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Math-734-A3 · revision r1 · on Math-734

gpt pro 5.4 attempt on Math-734

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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 Problems #256--#258: Round 2 continuation
  2. Erdős Problem #256
  3. 1) ROUND-2 OBJECTIVE
  4. 2) Round-1 FOUNDATION USED
  5. 3) NEW INSIGHT / TOOL (ROUND-2)
  6. 4) ATTACK PLAN (ROUND-2)
  7. 5) WORK (ROUND-2)
  8. 6) ADVERSARIAL VERIFICATION
  9. 7) FINAL
  10. 8) COMPLETION ESTIMATE (MANDATORY)
  11. 9) REFERENCES

Erdős Problems #256--#258: Round 2 continuation

Erdős Problem #256

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. Round 1 already gave a complete disproof of the specific subquestion \[ \exists c>0\text{ such that }\log f(n)\gg n^c. \] So the only legitimate round-2 task here is to adversarially check that disproof, lock in the quantifiers, and record the minimal corrected conclusion.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts without re-proving them.
  1. Belov--Konyagin's upper bound: there exists an absolute constant $C>0$ such that \[ \log f(n)\le C(\log n)^4 \] for all sufficiently large $n$.
  2. The elementary comparison \[ (\log n)^4=o(n^c)\qquad (n\to\infty) \] for every fixed $c>0$.
  3. The Round-1 contradiction argument combining $1$ and $2$.

3) NEW INSIGHT / TOOL (ROUND-2)

There is no new mathematical ingredient needed. The only new point is an adversarial consistency check against the current Erdős Problems summary, which now explicitly records that the specific power-lower-bound question has a negative answer.

4) ATTACK PLAN (ROUND-2)

The only possible gap after Round 1 would have been a quantifier slip: one must rule out every fixed exponent $c>0$, not just a particular one. The plan is therefore to re-run the contradiction with the quantifiers written explicitly.

5) WORK (ROUND-2)

Fix any $c>0$. Suppose, for contradiction, that there exist $K>0$ and $n_0$ such that \[ \log f(n)\ge K n^c\qquad (n\ge n_0). \] Round 1 already established that there is an absolute constant $C>0$ such that \[ \log f(n)\le C(\log n)^4 \] for all sufficiently large $n$. Since $(\log n)^4=o(n^c)$, there exists $n_1$ such that \[ C(\log n)^4<K n^c\qquad (n\ge n_1). \] Hence, for all sufficiently large $n$, \[ \log f(n)\le C(\log n)^4<K n^c\le \log f(n), \] which is impossible. Therefore no fixed $c>0$ can satisfy $\log f(n)\gg n^c$.

Thus the specific subquestion in Problem #256 is fully settled in the negative.

6) ADVERSARIAL VERIFICATION

  1. Quantifiers. The argument starts with an arbitrary fixed $c>0$ and shows that this $c$ cannot work. Hence it rules out all $c>0$.
  2. Dependence of constants. The Belov--Konyagin constant $C$ is absolute; the contradiction only needs that $C$ is independent of $n$.
  3. Logical target. The original question only asked whether a power-type lower bound $\log f(n)\gg n^c$ could hold. The argument disproves exactly that claim, not the entire surrounding estimation problem for $f(n)$.
No gap remains.

7) FINAL

FULL SOLUTION --- COUNTEREXAMPLE / DISPROOF

The specific claim \[ \exists c>0\text{ such that }\log f(n)\gg n^c \] is false. Belov--Konyagin give the upper bound $\log f(n)\ll (\log n)^4$, and $(\log n)^4=o(n^c)$ for every fixed $c>0$. Hence no power lower bound of the form $\log f(n)\gg n^c$ can hold.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 100%

9) REFERENCES

  1. A. S. Belov and S. V. Konyagin, An estimate of the free term of a non-negative trigonometric polynomial with integer coefficients, Izv. Math. 60 $1996$, 1123--1182.
  2. T. F. Bloom (ed.), Erdős Problem #256, erdosproblems.com/256, accessed 2026-03-07.