gpt pro 5.4 attempt on Math-5969, version 9

Erdős Problem #884 --- continuation

ROUND-2 OBJECTIVE

I pursue path (C): sharpen the first genuinely unresolved obstruction in the three-prime family.

Round-(round_no-1) proved two key facts:

1. the whole family $\omega(n)\le 2$ is already harmless, with a uniform absolute bound \[ S_{\mathrm{all}}(n)<147; \]
2. for distinct primes $p<q<r$, the boundedness of \[ L_{p,q,r}:=\sup_{a,b,c\ge 0} S_{\mathrm{all}}(p^a q^b r^c) \] is equivalent to the boundedness of the mixed one-vs-two sum \[ C(p,q,r):= \sum_{u,v,w\ge 1} \left( \frac{1}{|p^u-q^v r^w|} +\frac{1}{|q^v-p^u r^w|} +\frac{1}{|r^w-p^u q^v|} \right). \]

The exact remaining gap was that $C(p,q,r)$ still allowed all mixed differences, large and small. In this round I prove that, up to a uniform absolute error, only subpower one-vs-two collisions matter.

For $0<\theta<1$, define \[ C_\theta(p,q,r):= \sum_{u,v,w\ge 1} \left( \frac{\mathbf 1_{|p^u-q^v r^w|<p^{\theta u}}}{|p^u-q^v r^w|} + \frac{\mathbf 1_{|q^v-p^u r^w|<q^{\theta v}}}{|q^v-p^u r^w|} + \frac{\mathbf 1_{|r^w-p^u q^v|<r^{\theta w}}}{|r^w-p^u q^v|} \right). \] Also set \[ \Gamma_3:=\sup_{p<q<r} C(p,q,r), \qquad \Gamma_3^{(\theta)}:=\sup_{p<q<r} C_\theta(p,q,r). \]

The main new theorem proved in this round is:

Subpower obstruction theorem. For every fixed $0<\theta<1$ and every distinct primes $p<q<r$, \[ 0\le C(p,q,r)-C_\theta(p,q,r)\le \frac{3300}{2^{\theta/2}-1}. \] Consequently, \[ \Gamma_3<\infty \iff \Gamma_3^{(\theta)}<\infty. \] Combining this with the previous-round equivalence \[ L_{p,q,r}<\infty \iff C(p,q,r)<\infty, \] one obtains \[ \sup_{\omega(n)\le 3} S_{\mathrm{all}}(n)<\infty \iff \Gamma_3^{(\theta)}<\infty \qquad (0<\theta<1). \]

Thus the entire unresolved three-prime problem is reduced further: only subpower one-vs-two near-collisions can matter.

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 interval-energy lower bound proved in earlier rounds: if $x_1<\cdots<x_k$ are $k\ge 2$ points in an interval of length $L>0$, then \[ \sum_{1\le i<j\le k}\frac{1}{x_j-x_i}\ge \frac{1}{32}\,\frac{k^2\log k}{L}. \tag{F1} \]

3. The previously proved uniform bound for the entire family $\omega(n)\le 2$: \[ S_{\mathrm{all}}(n)<147 \qquad\text{for every }n\text{ with }\omega(n)\le 2. \tag{F2} \]

4. The previously proved three-prime reduction: for distinct primes $p<q<r$, \[ C(p,q,r)\le L_{p,q,r}<527+4\,C(p,q,r), \tag{F3} \] where \[ L_{p,q,r}:=\sup_{a,b,c\ge 0}S_{\mathrm{all}}(p^a q^b r^c). \] In particular, \[ \sup_{\omega(n)\le 3} S_{\mathrm{all}}(n)<\infty \iff \Gamma_3<\infty. \tag{F4} \]

I do not re-prove these earlier facts here.

NEW INSIGHT / TOOL (ROUND-2)

The new tool is a local packing estimate for two-prime products, obtained by combining the already-settled $\omega(n)\le 2$ case with the interval-energy lower bound.

For distinct primes $q<r$, define \[ \mathcal A_{q,r}:=\{q^v r^w : v,w\in \mathbf N\}. \] The key observation is:

Any interval of length $L$ can contain only $O(\sqrt L)$ elements of $\mathcal A_{q,r}$, uniformly in $q,r$.

Indeed, if an interval contained $k$ such elements, then these $k$ integers form a subset of the divisor set of $q^V r^W$ for large enough $V,W$, so by $(\mathrm{F2})$ their internal reciprocal energy is $<147$; but by $(\mathrm{F1})$, $k$ points packed into length $L$ force energy $\gg k^2/L$.

A dyadic shell decomposition then shows that, for any real $x$ and any threshold $D\ge 1$, \[ \sum_{\substack{a\in \mathcal A_{q,r}\\ |a-x|\ge D}}\frac{1}{|a-x|} \ll D^{-1/2}, \] uniformly in $q,r,x$. Applying this with $x=p^u$ and $D=p^{\theta u}$ yields a convergent geometric sum over $u$.

