Erdős Problem #884

Erdős Problem 884: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

Statement and current resolution

For a finite increasing set $D$ define \[ S(D)=\sum_{a,b\text{ consecutive in }D}\frac1{b-a},\qquad T(D)=\sum_{a<b\text{ in }D}\frac1{b-a}. \] Is $T(\operatorname{div}n)/(1+S(\operatorname{div}n))$ bounded independently of $n$? No. We reconstruct the multiscale part of Daniel Larsen's unconditional proof, https://raw.githubusercontent.com/Larsen-Daniel/Erdos-884/main/884.pdf. The one-scale prime-packet estimate is explicitly imported below. This distinguishes the current result from Tao's earlier conditional proof, https://terrytao.wordpress.com/wp-content/uploads/2025/09/erdos-884.pdf.

Precisely stated prime-packet input

Write $f(t)=\log_*t$, the number of iterated natural logarithms needed to reach at most one. Corollary 2.4 and its proof in Larsen's paper supply, for every sufficiently large $t$, a set $B(t)$ of primes in $[t,2t]$, with product $m(t)$, such that \[ k(t):=|B(t)|=o\left(\frac{\log t}{\log\log t}\right),\quad \operatorname{diam} B(t)\le\log^2t,\quad T(B(t))\gg\frac1{f(t)},\quad S(\operatorname{div}m(t))\ll\frac1{\log\log t}.\tag{1} \] The constants are absolute. This input is unconditional: its source uses the prime number theorem, an upper-bound Brun sieve, and Tao's real-rooted-polynomial gap lemma. We do not import the multiscale conclusion.

For context, the prime selection has $k\asymp\log t/(f(t)\log\log t)$ points separated by $\gg\log t$ in an interval of length $H\ll k\log t$. Its neighboring-prime contribution is $O(k/\log t)$; the nonprime-divisor contribution is $O(1/\log\log t)$ by the source's polynomial lemma. Its energy is at least $\gg k^2\log k/H$. This last assertion has a short elementary proof: for each $1\le d\le k/2$, the sum of the $k-d$ distances $b_{i+d}-b_i$ is at most $dH$, so Cauchy--Schwarz gives reciprocal sum at least $(k-d)^2/(dH)$. Summing in $d$ gives the claimed logarithm. These facts explain both bounds in $1$; the polynomial gap estimate itself remains a declared external input.

Products across widely separated scales

Choose large $t_0$ and put $t_{j+1}=\exp(t_j)$. Let $B_j=B(t_j)$, $m_j=\prod_{p\in B_j}p$, and $n_R=\prod_{j=0}^R m_j$. Prime packets are disjoint. A divisor $d$ of $n_R$ has unique components $d^{(j)}\mid m_j$; let $v_j(d)$ be the number of primes used in that component. Put $k_j=|B_j|$ and $w_j=\operatorname{diam}B_j$.

We record uniform scale bounds, all following from $1$ and the exponential recursion: \[ \sum_{j< i}k_j\log(2t_j)=o(\log t_i),\quad \sum_{j>i}\frac{k_jw_j}{t_j}=o((2t_i)^{-k_i}),\quad 4^{\sum_{j\le\ell}k_j}=t_\ell^{o(1)}.\tag{2} \] For example the first sum is dominated by $O((\log t_{i-1})^2)$, whereas $\log t_i=t_{i-1}$. For the second, its first term is at most a fixed power of $t_i$ times $\exp(-t_i)$, whereas $(2t_i)^{-k_i}=\exp(-o((\log t_i)^2))$; later terms decay still faster. The third uses $k_\ell=o(\log t_\ell)$. Increasing $t_0$ absorbs all finite initial qualifications.

Consider consecutive divisors $d_1<d_2$ of $n_R$, and let $\ell$ be the largest active packet, meaning $d_1^{(\ell)}d_2^{(\ell)}>1$. There are two types of exceptional pairs.

First suppose the vectors $v(d_1),v(d_2)$ differ, and let $i$ be the largest coordinate where they differ. Up to exchanging the two divisors, the logarithm of the component ratio at $i$ is at least $\log t_i-O(k_iw_i/t_i)$. At larger coordinates equal counts give logarithmic ratios of absolute value at most $k_jw_j/t_j$. At smaller coordinates use the first bound of $2$. Thus the total logarithmic ratio has absolute value at least $\frac12\log t_i$. The larger divisor is at least $t_\ell$, so $d_2-d_1\gg t_\ell$.

Second suppose all counts agree, let $i$ be the smallest coordinate with different components, and suppose $i<\ell$. The two integer components at $i$ are distinct and at most $(2t_i)^{k_i}$, so their logarithmic ratio has absolute value at least $c(2t_i)^{-k_i}$. The smaller-coordinate ratios are zero and the higher-coordinate errors are negligible by $2$. Both divisors are at least $t_\ell$, giving \[ d_2-d_1\gg t_\ell(2t_i)^{-k_i}. \] For $i<\ell$, the reciprocal extra factor $(2t_i)^{k_i}$ is $t_\ell^{o(1)}$, since $t_i\le\log t_\ell$.

There are at most $4^{\sum_{j\le\ell}k_j}=t_\ell^{o(1)}$ ordered divisor pairs whose largest active packet is $\ell$. The total contribution of both exceptional types is therefore at most \[ \sum_{\ell=0}^R t_\ell^{-1+o(1)}=O(1),\tag{3} \] with a bound independent of $R$.

