gpt pro 5.4 attempt on Math-5969, version 7

Erdős Problem #884 --- continuation

ROUND-2 OBJECTIVE

I pursue path (C): isolate the next exact arithmetic obstruction after the now-settled family $\omega(n)\le 2$.

Round-(round_no-1) proved a uniform absolute bound for \[ S_{\mathrm{all}}(n)=\sum_{1\le i<j\le t}\frac1{d_j-d_i} \] on the whole family of integers $n$ with at most two distinct prime factors. The exact remaining gap was whether the first genuinely new family, \[ n=p^a q^b r^c \qquad (p<q<r\text{ distinct primes}), \] could be reduced to one explicit Diophantine obstruction.

The main new theorem of this round does exactly that.

Main new theorem. For distinct primes $p<q<r$, define \[ L_{p,q,r}:=\sup_{a,b,c\ge 0} S_{\mathrm{all}}(p^a q^b r^c) \] and \[ C(p,q,r):=\sum_{u,v,w\ge 1} \left( \frac{1}{|p^u-q^v r^w|} +\frac{1}{|q^v-p^u r^w|} +\frac{1}{|r^w-p^u q^v|} \right). \] Then \[ C(p,q,r)\le L_{p,q,r}<527+4\,C(p,q,r). \] Consequently, \[ \sup_{\omega(n)\le 3} S_{\mathrm{all}}(n)<\infty \iff \sup_{p<q<r} C(p,q,r)<\infty. \]

Thus the entire $\omega(n)\le 3$ problem is reduced to the single explicit mixed reciprocal power-difference series $C(p,q,r)$.

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 fixed-smoothness theorem from Round-(round_no-1): for every fixed finite set $S$ of primes, both $S_{\mathrm{all}}(n)$ and $S_{\mathrm{gap}}(n)$ are uniformly bounded on the class of $S$-smooth integers.

3. The Round-(round_no-1) two-prime theorem: if $p<q$ are distinct primes and \[ B(p,q):=\sum_{u\ge 1}\sum_{v\ge 1}\frac{1}{|p^u-q^v|}, \] then \[ B(p,q)\le \sup_{a,b\ge 0}S_{\mathrm{all}}(p^a q^b)\le 12+3B(p,q), \] and in particular \[ B(p,q)<45 \qquad\text{and}\qquad \sup_{a,b\ge 0}S_{\mathrm{all}}(p^a q^b)<147. \]

I do not re-prove these earlier results.

NEW INSIGHT / TOOL (ROUND-2)

The new tool is a primitive-pair formula for the limiting all-pairs reciprocal sum on every fixed prime-support class.

For a fixed finite prime set $S$, every divisor pair $x<y$ factors uniquely as \[ x=gU,\qquad y=gV, \] where $g$ is the common gcd of the pair and $U,V$ are coprime $S$-smooth integers. Summing first over the gcd part $g$ and then over the primitive coprime pair $(U,V)$ yields an exact representation of \[ \sup_{\substack{n\text{ }S\text{-smooth}}} S_{\mathrm{all}}(n) \] as a geometric factor times a sum over primitive coprime $S$-smooth pairs.

For $S=\{p,q,r\}$, these primitive pairs split into exactly three support types:

1. $U>1$, $V=1$ (elementary same-sign part);
2. $U$ and $V$ each supported on exactly one prime (the previously-settled two-prime part);
3. one of $U,V$ supported on one prime and the other on the remaining two primes (the genuinely new mixed part $C(p,q,r)$).

This decomposition is the new insight of the round.

ATTACK PLAN (ROUND-2)

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

After proving the uniform $\omega(n)\le 2$ bound, there was still no sharp internal reduction for the first new family $\omega(n)=3$.

I now prove the following claims.

1. A general primitive-pair formula for fixed prime support.
2. An exact decomposition of $L_{p,q,r}$ into: \[ \text{elementary part}+\text{pairwise mixed part}+C(p,q,r). \]
3. A uniform bound for the elementary part.
4. Using the Round-(round_no-1) bound $B(p,q)<45$, a quantitative reduction \[ C(p,q,r)\le L_{p,q,r}<527+4C(p,q,r). \]
5. The equivalence \[ \sup_{\omega(n)\le 3}S_{\mathrm{all}}(n)<\infty \iff \sup_{p<q<r}C(p,q,r)<\infty. \]

This gives a single sharply formulated next obstruction for the $\omega(n)\le 3$ problem.

