GPT 6 Astra Ultra attempt on Math-657

Erdős Problem #50: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

Let $G(c)$ be the limiting distribution of $\varphi(n)/n$. Must there be no point of $[0,1]$ with an existing finite positive derivative (one-sided at endpoints)?

All supplied versions considered: 50.tex, 50_v2.tex, 50_v3.tex, 50_v4.tex, 50_v5.tex, 50_v6.tex, 50_v7.tex. There is a numbering collision: 50.tex treats minimal inverse totients, a different question. Its compression proof cannot advance this derivative problem. Versions 2--4 concern residual flatness, one-sided oscillation, and intermediate secant slopes; the intermediate-value step is elementary, not evidence that resolution is close. Version 5 and the consolidation in version 7 use a one-sided covering assertion requiring correction below. Version 6 concerns a logarithmic modulus, whose scale is much larger than the linear derivative scale. All seven files were read in full.

2) WORK

Repair of the measure-theoretic step.

Let $\mu$ be a finite singular Borel measure on $\mathbb R$, supported on a Lebesgue-null Borel set $S$. For positive integers $M,j$ define \[ E_{M,j}=\{x\in S:\mu((x-h,x])\le Mh\text{ for every }0<h<1/j\}. \] I prove $\mu(E_{M,j})=0$. Choose an open set $O\supset S$ with $|O|<\eta<1/j$. Each component $(a,b)$ of $O$ has length below $1/j$. For $x\in E_{M,j}\cap(a,b)$, \[ \mu((a,x])\le M(x-a)\le M(b-a). \] Take a sequence of such $x$ increasing to the supremum of the set in this component; if a maximum exists use it. Continuity from below shows $\mu(E_{M,j}\cap(a,b))\le M(b-a)$. Summing over the countably many components gives $\mu(E_{M,j})\le M\eta$. Let $\eta\downarrow0$. Every point with finite left upper density belongs to some $E_{M,j}$, apart from the $\mu$-null complement of $S$. Consequently the left upper density is infinite $\mu$-almost everywhere. Reflection gives the right-hand result. This proves the intended conclusion without the claimed one-sided Besicovitch selection.

Why the stated selection rule is unsafe.

The family $I_x=(x-1,x]$, $0<x<1$, cannot have a bounded-overlap subfamily covering all $x\in(0,1)$. Such a subfamily needs right endpoints approaching $1$ and hence infinitely many intervals containing $1/2$. This refutes the general endpoint-based selection assertion as it was stated. It does not refute the density theorem, repaired above.

For continuous $G$ with singular $dG$, finite positive differentiability points are therefore $dG$-null. This conclusion still permits an exceptional point.

3) ADVERSARIAL VERIFICATION

The sets can be defined using rational $h$ by continuity from below of interval masses, ensuring measurability. Finite-density points cannot be atoms. The component proof handles its supremum without assuming it belongs to $E_{M,j}$. Singularity and residual irregularity do not turn a null exceptional set into the empty set; endpoints also require separate analysis.

4) FINAL

UNRESOLVED. No pointwise exclusion of every nonspecial point, nor a point with a positive derivative, has been found. The inverse-totient branch was considered but is unrelated to the target. The corrected proof repairs a local gap rather than increasing an asserted 95% completion.

5) SOURCES CHECKED

Checked the actual statement in https://raw.githubusercontent.com/google-deepmind/formal-conjectures/main/FormalConjectures/ErdosProblems/50.lean on 2026-09-23 (statement reference, not a proof; the file contains sorry). Also checked the publisher record for Toulmonde, Acta Arith. 121 $2006$, 367--402, DOI 10.4064/aa121-4-6; its detailed modulus theorem is not needed in the repair.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 12%