This is what turns the general mixed sum $C(p,q,r)$ into the subpower-restricted sum $C_\theta(p,q,r)$ up to an absolute error.

ATTACK PLAN (ROUND-2)

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

The three-prime obstruction $C(p,q,r)$ still included every mixed difference, even those that are polynomially large compared with the one-prime term. It was not yet known whether those polynomial-scale terms could contribute to a uniform blowup.

I prove that they cannot.

The precise claims are:

1. A uniform interval-packing lemma for the sets $\mathcal A_{q,r}=\{q^v r^w:v,w\ge 1\}$.
2. A uniform tail-potential estimate \[ \sum_{\substack{a\in \mathcal A_{q,r}\\ |a-x|\ge D}}\frac{1}{|a-x|}\ll D^{-1/2}. \]
3. Hence, for every fixed $0<\theta<1$, the part of $C(p,q,r)$ coming from terms with \[ |p^u-q^v r^w|\ge p^{\theta u}, \quad |q^v-p^u r^w|\ge q^{\theta v}, \quad |r^w-p^u q^v|\ge r^{\theta w} \] is uniformly bounded.
4. Therefore the entire unresolved three-prime problem is equivalent to the boundedness of the subpower-collision sums $C_\theta(p,q,r)$.

This is strict progress because it rules out all proof strategies that only control mixed differences on polynomial scales: such terms are automatically harmless.

WORK (ROUND-2)

1. A uniform packing bound for two-prime products

Lemma 1.

Let $q<r$ be distinct primes, and let $I\subset \mathbf R$ be an interval of length $L\ge 0$. Then \[ \#(\mathcal A_{q,r}\cap I)\le 100\sqrt{L+1}. \]

Proof. Set \[ k:=\#(\mathcal A_{q,r}\cap I). \] If $k\le 1$, the claim is immediate. Assume now that $k\ge 2$, and list the points of $\mathcal A_{q,r}\cap I$ as \[ x_1<\cdots<x_k. \] Choose integers \[ V:=\max\{v: q^v r^w\in \mathcal A_{q,r}\cap I\}, \qquad W:=\max\{w: q^v r^w\in \mathcal A_{q,r}\cap I\}. \] Then every $x_i$ is a divisor of $q^V r^W$. Therefore the pairwise reciprocal sum over the set $\{x_1,\dots,x_k\}$ is bounded by the full divisor energy: \[ \sum_{1\le i<j\le k}\frac{1}{x_j-x_i} \le S_{\mathrm{all}}(q^V r^W) <147 \] by $(\mathrm{F2})$.

On the other hand, since $x_1,\dots,x_k$ lie in an interval of length at most $L+1$, the interval-energy lower bound $(\mathrm{F1})$ gives \[ \sum_{1\le i<j\le k}\frac{1}{x_j-x_i} \ge \frac{1}{32}\,\frac{k^2\log k}{L+1}. \] Combining the two inequalities yields \[ \frac{1}{32}\,\frac{k^2\log k}{L+1}<147. \] Since $k\ge 2$, one has $\log k\ge \log 2$, and therefore \[ k^2<\frac{32\cdot 147}{\log 2}(L+1)<10000(L+1). \] Hence \[ k<100\sqrt{L+1}. \] This proves the lemma. $\Box$

2. A uniform tail-potential estimate

Lemma 2.

Let $q<r$ be distinct primes. For every real $x$ and every $D\ge 1$, \[ \sum_{\substack{a\in\mathcal A_{q,r}\\ |a-x|\ge D}}\frac{1}{|a-x|} \le 1100\,D^{-1/2}. \]

Proof. For each integer $m\ge 0$, define the dyadic shell \[ \mathcal S_m:=\{a\in \mathcal A_{q,r}: 2^m D\le |a-x|<2^{m+1}D\}. \] Then \[ \sum_{\substack{a\in\mathcal A_{q,r}\\ |a-x|\ge D}}\frac{1}{|a-x|} = \sum_{m=0}^{\infty}\sum_{a\in \mathcal S_m}\frac{1}{|a-x|} \le \sum_{m=0}^{\infty}\frac{\#\mathcal S_m}{2^m D}. \tag{2.1} \] Now $\mathcal S_m$ is contained in the interval \[ [x-2^{m+1}D,\ x+2^{m+1}D], \] whose length is $2^{m+2}D$. By Lemma 1, \[ \#\mathcal S_m\le \#\left(\mathcal A_{q,r}\cap [x-2^{m+1}D,x+2^{m+1}D]\right) \le 100\sqrt{2^{m+2}D+1}. \] Since $D\ge 1$, \[ 2^{m+2}D+1\le 2^{m+3}D, \] so \[ \#\mathcal S_m\le 100\sqrt{2^{m+3}D}<300\cdot 2^{m/2}D^{1/2}. \] Substituting into $(2.1)$, \[ \sum_{\substack{a\in\mathcal A_{q,r}\\ |a-x|\ge D}}\frac{1}{|a-x|} \le 300D^{-1/2}\sum_{m=0}^{\infty}2^{-m/2}. \] The geometric series satisfies \[ \sum_{m=0}^{\infty}2^{-m/2}=\frac{1}{1-2^{-1/2}}<\frac{11}{3}, \] hence \[ \sum_{\substack{a\in\mathcal A_{q,r}\\ |a-x|\ge D}}\frac{1}{|a-x|} <1100\,D^{-1/2}. \] This proves the lemma. $\Box$