All remaining consecutive pairs have identical components below some $i$, different components at $i$, and no higher components. They are $au,av$, where $a\mid n_{i-1}$ and $u,v\mid m_i$. Necessarily $u,v$ are consecutive divisors of $m_i$, since any intermediate divisor would give an intermediate divisor of $n_R$. Their contribution is at most \[ \left(\sum_{a\mid n_{i-1}}\frac1a\right)S(\operatorname{div}m_i). \] The factor is uniformly bounded, even by two after increasing $t_0$: it is $\prod_{j<i}\prod_{p\in B_j}(1+1/p)\le\exp(\sum_j k_j/t_j)$. Together with $3$, this proves \[ S(\operatorname{div}n_R)\le O(1)+2\sum_{i=0}^R S(\operatorname{div}m_i).\tag{4} \]

Accumulating enough energy

Prime pairs from different packets are distinct pairs of divisors, so \[ T(\operatorname{div}n_R)\ge\sum_{i=0}^R T(B_i)=:U_R. \] Since $f(t_i)=f(t_0)+i$, $1$ shows $U_R\gg\sum_{i\le R}(f(t_0)+i)^{-1}\to\infty$. Moreover \[ \frac{S(\operatorname{div}m_i)}{T(B_i)} \ll\frac{f(t_i)}{\log\log t_i}\longrightarrow0. \] For any $\varepsilon>0$, split the sum at a fixed index beyond which this ratio is at most $\varepsilon$. The finite initial part divided by $U_R$ vanishes, and the tail is at most $\varepsilon$. Thus $4$ implies $S(\operatorname{div}n_R)=o(U_R)$, and \[ \frac{T(\operatorname{div}n_R)}{1+S(\operatorname{div}n_R)}\ge \frac{U_R}{1+o(U_R)}\longrightarrow\infty. \]

Assessment

The negative answer is fully deduced from the exact unconditional packet estimate $1$. The proof above supplies the cross-packet gap control, the uniform error sum, and the divergent energy needed to overcome the denominator's constant one. The packet's polynomial-divisor estimate is credited and explicitly external; no prime-tuples conjecture is assumed.

Completion rate estimate: 100%.

Erdős Problem #884 --- continuation

ROUND-2 OBJECTIVE

I pursue path (C) in the following precise sense: I isolate the first concrete arithmetic obstruction inside the genuine divisor family with exactly two distinct prime factors.

Round-(round_no-1) proved general positive results for fixed smoothness classes and for several structured subfamilies, but it still left the basic two-prime-power family \[ n=p^a q^b \qquad (p<q\text{ distinct primes},\ a,b\ge 0) \] without a sharp internal reduction. In this round I prove that the full reciprocal-gap sum on this family has an exact decomposition into:

1. an elementary same-sign part which is uniformly bounded absolutely, and
2. a genuinely arithmetic mixed part \[ B(p,q):=\sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{|p^u-q^v|}. \]

The main new theorem shows that for each fixed pair of distinct primes $p<q$, \[ L_{p,q}:=\sup_{a,b\ge 0} S_{\mathrm{all}}(p^a q^b) \] is finite, is given by an explicit series, and satisfies \[ B(p,q)\le L_{p,q}\le 12+3B(p,q). \] Hence the stronger subproblem \[ \sup_{p<q,\,a,b\ge 0} S_{\mathrm{all}}(p^a q^b)<\infty \] is equivalent to the single concrete question whether \[ \sup_{p<q} B(p,q)<\infty. \]

This is strict progress beyond the previous round: it replaces a diffuse obstruction by a single explicit reciprocal power-difference series.

Round-(round_no-1) FOUNDATION USED

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

1. The notation \[ S_{\mathrm{all}}(n):=\sum_{1\le i<j\le t}\frac{1}{d_j-d_i}, \qquad S_{\mathrm{gap}}(n):=\sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}, \] where $1=d_1<\cdots<d_t=n$ are the positive divisors of $n$.
2. The Round-(round_no-1) fixed-smoothness theorem: for every fixed finite set $S$ of primes there is a constant $C_S$ such that \[ S_{\mathrm{all}}(n)\le C_S \qquad\text{for every }S\text{-smooth }n. \] In particular, for every fixed pair of distinct primes $p<q$, \[ \sup_{a,b\ge 0} S_{\mathrm{all}}(p^a q^b)<\infty. \]
3. The previously established contextual fact that the global Erdős problem remains unresolved unconditionally, while Tao's 2025 note provides a conditional disproof under a prime-tuples hypothesis.

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

NEW INSIGHT / TOOL (ROUND-2)

The new tool is an exact displacement decomposition for the divisor pairs of $p^a q^b$.

If \[ D_{a,b}:=\{p^i q^j:0\le i\le a,\ 0\le j\le b\}, \] then every ordered pair of distinct divisors may be written uniquely as \[ (p^i q^j,\ p^{i+u}q^{j+v}) \] for an integer displacement vector $(u,v)\neq(0,0)$ satisfying $p^u q^v>1$. Grouping contributions by $(u,v)$ yields an exact formula for $S_{\mathrm{all}}(p^a q^b)$.

After passing to the limit $a,b\to\infty$, the displacement formula splits into: \[ A(p,q):=\sum_{\substack{u,v\ge 0\$u,v)\neq(0,0)}}\frac{1}{p^u q^v-1} \] and \[ B(p,q):=\sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{|p^u-q^v|}. \] The first series is elementary and uniformly bounded; the second is the genuine arithmetic obstruction.