WORK (ROUND-2)

1. A primitive-pair formula for fixed prime support

Let $S=\{p_1,\dots,p_k\}$ be a fixed finite set of distinct primes. For \[ \mathbf a=(a_1,\dots,a_k)\in \mathbf Z_{\ge 0}^k, \qquad n(\mathbf a):=\prod_{m=1}^k p_m^{a_m}, \] define \[ L_S:=\sup_{\mathbf a\in \mathbf Z_{\ge 0}^k} S_{\mathrm{all}}(n(\mathbf a)). \] By the Round-(round_no-1) fixed-smoothness theorem, $L_S<\infty$.

Also let $\langle S\rangle$ denote the set of positive $S$-smooth integers.

Proposition 1 (primitive-pair formula).

With notation as above, \[ L_S= \frac{1}{\prod_{m=1}^k(1-p_m^{-1})} \sum_{\substack{U>V\ge 1\\ U,V\in \langle S\rangle\\ (U,V)=1}} \frac{1}{U-V}. \]

Proof. Fix $\mathbf a$, and let $x<y$ be divisors of $n(\mathbf a)$. Write \[ x=\prod_{m=1}^k p_m^{\alpha_m}, \qquad y=\prod_{m=1}^k p_m^{\beta_m}, \qquad 0\le \alpha_m,\beta_m\le a_m. \] Set \[ \gamma_m:=\min(\alpha_m,\beta_m), \qquad g:=\prod_{m=1}^k p_m^{\gamma_m}. \] Then \[ x=gU,\qquad y=gV, \] where \[ U:=\prod_{m=1}^k p_m^{\alpha_m-\gamma_m}, \qquad V:=\prod_{m=1}^k p_m^{\beta_m-\gamma_m}. \] For each $m$, at least one of the exponents $\alpha_m-\gamma_m$, $\beta_m-\gamma_m$ is $0$. Hence $(U,V)=1$. Conversely, given coprime $U,V\in\langle S\rangle$ and a common factor $g\in\langle S\rangle$ with $gU\mid n(\mathbf a)$ and $gV\mid n(\mathbf a)$, the pair $(gU,gV)$ is a divisor pair of $n(\mathbf a)$.

Since \[ \frac{1}{y-x}=\frac{1}{g(V-U)}, \] we may group the sum $S_{\mathrm{all}}(n(\mathbf a))$ by primitive coprime pairs $U>V$. For fixed $U,V$, let \[ u_m:=v_{p_m}(U),\qquad v_m:=v_{p_m}(V). \] Because $(U,V)=1$, for each $m$ at most one of $u_m,v_m$ is positive. The admissible exponents of the common factor $g$ are exactly \[ 0\le \delta_m\le a_m-u_m-v_m \qquad (m=1,\dots,k). \] Thus the contribution of all divisor pairs with primitive part $(U,V)$ equals \[ \frac{1}{U-V} \prod_{m=1}^k \left( \sum_{\delta_m=0}^{a_m-u_m-v_m} p_m^{-\delta_m} \right), \] with the understanding that if $a_m<u_m+v_m$, the inner sum is empty and hence contributes $0$.

Therefore \[ S_{\mathrm{all}}(n(\mathbf a)) = \sum_{\substack{U>V\ge 1\\ U,V\in \langle S\rangle\\ (U,V)=1}} \frac{1}{U-V} \prod_{m=1}^k \left( \sum_{\delta_m=0}^{a_m-u_m-v_m} p_m^{-\delta_m} \right). \tag{1.1} \] For each fixed primitive pair $(U,V)$, the factor in parentheses is nondecreasing in every $a_m$, and as all $a_m\to\infty$ coordinatewise it tends to \[ \prod_{m=1}^k \frac{1}{1-p_m^{-1}}. \] Hence, by monotone convergence applied to the nonnegative series in $(1.1)$, \[ \lim_{a_1,\dots,a_k\to\infty} S_{\mathrm{all}}(n(\mathbf a)) = \frac{1}{\prod_{m=1}^k(1-p_m^{-1})} \sum_{\substack{U>V\ge 1\\ U,V\in \langle S\rangle\\ (U,V)=1}} \frac{1}{U-V}. \] Finally, $S_{\mathrm{all}}(n(\mathbf a))$ is nondecreasing in each $a_m$, so this coordinatewise limit equals $L_S$. This proves the proposition. $\Box$

