gpt pro 5.4 attempt on Math-5615

Problem #380 (Round 2)

1) ROUND-2 OBJECTIVE

Path C (obstruction / structural reduction). In Round 1 the main unresolved gap was the lack of a rigid description of a bad interval beyond "it contains no primes". The new goal is to prove a sharper internal structure theorem for bad intervals, then use it to isolate exactly what would have to happen in any counterexample to the conjectural behaviour of very bad intervals.

2) Round-1 FOUNDATION USED

I use only the following Round-1 results.

1. If $[u,v]$ is bad, then it contains no primes.
2. Consequently, every bad interval satisfies $v<2u$.
3. Hence if $n\le x$ lies in a bad interval, then the whole interval lies below $2x$.
4. For singletons, $[n,n]$ is bad if and only if $P(n)^2\mid n$, and $[n,n]$ is very bad if and only if $n$ is powerful.

I do not re-prove these.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is the following "anchor-smooth reduction".

Anchor-smooth reduction. If $[u,v]$ is bad, write \[ L=v-u,\qquad N(u,v)=\prod_{m=u}^{v}m,\qquad Q=P(N(u,v)). \] Then:

1. $Q>L$;
2. there is a unique $t\in [u,v]$ divisible by $Q$, and in fact $Q^2\mid t$;
3. every element of $[u,v]$ is $Q$-smooth.

So every bad interval is a run of consecutive $Q$-smooth integers of length $L+1<Q+1$ containing an "anchor" $t$ with $P(t)=Q$ and $Q^2\mid t$.

This is strictly stronger than the Round-1 information "bad $\Rightarrow$ prime-free".

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was this: even though bad intervals contain no primes, that alone did not explain how they are tied to the set \[ \{n:\ P(n)^2\mid n\}. \]

The new plan is:

1. prove the anchor-smooth reduction;
2. deduce a covering lemma showing that every integer counted by $B(x)$ lies within distance $<P(t)$ of an anchor $t$ with $P(t)^2\mid t$;
3. analyse the first genuinely new case for very bad intervals, namely intervals of length $1$ and $2$, to isolate the precise obstruction to proving $V(x)\sim$ (powerful numbers).

5) WORK (ROUND-2)

Theorem 1 (anchor-smooth reduction).

Let $[u,v]$ be bad. Put $L=v-u$ and $Q=P\!\left(\prod_{m=u}^{v}m\right)$. Then for $L\ge 1$ one has:

1. $Q>L$;
2. exactly one element $t\in [u,v]$ is divisible by $Q$;
3. $Q^2\mid t$;
4. every $m\in [u,v]$ has all prime factors at most $Q$.

Proof.

By the Round-1 lemma, $[u,v]$ contains no primes, hence $v<2u$. Therefore \[ L=v-u<u, \] so every integer in the block \[ u,u+1,\dots,v-1 \] is strictly greater than $L$.

Now $u(u+1)\cdots (v-1)$ is a product of $L$ consecutive integers, each greater than $L$. By the Sylvester--Schur theorem, this product has a prime divisor exceeding $L$. Since $Q$ is the largest prime divisor of the larger product $\prod_{m=u}^{v}m$, we get \[ Q>L. \]

Because the interval length is $L$, two distinct multiples of $Q$ cannot both lie in $[u,v]$: their difference would be at least $Q>L$. Thus at most one element of $[u,v]$ is divisible by $Q$.

On the other hand, the interval is bad, so the exponent of $Q$ in $\prod_{m=u}^{v}m$ is at least $2$. Hence the unique multiple $t$ of $Q$ in the interval must actually satisfy \[ Q^2\mid t. \]

Finally, $Q=P\!\left(\prod_{m=u}^{v}m\right)$ by definition, so no element of the interval can have a prime factor exceeding $Q$. Thus every element of $[u,v]$ is $Q$-smooth. This proves the theorem.

The degenerate case $L=0$ is even easier: then $[u,v]=[n,n]$, and badness means exactly $P(n)^2\mid n$.

Corollary 2 (covering by anchor neighbourhoods).

Let $n\le x$ lie in some bad interval. Then there exists an integer $t\le 2x$ such that \[ P(t)^2\mid t ,\qquad |n-t|<P(t). \]

Proof.

