gpt pro 5.2 attempt on Math-991

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.