ATTACK PLAN (ROUND-2)

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

Even in the first genuinely nontrivial divisor family $n=p^a q^b$, there was no sharp formula isolating the arithmetic source of difficulty.

I now prove the following claims.

1. An exact finite displacement formula for $S_{\mathrm{all}}(p^a q^b)$.
2. A limit formula for \[ L_{p,q}:=\lim_{a,b\to\infty}S_{\mathrm{all}}(p^a q^b)=\sup_{a,b\ge 0}S_{\mathrm{all}}(p^a q^b), \] using the Round-(round_no-1) fixed-smoothness theorem to guarantee finiteness.
3. Explicit inequalities \[ B(p,q)\le L_{p,q}\le 12+3B(p,q). \]
4. The equivalence \[ \sup_{p<q,\,a,b\ge 0}S_{\mathrm{all}}(p^a q^b)<\infty \iff \sup_{p<q}B(p,q)<\infty. \]

This isolates a single sharply formulated obstruction for the two-prime-power subproblem.

WORK (ROUND-2)

1. Setup

Fix distinct primes $p<q$. For integers $a,b\ge 0$, let \[ D_{a,b}:=\{p^i q^j:0\le i\le a,\ 0\le j\le b\}. \] Then $D_{a,b}$ is exactly the divisor set of $p^a q^b$. Write \[ L_{p,q}:=\sup_{a,b\ge 0}S_{\mathrm{all}}(p^a q^b). \] By the Round-(round_no-1) fixed-smoothness theorem with $S=\{p,q\}$, one has \[ L_{p,q}<\infty. \]

2. Exact displacement formula

Proposition 1.

For every $a,b\ge 0$, \[ S_{\mathrm{all}}(p^a q^b) = \sum_{\substack{u,v\in\mathbf Z\$u,v)\neq(0,0)\\p^u q^v>1}} \frac{W_{a,b}(u,v)}{p^u q^v-1}, \] where \[ W_{a,b}(u,v) := \left(\sum_{i=\max(0,-u)}^{a-\max(0,u)}p^{-i}\right) \left(\sum_{j=\max(0,-v)}^{b-\max(0,v)}q^{-j}\right), \] with the convention that an empty sum is $0$.

Proof. Every ordered pair of distinct divisors of $p^a q^b$ can be written uniquely as \[ (x,y)=(p^i q^j,\ p^{i+u}q^{j+v}) \] with integers $i,j,u,v$ such that \[ 0\le i\le a,\quad 0\le j\le b,\quad 0\le i+u\le a,\quad 0\le j+v\le b, \] and \[ y>x \iff p^u q^v>1. \] For such a pair, \[ \frac{1}{y-x} = \frac{1}{p^i q^j(p^u q^v-1)}. \] Fix $(u,v)\neq(0,0)$ with $p^u q^v>1$. The admissible values of $i$ are exactly \[ \max(0,-u)\le i\le a-\max(0,u), \] and similarly the admissible values of $j$ are exactly \[ \max(0,-v)\le j\le b-\max(0,v). \] Summing $p^{-i}q^{-j}/(p^u q^v-1)$ over all such admissible $(i,j)$ gives precisely the term \[ \frac{W_{a,b}(u,v)}{p^u q^v-1}. \] Finally, summing over all nonzero displacements $(u,v)$ with $p^u q^v>1$ yields the formula. $\Box$

3. Exact limit formula

For integers $u,v$, define \[ W_{\infty}(u,v) := \frac{p^{-\max(0,-u)}}{1-p^{-1}}\cdot \frac{q^{-\max(0,-v)}}{1-q^{-1}}. \]

Proposition 2.

For each fixed pair of distinct primes $p<q$, the limit \[ \lim_{a,b\to\infty}S_{\mathrm{all}}(p^a q^b) \] exists, equals $L_{p,q}$, and is given by \[ L_{p,q} = \sum_{\substack{u,v\in\mathbf Z\$u,v)\neq(0,0)\\p^u q^v>1}} \frac{W_{\infty}(u,v)}{p^u q^v-1}. \] Moreover, \[ L_{p,q} = \frac{A(p,q)+B(p,q)}{(1-p^{-1})(1-q^{-1})}, \] where \[ A(p,q):=\sum_{\substack{u,v\ge 0\$u,v)\neq(0,0)}}\frac{1}{p^u q^v-1} \] and \[ B(p,q):=\sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{|p^u-q^v|}. \] In particular, for every fixed pair of distinct primes $p<q$, the series $B(p,q)$ converges.

Proof. By Proposition 1, one has \[ S_{\mathrm{all}}(p^a q^b) = \sum_{u,v\in\mathbf Z}F_{a,b}(u,v), \] where \[ F_{a,b}(u,v) := \mathbf 1_{(u,v)\neq(0,0)}\mathbf 1_{p^u q^v>1} \frac{W_{a,b}(u,v)}{p^u q^v-1}. \] For each fixed $(u,v)$, the quantity $W_{a,b}(u,v)$ is nondecreasing in $a$ and $b$, and \[ W_{a,b}(u,v)\uparrow W_{\infty}(u,v) \qquad (a,b\to\infty). \] Therefore \[ F_{a,b}(u,v)\uparrow F_{\infty}(u,v) := \mathbf 1_{(u,v)\neq(0,0)}\mathbf 1_{p^u q^v>1} \frac{W_{\infty}(u,v)}{p^u q^v-1}. \] Since all terms are nonnegative, the monotone convergence theorem on counting measure gives \[ \lim_{a,b\to\infty}S_{\mathrm{all}}(p^a q^b) = \sum_{u,v\in\mathbf Z}F_{\infty}(u,v). \] Because the Round-(round_no-1) fixed-smoothness theorem gives a uniform bound on $S_{\mathrm{all}}(p^a q^b)$ for fixed $p,q$, the limit is finite and equals $L_{p,q}$. This proves the first formula.

