Erdős Problem #382

Problem #382 (Round 2)

1) ROUND-2 OBJECTIVE

Path C (obstruction / structural reduction). The Round-1 route gave only the conditional-style philosophy "short prime gaps force short bad intervals" and an unconditional bound of order $v^{0.525}$. The new goal is to remove that dependence entirely and force a square-divisibility anchor inside every bad interval, obtaining a sharper unconditional upper bound and a cleaner description of the intervals themselves.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts.

1. Every bad interval contains no primes.
2. Therefore every bad interval satisfies $v<2u$.
3. Round 1 recorded explicit examples with $v-u=3,4,5$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is the anchor-smooth reduction from Problem #380. Applied here, it says:

1. if $[u,v]$ is bad and $L=v-u$, then $Q:=P\!\left(\prod_{m=u}^{v}m\right)$ satisfies $Q>L$;
2. the interval contains a unique multiple of $Q$, and that multiple is divisible by $Q^2$;
3. every term in the interval is $Q$-smooth.

This immediately turns a bad interval into a short run of consecutive $Q$-smooth numbers with a built-in square anchor.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was that the $v^{0.525}$ bound still came from general theorems on primes in short intervals. The new plan is:

1. use the Round-2 structure theorem instead of prime-gap technology;
2. deduce a direct unconditional inequality of square-root type;
3. verify a much longer explicit example than the Round-1 examples.

5) WORK (ROUND-2)

Theorem 1 (square-root bound).

Let $[u,v]$ be bad and put \[ L=v-u,\qquad Q=P\!\left(\prod_{m=u}^{v}m\right). \] Then \[ L<Q\le \sqrt{v}. \] In particular, \[ v-u<\sqrt{v}. \]

Proof.

By the anchor-smooth reduction from Problem #380, we already know that $Q>L$ and that there is a unique $t\in [u,v]$ with $Q\mid t$, necessarily satisfying $Q^2\mid t$. Since $t\le v$, we get \[ Q^2\le t\le v, \] hence $Q\le \sqrt{v}$. Combining this with $L<Q$ gives \[ v-u=L<\sqrt{v}. \] This proves the theorem.

Remark (optional deeper strengthening).

A survey of Shorey and Tijdeman records that if $n,n+1,\dots,n+k-1$ are all composite and $n>100$, then the greatest prime factor of their product exceeds $4.42k$. Since every bad interval is prime-free, this gives for all sufficiently large bad intervals \[ Q>4.42(L+1). \] Together with $Q^2\mid t\le v$, one gets \[ L<\frac{\sqrt{v}}{4.42}. \] I do not need this stronger constant below; the elementary square-root bound already strictly improves Round 1.

Structural reformulation.

Every bad interval is now seen to be a block of consecutive $Q$-smooth integers of length $L+1<Q+1$ containing a $Q^2$-divisible term. Therefore the second question in Problem #382 can be rephrased as follows:

Can one find arbitrarily long runs of consecutive $Q$-smooth integers, of length still $<Q$, containing a term divisible by $Q^2$, with $Q$ prime?

This is more precise than the Round-1 heuristic description.

A verified example with $v-u=13$.

The Round-1 notes mentioned an informal forum claim of a length-$13$ example. I verified it exactly.

Let \[ q=4237033,\qquad q^2=17952448643089. \] Consider the interval \[ [q^2-11,\ q^2+2] = [17952448643078,\ 17952448643091]. \] Direct exact factorisation gives:

1. $q^2-11=2\cdot 89\cdot 30253\cdot 3333767$;
2. $q^2-10=3\cdot 43\cdot 153719\cdot 905329$;
3. $q^2-9=2^3\cdot 5\cdot 7^2\cdot 8647\cdot 1059259$;
4. $q^2-8=31\cdot 193\cdot 239\cdot 2129\cdot 5897$;
5. $q^2-7=2\cdot 3^2\cdot 47\cdot 103\cdot 109\cdot 1890121$;
6. $q^2-6=19\cdot 431\cdot 18287\cdot 119881$;
7. $q^2-5=2^2\cdot 331\cdot 80209\cdot 169049$;
8. $q^2-4=3\cdot 5\cdot 11\cdot 25679\cdot 4237031$;
9. $q^2-3=2\cdot 37\cdot 877\cdot 2063\cdot 134089$;
10. $q^2-2=7\cdot 937\cdot 5711\cdot 479263$;
11. $q^2-1=2^4\cdot 3\cdot 53\cdot 3331\cdot 2118517$;
12. $q^2=q^2$;
13. $q^2+1=2\cdot 5\cdot 13\cdot 17\cdot 29\cdot 113\cdot 2478877$;
14. $q^2+2=3^2\cdot 41\cdot 67\cdot 739\cdot 982603$.

Every prime shown above is at most $q$, and the middle term contributes $q^2$. Hence the largest prime factor of \[ \prod_{m=q^2-11}^{q^2+2} m \] is $q$, with exponent at least $2$. Therefore \[ [17952448643078,\ 17952448643091] \] is a bad interval, and \[ v-u=13. \]

By-product from Problem #383.

