Reused writeup on Math-953 (gpt pro 5.2)

Erdős Problem 829 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/829

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_5.2/829.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/829.tex.

FORMAL RESTATEMENT

Let $A\subset\mathbb{N}_0$ be the set of perfect cubes $\{0^3,1^3,2^3,\dots\}$. Let $1_A$ be its indicator function on $\mathbb{Z}$ (or on $\mathbb{N}_0$), and define the additive convolution \[ (1_A*1_A)(n)=\sum_{m\in\mathbb{Z}} 1_A(m)\,1_A(n-m). \] Equivalently, for $n\ge 0$, $(1_A*1_A)(n)$ counts the number of ordered pairs $(x,y)\in\mathbb{N}_0^2$ such that \[ x^3+y^3=n. \] Question: is it true that for all $n\ge 1$, \[ (1_A*1_A)(n) \ll (\log n)^{O(1)}? \] I.e. does there exist a constant $C>0$ such that $(1_A*1_A)(n)\le (\log n)^C$ for all sufficiently large $n$? (If $A$ is taken as positive cubes $\{1^3,2^3,\dots\}$, the count differs by at most $O(1)$ from allowing $0^3$; this does not affect polylogarithmic bounds.)

QUICK LITERATURE/CONTEXT CHECK

The problem statement records: Mordell proved $\limsup_{n\to\infty} (1_A*1_A)(n)=\infty$. It also records lower bounds for infinitely many $n$: Mahler proved $(1_A*1_A)(n)\gg (\log n)^{1/4}$ and Stewart improved to $(\log n)^{11/13}$. No upper bounds beyond what is proved below are assumed.

ATTACK PLAN

Proof track: $1$ Relate representations $x^3+y^3=n$ to factorizations $n=(x+y)(x^2-xy+y^2)$ to get divisor-type upper bounds. $2$ Try to sharpen using restrictions on the size of $x+y$ and the structure of the discriminant.

Disproof track: $1$ Attempt to build explicit $n$ with many representations by composing identities or multiplicative structures. $2$ Use computation to search for unusually large $(1_A*1_A)(n)$.

WORK

Lemma 1 (divisor bound via $x^3+y^3$ factorization). For every integer $n\ge 1$, \[ (1_A*1_A)(n) \le 2\,\tau(n), \] where $\tau(n)$ is the number of positive divisors of $n$.

Proof. Let $r(n)=(1_A*1_A)(n)$, the number of ordered pairs $(x,y)\in\mathbb{N}_0^2$ with $x^3+y^3=n$. For any such pair, \[ n=x^3+y^3=(x+y)(x^2-xy+y^2). \] Define $u=x+y$ and $v=x^2-xy+y^2$. Then $u,v\in\mathbb{N}$ and $uv=n$, so $u$ is a positive divisor of $n$.

Fix a divisor $u\mid n$ and set $v=n/u$. If there exists a representation with this $u$, then $x,y$ satisfy \[ x+y=u,\qquad x^2-xy+y^2=v. \] Using $v=(x+y)^2-3xy=u^2-3xy$, we get \[ xy=\frac{u^2-v}{3}. \] Thus $(x,y)$ must be a pair of roots (in $\mathbb{R}$, hence in $\mathbb{Z}$ if they exist as integers) of the quadratic equation \[ t^2-u t + \frac{u^2-v}{3}=0. \] A quadratic equation has at most two real roots, and therefore yields at most two ordered integer pairs $(x,y)$. Hence, for each divisor $u$ of $n$, there are at most two ordered pairs $(x,y)$ with $x+y=u$ and $x^3+y^3=n$. Summing over all $u\mid n$ gives $r(n)\le 2\tau(n)$. $\square$

Lemma 2 (size restriction on the divisor $u=x+y$). If $n=x^3+y^3$ with $x,y\in\mathbb{N}_0$ and $u=x+y$, then \[ n^{1/3}\le u \le (4n)^{1/3}. \] Consequently, \[ (1_A*1_A)(n) \le 2\,\#\left\{u\mid n:\ n^{1/3}\le u\le (4n)^{1/3}\right\}. \]

