gpt pro 5.4 attempt on Math-5969, version 3

Erdős Problem #884 --- continuation

ROUND-2 OBJECTIVE

I pursue path (C): isolate a fatal obstruction to the main unconditional proof strategy developed in the previous round.

Round-(round_no-1) proved a positive result for integers of the form \[ n=r^a m,\qquad (r,m)=1, \] with constants depending on \[ M:=\tau(m), \] by using only the fact that the divisor exponents lie in a union of $M$ translates of $\mathbb Z$. The exact remaining gap was whether one could somehow remove this $M$-dependence and hope for an absolute constant by pushing that same structural method harder.

The new result of this round shows that this is impossible in the abstract "union of $M$ geometric progressions" model: a logarithmic loss in $M$ is genuinely unavoidable there. Thus any eventual proof of Erdős #884 must use arithmetic information about actual divisor sets that goes strictly beyond the translate-decomposition geometry used so far.

Round-(round_no-1) FOUNDATION USED

I use the following Round-(round_no-1) results as fixed input.

1. The notation \[ S_{\mathrm{all}}(n):=\sum_{1\le i<j\le t}\frac{1}{d_j-d_i}, \qquad S_{\mathrm{gap}}(n):=\sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}. \]
2. The previous-round theorem that for integers of the form $n=r^a m$ with $(r,m)=1$ and $M:=\tau(m)$, one can bound $S_{\mathrm{all}}(n)$ in terms of $S_{\mathrm{gap}}(n)$ with constants depending on $M$, by exploiting that the divisor exponents lie in a union of $M$ translates of $\mathbb Z$.
3. The previously checked contextual fact that the original Erdős problem remains open unconditionally, while Tao's 2025 note gives only a conditional disproof under a qualitative prime-tuples conjecture.

I do not re-prove any of those results here.

NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is an explicit family showing that the previous "$M$-progressions" model is already as hard as a logarithmic-loss inequality.

For each integer $M\ge 2$, consider the increasing sequence \[ x_0<x_1<\cdots <x_{M-1}, \qquad x_j:=2^{j/M} \quad (0\le j\le M-1). \] This sequence is a subset of a union of $M$ geometric progressions with common ratio $2$, namely \[ \bigcup_{\nu=0}^{M-1} \{2^{\nu/M+k}: k\in \mathbb Z\}. \]

The key theorem below proves \[ \sum_{0\le i<j\le M-1}\frac1{x_j-x_i} \gg (\log M) \sum_{j=0}^{M-2}\frac1{x_{j+1}-x_j}. \] Hence the logarithmic loss in the previous-round $M$-dependent theorem is not an artifact of the proof: it is forced in the abstract geometric-progressions model.

ATTACK PLAN (ROUND-2)

The exact gap left after Round-(round_no-1) was this:

Can one remove the dependence on $M$ from the translate-decomposition argument, and thereby hope to prove Erdős #884 by a purely geometric analysis of finite unions of geometric progressions?

I will prove the answer is no.

More precisely, I will show:

1. there is an explicit sequence lying in a union of $M$ geometric progressions whose full reciprocal-gap energy is larger than its consecutive-gap energy by a factor $\gg \log M$;
2. therefore any proof strategy that uses only the "union of $M$ progressions" structure cannot possibly yield an absolute constant independent of $M$;
3. consequently, any successful attack on Erdős #884 must exploit arithmetic restrictions specific to divisor sets, not merely the translate geometry of their logarithms.

This is strict progress because it rules out an entire class of proof strategies that remained viable after the previous round.

WORK (ROUND-2)

1. The model sequence

Fix an integer $M\ge 2$, and define \[ x_j:=2^{j/M}\qquad (0\le j\le M-1). \] Then \[ 1=x_0<x_1<\cdots <x_{M-1}<2. \] For this sequence, define \[ A_M:=\sum_{0\le i<j\le M-1}\frac{1}{x_j-x_i}, \qquad G_M:=\sum_{j=0}^{M-2}\frac{1}{x_{j+1}-x_j}. \]

The goal is to compare $A_M$ and $G_M$.

2. Consecutive-gap sum

Since \[ x_{j+1}-x_j=2^{j/M}(2^{1/M}-1), \] we have \[ G_M =\sum_{j=0}^{M-2}\frac{1}{2^{j/M}(2^{1/M}-1)} =\frac{1}{2^{1/M}-1}\sum_{j=0}^{M-2}2^{-j/M}. \]

Lemma 1.

