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Source. The theorem on printed p. 74 and the cyclic-group corollary on p. 79 (PDF pp. 2 and 4). This is a complete rewritten deduction. The full residue-class and prime-adic correspondence is supplied in Part I, instead of leaving it as the source's “exactly as in [1]” reference.

Statement

Let $\{a_j\pmod{m_j}:1\le j\le k\}$ be a finite cover of the integers with distinct odd moduli $m_j>1$. Write

$$N=\operatorname{lcm}(m_1,\ldots,m_k)=\prod_{i=1}^n p_i^{s_i}, \qquad s_i\ge1,$$

where the $p_i$ are distinct odd primes. Part I implies $n\ge5$. For any labeling of these primes define

$$\bar w=\frac{p_1^{s_1}-1}{(p_1-2)p_1^{s_1}+1},\qquad \bar z_1=\frac{p_1^{s_1-1}-1}{(p_1-2)p_1^{s_1}+1},$$ $$\bar z_i=\frac{p_i^{s_i}-1}{(p_i-3)p_i^{s_i}+2} \qquad(2\le i\le n).$$

With the polynomial

$$\begin{aligned} g(w,z)={}&(1+w)\prod_{i=2}^n(1+z_i)-w -(1+w-z_1)\sum_{i=2}^n z_i\\ &-z_1(z_3z_4z_5+2z_2z_4z_5+3z_2z_3z_5+3z_2z_3z_4), \end{aligned} \tag{1}$$

the necessary condition is $g(\bar w,\bar z)\ge2$.

Equivalently, let $C$ be a finite cyclic group of odd order $\prod_i p_i^{s_i}$ with $n\ge5$. If proper cosets cover $C$ and $g(\bar w,\bar z)<2$, then two cosets in the cover have the same cardinality. For $1\le n<5$, the repeated-cardinality conclusion holds already by Part I, without evaluating (1).

Proof

Choose a generator of $C$. It identifies $C$ with $\mathbb Z/N\mathbb Z$. The prime-adic correspondence maps every coset of index $m\mid N$ to a box in $\prod_i\{0,\ldots,p_i^{s_i}-1\}$ with cardinality $N/m$. It is a bijection of the underlying point sets, preserves unions, and maps proper cosets to proper boxes. If all coset cardinalities were distinct, the geometric proposition would give $g(\bar w,\bar z)\ge2$. This proves the group assertion.

For a covering system of integers, every $m_j$ divides $N$. Its classes cover the integers if and only if their images cover $\mathbb Z/N\mathbb Z$: membership is periodic modulo $N$. These images have sizes $N/m_j$, which are distinct because the $m_j$ are distinct. Their properness follows from $m_j>1$. Apply the group result. The bound $n\ge5$ needed to write (1) comes from Part I's prime-factor corollary.

Scope

The formula is a necessary condition, not a construction or a sufficient condition. In particular $s_1=1$ is allowed: then $\bar z_1=0$, and (1) remains a well-defined polynomial. The six-prime corollary uses the limit of these parameters to exclude $n=5$ as well.

Bears on. Problem 7, through historical necessary conditions on a hypothetical distinct odd cover.