gpt pro 5.2 attempt on Math-5821

Erdős Problem 688

1) FORMAL RESTATEMENT.

Fix an integer $n\ge 2$. For a real parameter $\epsilon\ge 0$, let \[ \mathcal P_\epsilon(n):=\{\,p:\ p\text{ prime},\ n^\epsilon<p\le n\,\}. \] We say $\epsilon$ is admissible if there exist residue classes $a_p\pmod p$ for every prime $p\in\mathcal P_\epsilon(n)$ such that every integer $m\in[1,n]$ satisfies at least one congruence $m\equiv a_p\pmod p$ with $p\in\mathcal P_\epsilon(n)$. Define \[ \epsilon_n:=\sup\{\epsilon\ge 0:\ \epsilon\text{ is admissible}\}. \] (The strict inequality $n^\epsilon<p$ means the supremum need not be attained; the problem asks for estimates as $n\to\infty$.)

Question: estimate $\epsilon_n$; in particular, is $\epsilon_n=o(1)$?

2) QUICK LITERATURE/CONTEXT CHECK.

The problem statement records Erdős's lower bound \[ \epsilon_n \gg \frac{\log\log\log n}{\log\log n}. \] No other external results are used or claimed here.

3) ATTACK PLAN.

• Reformulate admissibility in terms of a shift $t$ modulo the product of the primes in $\mathcal P_\epsilon(n)$ (Lemma 688.1).
• Derive simple necessary conditions from counting/coverage constraints (Lemma 688.2).
• Use small-$n$ computations to sanity check the definition and see when admissibility first appears.

4) WORK.

Lemma 688.1 (shift reformulation for a fixed prime set). Fix $n\ge 2$ and a set of primes $S\subseteq\{p:\ p\le n\}$, and let $P_S:=\prod_{p\in S}p$. Choosing residue classes $\{a_p\pmod p\}_{p\in S}$ is equivalent (by CRT) to choosing a residue class $t\pmod{P_S}$ such that $t\equiv -a_p\pmod p$ for all $p\in S$. With this $t$, an integer $m\in[1,n]$ is covered (i.e. $m\equiv a_p\pmod p$ for some $p\in S$) if and only if $m+t$ is divisible by some prime in $S$, equivalently \[ \gcd(m+t,P_S)>1. \] In particular, $\epsilon$ is admissible (for the set $S=\mathcal P_\epsilon(n)$) if and only if there exists $t$ such that \[ \gcd(t+1,P_S)>1,\ \gcd(t+2,P_S)>1,\ \dots,\ \gcd(t+n,P_S)>1. \]

Proof. Identical to Lemma 687.1, but with the prime set $S$ in place of $\{p\le x\}$. $\square$

Lemma 688.2 (simple counting necessary condition). Let $S=\mathcal P_\epsilon(n)$ and suppose $\epsilon$ is admissible, witnessed by residue classes $\{a_p\pmod p\}_{p\in S}$. Then \[ \sum_{p\in S}\left\lceil\frac{n}{p}\right\rceil \ge n. \]

Proof. For each $p\in S$, the chosen residue class $a_p\pmod p$ hits at most $\lceil n/p\rceil$ integers in $[1,n]$, because among any $p$ consecutive integers there is at most one representative of a fixed residue class, and the interval $[1,n]$ has length $n$. Let \[ C_p:=\#\{\,m\in[1,n]:\ m\equiv a_p\pmod p\,\}. \] Then $C_p\le \lceil n/p\rceil$ for each $p$. If every $m\in[1,n]$ is covered by at least one of the residue classes, then counting incidences gives \[ n \le \sum_{p\in S} C_p \le \sum_{p\in S}\left\lceil\frac{n}{p}\right\rceil, \] which is the stated inequality. $\square$

FAST REALITY CHECK (local computation for small $n$). For each $n\le 20$, I tested admissibility for prime sets of the form $S=\{p:\ p_{\min}\le p\le n\}$ (all primes in an upper tail), using the shift formulation (Lemma 688.1) and scanning residues modulo $P_S$ to find the maximum run of consecutive integers with gcd$>1$ with $P_S$. Results:

• For $n=4,6,10$, no choice of residue classes for all primes $p\le n$ covers $[1,n]$ (verified by brute force over all residue choices). Equivalently, even with $S=\{p\le n\}$, the maximum run length is $<n$.
• For $n=3,5,7,8,9,11,\dots,20$, covering is possible, but in all feasible cases found, the smallest prime in $S$ must be $2$ (i.e. excluding $p=2$ makes the maximum run $<n$). For example:

  n=17: S={3,5,7,11,13,17} gives maxrun=12 < 17  (so p_min=3 fails)
  n=17: S={2,3,5,7,11,13,17} gives maxrun=25 >=17 (so p_min=2 works)
  

These computations are only for small $n$ and do not address asymptotics.

5) VERIFICATION.

• Lemma 688.1 is an exact CRT equivalence.
• Lemma 688.2 is a necessary condition only (it ignores overlaps between the classes), so it cannot by itself prove admissibility.
• The computed "no" instances ($n=4,6,10$) were verified by exhaustive search over all residue choices for all primes $\le n$.

FINAL.

UNRESOLVED

(i) Strongest proved partial result. Admissibility for $\epsilon$ is equivalent to the existence of a shift $t$ such that $t+1,\dots,t+n$ are all divisible by at least one prime in $\mathcal P_\epsilon(n)$ (Lemma 688.1). Any admissible $\epsilon$ must satisfy the counting constraint $\sum_{n^\epsilon<p\le n}\lceil n/p\rceil\ge n$ (Lemma 688.2).

(ii) First gap (crisp statement). Prove any asymptotic upper bound on $\epsilon_n$ (e.g. $\epsilon_n=o(1)$) without invoking literature beyond what is stated in the problem file. Equivalently: show that if one excludes all primes $\le n^\epsilon$ for fixed $\epsilon>0$, then for $n$ large, some $m\in[1,n]$ avoids all chosen classes modulo primes $>n^\epsilon$.

(iii) Top 3 next moves.

1. Use Lemma 688.2 together with explicit estimates for $\sum_{p>n^\epsilon}1/p$ (if allowed) to obtain a quantitative upper bound on admissible $\epsilon$.
2. In the shift model, study runs of length $n$ of integers all having a prime factor $>n^\epsilon$ and attempt to upper-bound their length using sieve methods on $y$-rough numbers.
3. Extend computation to moderately larger $n$ (where $P_S$ remains tractable) to guess the growth of $\epsilon_n$ and identify extremal shifts $t$.

(iv) What a minimal counterexample would likely look like. If $\epsilon_n$ does not tend to $0$, there would exist a fixed $\epsilon_0>0$ and infinitely many $n$ with a shift $t$ such that every one of $t+1,\dots,t+n$ has a prime divisor in $(n^{\epsilon_0},n]$; this would force unusually long intervals consisting entirely of integers with a relatively large prime factor.