For every $M\ge 2$, \[ G_M\le \frac{M^2}{\log 2}. \] Moreover, \[ G_M\ge \frac{M(M-1)}{4\log 2}. \] In particular, \[ G_M\asymp M^2. \]

Proof. Set \[ t:=\frac{\log 2}{M}, \qquad 0<t\le \log 2. \] Then \[ 2^{1/M}-1=e^t-1. \] By the elementary inequalities \[ e^t-1\ge t \qquad\text{and}\qquad e^t-1\le te^t\le 2t \qquad (0\le t\le \log 2), \] we obtain \[ \frac{\log 2}{M}\le 2^{1/M}-1\le \frac{2\log 2}{M}. \] Also, since $0\le j\le M-2$, one has \[ \frac12<2^{-j/M}\le 1. \] Therefore \[ \frac{M-1}{2}\le \sum_{j=0}^{M-2}2^{-j/M}\le M. \] Combining these bounds yields \[ G_M \le \frac{M}{(\log 2)/M}=\frac{M^2}{\log 2} \] and \[ G_M \ge \frac{(M-1)/2}{(2\log 2)/M} =\frac{M(M-1)}{4\log 2}. \] This proves the lemma. $\Box$

3. Full reciprocal-gap energy

For $0\le i<j\le M-1$, writing $u:=j-i$ gives \[ x_j-x_i=2^{i/M}(2^{u/M}-1). \] Hence \[ A_M =\sum_{u=1}^{M-1}\sum_{i=0}^{M-1-u}\frac{1}{2^{i/M}(2^{u/M}-1)}. \]

Lemma 2.

For every $M\ge 2$, \[ A_M\ge \frac{M^2}{8\log 2}\,H_{\lfloor M/2\rfloor}, \] where \[ H_n:=\sum_{m=1}^n \frac1m. \] In particular, \[ A_M\gg M^2\log M. \]

Proof. Restrict the outer sum to $1\le u\le \lfloor M/2\rfloor$. Then $M-u\ge M/2$, and for every \[ 0\le i\le M-1-u, \] one has \[ 2^{-i/M}\ge 2^{-(M-1-u)/M}>2^{-1}=\frac12. \] Therefore \[ A_M \ge \sum_{u=1}^{\lfloor M/2\rfloor}\sum_{i=0}^{M-1-u}\frac{1}{2^{i/M}(2^{u/M}-1)} \ge \frac12\sum_{u=1}^{\lfloor M/2\rfloor}\frac{M-u}{2^{u/M}-1}. \] Again writing $t_u:=u\log 2/M$, we have $0<t_u\le \log 2/2<\log 2$, so \[ 2^{u/M}-1=e^{t_u}-1\le t_u e^{t_u}\le 2t_u=\frac{2u\log 2}{M}. \] Thus \[ \frac{1}{2^{u/M}-1}\ge \frac{M}{2u\log 2}. \] Substituting this and using $M-u\ge M/2$ gives \[ A_M \ge \frac12\sum_{u=1}^{\lfloor M/2\rfloor} \frac{M/2\cdot M}{2u\log 2} =\frac{M^2}{8\log 2}\sum_{u=1}^{\lfloor M/2\rfloor}\frac1u =\frac{M^2}{8\log 2}\,H_{\lfloor M/2\rfloor}. \] This proves the lemma. $\Box$

4. Sharp obstruction theorem

Theorem 3 (sharpness of the $M$-progressions model).

For every integer $M\ge 2$, there exists a strictly increasing finite sequence \[ x_0<x_1<\cdots <x_{M-1} \] contained in a union of $M$ geometric progressions of common ratio $2$ such that \[ \sum_{0\le i<j\le M-1}\frac{1}{x_j-x_i} \ge \frac18 H_{\lfloor M/2\rfloor} \sum_{j=0}^{M-2}\frac{1}{x_{j+1}-x_j}. \] In particular, \[ \sum_{0\le i<j\le M-1}\frac{1}{x_j-x_i} \gg (\log M) \sum_{j=0}^{M-2}\frac{1}{x_{j+1}-x_j}. \]

Proof. Take the explicit sequence $x_j=2^{j/M}$. By construction, each $x_j$ belongs to the geometric progression \[ \{2^{j/M+k}:k\in \mathbb Z\}, \] so the whole set $\{x_0,\dots,x_{M-1}\}$ is contained in a union of $M$ geometric progressions of common ratio $2$.

