^{N}-1)\end{align*} $$

is the mask polynomial of a set, say $\overline {A}$ ; this follows from the fact that $(A-A)\cap \frac {N}{p^{m_{p}(A)+1}}\mathbb {Z}_{N}^{\star }=\varnothing $ , which implies that $A-A$ and the difference set whose mask polynomial is $\Phi _{p}(X^{p^{m-m_{p}(A)-1}q^{n}})$ intersect trivially. Consider now the multiset $p^{m_{p}(A)}\cdot A$ . By definition, its mask polynomial does not vanish on $\zeta _{N}$ , while on the other hand, $p^{m_{p}(A)+1}\cdot A$ is a disjoint union of q-cycles by Lemma 10.2. By Lemma 5.6 and Theorem 3.3(i), we obtain $B(\zeta _{p^{m_{p}(A)+1}q})=0$ and, similarly, $B(\zeta _{pq^{m_{q}(A)+1}})=0$ , as desired.

Proof. We first note, that $x=m_{p}(A)+1$ satisfies $B(\zeta _{p^{x}})\neq 0=B(\zeta _{p^{x}q})$ by Lemma 10.3 and Proposition 10.4, therefore $U_{B}^{N}(p)\neq \varnothing $ . By (10.1), we have $B(\zeta _{p})=0$ or, equivalently, $\lvert B_{i\ \bmod\ p} \rvert =\frac {1}{p}\lvert B \rvert $ for all i. At every step, we will estimate

$$ \begin{align*}\min{\left\{{i:v_{p}(\lvert B_{i\ \bmod\ p^{x}} \rvert)}\right\}}\end{align*} $$

or, in other words, how much the power of p drops from $\lvert B \rvert $ down to one of the subsets $\lvert B_{i\ \bmod\ p^{x}} \rvert $ . If $x\in S_{B}^{N}(p)$ , then each $B_{i\ \bmod\ p^{x-1}}$ is equidistributed $\ \bmod\ p^{x}$ , therefore

$$ \begin{align*}v_{p}(\lvert B_{i\ \bmod\ p^{x-1}} \rvert)-1=v_{p}(\lvert B_{i\ \bmod\ p^{x}} \rvert),\end{align*} $$

for all i. If $x\notin S_{B}^{N}(p)$ , we have in general

$$ \begin{align*}v_{p}(\lvert B_{i\ \bmod\ p^{x-1}} \rvert)\geq \min{\left\{{v_{p}(\lvert B_{i+kp^{x-1}\ \bmod\ p^{x}} \rvert):0\leq k\leq p-1}\right\}}.\end{align*} $$

Next, consider the minimum element of $U_{B}^{N}(p)$ , say x, and suppose that $B_{i\ \bmod\ p^{x-1}}$ is not absorbed $\ \bmod\ p^{x}$ ; such an i exists due to hypothesis and equation (9.4) since B is absorption-free. Each nonempty class $B_{i+kp^{x-1}\ \bmod\ p^{x}}$ would then be absorbed $\ \bmod\ p^{x}q$ . Otherwise, $\frac {N}{p^{x}q}B\cap (j+\frac {N}{pq}\mathbb {Z}_{N})$ would neither be supported on a p-cycle nor on a q-cycle for some j. For the same reason, each element in $B_{i+kp^{x-1}\ \bmod\ p^{x}}$ must have the same residue $\ \bmod\ q$ for all k, showing that $B_{i\ \bmod\ p^{x-1}}$ is absorbed $\ \bmod\ p^{x-1}q$ ; furthermore, since $B(\zeta _{p^{x}q})=0$ , $B_{i\ \bmod\ p^{x-1}}$ and $B_{i\ \bmod\ p^{x-1}q}$ are equidistributed $\ \bmod\ p^{x}$ and $\ \bmod\ p^{x}q$ , respectively. Therefore,

$$ \begin{align*}v_{p}(\lvert B_{i\ \bmod\ p^{x-1}} \rvert)-1=v_{p}(\lvert B_{i\ \bmod\ p^{x}} \rvert).\end{align*} $$

