Divisibility selections and simultaneous representatives

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These are elementary formulation checks for Problem 708, Problem 709 and Problem 711. They are author-recorded; the cited external bounds retain the standing on their claim pages.

Signs in the interval-product problem

Let $A$ be a nonempty set of positive integers, $n=|A|$, $M=\max A$, and $Q=\prod_{a\in A}a$. Suppose every interval of $M$ consecutive positive integers has a subset of at most $h(A)\ge1$ members whose product is divisible by $Q$. The same assertion holds for intervals of arbitrary integers.

An interval containing zero has the witness $\{0\}$, since $Q\mid0$. An interval containing only negative integers becomes a positive interval after negation. Reflect a positive witness back: the product changes only by the factor $(-1)^{|B|}$, which does not affect divisibility. An interval with both signs contains zero. The reverse implication is immediate by restriction.

Thus any positive-interval theorem with an at-most bound transfers unchanged, including all its restrictions on $n$ and $M$. In particular, the claimed bounds on [[problems/integer_sequences/E0708/claims/2026_09_05_chen|Chen's claim page]] have no additional sign gap. This observation does not prove their external premise or the desired $2n$ bound in the unrestricted range. It also does not fix a formulation requiring exactly $2n$ distinct choices when the interval has fewer than $2n$ elements.

What a finite residue test would establish

For fixed $A$, write $I_s=\{s+1,\ldots,s+M\}$. A subset of positions $J\subseteq\{1,\ldots,M\}$ works precisely when

$$|J|\le2n,\qquad Q\mid\prod_{j\in J}(s+j).$$

This predicate depends only on $s\pmod Q$. Therefore checking all $Q$ start residues and all subsets of at most $2n$ positions is a complete finite test for this fixed $A$. The modulus $Q$ is sufficient without any claim that it is minimal. Replacing it by $\operatorname{lcm}(A)$ without further proof would discard the multiplicities in the required product.

Equivalently, for every prime $p\mid Q$, put $S_p=\sum_{a\in A}v_p(a)$. A nonzero product works exactly when $\sum_{j\in J}v_p(s+j)\ge S_p$ for all such $p$. A zero factor is handled directly as a successful divisibility witness. Separate choices for separate primes do not suffice: the same set $J$ must satisfy every inequality.

The large-maximum $2n$ result recorded on Chen's claim page applies for $M\ge8n^3$. Conditional on that result, each fixed $n$ would leave finitely many $A\subseteq\{2,\ldots,8n^3-1\}$ with $|A|=n$, and finitely many start residues for each. The value of $n$ remains unbounded. This is not a finite complement for Problem 708 and supplies no search cutoff. A finite-verification route should stop until it has a uniform reduction bounding $n$, or a theorem covering the remaining maxima for every $n$.

Matching is a simultaneous-choice condition

For a finite divisor set $A$ and interval $I$, form the bipartite graph with left vertices $A$, right vertices $I$, and an edge $a\sim b$ when $a\mid b$. Distinct representatives require a matching saturating $A$. By Hall's criterion this is equivalent to $|\bigcup_{a\in S}\{b\in I:a\mid b\}|\ge|S|$ for every $S\subseteq A$. The assertion that every single divisor has a multiple checks only singleton sets $S$.

The initial-segment problem transfers to the arbitrary-set problem in one direction. Set $A=\{2,\ldots,n+1\}$. In any interval with at least $n+1$ members, distinct representatives for $A$ leave a spare integer, which serves as a representative for $1$. Hence a counterexample interval for $\{1,\ldots,n+1\}$ of that length is also a counterexample for $A$. An upper bound for initial segments does not reverse this implication and does not handle every set of $n$ divisors.

The counterexample intervals in [[../library/integer_sequences/doorn_2026_length_interval_distinct_multiples/_index|van Doorn's paper]] give a lower bound for Problem 709 by this transfer. The paper also proves the divergence of the difference asked for in the second part of Problem 711. Neither consequence removes the gap between the claimed $n^{4/3+o(1)}$ upper bound and the requested $n^{1+o(1)}$ estimate in its first part. The useful test of a proposed imported theorem is its Hall condition for arbitrary $A$, not merely its specialization to an initial segment. No new matching search is needed to expose this missing implication.