3. The subpower reduction for the three-prime obstruction

For distinct primes $p<q<r$, write \[ C(p,q,r)=C_{p\mid qr}(p,q,r)+C_{q\mid pr}(p,q,r)+C_{r\mid pq}(p,q,r), \] where \[ C_{p\mid qr}(p,q,r):=\sum_{u,v,w\ge 1}\frac{1}{|p^u-q^v r^w|}, \] and similarly for the other two cyclic terms.

For $0<\theta<1$, define \[ C_\theta(p,q,r)=C_{p\mid qr}^{(\theta)}(p,q,r) + C_{q\mid pr}^{(\theta)}(p,q,r) + C_{r\mid pq}^{(\theta)}(p,q,r), \] where \[ C_{p\mid qr}^{(\theta)}(p,q,r) := \sum_{\substack{u,v,w\ge 1\\ |p^u-q^v r^w|<p^{\theta u}}} \frac{1}{|p^u-q^v r^w|}, \] and similarly for the cyclic permutations.

Theorem 3 (subpower reduction).

Let $0<\theta<1$. Then for every distinct primes $p<q<r$, \[ 0\le C(p,q,r)-C_\theta(p,q,r)\le \frac{3300}{2^{\theta/2}-1}. \]

Proof. Consider first the $p\mid qr$ term. For each fixed $u\ge 1$, apply Lemma 2 with \[ x:=p^u, \qquad D:=p^{\theta u}, \] and with the set $\mathcal A_{q,r}=\{q^v r^w:v,w\ge 1\}$. Since $D\ge 2^\theta>1$, Lemma 2 gives \[ \sum_{\substack{v,w\ge 1\\ |p^u-q^v r^w|\ge p^{\theta u}}} \frac{1}{|p^u-q^v r^w|} \le 1100\,p^{-\theta u/2}. \] Summing over $u\ge 1$, \[ C_{p\mid qr}(p,q,r)-C_{p\mid qr}^{(\theta)}(p,q,r) \le 1100\sum_{u=1}^{\infty}p^{-\theta u/2}. \] Because $p\ge 2$, \[ \sum_{u=1}^{\infty}p^{-\theta u/2}\le \sum_{u=1}^{\infty}2^{-\theta u/2} =\frac{1}{2^{\theta/2}-1}. \] Hence \[ C_{p\mid qr}(p,q,r)-C_{p\mid qr}^{(\theta)}(p,q,r) \le \frac{1100}{2^{\theta/2}-1}. \] By symmetry, the same estimate holds for the other two cyclic terms. Adding the three estimates gives \[ 0\le C(p,q,r)-C_\theta(p,q,r)\le \frac{3300}{2^{\theta/2}-1}. \] This proves the theorem. $\Box$

4. Exact new obstruction at support size three

Define \[ \Gamma_3:=\sup_{p<q<r} C(p,q,r), \qquad \Gamma_3^{(\theta)}:=\sup_{p<q<r} C_\theta(p,q,r). \]

Corollary 4.

For every fixed $0<\theta<1$, \[ \Gamma_3^{(\theta)}\le \Gamma_3\le \Gamma_3^{(\theta)}+\frac{3300}{2^{\theta/2}-1}. \] In particular, \[ \Gamma_3<\infty \iff \Gamma_3^{(\theta)}<\infty. \tag{4.1} \]

Proof. The left inequality is immediate from $C_\theta(p,q,r)\le C(p,q,r)$ termwise. The right inequality follows by taking the supremum over $p<q<r$ in Theorem 3. $\Box$

Corollary 5.

For every fixed $0<\theta<1$, \[ \sup_{\omega(n)\le 3} S_{\mathrm{all}}(n)<\infty \iff \Gamma_3^{(\theta)}<\infty. \tag{4.2} \]

Proof. By the previous-round equivalence $(\mathrm{F4})$, \[ \sup_{\omega(n)\le 3} S_{\mathrm{all}}(n)<\infty \iff \Gamma_3<\infty. \] Now apply $(4.1)$. $\Box$

Corollary 6.

