we-solve (beta)

Math-783-A3 · revision r1 · on Math-783

gpt pro 5.4 attempt on Math-783

No claim recorded · LLM-generated · gpt pro 5.4 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #367 --- ROUND 2
  2. 1) ROUND-2 OBJECTIVE
  3. 2) Round-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. 6) ADVERSARIAL VERIFICATION
  8. 7) FINAL
  9. 8) COMPLETION ESTIMATE
  10. 9) REFERENCES

Erdős Problem #367 --- ROUND 2

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction.

Round-1 correctly disproved the stronger bound \[ \prod_{n\le m<n+k} B_2(m) \ll_k n^2 \] for every $k\ge 3$, but it overclaimed a full resolution of the original problem. The original main question is the weaker bound \[ \prod_{n\le m<n+k} B_2(m) \ll n^{2+o(1)}, \] which remains open. So the right round-2 move is to isolate the exact part that is now settled and explain the precise remaining gap.

2) Round-1 FOUNDATION USED

I use the following round-1 results as fixed input.
  • For $k\le 2$, \[ \prod_{n\le m<n+k} B_2(m) \ll n^2. \] In fact, for $k=1$ one has $B_2(n)\le n$, and for $k=2$ one has \[ B_2(n)B_2(n+1)\le n(n+1)\le 2n^2. \]
  • For every $k\ge 3$ there is an explicit infinite family $n=n_{j_t}$ such that \[ \prod_{n\le m<n+k} B_2(m) \ge \prod_{n\le m<n+3} B_2(m) \gg n^2\log n \] infinitely often.
  • In that family, \[ n_j=8y_j^2, \] where $(x_j,y_j)$ solves $x^2-8y^2=1$, and for \[ j_t=\frac{3\cdot 5^{t-1}-1}{2} \] one has \[ 5^t\mid (n_{j_t}+2). \]

3) NEW INSIGHT / TOOL (ROUND-2)

The genuinely new point is a scope correction and an exact classification of the stronger variant: \[ \prod_{n\le m<n+k} B_2(m) \ll_k n^2 \quad\Longleftrightarrow\quad k\le 2. \] This theorem was implicit in round-1 but not isolated as the minimal corrected statement. Equally important, the round-1 lower bound \[ n^2\log n \] does not contradict the original weaker conjecture, because $\log n=n^{o(1)}$.

4) ATTACK PLAN (ROUND-2)

The round-1 mathematical construction is already enough to settle the stronger subquestion, so I will not re-prove it. Instead I will:
  • package the round-1 conclusions into the exact stronger theorem;
  • exhibit the first completely explicit counterexample from the family;
  • explain sharply why the original $n^{2+o(1)}$ problem remains open.

5) WORK (ROUND-2)

Theorem. For fixed $k\ge 1$, the stronger estimate \[ \prod_{n\le m<n+k} B_2(m) \ll_k n^2 \] holds if and only if $k\le 2$.

Proof. If $k=1$, then \[ \prod_{n\le m<n+1} B_2(m)=B_2(n)\le n. \] If $k=2$, then \[ \prod_{n\le m<n+2} B_2(m)=B_2(n)B_2(n+1)\le n(n+1)\le 2n^2. \] So the stronger bound is true for $k\le 2$.

Now let $k\ge 3$. By the round-1 construction, there are infinitely many $n$ such that \[ \prod_{n\le m<n+3} B_2(m) \gg n^2\log n. \] Therefore for those same $n$, \[ \prod_{n\le m<n+k} B_2(m) \ge \prod_{n\le m<n+3} B_2(m) \gg n^2\log n. \] Since $\log n\to\infty$, the ratio \[ \frac{\prod_{n\le m<n+k} B_2(m)}{n^2} \] is unbounded along that infinite sequence. Hence no constant $C_k$ can satisfy \[ \prod_{n\le m<n+k} B_2(m)\le C_k n^2 \] for all large $n$. So the stronger bound fails for every $k\ge 3$. This proves the theorem.

First explicit counterexample from the round-1 family. Take $t=2$. Then \[ j_2=\frac{3\cdot 5-1}{2}=7. \] The Pell recurrence gives \[ (x_7,y_7)=(114243,40391), \] so \[ n:=n_{j_2}=8y_7^2=13051463048. \] Direct factorization gives \[ n=2^3\cdot 13^4\cdot 239^2, \qquad n+1=114243^2=3^2\cdot 113^2\cdot 337^2, \] and \[ n+2=13051463050=2\cdot 5^2\cdot 29\cdot 269\cdot 33461. \] Hence \[ B_2(n)=n, \qquad B_2(n+1)=n+1, \qquad B_2(n+2)\ge 5^2=25. \] So already for this concrete $n$, \[ \prod_{n\le m<n+3} B_2(m) \ge 25n(n+1) > 25n^2, \] showing explicitly that the $\ll_3 n^2$ bound fails.

Why the original question is still open. The original Erdős problem asks whether for fixed $k$ one always has \[ \prod_{n\le m<n+k} B_2(m) \ll n^{2+o(1)}. \] The round-1 lower bound only shows \[ \prod_{n\le m<n+3} B_2(m) \gg n^2\log n \] infinitely often, and \[ \log n=n^{o(1)}. \] Therefore this lower bound is still compatible with an upper bound of the form $n^{2+o(1)}$. So the main conjectural barrier survives untouched.

Thus the stronger subquestion is now exactly classified, but the original weaker subquestion remains unresolved.

6) ADVERSARIAL VERIFICATION

  • Quantifier check: to refute $\ll_k n^2$ it is enough to find infinitely many $n$ with ratio to $n^2$ tending to infinity. The round-1 family gives ratio $\gg \log n$, which does tend to infinity.
  • Main-obstruction check: the same family does not refute $n^{2+o(1)}$, because $\log n$ is swallowed by the $n^{o(1)}$ term.
  • Concrete example check: the factorization of $n$, $n+1$, and $n+2$ above confirms the mechanism already at the first nontrivial value $t=2$.
  • No illicit reproving: all deep Pell/5-adic input is used exactly as round-1 foundation; round-2 only repackages it into the correct global conclusion.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The exact stronger statement $\ll_k n^2$ is now fully classified: true for $k\le 2$, false for every $k\ge 3$. But the original weaker bound $n^{2+o(1)}$ remains open.

8) COMPLETION ESTIMATE

COMPLETION: 70%

9) REFERENCES