Choose a bad interval $[u,v]$ containing $n$, and let $t$ be the anchor supplied by Theorem 1. Then \[ P(t)=Q,\qquad P(t)^2\mid t,\qquad |n-t|\le v-u=L<Q=P(t). \] Also, by the Round-1 bound $v<2u$ and the fact that $u\le n\le v$, \[ t\le v<2u\le 2n\le 2x. \] So $t\le 2x$, as claimed. This proves the corollary.

Corollary 3 (set-theoretic reduction of Problem #380(a)).

For every $x\ge 1$, \[ \{n\le x:\ n \mbox{ lies in a bad interval}\} \subseteq \bigcup_{t\le 2x,\ P(t)^2\mid t} \{m:\ |m-t|<P(t)\}. \]

Equivalently, every integer counted by $B(x)$ is captured by a neighbourhood of radius $P(t)-1$ around an anchor $t$ counted by the singleton set from Round 1.

Short very-bad intervals.

The anchor-smooth theorem does not settle the powerful case, but it does sharply reduce the first unsolved lengths.

Length $1$ (two consecutive integers). If $[n,n+1]$ is very bad, then \[ n(n+1) \] is powerful and $\gcd(n,n+1)=1$. Therefore every prime divisor of the product comes from exactly one factor, with the same exponent as in that factor. Hence both $n$ and $n+1$ are powerful.

So a very bad interval of length $1$ is exactly a pair of consecutive powerful numbers.

Length $2$ (three consecutive integers), odd start. If $[n,n+2]$ is very bad and $n$ is odd, then $n,n+1,n+2$ are pairwise coprime. Therefore the same reasoning shows that each of \[ n,\ n+1,\ n+2 \] must itself be powerful.

Thus any odd-start very bad interval of length $2$ produces three consecutive powerful numbers.

Length $2$ (three consecutive integers), even start. Write the interval as \[ [2a,2a+2]. \] Then \[ (2a)(2a+1)(2a+2)=2^{2+\alpha+\beta}AB(2a+1), \] where \[ a=2^{\alpha}A,\qquad a+1=2^{\beta}B, \] with $A,B$ odd. The odd integers $A$, $B$, and $2a+1$ are pairwise coprime. Therefore, if the product is powerful, then each of \[ A,\ B,\ 2a+1 \] must be powerful.

So any even-start very bad interval of length $2$ forces a very restricted configuration: the odd part of $a$ is powerful, the odd part of $a+1$ is powerful, and $2a+1$ is powerful.

This is the first place where the Round-1 heuristic "maybe every very bad interval consists only of powerful numbers" meets a genuinely arithmetic obstruction.

6) ADVERSARIAL VERIFICATION

I checked the following possible failure points.

1. Does Theorem 1 accidentally use a Round-1 proof again? No. It only cites the Round-1 conclusion $v<2u$ and then adds a genuinely new Sylvester--Schur step.
2. Could the anchor be non-unique? No, because $Q>L$ and two multiples of $Q$ differ by at least $Q$.
3. Does the theorem claim the interval contains $Q^2$ itself? No. It only claims that the unique multiple of $Q$ in the interval is divisible by $Q^2$. This distinction matters: for example, $[242,243]$ is bad with $Q=11$, and the unique multiple is $242=2\cdot 11^2$, not $11^2$ itself.
4. In the even-start very-bad case, do we really get that $a$ and $a+1$ are powerful? No, and I have deliberately not claimed that. The extra factor $4$ can repair the $2$-adic defect. The correct statement is only about the odd parts of $a$ and $a+1$.
5. Boundary case $L=0$. This was separated off explicitly.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The new Round-2 progress is:

1. a rigorous anchor-smooth reduction for every bad interval;
2. a covering lemma reducing Problem #380(a) to neighbourhoods of singleton anchors $t$ with $P(t)^2\mid t$;
3. an exact classification of very bad intervals of length $1$;
4. a sharp reduction of the first new three-term cases to highly constrained powerful-number configurations.

This is a real advance over Round 1, but it does not yet prove either \[ B(x)\sim \#\{n\le x:\ P(n)^2\mid n\} \] or \[ V(x)\sim \#\{n\le x:\ n \mbox{ is powerful}\}. \]

8) COMPLETION ESTIMATE

COMPLETION: 45%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #380, accessed 2026-03-08, https://www.erdosproblems.com/380.
2. T. N. Shorey and R. Tijdeman, Arithmetic properties of blocks of consecutive integers, especially the discussion of Sylvester's theorem in Section 3.1.