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Source. R. J. Simpson, Regular coverings of the integers by arithmetic progressions, Acta Arithmetica 45 (1985), 145--152, published version, Theorem 1 on printed p. 147, proved on pp. 147--148 (canonical PDF, pp. 2--3).

Conventions

For $a\in\mathbb Z$ and a positive integer $d$, write

$$C(a,d)=a+d\mathbb Z.$$

This is Simpson's progression $\langle a,d\rangle$. A regular covering is a finite family $\mathcal A$ of such progressions whose union is $\mathbb Z$, with no proper subfamily still covering $\mathbb Z$. Thus regular means irredundant; the members need not be disjoint and their moduli need not be distinct. The definitions are on printed p. 145.

For a prime $p\mid d$, let $v_p(d)$ be the unique positive integer $\alpha$ such that $p^\alpha\mid d$ and $p^{\alpha+1}\nmid d$.

Statement

Let $\mathcal A$ be a regular covering. Fix a member $C(a,d)\in\mathcal A$ and a prime $p\mid d$, and put $\alpha=v_p(d)$. There are members

$$C\bigl(a_i^{(k)},d_i^{(k)}\bigr)\in\mathcal A, \qquad 1\le k\le\alpha,\quad 1\le i\le p-1,$$

such that

$$p^k\mid d_i^{(k)},\qquad a_i^{(k)}\equiv a\pmod{p^{k-1}},\qquad \frac{a_i^{(k)}-a}{p^{k-1}}\equiv i\pmod p.$$

The $\alpha(p-1)$ selected progressions are mutually disjoint, and none of them meets $C(a,d)$.

The selection is simultaneous across all depths $k$ for the fixed member $C(a,d)$ and fixed prime $p$. The divisibility condition permits $v_p(d_i^{(k)})>k$; it does not assert equality. There is no oddness or distinct-modulus hypothesis. The theorem does not assert that selections made for different fixed members or different primes form one disjoint family.

Proof route and dependencies

The source proves the theorem on printed pp. 147--148 using its Lemmas 1--3 on pp. 146--147. The following is a route sketch.

For each depth $k$, take a minimal subfamily covering $C(a,p^{k-1})$. Regularity of $\mathcal A$ forces that subfamily to contain $C(a,d)$. Lemma 3 reduces this subfamily, via $C(a,p^{k-1})$, to a regular covering of $\mathbb Z$. The distinguished member reduces to $C(0,d/p^{k-1})$, whose modulus is divisible by $p$. Lemma 2 then supplies reduced members with moduli divisible by $p$ and representatives in each nonzero residue class modulo $p$. Lifting them through Lemma 3 gives the displayed divisibility and residue conditions.

Lemma 1 supplies the intersection criterion

$$C(b,m)\cap C(c,n)\ne\varnothing \quad\Longleftrightarrow\quad b\equiv c\pmod{\gcd(m,n)}.$$

Applied to two selected members, it makes an intersection at different depths incompatible with the nonzero residue condition, and an intersection at the same depth incompatible with different indices $i$. The same criterion excludes intersection with $C(a,d)$. Lemma 1 is derived from the classical Chinese remainder theorem; Lemmas 2 and 3 are internal source steps.

Reading and proof coverage

The definitions on printed p. 145, Lemmas 1--3 and their proofs on pp. 146--147, and Theorem 1 and its proof on pp. 147--148 were visually read against the canonical PDF. Its text layer is empty. The adjacent Corollary 1 on pp. 148--149 and Theorem 2's statement and opening proof on p. 149 were read as context; the remainder of the paper was not inspected for this page.

This page records the exact statement and a proof-route sketch. It is not a complete proof reconstruction or independently accepted compilation proof coverage. The paper's general cardinality bound belongs to Theorem 2 and its corollary, rather than to this extracted statement alone.

Bears on

• Problem 1189: every covering realization of an irreducible covering set of distinct moduli is regular, since deleting a progression while preserving coverage would give a covering proper subset of the moduli. The theorem therefore constrains each such realization.