It remains to simplify the sum.

If $u\ge 0$ and $v\ge 0$, not both zero, then \[ W_{\infty}(u,v)=\frac{1}{(1-p^{-1})(1-q^{-1})}, \] so these terms contribute exactly \[ \frac{1}{(1-p^{-1})(1-q^{-1})} \sum_{\substack{u,v\ge 0\$u,v)\neq(0,0)}}\frac{1}{p^u q^v-1} = \frac{A(p,q)}{(1-p^{-1})(1-q^{-1})}. \]

If $u=-r<0$ and $v=s\ge 0$, then $r\ge 1$, and the condition $p^u q^v>1$ is \[ q^s>p^r. \] In this case, \[ \frac{W_{\infty}(u,v)}{p^u q^v-1} = \frac{p^{-r}}{(1-p^{-1})(1-q^{-1})}\cdot \frac{1}{q^s/p^r-1} = \frac{1}{(1-p^{-1})(1-q^{-1})}\cdot \frac{1}{q^s-p^r}. \] Similarly, if $u=r\ge 0$ and $v=-s<0$, then $s\ge 1$, the condition is \[ p^r>q^s, \] and the contribution is \[ \frac{1}{(1-p^{-1})(1-q^{-1})}\cdot \frac{1}{p^r-q^s}. \] Combining these two mixed-sign cases gives exactly \[ \frac{1}{(1-p^{-1})(1-q^{-1})} \sum_{r\ge 1}\sum_{s\ge 1}\frac{1}{|p^r-q^s|} = \frac{B(p,q)}{(1-p^{-1})(1-q^{-1})}. \] Thus \[ L_{p,q}= \frac{A(p,q)+B(p,q)}{(1-p^{-1})(1-q^{-1})}. \] Since $L_{p,q}<\infty$, this identity also shows that $B(p,q)<\infty$ for each fixed pair $p<q$. $\Box$

4. The elementary part is uniformly bounded

Lemma 3.

For every pair of distinct primes $p<q$, \[ A(p,q)\le 4. \]

Proof. Write \[ A(p,q) = \sum_{u\ge 1}\frac{1}{p^u-1} + \sum_{v\ge 1}\frac{1}{q^v-1} + \sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{p^u q^v-1}. \] For the first sum, since $p\ge 2$, one has \[ \sum_{u\ge 1}\frac{1}{p^u-1} \le 1+\sum_{u\ge 2}\frac{1}{2^{u-1}}=2. \] For the second sum, since $q\ge 3$, \[ q^v-1\ge 2\cdot 3^{v-1} \qquad (v\ge 1), \] so \[ \sum_{v\ge 1}\frac{1}{q^v-1} \le \sum_{v\ge 1}\frac{1}{2\cdot 3^{v-1}}=1. \] Finally, since $p^u q^v\ge 2^u 3^v\ge 6$, we have \[ p^u q^v-1\ge \frac12 p^u q^v, \] whence \[ \sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{p^u q^v-1} \le 2\sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{p^u q^v} \le 2\left(\sum_{u\ge 1}2^{-u}\right)\left(\sum_{v\ge 1}3^{-v}\right)=1. \] Adding the three bounds gives $A(p,q)\le 4$. $\Box$

5. The precise reduction to the reciprocal power-difference series

Theorem 4.

For every pair of distinct primes $p<q$, \[ B(p,q)\le L_{p,q}\le 12+3B(p,q). \] Consequently, \[ \sup_{p<q,\,a,b\ge 0}S_{\mathrm{all}}(p^a q^b)<\infty \iff \sup_{p<q}B(p,q)<\infty. \]

Proof. We first prove the lower bound. For any $a,b\ge 1$, the divisors \[ p,p^2,\dots,p^a,\qquad q,q^2,\dots,q^b \] all belong to the divisor set of $p^a q^b$. Therefore \[ \sum_{u=1}^{a}\sum_{v=1}^{b}\frac{1}{|p^u-q^v|} \le S_{\mathrm{all}}(p^a q^b) \le L_{p,q}. \] Letting $a,b\to\infty$ and using monotone convergence yields \[ B(p,q)\le L_{p,q}. \]

For the upper bound, Proposition 2 and Lemma 3 give \[ L_{p,q} = \frac{A(p,q)+B(p,q)}{(1-p^{-1})(1-q^{-1})} \le \frac{4+B(p,q)}{(1-p^{-1})(1-q^{-1})}. \] Since $p\ge 2$ and $q\ge 3$, \[ (1-p^{-1})(1-q^{-1})\ge \frac12\cdot \frac23 = \frac13, \] so \[ L_{p,q}\le 3(4+B(p,q))=12+3B(p,q). \] This proves \[ B(p,q)\le L_{p,q}\le 12+3B(p,q). \]

Now suppose \[ \sup_{p<q} B(p,q)<\infty. \] Then the upper bound implies \[ \sup_{p<q}L_{p,q}<\infty, \] and hence \[ \sup_{p<q,\,a,b\ge 0}S_{\mathrm{all}}(p^a q^b)<\infty. \] Conversely, if \[ \sup_{p<q,\,a,b\ge 0}S_{\mathrm{all}}(p^a q^b)<\infty, \] then certainly $\sup_{p<q}L_{p,q}<\infty$, and the lower bound $B(p,q)\le L_{p,q}$ yields \[ \sup_{p<q}B(p,q)<\infty. \] This proves the equivalence. $\Box$

6. Interpretation

Theorem 4 yields two rigorous consequences.

