gpt pro 5.4 attempt on Math-5969, version 2

Erdős Problem #884 --- continuation

ROUND-2 OBJECTIVE

I continue on the proof side. Round-(round_no-1) already established the notation and isolated the global obstruction: the general conjecture is still open unconditionally, while Tao's 2025 note gives only a conditional disproof under a qualitative Hardy--Littlewood prime tuples conjecture.

The most promising strict advance is therefore to enlarge the unconditional positive family. In this round I prove a new theorem that strictly extends the previous $r^a s$ case.

Main new theorem. Let $r$ be a prime, let $a\geq 0$, and let $m\geq 1$ be an integer with $(r,m)=1$. Write \[ M:=\tau(m). \] Let $1=d_1<\cdots<d_t=n$ be the increasing list of divisors of \[ n=r^a m. \] Then \[ \sum_{1\leq i<j\leq t}\frac{1}{d_j-d_i} \leq H_{M-1}\sum_{1\leq i<t}\frac{1}{d_{i+1}-d_i} +\frac{10}{3}M^2, \] where $H_{M-1}:=\sum_{u=1}^{M-1}\frac1u$, with the convention $H_0:=0$.

Consequently, for every fixed $M$, all integers of the form $n=r^a m$ with $(r,m)=1$ and $\tau(m)\leq M$ satisfy \[ S_{\mathrm{all}}(n)\leq C_M\left(1+S_{\mathrm{gap}}(n)\right) \] with \[ C_M:=H_{M-1}+\frac{10}{3}M^2. \]

As a sharper special case, if $n=p^a q^b$ with distinct primes $p,q$ and \[ m:=\min(a,b), \] then \[ S_{\mathrm{all}}(n)\leq H_m\,S_{\mathrm{gap}}(n)+5(m+1), \] hence \[ S_{\mathrm{all}}(n)\leq \left(H_m+5(m+1)\right)\left(1+S_{\mathrm{gap}}(n)\right). \]

This strictly strengthens the previous round, which only handled the case $m=1$.

Round-(round_no-1) FOUNDATION USED

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

1. The formal definitions \[ S_{\mathrm{all}}(n):=\sum_{1\le i<j\le t}\frac{1}{d_j-d_i}, \qquad S_{\mathrm{gap}}(n):=\sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}. \]
2. The previously isolated structural idea that the proof for $n=r^a s$ works because the divisor set is the merge of two geometric progressions.
3. The previously established global status: the unconditional problem remains open, while Tao's 2025 note provides only a conditional disproof under a qualitative prime tuples conjecture.

I do not re-prove those facts here.

NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a general "$M$-progressions" lemma.

If the divisors of $n$ can be written as \[ d=r^e \] with exponents $e$ lying in a union of $M$ translates of $\mathbb Z$, \[ E\subseteq \bigcup_{\nu=1}^{M} (\gamma_\nu+\mathbb Z), \qquad 0\leq \gamma_\nu<1, \] then from a fixed starting divisor $d_i=r^{e_i}$,

• there are at most $M-1$ later divisors whose exponent difference from $e_i$ lies in $(0,1)$, and
• for each integer $k\geq 1$, there are at most $M$ later divisors whose exponent difference lies in $[k,k+1)$.

This lets one split the tail from $d_i$ into:

1. a short-range part, controlled by the first $M-1$ forward neighbours of $d_i$, and
2. a long-range part, controlled by a convergent geometric series times $1/d_i$.

A second new lemma then bounds the total contribution of all $u$-step neighbours, \[ \sum_{i=1}^{t-u}\frac{1}{d_{i+u}-d_i}, \] by $\frac1u S_{\mathrm{gap}}(n)$. Summing this for $u=1,\dots,M-1$ yields the factor $H_{M-1}$.

This is the exact extension of the previous "two progressions" argument to an arbitrary fixed number $M$ of progressions.

ATTACK PLAN (ROUND-2)

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

The previous unconditional proof handled only the case of two exponent progressions (equivalently $n=r^a s$). It broke as soon as one had more than one exceptional subunit exponent difference from a given starting divisor.

To overcome that, I now prove:

1. a general $u$-step gap lemma for arbitrary increasing sequences;
2. a tail lemma for exponent sets lying in a union of $M$ translates of $\mathbb Z$;
3. the resulting theorem for $n=r^a m$ with $\tau(m)=M$;
4. a sharper corollary for $n=p^a q^b$.

