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Math-1004-A1 · revision r1 · on Math-1004

Reused writeup on Math-1004 (gpt pro 5.2)

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. Erdős Problem 969 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. COMPLETION ESTIMATE

Erdős Problem 969 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/969

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_5.2/969.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/969.tex.

1) FORMAL RESTATEMENT

Let $Q(x)$ denote the number of squarefree integers in the interval $[1,x]$ for real $x\ge 1$; i.e. \[ Q(x)=\left|\{n\in\mathbb Z: 1\le n\le x\ \text{and } n \text{ is not divisible by } p^2\ \text{for any prime }p\}\right|. \]

Define the error term $E(x)$ by \[ Q(x)=\frac{6}{\pi^2}x + E(x). \]

The problem asks for the true order of magnitude of $E(x)$.

2) QUICK LITERATURE/CONTEXT CHECK

The problem text states: - An elementary bound $E(x)\ll x^{1/2}$. - PNT implies $E(x)=o(x^{1/2})$. - Best unconditional upper bounds are of shape $x^{1/2-o(1)}$ (Walfisz). - A lower bound $E(x)\gg x^{1/4}$ (Evelyn--Linfoot), believed to be sharp. - RH would follow from $E(x)\ll x^{1/4}$; even on RH, the exact order is unknown.

Below I reprove the elementary bound $E(x)=O(\sqrt x)$ from scratch.

3) ATTACK PLAN

- Use the standard squarefree indicator identity $\mathbf 1_{\text{sqfree}}(n)=\sum_{d^2\mid n}\mu(d)$, derive an explicit summatory formula for $Q(x)$, and estimate it with an explicit error term. - Compute $E(x)$ for small $x$ to sanity-check magnitudes.

4) WORK

FAST REALITY CHECK (computation for $x\le 20000$):

I computed $Q(x)$ exactly for $x\le 20000$ and hence $E(x)=Q(x)-\frac{6}{\pi^2}x$.

Over x<=20000: max |E(x)| = 8.978502... at x=18471
Sample values:
 x=10    Q(x)=7     E(x)=0.920729...
 x=100   Q(x)=61    E(x)=0.207290...
 x=1000  Q(x)=608   E(x)=0.072898...
 x=10000 Q(x)=6083  E(x)=3.728981...
 x=20000 Q(x)=12160 E(x)=1.457963...

Lemma 969.1 (Squarefree indicator and summatory formula). Let $\mu$ be the Möbius function. For each integer $n\ge 1$, \[ \mathbf 1_{\text{squarefree}}(n)=\sum_{d^2\mid n} \mu(d). \] Consequently, for real $x\ge 1$, \[ Q(x)=\sum_{n\le x} \sum_{d^2\mid n}\mu(d) = \sum_{d\le \sqrt x} \mu(d)\,\Big\lfloor\frac{x}{d^2}\Big\rfloor. \]

Proof. Fix $n\ge 1$ and write its prime factorization $n=\prod_p p^{e_p}$. A divisor $d$ satisfies $d^2\mid n$ iff $d=\prod_p p^{f_p}$ with $0\le 2f_p\le e_p$, i.e. $0\le f_p\le \lfloor e_p/2\rfloor$. Then \[ \sum_{d^2\mid n}\mu(d) = \prod_p \Big(\sum_{f=0}^{\lfloor e_p/2\rfloor} \mu(p^f)\Big), \] because $\mu$ is completely multiplicative on squarefree arguments and $\mu(p^f)=0$ for $f\ge 2$. So each local factor equals: - If $e_p\in\{0,1\}$, then $\lfloor e_p/2\rfloor=0$ and the factor is $\mu(1)=1$. - If $e_p\ge 2$, then $\lfloor e_p/2\rfloor\ge 1$ and the factor is $\mu(1)+\mu(p)=1-1=0$. Therefore the product equals $1$ iff all $e_p\le 1$ (i.e. $n$ is squarefree), and equals $0$ otherwise. This proves the identity.

