Reused writeup on Math-6161 (gpt pro 5.2)

Erdős Problem 1107 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1107

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_5.2/1107.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1107.tex.

FORMAL RESTATEMENT

Fix an integer $r\ge 2$. A positive integer $n$ is called $r$-powerful if for every prime $p$ dividing $n$ we have $p^r\mid n$. Equivalently, writing $n=\prod_{p} p^{e_p}$ with $e_p\in\mathbb{Z}_{\ge 0}$, we require that for every prime $p$, \[e_p\in\{0\}\cup\{r,r+1,r+2,\dots\}.\] (We regard $1$ as $r$-powerful since it has no prime divisors.)

The problem asks whether the following is true:

For each fixed $r\ge 2$ there exists an integer $N_r$ such that every integer $n\ge N_r$ can be written as \[n=a_1+\cdots+a_m\] with $m\le r+1$ and each $a_i$ a positive $r$-powerful integer.

QUICK LITERATURE/CONTEXT CHECK

The problem file states that the claim is true for $r=2$ (squarefull numbers): Heath--Brown proved that every sufficiently large integer is the sum of three squarefull numbers. No general result for $r\ge 3$ is stated in the problem file, so I treat the general case as open here and I do not use any external literature beyond what is explicitly stated in the problem text.

ATTACK PLAN

Proof-track strategies.

• Try a local-to-global approach: understand the set of residues of $r$-powerful numbers modulo $M$, and attempt to show that the $(r+1)$-fold sumset covers all residues for sufficiently large $M$, then lift to integers.
• Exploit that $r$th powers are $r$-powerful: attempt to reduce the problem to a Waring-type statement plus a bounded number of extra $r$-powerful "adjustment" terms.
• Attempt an interval covering argument: show that sums of $r$-powerful numbers form long consecutive intervals once the scale is large enough.

Disproof-track strategies.

• Search for congruence obstructions: find a modulus $M$ such that $\mathcal{P}_r+\cdots+\mathcal{P}_r$ (with $r+1$ summands) misses some residue class, where $\mathcal{P}_r$ denotes the set of $r$-powerful numbers.
• Try to build an infinite family of integers forced to have small $p$-adic valuations after subtracting any bounded number of $r$-powerful numbers.
• Computationally hunt for a growing sequence of counterexamples requiring $>r+1$ terms.

WORK

FAST REALITY CHECK (computation).

I brute-forced representations $n=a_1+\cdots+a_{r+1}$ with each $a_i$ $r$-powerful (repetitions allowed) for small $r$ and ranges of $n$.

• For $r=2$ and $1\le n\le 100000$, the integers not representable as a sum of $\le 3$ squarefull numbers were exactly \[\{7,15,23,87,111,119\},\] and every $n\ge 120$ up to $100000$ was representable. Moreover, among $1\le n\le 20000$, the counts of minimal number of squarefull summands were: 1 term: $267$; 2 terms: $11571$; 3 terms: $8156$; non-representable: $6$.

• For $r=3$ and $1\le n\le 200000$, the integers not representable as a sum of $\le 4$ $3$-powerful numbers were exactly the following $45$ numbers (all $\le 2039$): \[\begin{aligned} &5,6,7,12,13,14,15,20,21,22,23,31,38,39,46,47,53,58,69,77,79,85,95,101,103,111,\\ &175,196,212,228,231,247,327,444,458,490,606,662,860,975,1167,1470,1821,1967,2039. \end{aligned}\] Every integer $n\ge 2040$ up to $200000$ was representable as a sum of $\le 4$ $3$-powerful numbers. Among $1\le n\le 50000$, the counts of minimal number of $3$-powerful summands were: 1 term: $97$; 2 terms: $3601$; 3 terms: $34484$; 4 terms: $11773$; non-representable: $45$.

These computations are evidence only; they do not constitute a proof that the exceptional set is finite for any $r\ge 3$.

Lemma 1107.1 (radical characterization).

Let $r\ge 2$ and let $\operatorname{rad}(n)=\prod_{p\mid n} p$ be the radical of $n$. Then a positive integer $n$ is $r$-powerful if and only if \[\operatorname{rad}(n)^r\mid n.\]

Proof.

Assume $n$ is $r$-powerful. For every prime $p\mid n$ we have $p^r\mid n$. Multiplying these divisibilities over all primes dividing $n$ gives \[\prod_{p\mid n} p^r\mid n,\] i.e. $\operatorname{rad}(n)^r\mid n$.

Conversely, assume $\operatorname{rad}(n)^r\mid n$. Let $p$ be a prime dividing $n$. Then $p\mid \operatorname{rad}(n)$, hence $p^r\mid \operatorname{rad}(n)^r$, hence $p^r\mid n$. Since this holds for every prime $p\mid n$, $n$ is $r$-powerful by definition. $\square$

Lemma 1107.2 (unique factorization $n=m^r s$ with $s\mid m^{r-1}$).

Fix $r\ge 2$. A positive integer $n$ is $r$-powerful if and only if there exist integers $m\ge 1$ and $s\ge 1$ such that \[n=m^r s \quad\text{and}\quad s\mid m^{r-1}.\] Moreover, if such a representation exists then the pair $(m,s)$ is uniquely determined by $n$.