Next, let x be the minimum element not belonging to $S_{B}^{N}(p)$ , that is,

$$ \begin{align*}B(\zeta_{p})=\dotsb=B(\zeta_{p^{x-1}})=0\neq B(\zeta_{p^{x}}).\end{align*} $$

Suppose that $B_{i\ \bmod\ p^{x-1}}$ is not absorbed $\ \bmod\ p^{x}$ ; such an i exists due to hypothesis and equation (9.4). Let then $B_{i+kp^{x-1}\ \bmod\ p^{x}}$ and $B_{i+k^{\prime }p^{x-1}\ \bmod\ p^{x}}$ be nonempty with $p\nmid k-k^{\prime }$ . If there are $b\in B_{i+kp^{x-1}\ \bmod\ p^{x}}$ and $b^{\prime }\in B_{i+k^{\prime }p^{x-1}\ \bmod\ p^{x}}$ with $q\nmid b-b^{\prime }$ , then $b-b^{\prime }\in p^{x-1}\mathbb {Z}_{N}^{\star }$ , yielding $A(\zeta _{N}^{p^{x-1}})=0$ by Theorem 3.3(i), and then $B(\zeta _{p^{x}})=0$ by Lemma 10.3, contradicting the definition of x. Thus, we must have $q\mid b-b^{\prime }$ for every such selection of $b,b^{\prime }$ , implying eventually that every element of $B_{i\ \bmod\ p^{x-1}}$ has the same residue $\ \bmod\ q$ .

For every $y\geq x$ , we choose an integer $i_{y}\equiv i\ \bmod\ p^{x-1}$ as follows (we denote $i_{j}=i$ for $1\leq j\leq x-1$ ):

• • If $y\notin S_{B}^{N}(p)\cup U_{B}^{N}(p)$ , choose $i_{y}$ satisfying

  $$ \begin{align*}v_{p}(\lvert B_{i_{y-1}\ \bmod\ p^{y-1}} \rvert)\geq v_{p}(\lvert B_{i_{y}\ \bmod\ p^{y}} \rvert).\end{align*} $$

• • If $y\in S_{B}^{N}(p)\cup U_{B}^{N}(p)$ , then $i_{y}=i_{y-1}$ .

We have $\lvert B_{i\ \bmod\ p^{x-1}} \rvert =\frac {1}{p^{x-1}}\lvert B \rvert $ . If $y\in S_{B}^{N}(p)$ , then $B_{i_{y-1}\ \bmod\ p^{y-1}}$ is equidistributed $\ \bmod\ p^{y}$ , so

(10.7) $$ \begin{align} \lvert B_{i_{y}\ \bmod\ p^{y}} \rvert=\frac{1}{p}\lvert B_{i_{y-1}\ \bmod\ p^{y-1}} \rvert. \end{align} $$

If $y\in U_{B}^{N}(p)$ , we have $B(\zeta _{p^{y}q})=0$ . Therefore, the multiset $\frac {N}{p^{y}q}\cdot B$ is a disjoint union of p- and q-cycles by Theorem 5.2. Every such cycle consists of elements $p^{y}qb$ for $b\in B$ that have the same residue $\ \bmod\ p^{y-1}$ ; indeed, if $\frac {N}{p^{y}q}(b-b^{\prime })\in \frac {N}{p}\mathbb {Z}_{N}$ , then $b-b^{\prime }\equiv kp^{y-1}q\ \bmod\ p^{y}q$ for some k, and if $\frac {N}{p^{y}q}(b-b^{\prime })\in \frac {N}{q}\mathbb {Z}_{N}$ , then $b-b^{\prime }\equiv kp^{y}\ \bmod\ p^{y}q$ for some k, proving our claim. This shows that $\frac {N}{p^{y}q}\cdot B_{i_{y-1}\ \bmod\ p^{y-1}}$ is a disjoint union of p- and q-cycles; however, a q-cycle in $\frac {N}{p^{y}q}\cdot B_{i_{y-1}\ \bmod\ p^{y-1}}\subseteq \frac {N}{p^{y}q}B$ consists of elements having all possible residues $\ \bmod\ q$ , a contradiction, since every element of $B_{i\ \bmod\ p^{x-1}}$ has the same residue $\ \bmod\ q$ . This shows that $\frac {N}{p^{y}q}\cdot B_{i_{y-1}\ \bmod\ p^{y-1}}$ is a union of p-cycles, yielding the equidistribution of $B_{i_{y-1}\ \bmod\ p^{y-1}}$ modulo $p^{y}$ . Thus, (10.7) holds in this case as well.

