gpt pro 5.4 attempt on Math-927

Problem 768 (Round 2)

1. ROUND-2 OBJECTIVE

Path (C): isolate a genuine quantitative obstruction and strengthen the Round-1 result from mere density $0$ to an explicit subexponential upper bound. Round 1 only proved that \[ \frac{|A\cap[1,x]|}{x}\to 0. \] The exact Erdős asymptotic \[ \frac{|A\cap[1,x]|}{x}=\exp\!\left(-(c+o(1))\sqrt{\log x}\,\log\log x\right) \] remains out of reach, so the most promising Round-2 target is to close the first major gap: obtain a nontrivial upper bound on the natural scale \[ \exp\!\left(-c\sqrt{\log x\,\log\log x}\right). \]

2. Round-1 FOUNDATION USED

I use exactly two Round-1 ingredients.

1. The definition of \[ A=\left\{n\ge 1:\ \forall p\mid n \text{ prime},\ \exists d\mid n,\ d>1,\ d\equiv 1 \pmod p\right\}. \]
2. The Round-1 rough/smooth decomposition idea: if $p=P(n)$ is the largest prime factor of $n\in A$, then there is a divisor of the form $1+ap$ dividing $n$, and this can be counted by summing over $(p,a)$.

I do not reprove the Round-1 density-$0$ theorem; I strengthen it quantitatively.

3. NEW INSIGHT / TOOL (ROUND 2)

The new step is to let the smoothness threshold depend on $x$ and optimize it.

Write $\Psi(x,y)$ for the number of $y$-smooth integers $\le x$. Then the Round-1 argument actually gives, for every $2\le y\le x$, \[ |A\cap[1,x]| \le \Psi(x,y) + x\sum_{\substack{p>y\\ p\ \mathrm{prime}}}\ \sum_{1\le a\le x/p}\frac{1}{p(1+ap)}. \] Using a standard smooth-number estimate in the Canfield--Erdős--Pomerance range, and then optimizing $y$, yields a genuine subexponential decay.

4. ATTACK PLAN (ROUND 2)

The exact Round-1 gap was this: density $0$ is much too weak compared to the problem's conjectural scale. So I now prove the following explicit theorem.

Theorem. Let \[ A(x)=|A\cap[1,x]|. \] Then, as $x\to\infty$, \[ A(x)\le x\exp\!\left(-\left(\frac{1}{\sqrt 2}+o(1)\right)\sqrt{\log x\,\log\log x}\right). \]

This does not settle Erdős's conjectured $\sqrt{\log x}\,\log\log x$ scale, but it is a strict Round-2 advance beyond density $0$.

5. WORK (ROUND 2)

Let \[ L=\log x,\qquad L_2=\log\log x. \]

Step 1: a uniform counting inequality.

Fix $y$ with $2\le y\le x$. Split \[ A\cap[1,x] = A_{\le y}(x)\cup A_{>y}(x), \] where $A_{\le y}(x)$ consists of those $n\in A\cap[1,x]$ with largest prime factor at most $y$, and $A_{>y}(x)$ those with largest prime factor exceeding $y$.

Clearly, \[ |A_{\le y}(x)|\le \Psi(x,y). \]

Now let $n\in A_{>y}(x)$, and let $p=P(n)>y$ be the largest prime factor of $n$. By the defining property of $A$, applied to this prime $p$, there exists a divisor $d\mid n$ with $d>1$ and $d\equiv 1\pmod p$. Write \[ d=1+ap,\qquad a\ge 1. \] Because $d\equiv 1\pmod p$, we have $p\nmid d$, so $p\,d\mid n$. Hence $n$ is a multiple of \[ p(1+ap). \] Therefore, for fixed $(p,a)$, the number of such $n\le x$ is at most \[ \frac{x}{p(1+ap)}. \] Summing over all admissible $p$ and $a$ gives \[ |A_{>y}(x)| \le x\sum_{\substack{p>y\\ p\ \mathrm{prime}}}\ \sum_{1\le a\le x/p}\frac{1}{p(1+ap)}. \] Since \[ \frac{1}{p(1+ap)}\le \frac{1}{a p^2}, \] we obtain \[ |A_{>y}(x)| \le x\sum_{\substack{p>y\\ p\ \mathrm{prime}}}\frac{1}{p^2}\sum_{1\le a\le x/p}\frac{1}{a} \le x(1+\log x)\sum_{p>y}\frac{1}{p^2}. \] Finally, \[ \sum_{p>y}\frac{1}{p^2}\le \sum_{m>y}\frac{1}{m^2}\le \frac{1}{y-1}, \] so \[ |A_{>y}(x)|\le \frac{x(1+\log x)}{y-1}. \]

Combining the two pieces: \[ A(x)\le \Psi(x,y)+\frac{x(1+\log x)}{y-1}. \tag{5.1} \]

Step 2: choose the threshold.