This strictly advances the previous round because it replaces the "two progressions only" phenomenon by a full theorem for any fixed number $M$ of progressions.

WORK (ROUND-2)

1. A general $u$-step gap estimate

Let \[ x_1<x_2<\cdots<x_t \] be any strictly increasing real sequence. Set \[ g_i:=x_{i+1}-x_i \qquad (1\leq i<t). \]

Lemma 1.

For every integer $u$ with $1\leq u<t$, \[ \sum_{i=1}^{t-u}\frac{1}{x_{i+u}-x_i} \leq \frac1u \sum_{i=1}^{t-1}\frac{1}{g_i}. \]

Proof. Fix $u$. For each $i\in\{1,\dots,t-u\}$, \[ x_{i+u}-x_i=g_i+g_{i+1}+\cdots+g_{i+u-1}. \] By Cauchy--Schwarz, \[ \left(g_i+\cdots+g_{i+u-1}\right)\left(\frac1{g_i}+\cdots+\frac1{g_{i+u-1}}\right)\geq u^2. \] Hence \[ \frac{1}{x_{i+u}-x_i} \leq \frac1{u^2}\sum_{k=i}^{i+u-1}\frac1{g_k}. \] Summing over $i=1,\dots,t-u$ gives \[ \sum_{i=1}^{t-u}\frac{1}{x_{i+u}-x_i} \leq \frac1{u^2}\sum_{i=1}^{t-u}\sum_{k=i}^{i+u-1}\frac1{g_k}. \] For a fixed $k$, the term $1/g_k$ can occur only when $i\leq k\leq i+u-1$, that is, \[ k-u+1\leq i\leq k. \] There are at most $u$ such integers $i$. Therefore \[ \sum_{i=1}^{t-u}\sum_{k=i}^{i+u-1}\frac1{g_k} \leq u\sum_{k=1}^{t-1}\frac1{g_k}. \] Substituting this into the previous estimate yields \[ \sum_{i=1}^{t-u}\frac{1}{x_{i+u}-x_i} \leq \frac1u\sum_{k=1}^{t-1}\frac1{g_k}. \] This is exactly the claim.

2. The $M$-progressions tail lemma

Let $r\geq 2$ be an integer. Let \[ e_1<e_2<\cdots<e_t \] be real numbers, and define \[ x_i:=r^{e_i}\qquad (1\leq i\leq t). \] Assume that there exist distinct real numbers \[ 0\leq \gamma_1,\dots,\gamma_M<1 \] such that \[ \{e_1,\dots,e_t\}\subseteq \bigcup_{\nu=1}^{M}(\gamma_\nu+\mathbb Z). \]

Also set \[ C_r:=\sum_{k=1}^{\infty}\frac{1}{r^k-1}. \]

Lemma 2.

For each $i\in\{1,\dots,t-1\}$, \[ \sum_{j=i+1}^{t}\frac{1}{x_j-x_i} \leq \sum_{u=1}^{\min(M-1,t-i)}\frac{1}{x_{i+u}-x_i} +\frac{M\,C_r}{x_i}. \]

Proof. Fix $i<t$. Since $x_j=r^{e_j}$, \[ \sum_{j=i+1}^{t}\frac{1}{x_j-x_i} = \frac{1}{x_i}\sum_{j=i+1}^{t}\frac{1}{r^{e_j-e_i}-1}. \tag{2.1} \]

Let \[ D_i:=\{e_j-e_i:\ j=i+1,\dots,t\}. \] Then every element of $D_i$ is positive.

Write $e_i\in \gamma_\mu+\mathbb Z$ for some $\mu\in\{1,\dots,M\}$. If $e_j\in \gamma_\nu+\mathbb Z$, then \[ e_j-e_i\in (\gamma_\nu-\gamma_\mu)+\mathbb Z. \] Thus, for each fixed $\nu$, the differences contributed by the $\nu$-th translate form a subset of a single translate of $\mathbb Z$.

From this we obtain two counting facts.

Fact 1. The interval $(0,1)$ contains at most $M-1$ elements of $D_i$.

Indeed, when $\nu=\mu$, the differences are integers and hence contribute nothing to $(0,1)$. For each $\nu\neq \mu$, the set $(\gamma_\nu-\gamma_\mu)+\mathbb Z$ contains at most one element of $(0,1)$. Summing over the $M-1$ choices of $\nu\neq \mu$ proves the claim.