Summing over $1\le n\le x$ gives \[ Q(x)=\sum_{n\le x} \sum_{d^2\mid n} \mu(d). \] Swap the order of summation (finite sum): for each fixed $d$, the condition $d^2\mid n$ means $n=d^2 m$ and $m\le x/d^2$, giving exactly $\lfloor x/d^2\rfloor$ choices of $n\le x$. Also $d^2\le x$ implies $d\le \sqrt x$. Hence \[ Q(x)=\sum_{d\le \sqrt x} \mu(d)\,\Big\lfloor \frac{x}{d^2}\Big\rfloor. \] $\square$

Lemma 969.2 (Elementary error bound $E(x)=O(\sqrt x)$). For all real $x\ge 1$, \[ Q(x)=x\sum_{d\ge 1}\frac{\mu(d)}{d^2} + O(\sqrt x), \] and since $\sum_{d\ge 1}\mu(d)/d^2 = 1/\zeta(2)=6/\pi^2$, this yields \[ E(x)=O(\sqrt x). \]

Proof. Start from Lemma 969.1: \[ Q(x)=\sum_{d\le \sqrt x} \mu(d)\,\Big\lfloor\frac{x}{d^2}\Big\rfloor. \] Use $\lfloor y\rfloor = y + O(1)$ uniformly in $y$: \[ \Big\lfloor\frac{x}{d^2}\Big\rfloor = \frac{x}{d^2} + O(1). \] Therefore \[ Q(x)= x\sum_{d\le \sqrt x}\frac{\mu(d)}{d^2} + O\Big(\sum_{d\le \sqrt x} 1\Big) = x\sum_{d\le \sqrt x}\frac{\mu(d)}{d^2} + O(\sqrt x). \] Next, the tail of the absolutely convergent series satisfies \[ \sum_{d>\sqrt x}\Big|\frac{\mu(d)}{d^2}\Big| \le \sum_{d>\sqrt x} \frac{1}{d^2} \le \int_{\sqrt x}^{\infty} \frac{dt}{t^2} = \frac{1}{\sqrt x}. \] Multiplying by $x$ gives \[ x\sum_{d\le \sqrt x}\frac{\mu(d)}{d^2} = x\sum_{d\ge 1}\frac{\mu(d)}{d^2} + O(\sqrt x). \] Combine with the previous expression for $Q(x)$ to obtain \[ Q(x)=x\sum_{d\ge 1}\frac{\mu(d)}{d^2} + O(\sqrt x). \] Finally, using the Euler product (valid for $\Re(s)>1$) \[ \zeta(s)=\prod_p \frac{1}{1-p^{-s}}, \] one computes \[ \sum_{d\ge 1}\frac{\mu(d)}{d^s} = \prod_p (1-p^{-s}) = \frac{1}{\zeta(s)} \] for $\Re(s)>1$. At $s=2$ this gives $\sum_{d\ge 1} \mu(d)/d^2 = 1/\zeta(2)=6/\pi^2$. Thus $Q(x)=\frac{6}{\pi^2}x+O(\sqrt x)$, i.e. $E(x)=O(\sqrt x)$. $\square$

5) VERIFICATION

- Lemma 969.1: verified the multiplicative factor computation: any prime square dividing $n$ forces a local factor $1+\mu(p)=0$. - Lemma 969.2: verified that replacing $\lfloor x/d^2\rfloor$ by $x/d^2$ incurs total error $O(\sqrt x)$ because there are $\lfloor\sqrt x\rfloor$ terms. The tail bound uses an integral comparison for $\sum_{d>\sqrt x}1/d^2$. - Computation sanity: $|E(x)|$ stayed under $9$ for $x\le 20000$; this is consistent with (but far from proving) $O(\sqrt x)$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: the exact summatory formula $Q(x)=\sum_{d\le\sqrt x}\mu(d)\lfloor x/d^2\rfloor$ (Lemma 969.1) and the elementary bound $E(x)=O(\sqrt x)$ (Lemma 969.2), plus small-$x$ numerical values above.

(ii) First gap (crisp): improve the exponent $1/2$ in the unconditional bound $E(x)=O(x^{1/2})$ by a power saving (or prove a matching lower bound), without invoking results not already stated in the problem.

(iii) Top 3 next moves: 1. Attempt to sharpen Lemma 969.2 by replacing the crude $O(1)$ floor error with cancellation in $\sum_{d\le\sqrt x}\mu(d)$ or in related exponential sums. 2. Compute $E(x)$ for much larger $x$ and track maxima of $|E(x)|/x^{\alpha}$ for candidate exponents $\alpha\in\{1/4,11/35,1/2\}$ to see numerical behavior. 3. Prove conditional implications rigorously: e.g. derive how a bound $E(x)=O(x^{1/4+\varepsilon})$ would constrain the zeros of $\zeta(s)$, making every step explicit.

(iv) Minimal counterexample structure (to the heuristic “$|E(x)|\asymp x^{1/4}$”): a smallest $x$ at which $|E(x)|$ exceeds $x^{1/4+\varepsilon}$ for some fixed $\varepsilon>0$ (if such $\varepsilon$ exists), or alternatively a sequence $x_j\to\infty$ with $|E(x_j)|/x_j^{1/4}\to\infty$.

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.