gpt pro 5.4 attempt on Math-789

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.