Proof.

Write the prime factorization $n=\prod_p p^{e_p}$.

($\Rightarrow$) Assume $n$ is $r$-powerful. Thus if $e_p>0$ then $e_p\ge r$. For each prime $p$, write $e_p=r f_p + g_p$ where $f_p=\lfloor e_p/r\rfloor\in\mathbb{Z}_{\ge 0}$ and $g_p\in\{0,1,\dots,r-1\}$. Define \[m:=\prod_p p^{f_p},\qquad s:=\prod_p p^{g_p}. \] Then $n=m^r s$ by construction. We show $s\mid m^{r-1}$. Fix a prime $p$. The exponent of $p$ in $s$ is $g_p\le r-1$. If $g_p>0$ then $e_p>0$, hence $e_p\ge r$, so $f_p\ge 1$ and the exponent of $p$ in $m^{r-1}$ is $(r-1)f_p\ge r-1\ge g_p$. If $g_p=0$ the claim is trivial for this prime. Therefore the $p$-adic exponent of $s$ is at most the $p$-adic exponent of $m^{r-1}$ for every $p$, hence $s\mid m^{r-1}$.

($\Leftarrow$) Assume $n=m^r s$ with $s\mid m^{r-1}$. Let $p$ be any prime dividing $n$. Then $p$ divides $m^r s$, so $p$ divides $m$ or $s$. If $p\mid s$, the condition $s\mid m^{r-1}$ implies $p\mid m$ as well. In all cases, $p\mid m$, so write $m=p^a u$ with $a\ge 1$. Then $m^r$ is divisible by $p^{ra}$, hence $n=m^r s$ is divisible by $p^r$. Thus every prime divisor $p$ of $n$ satisfies $p^r\mid n$, so $n$ is $r$-powerful.

Uniqueness. The decomposition $e_p=r f_p+g_p$ with $0\le g_p<r$ is unique for each prime $p$, so the exponent sequences $(f_p)_p$ and $(g_p)_p$ are uniquely determined. Hence $m$ and $s$ are uniquely determined. $\square$

Corollary 1107.3 (counting bound).

Let $P_r(x)$ be the number of $r$-powerful integers $n$ with $1\le n\le x$. Then for all $x\ge 1$, \[P_r(x)\le \sum_{m\le x^{1/r}} \tau\!\left(m^{r-1}\right),\] where $\tau(n)$ is the number of positive divisors of $n$.

Proof.

By Lemma 1107.2, each $r$-powerful $n$ corresponds to a unique pair $(m,s)$ with $n=m^r s$ and $s\mid m^{r-1}$. If $n\le x$ then $m^r\le n\le x$, so $m\le x^{1/r}$. For a fixed $m$, there are at most $\tau(m^{r-1})$ choices for $s$. Summing over $m\le x^{1/r}$ gives the bound. $\square$

VERIFICATION

• Edge cases: $n=1$ is $r$-powerful and Lemma 1107.1 holds since $\operatorname{rad}(1)=1$. The decomposition in Lemma 1107.2 gives $m=s=1$ for $n=1$, consistent with uniqueness.
• In Lemma 1107.2($\Rightarrow$), the step "$g_p>0\Rightarrow f_p\ge 1$" uses that $g_p>0\Rightarrow e_p>0$ and $n$ is $r$-powerful, hence $e_p\ge r$.
• Computations: for a given bound $N$, every summand in a representation of $n\le N$ is at most $n$, hence at most $N$. Thus restricting the candidate $r$-powerful summands to $\le N$ does not miss any representations of numbers $\le N$.

FINAL

UNRESOLVED

(i) Strongest proved partial result. Lemmas 1107.1--1107.2 give a complete structural description of $r$-powerful numbers, including the unique representation $n=m^r s$ with $s\mid m^{r-1}$, and Corollary 1107.3 gives an explicit (though weak) counting upper bound.

(ii) First gap. Show (or refute) that for each $r\ge 2$ there exists $N_r$ such that every integer $n\ge N_r$ is the sum of at most $r+1$ $r$-powerful integers.

(iii) Top 3 next moves. 1. For fixed $r$, analyze the set of residues of $r$-powerful numbers modulo $M$ and test whether their $(r+1)$-fold sumset equals $\mathbb{Z}/M\mathbb{Z}$ for all sufficiently large $M$. 2. Prove an interval-covering statement: find $\theta>0$ such that for large $X$ the set $(\mathcal{P}_r+\cdots+\mathcal{P}_r)\cap [1,X]$ contains an interval of length $X^\theta$. 3. Extend computation to $r=4,5$ and search for stabilization of the maximal nonrepresentable integer, to guide conjectures about the minimal obstruction.

(iv) What a minimal counterexample would likely look like. If the statement fails for some $r$, a minimal counterexample should likely be forced by a local obstruction: a residue class modulo some prime power $p^k$ that cannot be hit by a sum of $r+1$ $r$-powerful residues, reflecting the constraint that each prime dividing an $r$-powerful summand appears with exponent at least $r$.

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.