gpt pro 5.2 attempt on Math-789
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Problem #382
1) FORMAL RESTATEMENT
Let $1\le u\le v$ be integers, and set \[ N(u,v):=\prod_{m=u}^{v} m,\qquad Q:=P(N(u,v)). \] Assume the exponent of $Q$ in $N(u,v)$ is at least $2$, i.e. $\nu_Q(N(u,v))\ge 2$ (equivalently, $[u,v]$ is bad in the sense of Problem #380).
Question 382(a). Must one have \[ v-u \;=\; v^{o(1)}\quad\text{as }v\to\infty\text{ along such bad intervals?} \]
Question 382(b). Can $v-u$ be arbitrarily large (i.e. do bad intervals exist with unbounded length)?
2) QUICK LITERATURE/CONTEXT CHECK
The Erdős Problems forum thread for #382 records (from Erdős--Graham, 1980) that a result of Ramachandra implies $v-u\le v^{1/2+o(1)}$, and that Cramér's conjecture on prime gaps would imply $v-u=v^{o(1)}$.(https://www.erdosproblems.com/forum/thread/382 (accessed 2026-01-17).) The same thread notes an example with $v-u=13$ (for a very large $u$), though this is presented informally.(Same source as above; see the comment by Stijn Cambie.)
A general unconditional input one can combine with the "no primes in bad intervals" lemma is the existence of primes in short intervals. For example, Baker--Harman--Pintz proved a weak form of the prime number theorem in intervals $[x-x^{0.525},x]$ for large $x$.(https://arxiv.org/abs/1707.05437 (Alweiss--Luo, 2017, abstract cites Baker--Harman--Pintz).)
3) ATTACK PLAN
- Use the structural lemma from Problem #380 to obtain unconditional restrictions (prime-free $\Rightarrow$ $v<2u$).
- Combine prime-free-ness with known bounds for primes in short intervals to get an unconditional upper bound $v-u\ll v^{0.525}$ (still far from $v^{o(1)}$).
- Search for explicit families/examples of bad intervals to gauge how large $v-u$ can get, in particular by looking near squares $q^2$ of a prime $q$ (since $q^2$ supplies $\nu_q\ge 2$).
4) WORK
Unconditional prime-free restriction.
From Problem #380 we have:A further unconditional upper bound via primes in short intervals.
Assume a theorem of the following kind: for all sufficiently large $x$, the interval $[x-x^{\theta},x]$ contains a prime, where $\theta=0.525$ is admissible by Baker--Harman--Pintz (as cited in Alweiss--Luo).(https://arxiv.org/abs/1707.05437) Then any prime-free interval $[u,v]$ with $v$ large must satisfy \[ v-u < v^{\theta}, \] because otherwise $[v-v^{\theta},v]\subseteq [u,v]$ would contain a prime. Consequently, for all sufficiently large bad intervals $[u,v]$, \[ v-u < v^{0.525}. \] This improves the trivial bound $v-u < v$ but is still much weaker than the desired $v^{o(1)}$.Explicit examples with growing $v-u$.
One natural way to build bad intervals is to choose a prime $q$ and look for short runs of consecutive $q$-smooth integers (all prime factors $\le q$) that contain $q^2$. If $[u,v]$ is such a run containing $q^2$, then $P(N(u,v))=q$ and $\nu_q(N(u,v))\ge 2$ (because $q^2\mid N(u,v)$), hence $[u,v]$ is bad.By direct computation one finds, for example: \[ [1680,1683]\ \text{ is bad with }P(N)=41\text{ and }\nu_{41}(N)=2, \] so $v-u=3$. Further examples with larger length include \[ [76725,76729]\ \text{ bad with maximal prime }277,\quad (v-u=4), \] and \[ [332925,332930]\ \text{ bad with maximal prime }577,\quad (v-u=5). \] These examples show unconditionally that lengths $v-u$ can exceed $3$ and at least reach $5$ for moderately sized $v$.
5) VERIFICATION
- The deduction "bad $\Rightarrow$ prime-free $\Rightarrow v<2u$" is fully proved in Problem #380.
- The implication "prime in $[x-x^{0.525},x]$ for all large $x$ $\Rightarrow v-u<v^{0.525}$ for prime-free $[u,v]$" is a straightforward containment argument, conditional only on the short-interval prime theorem.
- The explicit examples above were verified by exact integer factorisations: in each interval, every prime factor of every term is $\le q$ and the term $q^2$ lies inside, forcing $\nu_q\ge 2$.
6) FINAL
UNRESOLVED.
- (i) Strongest partial result obtained. Unconditionally, any bad interval contains no primes and thus satisfies $v<2u$. Combining with known results on primes in short intervals (Baker--Harman--Pintz with exponent $0.525$) yields the unconditional bound $v-u < v^{0.525}$ for all sufficiently large bad intervals. We also exhibited explicit bad intervals with $v-u=5$.
- (ii) First hard gap. To upgrade $v-u < v^{0.525}$ to $v-u=v^{o(1)}$ would essentially require prime-free intervals to be much shorter than any fixed power of $v$, i.e. deep improvements in prime gap technology (or a different argument not mediated by primes).
- (iii) Top 3 next moves.
- Try to prove that in a bad interval the maximal prime must arise from a square $q^2$ inside the interval for "most" such intervals; then reduce to understanding runs of $q$-smooth integers near $q^2$.
- Improve the unconditional exponent $0.525$ by importing the best currently known prime-in-short-interval exponents (or conditional exponents under RH/EH/Cramér) and translate directly to $v-u$ bounds.
- For the "arbitrarily large" question: search systematically for long $q$-smooth runs containing $q^2$; develop heuristic predictions for their maximal length as $q\to\infty$ and attempt to turn them into rigorous existence via sieve bounds for smooth values.
- (iv) Why it might be true. If primes appear "often enough" in intervals $[x,x+x^{\varepsilon}]$ for every fixed $\varepsilon>0$, then prime-free-ness forces $v-u=x^{o(1)}$. Since bad intervals are prime-free, this would yield the desired $v^{o(1)}$ bound.