The new explicit primes found in the Round-2 work for Problem #383 give the following additional bad intervals: \[ [510569^2,\ 510569^2+7],\qquad [2869123^2,\ 2869123^2+8],\qquad [3092701^2,\ 3092701^2+9]. \] So there are rigorously verified bad intervals of lengths $7,8,9$, in addition to the verified length-$13$ example above.

6) ADVERSARIAL VERIFICATION

I checked the following possible issues.

1. Does $Q>L$ really follow for every non-singleton bad interval? Yes: this is exactly the new Sylvester--Schur step from Problem #380.
2. Could the $Q$-multiple occur twice anyway? No, because two distinct multiples of $Q$ differ by at least $Q>L$.
3. Did I overstate the structure by saying the interval contains $Q^2$ itself? No; the theorem only guarantees a $Q^2$-divisible term. The verified length-$13$ example happens to be centred at $q^2$, but this need not happen in general.
4. Is the length-$13$ example merely borrowed from the forum? No. The forum comment suggested it; the displayed exact factorisations above verify it independently.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The new Round-2 gains are:

1. a direct unconditional bound \[ v-u<\sqrt{v}, \] which is sharper than the Round-1 $v^{0.525}$ route;
2. a rigorous reformulation of every bad interval as a short run of consecutive $Q$-smooth integers with a $Q^2$-divisible anchor;
3. a fully verified explicit example with $v-u=13$;
4. further verified examples of lengths $7,8,9$ coming from the new Problem #383 computations.

The two original questions in Problem #382 remain open.

8) COMPLETION ESTIMATE

COMPLETION: 60%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #382, accessed 2026-03-08, https://www.erdosproblems.com/382.
2. T. F. Bloom, forum thread for Problem #382, accessed 2026-03-08, https://www.erdosproblems.com/forum/thread/382. The length-$13$ interval was mentioned there and independently verified above.
3. T. N. Shorey and R. Tijdeman, Arithmetic properties of blocks of consecutive integers, especially Section 3.1 for the Sylvester theorem discussion and the quoted Nair--Shorey bound for composite blocks.

OBJECTIVE AND SOURCE REVIEW

Erdős #382; independent attempt, 2026-09-23. If the largest prime factor $Q$ of an interval product $\prod_{m=u}^v m$ occurs with multiplicity at least two, must $v-u=v^{o(1)}$, and can these lengths be unbounded? I read all of 382.tex, including its smooth-number anchor and factorization example. I examine higher multiplicity and the exact logical scope of that reduction.

WORK: MULTIPLICITY-SENSITIVE ANCHOR BOUND

Put $L=v-u$. Whenever $Q>L$, the interval has a unique multiple $t$ of $Q$: two such multiples would differ by at least $Q$. If the total exponent of $Q$ in the interval product is $r$, this unique term satisfies $Q^r\mid t$. Hence \[ L<Q\le v^{1/r}. \] Thus a block with exponent at least three satisfies a cube-root bound, and a sequence of such blocks with $r\to\infty$ satisfies $L=v^{o(1)}$. The last statement follows by fixing $\varepsilon>0$ and eventually having $1/r<\varepsilon$. The supplied anchor theorem obtains $Q>L$ by a Sylvester--Schur step. The displayed argument above is explicitly conditional on that inequality; it verifies the subsequent valuation step without treating the imported theorem as an in-file proof. In particular it shows that, within that framework, the hardest unresolved case has bounded multiplicity, especially exponent two.

ADVERSARIAL VERIFICATION AND REMAINING GAP

Total exponent need not belong to one term unless $Q>L$ is first proved. The anchor is a multiple of $Q^r$, not necessarily $Q^r$ itself. Conversely, if every interval term is $Q$-smooth and a term is divisible by $Q^2$ with $Q$ prime, then the largest prime factor really is $Q$ and the interval is bad. This converse requires the full smoothness condition; a square anchor alone proves nothing. No unbounded family of smooth blocks has been constructed here. For fixed $r=2$, the power $1/r$ does not tend to zero, so this refinement does not settle the original subpower assertion.

SOURCES CHECKED

Complete local source gpt_pro_5.4/382.tex; official indexed statement checked at https://www.erdosproblems.com/382, 2026-09-23. No claim is made to have rerun the source's large factorization search.

FINAL

UNRESOLVED. The anchor bound is refined by exact multiplicity, while bounded-multiplicity blocks and unbounded lengths remain unresolved.

COMPLETION ESTIMATE

Conservative subjective progress toward both original questions. COMPLETION: 8%

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

1. Use the structural lemma from Problem #380 to obtain unconditional restrictions (prime-free $\Rightarrow$ $v<2u$).
2. 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 $v^{o(1)}$).
3. 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.

1. (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 $v-u=5$.
2. (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).
3. (iii) Top 3 next moves.
   1. 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 $q^2$.
   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.
   3. 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.
4. (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.

7) COMPLETION ESTIMATE

Estimated probability of a full resolution from these notes alone: 20%.