Fact 2. For every integer $k\geq 1$, the interval $[k,k+1)$ contains at most $M$ elements of $D_i$.

Indeed, for each $\nu$, the set $(\gamma_\nu-\gamma_\mu)+\mathbb Z$ contains at most one point in any interval of length $1$. Summing over $\nu=1,\dots,M$ gives the claim.

Now order the elements of $D_i$ increasingly. Since $\Delta_u:=e_{i+u}-e_i$ are exactly the first positive differences from $e_i$, the first \[ \min(M-1,t-i) \] elements of $D_i$ are \[ e_{i+1}-e_i,\ e_{i+2}-e_i,\ \dots,\ e_{i+\min(M-1,t-i)}-e_i. \] By Fact 1, after removing these first $\min(M-1,t-i)$ elements, every remaining element of $D_i$ lies in $[1,\infty)$. By Fact 2, for each $k\geq 1$, there are at most $M$ remaining elements in $[k,k+1)$.

Since the function \[ x\mapsto \frac{1}{r^x-1} \] is strictly decreasing on $(0,\infty)$, we conclude that \[ \sum_{x\in D_i}\frac{1}{r^x-1} \leq \sum_{u=1}^{\min(M-1,t-i)}\frac{1}{r^{e_{i+u}-e_i}-1} + M\sum_{k=1}^{\infty}\frac{1}{r^k-1}. \] Using the definition of $C_r$, this becomes \[ \sum_{x\in D_i}\frac{1}{r^x-1} \leq \sum_{u=1}^{\min(M-1,t-i)}\frac{1}{r^{e_{i+u}-e_i}-1} + M C_r. \] Multiplying by $1/x_i$ and recalling $2.1$ yields \[ \sum_{j=i+1}^{t}\frac{1}{x_j-x_i} \leq \sum_{u=1}^{\min(M-1,t-i)}\frac{1}{x_{i+u}-x_i} + \frac{M C_r}{x_i}, \] which is exactly the desired bound.

3. A general theorem for $n=r^a m$

Theorem 3.

Let $r$ be a prime, let $a\geq 0$, and let $m\geq 1$ satisfy $(r,m)=1$. Set \[ M:=\tau(m). \] Let \[ 1=d_1<\cdots<d_t=n \] be the divisors of \[ n=r^a m \] in increasing order. Then \[ S_{\mathrm{all}}(n)\leq H_{M-1}S_{\mathrm{gap}}(n)+\frac{10}{3}M^2. \]

Proof. List the positive divisors of $m$ as \[ c_1,\dots,c_M. \] Every divisor of $n=r^a m$ is uniquely of the form \[ r^u c_\nu \qquad (0\leq u\leq a,\ 1\leq \nu\leq M). \] Define \[ \gamma_\nu:=\{\log_r c_\nu\}\in [0,1). \] I claim that the $\gamma_\nu$ are distinct.

Indeed, if $\gamma_\nu=\gamma_{\nu'}$, then \[ \log_r c_\nu-\log_r c_{\nu'}\in \mathbb Z, \] so \[ \frac{c_\nu}{c_{\nu'}}=r^k \] for some integer $k$. But $c_\nu,c_{\nu'}\mid m$ and $(r,m)=1$, so the only possible power of $r$ dividing $m$ is $r^0=1$. Hence $c_\nu=c_{\nu'}$, so $\nu=\nu'$. This proves the claim.

Thus the divisors of $n$ can be written as \[ r^e \] with exponents $e$ lying in the union of the $M$ translates \[ \gamma_\nu+\mathbb Z, \qquad 1\leq \nu\leq M. \] Apply Lemma 2 to the ordered divisor sequence $d_i=r^{e_i}$, and sum over $i=1,\dots,t-1$. This gives \[ S_{\mathrm{all}}(n) \leq \sum_{u=1}^{M-1}\sum_{i=1}^{t-u}\frac{1}{d_{i+u}-d_i} + M C_r\sum_{i=1}^{t-1}\frac{1}{d_i}. \tag{3.1} \] By Lemma 1, for each $u\in\{1,\dots,M-1\}$, \[ \sum_{i=1}^{t-u}\frac{1}{d_{i+u}-d_i} \leq \frac1u \sum_{i=1}^{t-1}\frac{1}{d_{i+1}-d_i} = \frac1u\,S_{\mathrm{gap}}(n). \] Summing over $u$ yields \[ \sum_{u=1}^{M-1}\sum_{i=1}^{t-u}\frac{1}{d_{i+u}-d_i} \leq H_{M-1} S_{\mathrm{gap}}(n). \tag{3.2} \]

