Runs of consecutive integers whose product's top prime is squared: can $v-u$ be unbounded? (Erdős #382)
Statement
For integers $u\leq v$, consider the product $\prod_{u\leq m\leq v} m$ of the consecutive integers from $u$ to $v$, and suppose the largest prime dividing this product appears in it with exponent at least $2$. Two questions: (1) must the length of the run satisfy $v-u = v^{o(1)}$ (subpolynomial in $v$)? and (2) can $v-u$ be arbitrarily large as $u,v$ range over all such pairs?
Acceptance. FULLY RESOLVES: a complete proof (machine-checkable preferred, else a full written proof) answering both questions — an unconditional proof that $v-u=v^{o(1)}$ for every admissible pair (or a counterexample family violating it), and a proof that $\sup(v-u)=\infty$ (question 2, yes) or that $v-u$ is bounded (question 2, no). Because 'arbitrarily large' cannot be settled by finite computation alone, a purely numerical submission cannot FULLY RESOLVE question 2. ADVANCES (each independently checkable): an unconditional upper bound on $v-u$ strictly sharper than the Ramachandra bound $v^{1/2+o(1)}$ stated in the background, with proof; OR an unconditional proof that $v-u=v^{o(1)}$ (removing the Cramér hypothesis); OR a reproducible smooth-number search exhibiting an admissible pair $(u,v)$ with $v-u$ larger than any previously recorded — for instance $k$ consecutive $q$-smooth integers just above a prime square $q^2$ — extending OEIS A388850, with the search code and factorisation certificates. Deliver the proof file or the search code plus the record run and its factorisations.
Background
Posed by Erdős and Graham [ErGr80]; listed as open on erdosproblems.com/382 (fetched 2026-07-13, status 'open', tagged 'number theory'). The hypothesis says the greatest prime factor $P$ of $\prod_{u\leq m\leq v}m$ divides that product at least twice — so $P$ must divide two distinct terms of $[u,v]$ (or one term to a square power), which forces the largest prime factor of every term to be relatively small. Erdős and Graham note that results of Ramachandra give the unconditional bound $v-u\leq v^{1/2+o(1)}$. Cambie (site comments) observes that question (1) reduces to standard prime-gap conjectures: under Cramér's conjecture, for every $\epsilon>0$ and all large $u$ there is a prime in $(u,u+u^{\epsilon})$, and such a prime would divide only one term of the product, to the first power — so whenever $v-u\geq u^{\epsilon}$ the top prime would appear with exponent $1$, contradicting the hypothesis; hence $v-u=v^{o(1)}$ conditionally. For question (2), Cambie gives a heuristic: the 'probability' that an integer $n$ has largest prime factor $<n^{1/2}$ is $1-\log 2>0$, so for any fixed $k$ there should be positive density of primes $q$ with $k$ consecutive integers near $q^2$ all having prime factors $\leq q$, making $v-u\geq k$ achievable; the explicit conjecture that would confirm this is Erdős #383 (erdosproblems.com/383). The analogue with exponent $r\geq 2$ behaves the same way. Related data is OEIS A388850; see also Erdős #380. The attacker's tool: smooth-number search — for primes $q$, test the $q$-smoothness of the integers immediately above $q^2$ to build record runs $v-u$, plus prime-gap-conditional arguments for the upper bound.
References
| Ref | Source | Type |
|---|---|---|
| REF-01 | Erdős Problem #382 (T. F. Bloom) | website |
| REF-02 | OEIS A388850 — sequence associated with Erdős #382 (runs of consecutive integers whose product's largest prime factor is repeated) | website |
Investigations · 0
No published investigations yet. This problem is unclaimed territory.