Reused writeup on Math-991 (gpt pro 5.2)

Erdős Problem 940 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/940

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_5.2/940.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/940.tex.

1) FORMAL RESTATEMENT

Fix an integer $r\ge 3$. A positive integer $n$ is $r$-powerful if for every prime $p\mid n$ we have $p^r\mid n$. Let $\mathcal P_r$ be the set of $r$-powerful positive integers (note $1\in\mathcal P_r$ vacuously). Define the representable set \[\mathcal S_r := \Big\{m\in\mathbb N: \exists k\in\{1,2,\dots,r\},\ \exists a_1,\dots,a_k\in\mathcal P_r\ \text{with}\ m=a_1+\cdots+a_k\Big\}. \] Questions: (a) Are there infinitely many integers not in $\mathcal S_r$? (b) Does $\mathcal S_r$ have natural density 0, i.e. \[\lim_{X\to\infty} \frac{|\mathcal S_r\cap [1,X]|}{X}=0?\]

2) QUICK LITERATURE/CONTEXT CHECK

The statement reports: - For $r=2$, density 0 is "easy" and was first proved by Baker--Brüdern. - Erdős claimed a counting argument for infinitely many non-representable integers, but Schinzel noted an error. - Heath-Brown proved every sufficiently large integer is a sum of at most three 2-powerful numbers. I do not use any external results beyond these contextual remarks.

3) ATTACK PLAN

- Prove structural/counting lemmas about $r$-powerful numbers (how many are $\le X$, and how to decompose them). - Do a small computational sanity check for small $X$ and small $r$ to see how large $\mathcal S_r\cap[1,X]$ can be.

4) WORK

Fast reality check (explicit computation). For $X=200000$: - For $r=3$: there are 168 many 3-powerful numbers $\le X$, and $|\mathcal S_3\cap[1,X]|=161164$ (density $\approx 0.80582$). - For $r=4$: there are 72 many 4-powerful numbers $\le X$, and $|\mathcal S_4\cap[1,X]|=153107$ (density $\approx 0.765535$). These are only finite-range measurements and do not address the asymptotic density question.

Lemma 940.1 (Decomposition into an $r$-th power times an $(r+1)$-powerful number). Let $r\ge 2$. Every $r$-powerful integer $n\ge 1$ can be written as \[n=a^r\,b\] where $a\ge 1$ is an integer and $b$ is $(r+1)$-powerful.

Proof. Write $n=\prod_p p^{e_p}$ with all $e_p\ge r$. For each prime $p$, define integers $\alpha_p\ge 0$ and $\beta_p\ge 0$ as follows. - If $r\mid e_p$, set $\alpha_p=e_p/r$ and $\beta_p=0$. - If $r\nmid e_p$, write $e_p=rq_p+s_p$ with integers $q_p\ge 1$ and $1\le s_p\le r-1$. Set $\alpha_p=q_p-1$ and $\beta_p=r+s_p$. Then $\beta_p\ge r+1$ in the second case, and $\beta_p=0$ in the first. Define \[a:=\prod_p p^{\alpha_p},\qquad b:=\prod_p p^{\beta_p}. \] Then \[a^r b = \prod_p p^{r\alpha_p+\beta_p}. \] If $r\mid e_p$, then $r\alpha_p+\beta_p=r(e_p/r)+0=e_p$. If $r\nmid e_p$ and $e_p=rq_p+s_p$, then $r\alpha_p+\beta_p=r(q_p-1)+(r+s_p)=rq_p+s_p=e_p$. Thus $a^r b=n$. Finally, in $b$ each prime exponent $\beta_p$ is either 0 or at least $r+1$, so by definition $b$ is $(r+1)$-powerful. QED.

Lemma 940.2 (Counting $r$-powerful numbers). For fixed $r\ge 2$, let $F_r(X):=|\{n\le X: n\ \text{is $r$-powerful}\}|$. Then \[F_r(X)=O_r(X^{1/r}).\]