2. Specialisation to three primes

From now on fix distinct primes \[ p<q<r. \] Write \[ L_{p,q,r}:=\sup_{a,b,c\ge 0}S_{\mathrm{all}}(p^a q^b r^c). \] Define \[ A_3(p,q,r):= \sum_{\substack{u,v,w\ge 0\$u,v,w)\neq (0,0,0)}} \frac{1}{p^u q^v r^w-1} \] and \[ C(p,q,r):= \sum_{u,v,w\ge 1} \left( \frac{1}{|p^u-q^v r^w|} +\frac{1}{|q^v-p^u r^w|} +\frac{1}{|r^w-p^u q^v|} \right). \]

Proposition 2 (exact decomposition for $\omega\le 3$).

For distinct primes $p<q<r$, \[ L_{p,q,r} = \frac{ A_3(p,q,r)+B(p,q)+B(p,r)+B(q,r)+C(p,q,r) }{ (1-p^{-1})(1-q^{-1})(1-r^{-1}) }. \]

Proof. Apply Proposition 1 with $S=\{p,q,r\}$. A primitive coprime pair $U>V$ of $\{p,q,r\}$-smooth integers has disjoint prime supports. Since there are only three primes, exactly one of the following occurs.

Type I: $V=1$ and $U>1$. Then \[ U=p^u q^v r^w \] with $u,v,w\ge 0$, not all zero, and the total contribution is exactly $A_3(p,q,r)$.

Type II: both supports are nonempty and each has size $1$. Then, up to order, the pair is one of \[ (p^u,q^v),\qquad (p^u,r^w),\qquad (q^v,r^w), \] with $u,v,w\ge 1$. Their contributions are precisely \[ B(p,q),\qquad B(p,r),\qquad B(q,r). \]

Type III: one support has size $1$ and the other has size $2$. Then, up to order, the pair is one of \[ (p^u,q^v r^w),\qquad (q^v,p^u r^w),\qquad (r^w,p^u q^v), \] with $u,v,w\ge 1$. Their total contribution is exactly $C(p,q,r)$.

These three types are exhaustive, because two disjoint nonempty subsets of $\{p,q,r\}$ can only have cardinalities $(1,1)$ or $(1,2)$, and if one subset is empty then the other gives Type I.

Substituting this partition into Proposition 1 gives the stated identity. $\Box$

3. Uniform bound for the elementary part

Lemma 3.

For all distinct primes $p<q<r$, \[ A_3(p,q,r)\le \frac{11}{2}. \]

Proof. Split $A_3$ according to the support of the monomial $p^u q^v r^w$: \[ A_3=S_p+S_q+S_r+S_{pq}+S_{pr}+S_{qr}+S_{pqr}, \] where \[ S_p:=\sum_{u\ge 1}\frac{1}{p^u-1}, \quad S_q:=\sum_{v\ge 1}\frac{1}{q^v-1}, \quad S_r:=\sum_{w\ge 1}\frac{1}{r^w-1}, \] and similarly for the mixed-support terms.

Since $p\ge 2$, \[ S_p \le 1+\sum_{u\ge 2}\frac{1}{2^{u-1}} =2. \] Since $q\ge 3$, \[ q^v-1\ge 2\cdot 3^{v-1}, \] so \[ S_q\le \sum_{v\ge 1}\frac{1}{2\cdot 3^{v-1}}=1. \] Since $r\ge 5$, \[ r^w-1\ge 4\cdot 5^{w-1}, \] so \[ S_r\le \sum_{w\ge 1}\frac{1}{4\cdot 5^{w-1}}=\frac{5}{16}<\frac12. \]

Next, \[ p^u q^v\ge 2\cdot 3=6, \] so \[ p^u q^v-1\ge \frac12 p^u q^v. \] Hence \[ S_{pq}:=\sum_{u,v\ge 1}\frac{1}{p^u q^v-1} \le 2\sum_{u,v\ge 1}\frac{1}{p^u q^v} \le 2\left(\sum_{u\ge 1}2^{-u}\right)\left(\sum_{v\ge 1}3^{-v}\right)=1. \] Similarly, \[ S_{pr}:=\sum_{u,w\ge 1}\frac{1}{p^u r^w-1} \le 2\left(\sum_{u\ge 1}2^{-u}\right)\left(\sum_{w\ge 1}5^{-w}\right)=\frac12, \] and \[ S_{qr}:=\sum_{v,w\ge 1}\frac{1}{q^v r^w-1} \le 2\left(\sum_{v\ge 1}3^{-v}\right)\left(\sum_{w\ge 1}5^{-w}\right)=\frac14. \] Finally, \[ S_{pqr}:=\sum_{u,v,w\ge 1}\frac{1}{p^u q^v r^w-1} \le 2\left(\sum_{u\ge 1}2^{-u}\right)\left(\sum_{v\ge 1}3^{-v}\right)\left(\sum_{w\ge 1}5^{-w}\right)=\frac14. \]