Summarizing, we obtain the following: If $y\in S_{B}^{N}(p)\cup U_{B}^{N}(p)$ , then (10.7) is valid; otherwise,

$$ \begin{align*}v_{p}(\lvert B_{i_{y-1}\ \bmod\ p^{y-1}} \rvert)\geq v_{p}(\lvert B_{i_{y}\ \bmod\ p^{y}} \rvert)\end{align*} $$

holds. This obviously yields that

$$ \begin{align*}0\leq v_p(\lvert B_{i_m\ \bmod\ p^m} \rvert)\leq v_p(\lvert B \rvert)-\lvert S_B^N(p) \rvert-\lvert U_B^N(p) \rvert,\end{align*} $$

proving the first part.

For the second part of the lemma, we proceed by contradiction. If there is some j and $x\in U_{B}^{N}(p)$ with $x\leq m_{p}(A)+1$ such that $\frac {N}{p^{x}q}B\cap (j+\frac {N}{pq}\mathbb {Z}_{N})$ is supported neither on a p- nor on a q-cycle, then by Proposition 5.5 we get that

$$ \begin{align*}{\left({\frac{N}{p^{x}q}B-\frac{N}{p^{x}q}B}\right)}\cap\frac{N}{pq}\mathbb{Z}_{N}^{\star}\neq\varnothing.\end{align*} $$

This shows the existence of $b,b^{\prime }\in B$ such that

$$ \begin{align*}\frac{N}{p^{x}q}(b-b^{\prime})\in\frac{N}{pq}\mathbb{Z}_{N}^{\star}\end{align*} $$

or, equivalently,

$$ \begin{align*}\operatorname{\mathrm{ord}}(b-b^{\prime})=pq\frac{N}{p^{x}q}=\frac{N}{p^{x-1}}.\end{align*} $$

Hence, by Theorem 3.3(i), we obtain

$$ \begin{align*}A(\zeta_{N/p^{x-1}})=A(\zeta_{N}^{p^{x-1}})=0,\end{align*} $$

contradicting the definition of x. This completes the proof.

Proof. This is obvious by Lemma 10.5 since the sets $U_{B}^{N}(p)$ and $U_{B}^{N}(q)$ are nonempty.

Proof. If B was a union of (say) p-cycles only, then by Proposition 8.4, B satisfies (T1) and (T2). On the other hand, Corollary 10.7 gives us that (wT1) fails for B, a contradiction, as (T1) implies (wT1) .

Proof. Suppose on the contrary that there is such an N with $\mathscr {F}(N)\neq \varnothing $ , $A\in \mathscr {F}_{\max }(N)$ and B a spectrum. Corollary 6.3 implies that $n\geq 2$ . By Theorem 10.9, we obtain $A(\zeta _{q})=A(\zeta _{q^{n}})=0$ , so if $n=2$ we have $q^{n}\mid \lvert A \rvert $ , and so by Proposition 6.2 we get that A tiles, contradiction. If $n=3$ , we have $B(\zeta _{q})=B(\zeta _{q^{3}})=0$ again by Theorem 10.9 so that $\lvert S_{B}^{N}(q) \rvert \geq 2$ . Since $U_{B}^{N}(q)$ is nonempty, by Lemma 10.5 we obtain $q^{3}\mid \lvert B \rvert $ , and then Proposition 6.2 gives us again that A tiles $\mathbb {Z}_{N}$ , a contradiction. Thus, $\mathscr {S}_{p,q}$ cannot have elements of the form $p^{m}q^{n}$ with $n\leq 3$ , proving the desired result.