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Source. Pipeline-math, Erdős problem 477, commit 99d916ff32a90e77c98eb004537ccda409262346 (29 June 2026), Lemma 1.7, printed/PDF pp. 5-6 of the manuscript.

Statement

Take any $B\subseteq\mathbb Z$, with difference set $D=B-B$. Assume that each finite set $C$ of integers disjoint from $B$ has some $b\in B$ for which the translate $C-b$ misses $D$:

$$(C-b)\cap D=\varnothing. \tag{1}$$

Then for some $A\subseteq\mathbb Z$, each integer $n$ equals $a+b$ for exactly one pair $(a,b)\in A\times B$.

The hypothesis includes $C=\varnothing$, so it implies $B\ne\varnothing$. No sparseness, symmetry, or polynomial description of $B$ is assumed.

Proof

List the integers as $n_1,n_2,\ldots$ so that each one appears, for example in the order $0,1,-1,2,-2,\ldots$. By induction on $j$ we build finite sets $A_0\subseteq A_1\subseteq\cdots$ with two properties: no two of the sets $a+B$ with $a\in A_j$ meet, and together these sets contain $n_1,\ldots,n_j$.

The empty set serves as $A_0$. Given $A_{j-1}$, if $n_j$ already lies in $A_{j-1}+B$, keep $A_j=A_{j-1}$; both properties persist. Otherwise $n_j-a\notin B$ for every $a\in A_{j-1}$, so

$$C_j=\{n_j-a:a\in A_{j-1}\}$$

is a finite subset of $\mathbb Z\setminus B$. Apply (1) to obtain $b_j\in B$, and set

$$a_j=n_j-b_j,\qquad A_j=A_{j-1}\cup\{a_j\}.$$

The set $A_j$ is finite and contains $A_{j-1}$. Also $n_j=a_j+b_j\in a_j+B$, so all required integers are covered. The new element $a_j$ is not in $A_{j-1}$, since otherwise this equality would contradict that $n_j$ was uncovered.

For an old element $a\in A_{j-1}$, the avoidance condition gives

$$a_j-a=(n_j-a)-b_j\notin D.$$

If $(a_j+B)\cap(a+B)$ contained a point, there would be $b',b''\in B$ with $a_j+b'=a+b''$, forcing $a_j-a=b''-b'\in D$. This contradiction shows that the new translate is disjoint from every old one. Old translates remain pairwise disjoint by induction. Both properties are therefore maintained at every stage. If desired, select each available $b_j$ as the first in the fixed integer enumeration; no effectiveness assertion is needed for this existence construction.

Now put

$$A=\bigcup_{j\ge0}A_j.$$

Each integer appears as some $n_j$ and is covered by that stage, hence by $A+B$. If $a,a'\in A$ are distinct, each belongs to a finite stage; both belong to the later of those stages. Their translates are disjoint there, so they are disjoint in the final family as well.

Every integer thus belongs to exactly one translate $a+B$. Once $a$ is fixed, its summand in $B$ must be $b=n-a$, so the pair $(a,b)$ is unique. This proves the criterion.

Dependencies and current verification

This proof uses only the stated finite-avoidance hypothesis and the enumerability of $\mathbb Z$. It has no external theorem premise and does not use any earlier numbered result in the manuscript.

This complete reconstruction received [[diophantine_problems/pipeline_math_2026_tiling_complement/evidence/verify/compilation_review|independent compilation review]]. No material defect was found in its exact frozen statement or essential deductions. Both covering and pairwise disjointness are proved at finite stages and at the union. The source's pp. 5-6 were read in text and rendered images. Attack selection was partly pre-directed; the derivations were independently performed. The six-result review is relative to the Corvaja-Zannier-recalled unit bounds and Heath-Brown's journal Theorem 2, with the recorded nonconstant-family qualification. The external proofs were not independently reviewed; no formal verification is claimed. This lemma itself uses neither external premise. See the [[diophantine_problems/pipeline_math_2026_tiling_complement/_index|source digest]].

Bears on. Applied with the hypothesis proved in Proposition 1.8, the criterion yields Theorem 1.1 and answers Problem 477.