Reused writeup on Math-993 (gpt pro 5.2)

Erdős Problem 943 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/943

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/943.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/943.tex.

Let $A$ be the set of powerful numbers (if $p\mid n$ then $p^2\mid n$). Is it true that\[1_A\ast 1_A(n)=n^{o(1)}\]for every $n$?

FORMAL RESTATEMENT

Let $A\subseteq\mathbb{N}$ be the set of powerful positive integers: $m\in A$ iff for every prime $p$, $p\mid m\Rightarrow p^2\mid m$. Let $1_A$ be its indicator function on $\mathbb{Z}$. Define the additive convolution \[ (1_A*1_A)(n) := \sum_{m\in\mathbb{Z}} 1_A(m)\,1_A(n-m). \] Equivalently, for $n\ge 1$, \[ (1_A*1_A)(n)=\#\{(a,b)\in A\times A: a+b=n\} \] (counting ordered pairs). The question asks whether for every $n\to\infty$ we have \[ (1_A*1_A)(n)=n^{o(1)}. \]

QUICK LITERATURE/CONTEXT CHECK

I do not use external results beyond what is in the problem text (which provides no additional bounds here). Heuristically, since $\#(A\cap[1,N])\asymp N^{1/2}$, the average value of $(1_A*1_A)(n)$ for $n\le 2N$ should be $O(1)$, but the question is about worst-case $n$.

ATTACK PLAN

Proof-track ideas.

• Use the parametrization $m=a^2b^3$ (squarefree $b$) to reduce counting solutions of $a^2b^3+c^2d^3=n$.
• Seek a divisor-type upper bound: show that each representation forces a large square divisor of $n$ (or of some related quantity), then apply a divisor bound.
• Use average bounds (global counting) as a baseline and try to upgrade to pointwise bounds with additional structure.

Disproof-track ideas.

• Search for $n$ with many representations as sum of two powerful numbers (e.g. via many representations as sum of two squares, since squares are powerful).
• If spikes seem large computationally, attempt to build a parametric family producing polynomially many representations.

WORK

Fast reality check (explicit computation).

I computed $(1_A*1_A)(n)$ for $1\le n\le N$ for $N\in\{10^3,10^4,10^5\}$ by enumerating powerful numbers $\le N$ and counting ordered pairs summing to $n$. Exact maxima observed:

• For $N=10^3$: $\max_{n\le 1000}(1_A*1_A)(n)=12$, attained at $n\in\{657,684,873\}$.
• For $N=10^4$: $\max_{n\le 10000}(1_A*1_A)(n)=20$, attained at $n\in\{6156,7884,9225\}$.
• For $N=10^5$: $\max_{n\le 100000}(1_A*1_A)(n)=43$, attained at $n=88200$.

These are sanity checks only.

Lemma 943.1 (counting powerful numbers up to $X$).

Let $P(X)=\#(A\cap[1,X])$. Then for all $X\ge 1$, \[ P(X) \le \Big(\sum_{b=1}^{\infty} b^{-3/2}\Big)\, X^{1/2}. \]

Proof. This is exactly Lemma 942.2, since the definition of $A$ is the same. $\Box$

Lemma 943.2 (trivial pointwise upper bound on the convolution).

For every integer $n\ge 1$, \[ (1_A*1_A)(n) \le P(n) \le \Big(\sum_{b=1}^{\infty} b^{-3/2}\Big)\, n^{1/2}. \]

Proof. By definition, \[ (1_A*1_A)(n)=\sum_{a\in\mathbb{Z}} 1_A(a)1_A(n-a). \] All terms vanish unless $1\le a\le n-1$ and $a\in A$, so \[ (1_A*1_A)(n)\le \sum_{a=1}^{n-1}1_A(a)=\#(A\cap[1,n-1])\le P(n). \] Now apply Lemma 943.1. $\Box$

Proposition 943.3 (bounded average size of the convolution).

For every $N\ge 1$, \[ \sum_{n=1}^{2N} (1_A*1_A)(n) = \big(\#(A\cap[1,N])\big)^2 \le \Big(\sum_{b=1}^{\infty} b^{-3/2}\Big)^2 N. \] In particular, the average of $(1_A*1_A)(n)$ over $1\le n\le 2N$ is $O(1)$.

Proof. Each ordered pair $(a,b)\in (A\cap[1,N])^2$ contributes exactly once to the sum on the left, namely to $n=a+b\in[2,2N]$. Therefore \[ \sum_{n=1}^{2N} (1_A*1_A)(n)=\#\{(a,b)\in A\times A: 1\le a,b\le N\} = \big(\#(A\cap[1,N])\big)^2. \] The final inequality is Lemma 943.1. $\Box$

VERIFICATION

• Lemma 943.2: checked the support of $1_A(n-a)$ forces $1\le a\le n-1$ for $n\ge 2$; for $n=1$ the convolution is $0$ (no positive $a,b$ sum to $1$), consistent with the inequality.
• Proposition 943.3: verified the counting identity is exact because each pair contributes to exactly one sum.
• Computation sanity: for $n=2$, $(1_A*1_A)(2)=1$ because $2=1+1$ and $1\in A$. This matches the program output.

FINAL

UNRESOLVED (i) Strongest proved partial result: the pointwise bound $(1_A*1_A)(n)\le C\sqrt{n}$ (Lemma 943.2) and the exact average identity $\sum_{n\le 2N}(1_A*1_A)(n)=(\#(A\cap[1,N]))^2=O(N)$ (Proposition 943.3). (ii) First gap: improve the worst-case bound from $O(n^{1/2})$ to $n^{o(1)}$ (or find an $n$ giving a superpolynomial-in-$\log n$ lower bound that would disprove the statement). (iii) Top 3 next moves:

1. Study the Diophantine equation $a^2b^3+c^2d^3=n$ (with $b,d$ squarefree) via the determinant method / geometry-of-numbers to get pointwise upper bounds on the number of solutions.
2. Try to relate representations to large square divisors of $n$ or nearby integers and then apply sharp divisor bounds for those square divisors.
3. Extend computation to larger $n$ and inspect maximizing $n$ (e.g. $n=88200$ for $n\le 10^5$) to see whether large values correlate with special factorization patterns.

(iv) Minimal counterexample structure (if the conjecture is false): an integer $n$ with many decompositions $n=a+b$ where both $a$ and $b$ have all prime exponents $\ge 2$; heuristically such an $n$ may have many representations as sum of two squares plus additional squarefull perturbations.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.