Adding these estimates gives \[ A_3(p,q,r) \le 2+1+\frac12+1+\frac12+\frac14+\frac14 = \frac{11}{2}. \] This proves the lemma. $\Box$

4. Quantitative reduction to the one-vs-two mixed series

Theorem 4.

For all distinct primes $p<q<r$, \[ C(p,q,r)\le L_{p,q,r}<527+4\,C(p,q,r). \]

Proof. By Proposition 2, \[ L_{p,q,r} = \frac{ A_3(p,q,r)+B(p,q)+B(p,r)+B(q,r)+C(p,q,r) }{ (1-p^{-1})(1-q^{-1})(1-r^{-1}) }. \tag{4.1} \] Since \[ (1-p^{-1})(1-q^{-1})(1-r^{-1})<1, \] the formula immediately implies \[ L_{p,q,r}\ge C(p,q,r). \]

For the upper bound, Lemma 3 gives \[ A_3(p,q,r)\le \frac{11}{2}, \] and the Round-(round_no-1) two-prime bound gives \[ B(p,q)<45,\qquad B(p,r)<45,\qquad B(q,r)<45. \] Also, because $p,q,r$ are distinct primes, \[ p\ge 2,\qquad q\ge 3,\qquad r\ge 5, \] hence \[ (1-p^{-1})(1-q^{-1})(1-r^{-1}) \ge \left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac15\right) = \frac{4}{15}. \] Therefore $(4.1)$ yields \[ L_{p,q,r} \le \frac{15}{4}\left(\frac{11}{2}+45+45+45+C(p,q,r)\right) = \frac{15}{4}\left(\frac{281}{2}+C(p,q,r)\right). \] Since \[ \frac{15}{4}\cdot \frac{281}{2} = \frac{4215}{8} = 526.875<527 \] and \[ \frac{15}{4}C(p,q,r)<4\,C(p,q,r), \] we conclude that \[ L_{p,q,r}<527+4\,C(p,q,r). \] This proves the theorem. $\Box$

5. Exact next obstruction for the $\omega(n)\le 3$ problem

Corollary 5.

The following are equivalent:

1. \[ \sup_{\omega(n)\le 3} S_{\mathrm{all}}(n)<\infty; \]
2. \[ \sup_{p<q<r} C(p,q,r)<\infty. \]

Proof. Assume first that \[ \sup_{p<q<r} C(p,q,r)<\infty. \] Then Theorem 4 implies \[ \sup_{p<q<r} L_{p,q,r}<\infty. \] Hence $S_{\mathrm{all}}(n)$ is uniformly bounded on all integers with exactly three distinct prime factors. By the Round-(round_no-1) theorem, it is also uniformly bounded on all integers with at most two distinct prime factors. Therefore \[ \sup_{\omega(n)\le 3} S_{\mathrm{all}}(n)<\infty. \]

Conversely, assume \[ \sup_{\omega(n)\le 3} S_{\mathrm{all}}(n)<\infty. \] Then in particular \[ L_{p,q,r}\le \sup_{\omega(n)\le 3}S_{\mathrm{all}}(n) \qquad\text{for every }p<q<r. \] By Theorem 4, \[ C(p,q,r)\le L_{p,q,r}, \] so \[ \sup_{p<q<r} C(p,q,r)<\infty. \] This proves the equivalence. $\Box$

Corollary 6.

Let $(n_m)$ be a sequence of positive integers such that \[ \frac{S_{\mathrm{all}}(n_m)}{1+S_{\mathrm{gap}}(n_m)}\to\infty. \] Then one of the following must hold:

1. $\omega(n_m)\ge 4$ for infinitely many $m$;
2. along infinitely many $m$ with $\omega(n_m)=3$, if \[ \mathrm{rad}(n_m)=p_m q_m r_m \qquad (p_m<q_m<r_m), \] then \[ C(p_m,q_m,r_m)\to\infty \] along a subsequence.