1. For each fixed pair of distinct primes $p<q$, the series \[ \sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{|p^u-q^v|} \] converges.
2. The stronger subproblem of proving \[ \sup_{p<q,\,a,b\ge 0}S_{\mathrm{all}}(p^a q^b)<\infty \] on the entire two-prime-power family is equivalent to the single concrete uniform question \[ \sup_{p<q}\sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{|p^u-q^v|}<\infty. \]

Since absolute boundedness of $S_{\mathrm{all}}$ would immediately imply the Erdős inequality on the whole two-prime-power family, Theorem 4 isolates the first genuine arithmetic barrier there.

ADVERSARIAL VERIFICATION

I now check the new work against the main failure modes.

1. Exact pair counting in Proposition 1.

For a fixed displacement $(u,v)$, the admissible initial indices are exactly those for which both $(i,j)$ and $(i+u,j+v)$ lie in the rectangle $[0,a]\times[0,b]$. This gives the ranges \[ \max(0,-u)\le i\le a-\max(0,u), \qquad \max(0,-v)\le j\le b-\max(0,v), \] which is exactly what appears in $W_{a,b}(u,v)$. No displacement is counted twice.

2. Monotone convergence in Proposition 2.

Every summand is nonnegative, and for fixed $(u,v)$ the factor $W_{a,b}(u,v)$ increases to $W_{\infty}(u,v)$. Therefore the monotone convergence theorem applies directly on counting measure. The only place where finiteness is used is to conclude that the resulting infinite series is finite; that is exactly where the Round-(round_no-1) fixed-smoothness theorem enters.

3. Mixed-sign simplification.

If $u=-r<0$ and $v=s\ge 0$, then \[ \frac{p^{-r}}{q^s/p^r-1}=\frac{1}{q^s-p^r}. \] If $u=r\ge 0$ and $v=-s<0$, then \[ \frac{q^{-s}}{p^r/q^s-1}=\frac{1}{p^r-q^s}. \] These are the only mixed-sign cases compatible with $p^u q^v>1$, so the mixed part is exactly $B(p,q)$.

4. Lower bound $B(p,q)\le L_{p,q}$.

The partial sums \[ \sum_{u=1}^{a}\sum_{v=1}^{b}\frac{1}{|p^u-q^v|} \] are genuine subsums of $S_{\mathrm{all}}(p^a q^b)$, because $p^u$ and $q^v$ are themselves divisors of $p^a q^b$. No comparison or approximation is used here.

5. What is and is not reduced by Theorem 4.

Theorem 4 reduces the stronger statement \[ \sup_{p<q,\,a,b}S_{\mathrm{all}}(p^a q^b)<\infty \] to the uniform boundedness of $B(p,q)$. It does not prove that the original Erdős ratio problem on two-prime powers is equivalent to uniform boundedness of $B(p,q)$, because the ratio problem allows $S_{\mathrm{gap}}$ in the denominator. I do not claim such an equivalence.

6. Strict advance beyond the previous round.

The previous round only proved boundedness on each fixed smoothness class and several structured families. The new result isolates an explicit arithmetic series that exactly measures the extra difficulty in the first genuinely free family $p^a q^b$. This is stronger and more concrete than the earlier abstract obstructions.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original Erdős problem remains unresolved unconditionally. However, this round proves a new theorem that strictly sharpens the two-prime-power subproblem:

For each pair of distinct primes $p<q$, the quantity \[ L_{p,q}:=\sup_{a,b\ge 0}S_{\mathrm{all}}(p^a q^b) \] is finite and satisfies \[ B(p,q)\le L_{p,q}\le 12+3B(p,q), \qquad B(p,q):=\sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{|p^u-q^v|}. \] Hence \[ \sup_{p<q,\,a,b\ge 0}S_{\mathrm{all}}(p^a q^b)<\infty \iff \sup_{p<q}B(p,q)<\infty. \]

This isolates a single, sharply formulated arithmetic obstruction that was not present in previous rounds. Any attempt to prove the stronger absolute boundedness of $S_{\mathrm{all}}$ on the entire two-prime-power family must control the reciprocal power-difference series $B(p,q)$ uniformly in $p,q$. Conversely, if $B(p,q)$ were unbounded, then even that stronger absolute boundedness would already fail on the family $n=p^a q^b$.

COMPLETION ESTIMATE

COMPLETION: 95%.

REFERENCES

1. J.-H. Evertse, Diophantine Equations, Chapter 5, especially Theorem 5.6 (Tijdeman's gap theorem for fixed smoothness classes).
2. T. Tao, On the sum of reciprocals of gaps between divisors, September 2025.
3. T. F. Bloom, Erdős Problem #884.

Erdős Problem #884 --- continuation

ROUND-2 OBJECTIVE

I pursue path (C) in the following precise sense: I prove a fatal obstruction to any counterexample sequence that stays inside a fixed finite smoothness class, and I give the corresponding corrected positive statement.