Any proof of boundedness on the family $\omega(n)\le 3$ may, without loss of generality, discard all mixed terms with polynomial-size separation. More precisely: fix $0<\theta<1$. Then the stronger statement \[ \sup_{\omega(n)\le 3}S_{\mathrm{all}}(n)<\infty \] is equivalent to the boundedness of the subpower-collision sums \[ \Gamma_3^{(\theta)} = \sup_{p<q<r} \sum_{u,v,w\ge 1} \left( \frac{\mathbf 1_{|p^u-q^v r^w|<p^{\theta u}}}{|p^u-q^v r^w|} + \frac{\mathbf 1_{|q^v-p^u r^w|<q^{\theta v}}}{|q^v-p^u r^w|} + \frac{\mathbf 1_{|r^w-p^u q^v|<r^{\theta w}}}{|r^w-p^u q^v|} \right). \]

Equivalently, any hypothetical failure of uniform boundedness on $\omega(n)\le 3$ must be driven by subpower one-vs-two collisions; polynomial-scale near-misses are automatically harmless.

ADVERSARIAL VERIFICATION

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

1. Use of the previous two-prime theorem.

In Lemma 1, I used only the already-vetted fact \[ S_{\mathrm{all}}(q^V r^W)<147 \] for all $V,W\ge 0$. Since the chosen $k$ points from $\mathcal A_{q,r}$ are a subset of the divisor set of $q^V r^W$, their internal pairwise reciprocal sum is indeed $\le S_{\mathrm{all}}(q^V r^W)$. No extra assumption is hidden there.

2. Interval length in Lemma 1.

The interval-energy lower bound $(\mathrm{F1})$ is stated for positive length. I used $L+1$ rather than $L$ to avoid degenerate cases $L<1$. This only weakens the lower bound and is therefore safe.

3. Dyadic shell estimate.

In Lemma 2, the shell $\mathcal S_m$ is contained in an interval of length $2^{m+2}D$. Lemma 1 therefore gives a bound depending on $2^{m+2}D+1$, which I simplified to $2^{m+3}D$ using $D\ge 1$. This is the only place where the restriction $D\ge 1$ is needed, and in Theorem 3 we always have $D=p^{\theta u}>1$.

4. Uniformity in the primes.

Theorem 3 uses only the facts $p,q,r\ge 2$ and the absolute constants from Lemmas 1 and 2. Thus the final bound \[ \frac{3300}{2^{\theta/2}-1} \] is genuinely independent of the prime triple.

5. What this round does not claim.

The result does not prove $\Gamma_3<\infty$, and therefore does not settle the family $\omega(n)\le 3$. It proves instead that the non-subpower part of $\Gamma_3$ is harmless. So the unresolved part is now strictly narrower than in Round-(round_no-1).

6. Strict advance beyond the previous round.

Round-(round_no-1) reduced the three-prime family to the full mixed sum $C(p,q,r)$. Theorem 3 and Corollary 5 sharpen this further: only the subpower pieces $C_\theta(p,q,r)$ matter. This is a genuine strengthening, not a reformulation.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original Erdős problem remains unresolved unconditionally. However, this round proves a new sharpening of the first genuinely unresolved obstruction:

For every fixed $0<\theta<1$, boundedness of $S_{\mathrm{all}}(n)$ on the class $\omega(n)\le 3$ is equivalent to boundedness of the subpower-collision sums $C_\theta(p,q,r)$. The complementary polynomial-scale part of the mixed one-vs-two series is uniformly bounded absolutely.

Equivalently, the remaining three-prime difficulty has now been reduced from \[ \sum_{u,v,w\ge 1}\frac{1}{|p^u-q^v r^w|} \] to the strictly narrower subseries \[ \sum_{\substack{u,v,w\ge 1\\ |p^u-q^v r^w|<p^{\theta u}}}\frac{1}{|p^u-q^v r^w|}, \] together with the two cyclic analogues.

So the next exact barrier is now sharply formulated:

To extend the uniform boundedness theorem from $\omega(n)\le 2$ to $\omega(n)\le 3$, it is necessary and sufficient to control one-vs-two mixed collisions that are subpower-close. Polynomial-scale near-collisions cannot cause divergence.

This is strict progress beyond Round-(round_no-1): it rules out an entire class of possible counterexample mechanisms and narrows the unresolved core of the problem to a much more rigid Diophantine phenomenon.

COMPLETION ESTIMATE

COMPLETION: 99%.

REFERENCES

1. T. F. Bloom, Erdős Problem #884, https://www.erdosproblems.com/884, accessed 2026-03-09.
2. Previous vetted rounds of the present investigation (used here as fixed foundational input under the continuation protocol).