Proof. Let $u=x+y$ and $v=x^2-xy+y^2$ so that $uv=n$. We show $\frac{u^2}{4}\le v\le u^2$. First, since $xy\ge 0$ we have $v=u^2-3xy\le u^2$. Second, by AM--GM, $xy\le \left(\frac{x+y}{2}\right)^2=\frac{u^2}{4}$, hence \[ v=u^2-3xy\ge u^2-3\cdot \frac{u^2}{4}=\frac{u^2}{4}. \] Multiplying by $u>0$ gives \[ \frac{u^3}{4} \le uv=n \le u^3. \] Thus $u^3\ge n$ and $u^3\le 4n$, i.e. $n^{1/3}\le u\le (4n)^{1/3}$. Finally, Lemma 1 already shows that for each admissible divisor $u$ there are at most two ordered representations, yielding the stated bound. $\square$

Fast reality check (computation; exact search). We brute-forced $r(n)$ by enumerating cubes $x^3,y^3\le N$ and counting sums $x^3+y^3=n$. For each cutoff $N$ we recorded the maximum value of $r(n)$ for $n\le N$: \[ \begin{array}{r|r|l} N & \max_{n\le N} r(n) & \text{first $n$ achieving the maximum}\\\hline 10^6 & 4 & 1729\\ 10^7 & 4 & 1729\\ 10^8 & 6 & 87539319\\ 10^9 & 6 & 87539319 \end{array} \] For $N=10^9$, the exact set of $n\le 10^9$ with $r(n)=6$ found by the script was: \[ \{87539319,\ 119824488,\ 143604279,\ 175959000,\ 327763000,\ 700314552,\ 804360375,\ 958595904\}. \] These small-scale data show unboundedness is not visible at this range; they are only a sanity check.

VERIFICATION

$1$ Lemma 1: the only nontrivial point is that for fixed $u=x+y$, the pair $(x,y)$ is determined by $x+y$ and $xy$, hence by a quadratic with at most two solutions. This is correct. $2$ Lemma 2: the inequalities use only $xy\ge 0$ and $xy\le (x+y)^2/4$; both are valid for $x,y\in\mathbb{N}_0$. $3$ Edge cases: if $x=0$ then $u=y$ and $n=y^3$, giving $u=n^{1/3}$, consistent with the lower bound.

FINAL

UNRESOLVED

(i) Strongest proved partial result. We proved the unconditional upper bound $(1_A*1_A)(n)\le 2\tau(n)$ (Lemma 1), and the stronger restriction that only divisors $u\mid n$ with $n^{1/3}\le u\le (4n)^{1/3}$ can contribute (Lemma 2).

(ii) First gap (crisp statement). Prove or disprove: there exists $C>0$ such that for all sufficiently large $n$, $(1_A*1_A)(n)\le (\log n)^C$.

(iii) Top 3 next moves. 1. Improve Lemma 2 by bounding the number of divisors of $n$ in the short interval $[n^{1/3},(4n)^{1/3}]$ by a polylogarithm, or show this is impossible. 2. Use the elliptic curve viewpoint of $x^3+y^3=n$ to relate $r(n)$ to the number of integer points and seek unconditional upper bounds for integral points on this family. 3. Computation: extend searches for large $r(n)$ using hash tables for $x^3+y^3$ up to larger limits (e.g. $x\le 10^6$) to see empirical growth, and record $n$ with unusually large multiplicity.

(iv) Minimal counterexample structure. A counterexample to the polylog upper bound would be an $n$ with extremely many distinct pairs $(x,y)$, i.e. an integer whose corresponding curve $x^3+y^3=n$ has many integer points; heuristically this might arise from $n$ with unusually large arithmetic structure (e.g. related to high-rank curves or compositions of sum-of-two-cubes representations).

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.