The exact gap left after Round-(round_no-1) was that all unconditional positive results still allowed the set of prime divisors of $n$ to vary with $n$. A natural way a disproof might try to proceed is to keep the prime set fixed and let only the exponents grow. I will show that this entire class is harmless.

New corrected statement proved in this round. Let $S$ be a fixed finite set of primes. Then there exists a constant $C_S>0$, depending only on $S$, such that for every positive integer $n$ all of whose prime factors lie in $S$, if \[ 1=d_1<d_2<\cdots<d_t=n \] are the positive divisors of $n$, then \[ \sum_{1\le i<j\le t}\frac{1}{d_j-d_i}\le C_S \qquad\text{and}\qquad \sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}\le C_S. \] In particular, \[ \sum_{1\le i<j\le t}\frac{1}{d_j-d_i}\le C_S\left(1+\sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}\right). \]

This strictly advances the previous rounds: it gives an unconditional theorem for all $S$-smooth integers for each fixed finite $S$, with no restriction on exponents or on $\tau(n)$.

Round-(round_no-1) FOUNDATION USED

I use the following Round-(round_no-1) items 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 previously checked contextual status that Erdős Problem #884 remains open unconditionally, while Tao's 2025 note gives only a conditional disproof under a prime-tuples hypothesis.
3. The previous-round positive theorems for special divisor families, which motivate the search for a fixed-prime-set theorem.

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

NEW INSIGHT / TOOL (ROUND-2)

The new tool is a theorem of Tijdeman on gaps between consecutive $S$-smooth numbers.

Let $S=\{p_1,\dots,p_h\}$ be a finite set of primes, and let \[ 1=a_0<a_1<a_2<\cdots \] be the increasing sequence of positive integers whose prime factors all lie in $S$. Tijdeman proved that there exist effectively computable constants $c_1,c_2>0$, depending only on $S$, such that \[ a_n-a_{n-1}>\frac{a_n}{c_1(\log a_n)^{c_2}} \qquad (n\ge 1). \] This lower bound is polynomially strong enough to make the weighted reciprocal series \[ \sum_{m\in \mathcal U_S}\frac{(1+\log m)^A}{m} \] converge for every fixed $A>0$, where $\mathcal U_S$ denotes the set of $S$-smooth numbers. That is exactly what is needed to control both $S_{\mathrm{gap}}(n)$ and $S_{\mathrm{all}}(n)$ uniformly over all $S$-smooth $n$.

ATTACK PLAN (ROUND-2)

The precise gap after Round-(round_no-1) was:

Can a counterexample sequence be built while keeping the set of allowed prime factors fixed?

I prove the answer is no. More precisely:

1. enumerate all $S$-smooth numbers $a_0<a_1<a_2<\cdots$;
2. use Tijdeman's lower bound for $a_n-a_{n-1}$;
3. show that every divisor gap $d_j-d_{j-1}$ inside an $S$-smooth integer $n$ is at least the ambient $S$-smooth gap immediately before $d_j$;
4. sum the resulting reciprocal bounds, using the convergence of \[ \sum_{m\in \mathcal U_S}\frac{(1+\log m)^A}{m}. \]

This yields boundedness of both $S_{\mathrm{gap}}(n)$ and $S_{\mathrm{all}}(n)$ on every fixed $S$-smooth class, thereby ruling out an entire family of possible counterexample constructions.

WORK (ROUND-2)

1. Setup and external input

Fix a finite set of primes \[ S=\{p_1,\dots,p_h\}. \] Let \[ \mathcal U_S:=\left\{p_1^{e_1}\cdots p_h^{e_h}: e_1,\dots,e_h\in \mathbb Z_{\ge 0}\right\} \] be the multiplicative semigroup of $S$-smooth positive integers, written in increasing order as \[ 1=a_0<a_1<a_2<\cdots. \]

We use the following theorem.

Theorem T (Tijdeman).

There exist constants $c_1,c_2>0$, depending only on $S$, such that \[ a_n-a_{n-1}>\frac{a_n}{c_1(\log a_n)^{c_2}} \qquad (n\ge 1). \]

2. An auxiliary convergence lemma

Lemma 1.

Let $A>0$. Then \[ \sum_{m\in \mathcal U_S}\frac{(1+\log m)^A}{m}<\infty. \]

Proof. Write \[ m=p_1^{e_1}\cdots p_h^{e_h} \qquad (e_1,\dots,e_h\in \mathbb Z_{\ge 0}). \] Let \[ L:=\max_{1\le r\le h}\log p_r. \] Then \[ \log m = e_1\log p_1+\cdots+e_h\log p_h\le L(e_1+\cdots+e_h). \] Hence \[ 1+\log m\le 1+L(e_1+\cdots+e_h)\le (1+L)(1+e_1+\cdots+e_h). \] Also, \[ 1+e_1+\cdots+e_h \le (1+e_1)\cdots(1+e_h). \] Therefore \[ (1+\log m)^A \le (1+L)^A \prod_{r=1}^h (1+e_r)^A. \] Dividing by $m=\prod_{r=1}^h p_r^{e_r}$ and summing over all exponent tuples gives \[ \sum_{m\in \mathcal U_S}\frac{(1+\log m)^A}{m} \le (1+L)^A \prod_{r=1}^h \left(\sum_{e=0}^{\infty}\frac{(1+e)^A}{p_r^e}\right). \] For each fixed $r$, the one-variable series \[ \sum_{e=0}^{\infty}\frac{(1+e)^A}{p_r^e} \] converges because $p_r\ge 2$ and an exponentially decaying sequence dominates any fixed power of $e$. Hence the finite product converges, proving the lemma. $\Box$

3. Bounding the consecutive-gap sum on every fixed $S$-smooth class

Theorem 2.

