GPT 6 Astra Ultra attempt on Math-5391

Erdős Problem 47: the harmonic-mass criterion and its arithmetic limits

1) Problem and source status

Fix $\delta>0$. If $N$ is sufficiently large in terms of $\delta$ and $A\subset[1,N]$ satisfies \[ R(A):=\sum_{a\in A}\frac1a>\delta\log N, \] must there be a subset $S\subseteq A$ with $R(S)=1$? The current problem page, checked on 23 September 2026, records the affirmative answer due to Bloom, with a subsequently improved quantitative bound by Liu and Sawhney.

We use Bloom's explicitly stated quantitative theorem to deduce the original assertion and its stability under deleting small denominators. We then prove elementary arithmetic obstructions explaining why an exact unit sum does not follow from a merely large fixed reciprocal mass. The analytic theorem is an external dependency, not a proof newly established in this note.

2) The external quantitative theorem

Theorem 3 of Bloom's On a density conjecture about unit fractions states that an absolute constant $C>0$ suffices as follows: for all sufficiently large $N$, if $A\subset[1,N]$ and \[ R(A)\geq C\frac{\log\log\log N}{\log\log N}\log N, \] then some $S\subseteq A$ has $R(S)=1$. We use this theorem in exactly this form. Its proof involves an arithmetic selection argument and a Fourier estimate; those parts are not rederived here. The more recent improvements are unnecessary for the fixed-$\delta$ question, and no optimal threshold is claimed.

3) Completing the fixed-$\delta$ deduction

Put $t=\log\log N$. The factor in Bloom's threshold is \[ \frac{\log\log\log N}{\log\log N}=\frac{\log t}{t}, \] which tends to zero as $N$ tends to infinity. For a concrete sufficient condition in terms of the theorem's constant, use $\log t\leq\sqrt t$ for $t\geq1$ and choose \[ t\geq\max\{1,(2C/\delta)^2\}. \] Then \[ C\frac{\log t}{t}\leq\frac{C}{\sqrt t}\leq\frac\delta2. \] Choose $N$ also beyond the absolute threshold in Bloom's theorem. The assumed inequality $R(A)>\delta\log N$ exceeds the required threshold, so the desired subset exists.

This proves the assertion for every fixed $\delta>0$, with one threshold depending only on $\delta$ and the constants in the cited theorem. If $1\in A$, the conclusion is already given by $S=\{1\}$; otherwise the theorem supplies a set of denominators all at least $2$. We have not converted the proof into a numerical value of $N_0(\delta)$, because the absolute constants in the external input have not been numerically specified.

4) Deleting an initial segment and obtaining disjoint witnesses

Fix a cutoff $T\geq1$. Removing all denominators at most $T$ decreases the reciprocal mass by at most \[ H_T=\sum_{n=1}^T\frac1n. \] If $N$ is also large enough that $H_T<\delta\log N/2$, then \[ R(A\cap(T,N])>\frac\delta2\log N. \] Applying the fixed-density deduction with $\delta/2$ gives a witness using only denominators larger than $T$.

As a consequence, suppose an infinite set $A\subset\mathbb N$ has positive upper logarithmic density: \[ \limsup_{N\to\infty}\frac{R(A\cap[1,N])}{\log N}>0. \] Choose a fixed positive $\delta$ smaller than this upper limit. There are arbitrarily large $N$ satisfying the required mass bound, so the preceding argument gives a witness past every cutoff $T$. Repeating with the maximum denominator found so far produces infinitely many pairwise disjoint finite subsets of $A$, each of reciprocal sum $1$. This argument specifically uses logarithmic density; it does not replace an ordinary-density theorem by an unjustified comparison of finite cardinalities and reciprocal sums.

5) A prime-power obstruction and safe elimination rule

Let $S$ be a finite set of integers at least $2$ with $R(S)=1$. Fix a prime $p$ which divides some element of $S$, and let \[ a=\max_{n\in S}v_p(n)\geq1. \] At least two elements must attain this maximum. For if exactly one element $n_0$ attained it, take $L=\operatorname{lcm}(S)$ and multiply the unit-fraction equality by $L$: \[ \sum_{n\in S}L/n=L. \] Modulo $p$, every term except $L/n_0$ is zero, whereas $L/n_0$ is nonzero. The right side is zero. This contradiction proves the claim.

This gives a valid finite preprocessing rule. If an element $n_0$ of an available denominator set $A$ uniquely maximizes a positive $p$-adic valuation for some prime $p$, it cannot belong to any subset of $A$ with reciprocal sum $1$, and may be removed. Applying the same reasoning after each removal preserves all possible witnesses. The terminal set need not contain a witness: for instance $\{6,10,15\}$ has no uniquely attained positive maximal valuation, but its total reciprocal mass is only $1/3$.

In particular, no finite set of distinct prime denominators has a unit sum, since each prime appears to positive valuation only in its own denominator. Nevertheless, the sum of reciprocals of the primes diverges. Thus no fixed absolute reciprocal-mass threshold can replace the hypothesis in the problem. This does not contradict the theorem: prime reciprocal mass grows much more slowly than $\log N$.

6) Verification and outcome

The analytic dependence is confined to the explicitly stated theorem in Section 2. The limit estimate, deletion argument, and valuation obstruction are proved above. The finite verification script verification/solved_37_45_46_47_checks.py checks that the elimination rule never removes a denominator belonging to a unit subset for any set of available denominators in $\{2,\ldots,14\}$. It uses exact rational arithmetic. This validates a finite illustration of the elementary rule, not Bloom's infinite theorem.

Outcome: SOLVED using Bloom's quantitative theorem. The original fixed-$\delta$ assertion follows completely. No claim is made to prove the best possible harmonic-mass threshold, to reproduce the full analytic argument, or to rerun a formal proof.

7) Sources

• T. F. Bloom, On a density conjecture about unit fractions, arXiv:2112.03726v2, Theorem 3, https://arxiv.org/html/2112.03726v2; the theorem statement and surrounding discussion were checked on 23 September 2026.
• T. F. Bloom, Problem 47, https://www.erdosproblems.com/47, checked 23 September 2026, for the current status and the distinction between the proved bound and sharper threshold questions.

The estimate records coverage of the original question with the stated external theorem accepted, not a fresh proof of that theorem or a novelty assessment.

8) COMPLETION ESTIMATE

COMPLETION: 100%