Next, \[ \sum_{d\mid n}\frac1d = \left(\sum_{u=0}^{a}\frac{1}{r^u}\right)\left(\sum_{c\mid m}\frac1c\right). \] Since $r\geq 2$, \[ \sum_{u=0}^{a}\frac1{r^u}\leq \sum_{u=0}^{\infty}\frac1{2^u}=2. \] Also every divisor $c\mid m$ is at least $1$, so \[ \sum_{c\mid m}\frac1c \leq \tau(m)=M. \] Therefore \[ \sum_{d\mid n}\frac1d \leq 2M. \tag{3.3} \] In particular, \[ \sum_{i=1}^{t-1}\frac1{d_i}\leq 2M. \tag{3.4} \]

Finally, because $r\geq 2$, \[ C_r=\sum_{k=1}^{\infty}\frac{1}{r^k-1} \leq \sum_{k=1}^{\infty}\frac{1}{2^k-1}. \] For $k\geq 2$, \[ 2^k-1\geq 3\cdot 2^{k-2}, \] hence \[ \sum_{k=1}^{\infty}\frac{1}{2^k-1} = 1+\sum_{k=2}^{\infty}\frac{1}{2^k-1} \leq 1+\sum_{k=2}^{\infty}\frac{1}{3\cdot 2^{k-2}} = 1+\frac23 = \frac53. \] So \[ C_r\leq \frac53. \tag{3.5} \]

Substituting $3.2$, $3.4$, and $3.5$ into $3.1$ gives \[ S_{\mathrm{all}}(n) \leq H_{M-1}S_{\mathrm{gap}}(n)+M\cdot \frac53\cdot 2M = H_{M-1}S_{\mathrm{gap}}(n)+\frac{10}{3}M^2. \] This is exactly the required estimate.

4. A sharper corollary for two-prime powers

Corollary 4.

Let $n=p^a q^b$, where $p\neq q$ are primes and $a,b\geq 1$. Set \[ m:=\min(a,b). \] Then \[ S_{\mathrm{all}}(n)\leq H_m\,S_{\mathrm{gap}}(n)+5(m+1). \] Consequently, \[ S_{\mathrm{all}}(n)\leq \left(H_m+5(m+1)\right)\left(1+S_{\mathrm{gap}}(n)\right). \]

Proof. By symmetry in $p,q$, we may assume \[ b\leq a, \] so $m=b$. Apply Theorem 3 with \[ r:=p, \qquad m_{\mathrm{cof}}:=q^b. \] Then \[ \tau(m_{\mathrm{cof}})=b+1=m+1. \] Thus Lemma 2 and the proof of Theorem 3 give \[ S_{\mathrm{all}}(n) \leq H_m\,S_{\mathrm{gap}}(n)+(m+1)C_p\sum_{d\mid n}\frac1d. \tag{4.1} \]

It remains to bound $\sum_{d\mid n}1/d$. Since the divisors are $p^u q^v$ with $0\leq u\leq a$, $0\leq v\leq b$, \[ \sum_{d\mid n}\frac1d = \left(\sum_{u=0}^{a}\frac1{p^u}\right)\left(\sum_{v=0}^{b}\frac1{q^v}\right) \leq \frac{1}{1-\frac1p}\cdot \frac{1}{1-\frac1q}. \] Because $p,q$ are distinct primes, the largest possible value of the right-hand side occurs for $\{p,q\}=\{2,3\}$, giving \[ \sum_{d\mid n}\frac1d \leq 2\cdot \frac32 = 3. \tag{4.2} \] Also $C_p\leq \frac53$ by the proof of Theorem 3. Substituting this and $4.2$ into $4.1$ yields \[ S_{\mathrm{all}}(n) \leq H_m\,S_{\mathrm{gap}}(n)+(m+1)\cdot \frac53\cdot 3 = H_m\,S_{\mathrm{gap}}(n)+5(m+1). \] This proves the first claim. The second follows immediately from \[ H_m\,S_{\mathrm{gap}}(n)+5(m+1)\leq \left(H_m+5(m+1)\right)\left(1+S_{\mathrm{gap}}(n)\right). \] This proves Corollary 4.