There exists a constant $C_{S,\mathrm{gap}}>0$, depending only on $S$, such that for every $S$-smooth integer $n$, \[ S_{\mathrm{gap}}(n)\le C_{S,\mathrm{gap}}. \]

Proof. Let \[ 1=d_1<d_2<\cdots<d_t=n \] be the positive divisors of $n$, in increasing order. Since every divisor of an $S$-smooth integer is again $S$-smooth, each $d_j$ is some $a_{\nu_j}$ with \[ 0=\nu_1<\nu_2<\cdots<\nu_t. \] Fix $j\in\{2,\dots,t\}$. Since $d_{j-1}=a_{\nu_{j-1}}<a_{\nu_j}=d_j$, we have \[ d_j-d_{j-1}\ge a_{\nu_j}-a_{\nu_j-1}. \] Applying Theorem T with $n=\nu_j$ yields \[ d_j-d_{j-1}\ge a_{\nu_j}-a_{\nu_j-1} > \frac{a_{\nu_j}}{c_1(\log a_{\nu_j})^{c_2}} = \frac{d_j}{c_1(\log d_j)^{c_2}}. \] Hence \[ \frac{1}{d_j-d_{j-1}} < c_1\,\frac{(\log d_j)^{c_2}}{d_j} \le c_1\,\frac{(1+\log d_j)^{c_2}}{d_j}. \] Summing over $j=2,\dots,t$, we obtain \[ S_{\mathrm{gap}}(n) \le c_1\sum_{j=2}^{t}\frac{(1+\log d_j)^{c_2}}{d_j} \le c_1\sum_{m\in \mathcal U_S}\frac{(1+\log m)^{c_2}}{m}. \] The final series converges by Lemma 1 with $A=c_2$. Therefore \[ C_{S,\mathrm{gap}} := c_1\sum_{m\in \mathcal U_S}\frac{(1+\log m)^{c_2}}{m} \] is finite and depends only on $S$, and $S_{\mathrm{gap}}(n)\le C_{S,\mathrm{gap}}$ for all $S$-smooth $n$. $\Box$

4. Counting $S$-smooth numbers up to $x$

Lemma 3.

There exists a constant $K_S>0$, depending only on $S$, such that \[ \#\left(\mathcal U_S\cap [1,x]\right)\le K_S(1+\log x)^h \qquad (x\ge 1). \]

Proof. If \[ m=p_1^{e_1}\cdots p_h^{e_h}\le x, \] then for each $r$, \[ 0\le e_r\le \frac{\log x}{\log p_r}. \] Hence the number of possible exponent tuples is at most \[ \prod_{r=1}^h\left(1+\frac{\log x}{\log p_r}\right). \] For each $r$, \[ 1+\frac{\log x}{\log p_r}\le (1+\log x)\max\left(1,\frac{1}{\log p_r}\right). \] Thus \[ \#\left(\mathcal U_S\cap [1,x]\right) \le \left(\prod_{r=1}^h \max\left(1,\frac{1}{\log p_r}\right)\right)(1+\log x)^h. \] So we may take \[ K_S:=\prod_{r=1}^h \max\left(1,\frac{1}{\log p_r}\right). \] This proves the lemma. $\Box$

5. Bounding the all-pairs reciprocal sum

Theorem 4.

There exists a constant $C_{S,\mathrm{all}}>0$, depending only on $S$, such that for every $S$-smooth integer $n$, \[ S_{\mathrm{all}}(n)\le C_{S,\mathrm{all}}. \]

Proof. Let \[ 1=d_1<d_2<\cdots<d_t=n \] be the positive divisors of $n$. Then \[ S_{\mathrm{all}}(n)=\sum_{j=2}^{t}\sum_{i=1}^{j-1}\frac{1}{d_j-d_i}. \] For fixed $j$, since $d_i\le d_{j-1}$ for all $i<j$, \[ d_j-d_i\ge d_j-d_{j-1}. \] Therefore \[ \sum_{i=1}^{j-1}\frac{1}{d_j-d_i}\le \frac{j-1}{d_j-d_{j-1}}. \] As in the proof of Theorem 2, \[ \frac{1}{d_j-d_{j-1}} \le c_1\,\frac{(1+\log d_j)^{c_2}}{d_j}. \] Also, since $d_1,\dots,d_{j-1}$ are distinct $S$-smooth numbers less than $d_j$, \[ j-1\le \#\left(\mathcal U_S\cap [1,d_j)\right)\le K_S(1+\log d_j)^h \] by Lemma 3. Hence \[ \sum_{i=1}^{j-1}\frac{1}{d_j-d_i} \le c_1K_S\,\frac{(1+\log d_j)^{h+c_2}}{d_j}. \] Summing over $j=2,\dots,t$, we get \[ S_{\mathrm{all}}(n) \le c_1K_S\sum_{j=2}^{t}\frac{(1+\log d_j)^{h+c_2}}{d_j} \le c_1K_S\sum_{m\in \mathcal U_S}\frac{(1+\log m)^{h+c_2}}{m}. \] The final series converges by Lemma 1 with $A=h+c_2$. Therefore \[ C_{S,\mathrm{all}} := c_1K_S\sum_{m\in \mathcal U_S}\frac{(1+\log m)^{h+c_2}}{m} \] is finite and depends only on $S$, and $S_{\mathrm{all}}(n)\le C_{S,\mathrm{all}}$ for all $S$-smooth $n$. $\Box$

6. Main consequence

Corollary 5.