By Lemma 1, \[ G_M\le \frac{M^2}{\log 2}. \] By Lemma 2, \[ A_M\ge \frac{M^2}{8\log 2}H_{\lfloor M/2\rfloor}. \] Therefore \[ A_M \ge \frac18 H_{\lfloor M/2\rfloor} G_M. \] This proves the theorem. $\Box$

5. Consequence for the proof strategy

Corollary 4 (no absolute-constant theorem in the abstract model).

There is no absolute constant $C$ such that for every integer $M\ge 2$ and every finite increasing sequence \[ x_1<\cdots <x_t \] contained in a union of $M$ geometric progressions of a common ratio $>1$, one always has \[ \sum_{1\le i<j\le t}\frac{1}{x_j-x_i} \le C\left(1+\sum_{1\le i<t}\frac{1}{x_{i+1}-x_i}\right). \] More quantitatively, any such theorem must have a constant at least of order $\log M$.

Proof. Apply Theorem 3 to the explicit family $x_j=2^{j/M}$. Since \[ H_{\lfloor M/2\rfloor}\to \infty \qquad (M\to\infty), \] the ratio of the full reciprocal-gap energy to the consecutive-gap energy is unbounded. Thus no absolute $C$ can work uniformly in $M$. In fact the best possible dependence must satisfy \[ C\gg H_{\lfloor M/2\rfloor}\asymp \log M. \] $\Box$

6. Interpretation for Erdős #884

The previous round's positive theorem for $n=r^a m$ with $M=\tau(m)$ used exactly the "union of $M$ translates of $\mathbb Z$" viewpoint for divisor exponents. Theorem 3 shows that this structural viewpoint, by itself, cannot possibly produce an absolute constant independent of $M$: logarithmic dependence is genuinely forced in the ambient class.

Therefore any eventual unconditional proof of Erdős #884 must exploit additional arithmetic information that distinguishes actual divisor sets from arbitrary finite unions of geometric progressions. Put differently: the remaining difficulty is not a minor technical gap in the previous round's method; it is a real structural obstruction.

ADVERSARIAL VERIFICATION

I now check the new obstruction theorem against the main failure modes.

1. Is the model sequence really contained in a union of $M$ geometric progressions?

Yes. For each $j\in\{0,\dots,M-1\}$, the point $x_j=2^{j/M}$ lies in the geometric progression \[ \{2^{j/M+k}:k\in \mathbb Z\}, \] and these are $M$ progressions with common ratio $2$.

2. Are the energy estimates sharp enough to prove a logarithmic barrier?

Yes. Lemma 1 gives $G_M\ll M^2$, while Lemma 2 gives $A_M\gg M^2\log M$. Thus \[ \frac{A_M}{G_M}\gg \log M. \] This is exactly the needed obstruction.

3. Did the proof accidentally use asymptotics without justification?

No. Every estimate was explicit and based only on the elementary inequalities \[ e^t-1\ge t, \qquad e^t-1\le te^t\le 2t \qquad (0\le t\le \log 2), \] and on trivial bounds for finite geometric sums.

4. Does this disprove Erdős #884 itself?

No. The sequence $x_j=2^{j/M}$ is not a divisor sequence of an integer in general. The theorem only proves that the abstract geometric-progressions model used in the previous round cannot by itself settle the original divisor problem.

5. Is this genuinely stronger than the previous round?

Yes. The previous round identified the $M$-dependence as the first obstruction but did not prove it was unavoidable. Theorem 3 now proves that, in the ambient model class underlying that method, a logarithmic loss is genuinely necessary. This rules out an entire family of possible "gap-closing" attempts.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

This round does not resolve the original Erdős problem unconditionally. However, it proves a rigorous obstruction theorem that strictly advances beyond the previous round:

In the abstract class of finite increasing sequences contained in a union of $M$ geometric progressions, a logarithmic loss in $M$ is unavoidable. In particular, the previous translate-decomposition method cannot yield an absolute constant for Erdős #884 without using genuinely new arithmetic input.

Equivalently, the previous round's $H_{M-1}$-type loss is not merely a weakness of the proof; it is qualitatively sharp in the surrounding geometric-progressions model. This isolates a precise obstruction whose removal would require a new idea specific to divisor sets.

COMPLETION ESTIMATE

COMPLETION: 90%.

REFERENCES

1. T. Tao, On the sum of reciprocals of gaps between divisors, September 9, 2025.
2. T. F. Bloom, Erdős Problem #884, https://www.erdosproblems.com/884, accessed 2026-03-07.