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Source. Zhi-Wei Sun, Covering the integers by arithmetic sequences II, Trans. Amer. Math. Soc. 348 (1996), no. 11, 4279–4320, DOI. Theorem 1 is on pp. 10–11 of the 48-page author copy described on the source card, which does not carry the journal pagination; its proof is on p. 11.

Conventions

A system $A=\{a_s+n_s\mathbb Z\}_{s=1}^k$ has $a_s\in\mathbb Z$ and $n_s\in\mathbb Z^+$. For $m\in\mathbb Z^+$ and $S\subseteq\mathbb Z$, $A$ is an $m$-cover of $S$ when every $x\in S$ lies in at least $m$ of the sequences $a_s+n_s\mathbb Z$, and an exact $m$-cover when every $x\in S$ lies in exactly $m$ of them (p. 2). From Section 2 on, $m$ and $n$ are positive integers (p. 7). Write $(x,y)$ for the greatest common divisor and $\{x\}$ for the fractional part of a real $x$.

Statement

Let $a\in\mathbb Z$, and let $m_1,\dots,m_k$ be positive integers with

$$(m_s,n_s)=(n,n_s)\qquad(s=1,\dots,k).$$

(i) Put $J=\{1\le s\le k:\ (n,n_s)\mid a_s-a\}$ and

$$N=\Bigl|\Bigl\{\Bigl\{\sum_{s\in I}\frac{m_s}{n_s}\Bigr\}:\ I\subseteq J\Bigr\}\Bigr|.$$

If $A$ covers each of some $N$ consecutive integers congruent to $a$ modulo $n$ at least $m$ times, then $A$ is an $m$-cover of $a+n\mathbb Z$.

(ii) Let $1\le t\le k$ and let $d$ be a divisor of $n$. If $A$ is an $m$-cover of $a+d\mathbb Z$ but $A_t=\{a_s+n_s\mathbb Z\}_{s\ne t}$ is not, then

$$\Bigl|\Bigl\{\Bigl\{\sum_{s\in I}\frac{m_s}{n_s}\Bigr\}:\ I\subseteq\{1,\dots,k\}\setminus\{t\}\Bigr\}\Bigr| \ \ge\ \Bigl|\Bigl\{\Bigl\{\sum_{s\in I}\frac{m_s}{n_s}\Bigr\}:\ I\subseteq\{1\le s\le k:\ s\ne t,\ (d,n_s)\mid a_s-a\}\Bigr\}\Bigr| \ \ge\ \frac{n_t}{(n,n_t)}.$$

Special case

Take $n=1$ and $m_s=1$ for every $s$. The hypothesis $(m_s,n_s)=(n,n_s)$ holds, $J=\{1,\dots,k\}$, and $N$ is the number of distinct fractional parts of the sums $\sum_{s\in I}1/n_s$, so $N\le 2^k$. Part (i) then says: if $A$ covers each of some $N$ consecutive integers at least $m$ times, it is an $m$-cover of $\mathbb Z$. In particular, a system of $k$ sequences that covers $2^k$ consecutive integers at least $m$ times is an $m$-cover of $\mathbb Z$. For integer moduli this special case is the paper's Lemma 3 (p. 10), which the paper attributes to its predecessor Sun (1995), Lemma 3 itself is stated for real $\alpha_s$ and positive real $\beta_s$, and the paper presents it as stronger than the theorem of Crittenden and Vanden Eynden.

Proof route and dependencies

The paper proves (i) on p. 11 by translating, through its Lemma 1 (p. 8), the part of $A$ meeting $a+n\mathbb Z$ into a system of sequences with moduli $n_j/m_j$ and applying Lemma 3. Part (ii) splits $a+d\mathbb Z$ into the sequences $a+rd+n\mathbb Z$, picks one on which $A_t$ fails to be an $m$-cover, and applies (i) to $n_t/(n,n_t)-1$ consecutive terms of it that avoid $a_t+n_t\mathbb Z$. The proof rests on Lemma 3, which this paper quotes from Sun (1995) without proof. No proof is reconstructed here.

The paper's Corollary 4 (p. 12), Corollary 5 (pp. 12–13) and the remark on p. 14 apply the theorem; part (iii) of Theorem I uses part (ii) (p. 25).

Bears on

• Problem 275: the special case above with $m=1$ is the problem's statement, that $k$ congruences covering $2^k$ consecutive integers cover every integer. This derivation is the corpus's; the paper credits the $2^k$ form to Crittenden and Vanden Eynden and presents Lemma 3, quoted from Sun (1995), as a stronger result (p. 10).