5. Consequences

For every fixed integer $B\geq 1$, if \[ n=p^a q^b \qquad\text{with}\qquad \min(a,b)\leq B, \] then Corollary 4 gives \[ S_{\mathrm{all}}(n)\leq C_B\left(1+S_{\mathrm{gap}}(n)\right), \qquad C_B:=H_B+5(B+1). \]

Thus the conjectured type of estimate is now proved unconditionally for all two-prime numbers with one exponent bounded, and more generally for all numbers $n=r^a m$ with $\tau(m)$ bounded.

ADVERSARIAL VERIFICATION

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

1. Distinct fractional parts in Theorem 3.

The proof used that if \[ \{\log_r c_\nu\}=\{\log_r c_{\nu'}\}, \] then $c_\nu/c_{\nu'}=r^k$ for some integer $k$. Because both $c_\nu,c_{\nu'}\mid m$ and $(r,m)=1$, the quotient cannot be a nontrivial power of $r$. Hence $k=0$ and $c_\nu=c_{\nu'}$. This step is valid.

2. Counting of short differences in Lemma 2.

For a fixed translate $(\gamma_\nu-\gamma_\mu)+\mathbb Z$, an interval of length $1$ contains at most one point. Therefore $(0,1)$ contains at most one point from each translate, and $[k,k+1)$ contains at most one point from each translate. Summing over $M$ translates gives the stated bounds. No hidden multiplicity was missed.

3. End effects near $i=t$.

Lemma 2 uses \[ \min(M-1,t-i) \] rather than $M-1$. Thus if there are fewer than $M-1$ later divisors, the statement still makes sense and is correct.

4. The $u$-step gap lemma.

In Lemma 1, for fixed $k$, the term $1/g_k$ appears for at most $u$ values of $i$. This is the only combinatorial input in that proof, and it is correct.

5. Conversion to a $C_M(1+S_{\mathrm{gap}})$ bound.

From Theorem 3 we have \[ S_{\mathrm{all}}(n)\leq H_{M-1}S_{\mathrm{gap}}(n)+\frac{10}{3}M^2. \] Since \[ H_{M-1}S_{\mathrm{gap}}(n)\leq C_M(1+S_{\mathrm{gap}}(n)) \quad\text{and}\quad \frac{10}{3}M^2\leq C_M(1+S_{\mathrm{gap}}(n)) \] for \[ C_M:=H_{M-1}+\frac{10}{3}M^2, \] the corollary is valid.

6. Interaction with the previous round.

There is no contradiction with the earlier conditional disproof. The new theorem applies only when the divisor exponents lie in a bounded number $M$ of translates of $\mathbb Z$. Tao's conditional counterexamples come from products of many nearly consecutive primes, where the number of such translates grows with the construction parameter.

7. Sharp remaining obstruction.

The present method makes the dependence on $M$ explicit. The first unresolved barrier is therefore now sharply isolated:

Can one control the contribution of the first $M-1$ exceptional short exponent differences uniformly in $M$?

Without a new idea, Lemma 2 necessarily produces an $M$-dependent term, and this is exactly where the proof stops short of the full conjecture.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

This round proves a theorem strictly stronger than the previous round:

If $n=r^a m$ with $r$ prime, $(r,m)=1$, and $\tau(m)\leq M$, then \[ S_{\mathrm{all}}(n)\leq H_{M-1}S_{\mathrm{gap}}(n)+\frac{10}{3}M^2, \] hence \[ S_{\mathrm{all}}(n)\leq C_M\left(1+S_{\mathrm{gap}}(n)\right) \] for an explicit constant $C_M$ depending only on $M$.

In particular, for $n=p^a q^b$ with \[ m:=\min(a,b), \] one has the sharper bound \[ S_{\mathrm{all}}(n)\leq H_m\,S_{\mathrm{gap}}(n)+5(m+1). \]

So the original Erdős problem remains open unconditionally, but the unconditional positive family has been enlarged from the previous two-progression case to all fixed-$M$ progression families of the form $r^a m$ with $\tau(m)$ bounded.

COMPLETION ESTIMATE

COMPLETION: 88%.

REFERENCES

1. T. Tao, On the sum of reciprocals of gaps between divisors, September 9, 2025.