Proof. By Lemma 940.1, every $r$-powerful $n\le X$ can be written as $n=a^r b$ with $b$ $(r+1)$-powerful. For any fixed $(r+1)$-powerful $b\le X$, the number of integers $a\ge 1$ with $a^r b\le X$ is at most $\lfloor (X/b)^{1/r}\rfloor\le (X/b)^{1/r}$. Hence \[F_r(X) \le \sum_{\substack{b\le X\\ b\ \text{$(r+1)$-powerful}}} \Big(\frac{X}{b}\Big)^{1/r} = X^{1/r} \sum_{\substack{b\le X\\ b\ \text{$(r+1)$-powerful}}} b^{-1/r} \le X^{1/r} \sum_{\substack{b\ge 1\\ b\ \text{$(r+1)$-powerful}}} b^{-1/r}. \] It remains to show the infinite sum converges. Write each $(r+1)$-powerful $b$ as $b=\prod_p p^{f_p}$ where each $f_p\in\{0\}\cup\{r+1,r+2,\dots\}$. Then \[\sum_{b\ \text{$(r+1)$-powerful}} b^{-1/r} = \prod_{p}\Big(1+\sum_{f=r+1}^{\infty} p^{-f/r}\Big) = \prod_{p}\Big(1+\frac{p^{-(r+1)/r}}{1-p^{-1/r}}\Big). \] For each prime $p$, the factor is finite because $p^{-1/r}<1$. Moreover, for large $p$ we have $\frac{1}{1-p^{-1/r}}\le 2$, hence \[\log\Big(1+\frac{p^{-(r+1)/r}}{1-p^{-1/r}}\Big) \ll p^{-(r+1)/r}. \] Since $(r+1)/r=1+1/r>1$, the series $\sum_{p} p^{-(r+1)/r}$ converges by comparison with $\sum_{n\ge 2} n^{-(1+1/r)}$. Therefore the Euler product converges to a finite constant $C_r$, and the sum over $(r+1)$-powerful $b$ is finite. Thus $F_r(X)\le C_r X^{1/r}$, proving the claim. QED.

Lemma 940.3 (Squarefree kernel bound for the $r=2$ case). Let $n\ge 1$ be 2-powerful (squarefull). Write $n=u^2 v$ where $v$ is squarefree (so $u^2$ is the largest square dividing $n$). Then \[v\le n^{1/3}\qquad\text{and hence}\qquad u^2\ge n^{2/3}. \]

Proof. Since $v$ is the squarefree part, we can write $v=\prod_{p\in S} p$ where $S$ is the set of primes dividing $n$ to an odd exponent. Because $n$ is squarefull, every prime divisor appears with exponent at least 2; in particular, every $p\in S$ appears with exponent at least 3 (it is odd and $\ge2$). Thus $p^3\mid n$ for each $p\in S$. Since the primes in $S$ are distinct, their product satisfies \[v^3=\prod_{p\in S} p^3 \mid n. \] Therefore $v^3\le n$, so $v\le n^{1/3}$. Finally, $u^2=n/v\ge n/n^{1/3}=n^{2/3}$. QED.

5) VERIFICATION

- Lemma 940.1: checks $\alpha_p\ge0$ in the non-multiple case because $q_p\ge1$, so $q_p-1\ge0$. Also $\beta_p\ge r+1$ because $s_p\ge1$. - Lemma 940.2: the Euler product bound uses only that $\sum_{n\ge2} n^{-(1+1/r)}$ converges; since primes are a subset, $\sum_p p^{-(1+1/r)}$ also converges. - Lemma 940.3: uses that odd exponents in a squarefull number are $\ge3$. - Computations: the reported densities are finite-range; they do not establish any asymptotic claim.

6) FINAL

UNRESOLVED (i) Strongest proved partial result: Every $r$-powerful number factors as $a^r b$ with $b$ $(r+1)$-powerful (Lemma 940.1), implying the counting bound $F_r(X)=O_r(X^{1/r})$ (Lemma 940.2). For $r=2$, the squarefree kernel of a squarefull number is $\le n^{1/3}$ (Lemma 940.3). (ii) First gap: Prove or disprove that $|\mathcal S_r\cap[1,X]|=o(X)$ as $X\to\infty$ for each fixed $r\ge 3$, or even that infinitely many integers are not in $\mathcal S_r$. (iii) Top 3 next moves: 1. Improve from $F_r(X)=O(X^{1/r})$ to a structural description of residue classes modulo large $r$-th powers that $r$-powerful numbers occupy, then bound how many sums of $\le r$ such classes can cover. 2. For fixed $r$, compute $|\mathcal S_r\cap[1,X]|$ for growing $X$ (beyond $2\cdot 10^5$) using a generator for $r$-powerful numbers rather than factorising every integer, to see whether the apparent density decreases. 3. Attempt to create explicit infinite families of integers with a congruence obstruction to being a sum of $\le r$ $r$-powerful numbers (e.g. mod $p^r$ for carefully chosen primes $p$). (iv) Minimal counterexample structure: A hypothetical smallest integer $m\notin\mathcal S_r$ would have to avoid all representations by $\le r$ summands from the sparse set $\mathcal P_r$; given Lemma 940.2, one expects such an $m$ to lie in a residue class that is poorly approximated by sums of $r$-powerful residues modulo many primes $p^r$, so a promising obstruction would be simultaneous congruences modulo several $p^r$ forcing each summand to be 0 mod some $p^r$ while $m$ is not.

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.