Take \[ y=\exp\!\left(\frac{1}{\sqrt 2}\sqrt{L\,L_2}\right). \] Then \[ u:=\frac{\log x}{\log y}=\sqrt{\frac{2L}{L_2}}. \]

A standard smooth-number theorem of Canfield--Erdős--Pomerance (in the form quoted in Granville's survey) says that if $u\to\infty$ and $u\le y^{1-\varepsilon}$ for some fixed $\varepsilon>0$, then \[ \Psi(x,y)=x\,u^{-u+o(u)}=x\exp\left(-(1+o(1))u\log u\right). \] For the present choice of $y$, we have $u\to\infty$ and, for example, $u\le y^{1/2}$ for all large $x$, because \[ \log u=O(L_2)=o(\log y). \] Hence the theorem applies.

Now \[ u\log u = \sqrt{\frac{2L}{L_2}}\cdot \frac{1}{2}\log\!\left(\frac{2L}{L_2}\right) = \left(\frac{1}{\sqrt 2}+o(1)\right)\sqrt{L\,L_2}. \] Therefore \[ \Psi(x,y) \le x\exp\!\left(-\left(\frac{1}{\sqrt 2}+o(1)\right)\sqrt{L\,L_2}\right). \tag{5.2} \]

On the other hand, \[ \frac{x(1+\log x)}{y-1} = x\exp\!\left(-\frac{1}{\sqrt 2}\sqrt{L\,L_2}+O(L_2)\right) = x\exp\!\left(-\left(\frac{1}{\sqrt 2}+o(1)\right)\sqrt{L\,L_2}\right), \tag{5.3} \] since $L_2=o(\sqrt{L\,L_2})$.

Insert $5.2$ and $5.3$ into $5.1$: \[ A(x)\le x\exp\!\left(-\left(\frac{1}{\sqrt 2}+o(1)\right)\sqrt{L\,L_2}\right). \] This proves the theorem.

A simple structural check.

The proof above is consistent with the elementary fact that every $n\in A$ satisfies \[ P(n)\le \sqrt n. \] Indeed, if $p=P(n)>\sqrt n$, then $p^2>n$, so $p$ occurs only to the first power in $n$; any divisor $d\equiv 1\pmod p$ must be coprime to $p$, hence $d\mid n/p<p$, impossible because $d>1$ and $d\equiv 1\pmod p$ force $d\ge p+1$.

6. ADVERSARIAL VERIFICATION

There are three places where the new argument could fail.

1. Double counting of $n$. That is harmless: the bound for $A_{>y}(x)$ is an upper bound, so overcounting is allowed.

2. Range of the smooth-number theorem. For my choice \[ y=\exp\!\left(\frac{1}{\sqrt 2}\sqrt{L\,L_2}\right), \] we have $u=\sqrt{2L/L_2}\to\infty$ and \[ \log u=O(L_2)=o(\log y), \] hence $u\le y^{1/2}$ for large $x$. So the Canfield--Erdős--Pomerance estimate applies.

3. Possible loss in the rough part. The rough bound is extremely crude, but all I need is the exponent scale. The factor $1+\log x$ only contributes $O(L_2)$ in the exponent, which is negligible compared with $\sqrt{L\,L_2}$.

So the Round-2 theorem is gap-free.

7. FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 strengthens the Round-1 conclusion from \[ A(x)=o(x) \] to the explicit subexponential estimate \[ A(x)\le x\exp\!\left(-\left(\frac{1}{\sqrt 2}+o(1)\right)\sqrt{\log x\,\log\log x}\right). \] This is still weaker than the upper bound quoted on the problem page, and very far from the conjectured \[ \exp\!\left(-(c+o(1))\sqrt{\log x}\,\log\log x\right) \] law, but it is a real quantitative advance beyond Round 1.

8. COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 38%

9. REFERENCES

• T. F. Bloom, Erdős Problem #768, ErdosProblems.com, accessed 2026-03-07.
• A. Granville, Smooth numbers: computational number theory and beyond; used for the quoted Canfield--Erdős--Pomerance smooth-number estimate.