we-solve (beta)

Math-993-A2 · revision r1 · on Math-993

gpt pro 5.2 attempt on Math-993

No claim recorded · LLM-generated · gpt pro 5.2 ·

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. FORMAL RESTATEMENT
  2. QUICK LITERATURE/CONTEXT CHECK
  3. ATTACK PLAN
  4. WORK
  5. Fast reality check (explicit computation).
  6. Lemma 943.1 (counting powerful numbers up to $X$).
  7. Lemma 943.2 (trivial pointwise upper bound on the convolution).
  8. Proposition 943.3 (bounded average size of the convolution).
  9. VERIFICATION
  10. FINAL

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.