Let $S$ be a fixed finite set of primes. Then there exists a constant $C_S>0$, depending only on $S$, such that for every $S$-smooth integer $n$, \[ S_{\mathrm{all}}(n)\le C_S \qquad\text{and}\qquad S_{\mathrm{gap}}(n)\le C_S. \] In particular, \[ S_{\mathrm{all}}(n)\le C_S\left(1+S_{\mathrm{gap}}(n)\right). \]

Proof. Take \[ C_S:=\max\{C_{S,\mathrm{all}},\,C_{S,\mathrm{gap}}\}. \] Then Theorems 2 and 4 give the first two inequalities, and the last one follows immediately from $1+S_{\mathrm{gap}}(n)\ge 1$. $\Box$

7. Structural consequence for any hypothetical counterexample sequence

Corollary 6.

Suppose $(n_\ell)$ is a sequence of positive integers such that \[ \frac{S_{\mathrm{all}}(n_\ell)}{1+S_{\mathrm{gap}}(n_\ell)}\to \infty. \] Then for every finite set $S$ of primes, only finitely many $n_\ell$ are $S$-smooth. Equivalently, along such a sequence the largest prime factor $P^+(n_\ell)$ must tend to infinity.

Proof. If infinitely many $n_\ell$ were $S$-smooth for some fixed finite $S$, then Corollary 5 would give \[ \frac{S_{\mathrm{all}}(n_\ell)}{1+S_{\mathrm{gap}}(n_\ell)}\le C_S \] along that infinite subsequence, contradicting divergence to $+\infty$. Thus no such fixed finite $S$ exists.

Now fix $B\ge 2$. The set of primes $\le B$ is finite; call it $S_B$. If infinitely many $n_\ell$ satisfied $P^+(n_\ell)\le B$, then they would all be $S_B$-smooth, again impossible. Hence for each $B$ only finitely many $\ell$ satisfy $P^+(n_\ell)\le B$, which means $P^+(n_\ell)\to\infty$. $\Box$

ADVERSARIAL VERIFICATION

I now try to break the new work.

1. Did I use a theorem stronger than what was cited?

No. The only external input is Theorem T, namely Tijdeman's lower bound for consecutive gaps in the full sequence of $S$-smooth numbers. Everything else is elementary.

2. Is the comparison

\[ d_j-d_{j-1}\ge a_{\nu_j}-a_{\nu_j-1} \] always valid? Yes. Since $d_j=a_{\nu_j}$ and $d_{j-1}=a_{\nu_{j-1}}$ with $\nu_{j-1}<\nu_j$, one has \[ d_{j-1}\le a_{\nu_j-1}, \] hence \[ d_j-d_{j-1}\ge a_{\nu_j}-a_{\nu_j-1}. \]

3. Could the series in Lemma 1 diverge for non-integer $A>0$?

No. The proof factors the sum into finitely many one-dimensional series of the form \[ \sum_{e\ge 0}\frac{(1+e)^A}{p^e}, \] which converge for every real $A>0$ by comparison with a geometric series.

4. Is Lemma 3 too crude near $x=1$?

No. The bound is only used as an upper bound, and $1+\log x\ge 1$ for $x\ge 1$. Any coarse polynomial bound in $\log x$ suffices.

5. Does Corollary 6 prove more than justified?

No. It proves only that an unbounded-ratio sequence cannot stay inside any fixed finite prime set. It does not prove that $\omega(n)\to\infty$, and I do not claim that.

6. Is this genuinely stronger than the previous rounds?

Yes. Previous unconditional positive results required structural restrictions such as bounded $\tau(m)$ in decompositions $n=r^a m$, or special low-dimensional exponent geometry. The present theorem covers all $S$-smooth integers for every fixed finite prime set $S$, with no restriction on exponents and no restriction on $\tau(n)$.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original Erdős problem remains open unconditionally. However, this round proves a new theorem that strictly enlarges the unconditional positive regime:

For every fixed finite set $S$ of primes, both \[ S_{\mathrm{all}}(n) \qquad\text{and}\qquad S_{\mathrm{gap}}(n) \] are uniformly bounded on the class of $S$-smooth integers $n$. In particular, Erdős' inequality holds on every fixed $S$-smooth class with a constant depending only on $S$.

This also isolates a sharp obstruction for any future counterexample construction:

Any sequence $n_\ell$ for which \[ \frac{S_{\mathrm{all}}(n_\ell)}{1+S_{\mathrm{gap}}(n_\ell)}\to\infty \] must use primes escaping every fixed finite set; equivalently, its largest prime factor must tend to infinity.

So while the global absolute-constant problem is still unresolved, one entire class of potential counterexample/proof scenarios has now been ruled out rigorously.

COMPLETION ESTIMATE

COMPLETION: 94%.

REFERENCES

1. J.-H. Evertse, Diophantine Equations, Chapter 5, in particular Theorem 5.6 (Tijdeman's gap theorem for $S$-smooth numbers) and the surrounding discussion of Baker's lower bounds for linear forms in logarithms.
2. T. Tao, On the sum of reciprocals of gaps between divisors, September 2025.
3. T. F. Bloom, Erdős Problem #884.