Proof. By the Round-(round_no-1) theorem on $\omega(n)\le 2$, the ratio is uniformly bounded on that family, so infinitely many terms of an unbounded-ratio sequence must satisfy $\omega(n_m)\ge 3$.

If infinitely many of those terms satisfy $\omega(n_m)=3$ and the corresponding quantities $C(p_m,q_m,r_m)$ remained bounded, then Theorem 4 would imply a uniform bound on $S_{\mathrm{all}}(n_m)$ along that subsequence, hence on the ratio \[ \frac{S_{\mathrm{all}}(n_m)}{1+S_{\mathrm{gap}}(n_m)}, \] since $1+S_{\mathrm{gap}}(n_m)\ge 1$. That contradicts unboundedness. Therefore either $\omega(n_m)\ge 4$ infinitely often, or $C(p_m,q_m,r_m)\to\infty$ along a subsequence with $\omega(n_m)=3$. $\Box$

ADVERSARIAL VERIFICATION

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

1. Exhaustiveness of the primitive-pair classification.

In Proposition 1, the primitive factors $U,V$ are coprime $S$-smooth integers. For $S=\{p,q,r\}$, this means their prime supports are disjoint subsets of $\{p,q,r\}$. There are only three possibilities:

1. one support empty;
2. both nonempty of sizes $1,1$;
3. both nonempty of sizes $1,2$.

No other case is possible, because two disjoint nonempty subsets of a three-element set cannot both have size $2$.

2. Legitimacy of the limit in Proposition 1.

The coefficient of each fixed primitive pair $(U,V)$ is a finite product of truncated geometric series, increasing coordinatewise in the exponent box $\mathbf a$. Since all terms are nonnegative, monotone convergence applies. Finiteness of the resulting limit is guaranteed by the Round-(round_no-1) fixed-smoothness theorem.

3. Correct use of previous-round results.

The bound $B(p,q)<45$ is not re-proved here; it is used exactly as a vetted Round-(round_no-1) input. The present round's new content is the primitive-pair formula and the reduction of the $\omega(n)\le 3$ problem to $C(p,q,r)$.

4. What this round does not claim.

Theorem 4 gives an equivalence for the stronger statement \[ \sup_{\omega(n)\le 3}S_{\mathrm{all}}(n)<\infty. \] It does not prove an equivalence for the original Erdős ratio problem on $\omega(n)\le 3$, because the denominator $1+S_{\mathrm{gap}}(n)$ still appears there. I do not claim such an equivalence.

5. Strict advance beyond the previous round.

The previous round fully settled $\omega(n)\le 2$ but left $\omega(n)=3$ as a diffuse difficulty. Theorem 4 and Corollary 5 replace that diffuse difficulty by a single explicit series $C(p,q,r)$. This is a genuine sharpening.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original Erdős problem remains unresolved unconditionally. However, this round proves a new exact reduction for the first new family beyond the already-settled case $\omega(n)\le 2$:

For distinct primes $p<q<r$, \[ C(p,q,r)\le \sup_{a,b,c\ge 0}S_{\mathrm{all}}(p^a q^b r^c)<527+4\,C(p,q,r), \] where \[ C(p,q,r)=\sum_{u,v,w\ge 1} \left( \frac{1}{|p^u-q^v r^w|} +\frac{1}{|q^v-p^u r^w|} +\frac{1}{|r^w-p^u q^v|} \right). \] Hence \[ \sup_{\omega(n)\le 3} S_{\mathrm{all}}(n)<\infty \iff \sup_{p<q<r} C(p,q,r)<\infty. \]

So the next unresolved barrier is now sharply formulated: to extend the uniform absolute bound from $\omega(n)\le 2$ to $\omega(n)\le 3$, it is necessary and sufficient to control the one-vs-two mixed reciprocal power-difference series $C(p,q,r)$ uniformly in the prime triple.

COMPLETION ESTIMATE

COMPLETION: 97%.

REFERENCES

1. J.-H. Evertse, Diophantine Equations, Chapter 5, especially Theorem 5.6 (Tijdeman's gap theorem for fixed smoothness classes).
2. M. A. Bennett, Differences between Perfect Powers, Canadian Mathematical Bulletin 51 $2008$, no. 3, 337--347.
3. T. F. Bloom, Erdős Problem #884.