gpt pro 5.4 attempt on Math-642, version 3

Erdős Problem #12 --- Round 4 (Continuation & Gap-Closure Mode)

1) ROUND-4 OBJECTIVE

Path (C): obstruction/correction via reduction to the primitive case.

Round 3 pushed one important primitive construction paradigm (the square-based one) to its natural asymptotic limit. The next unresolved issue is whether nonprimitive behaviour could fundamentally change the problem: perhaps by adjoining carefully chosen larger multiples of existing elements one could obtain denser sets or a divergent reciprocal sum, even if the primitive core remains sparse.

The most promising Round-4 direction is therefore to determine whether the original questions are genuinely more general than their primitive analogues.

I will prove that they are not. More precisely, every infinite set $A$ satisfying $(*)$ has an infinite primitive kernel $K\subseteq A$ satisfying $(*)$ such that: \[ |A\cap[1,N]|\le |K\cap[1,N]|+|K\cap[1,N/2]|\le 2|K\cap[1,N]| \] for every $N$, and \[ \sum_{a\in A}\frac1a\le \frac32\sum_{k\in K}\frac1k, \qquad \sum_{a\in A}\frac1{a\log a}\le \frac32\sum_{k\in K}\frac1{k\log k}. \] Consequently:

• Q1 is exactly equivalent to its restriction to primitive sets;
• Q3 is exactly equivalent to its restriction to primitive sets;
• Q2 is equivalent at the yes/no level to its restriction to primitive sets (the exponent may shrink by an arbitrarily small amount when passing back to general sets).

Thus any eventual counterexample to Q1 or Q3, and any eventual proof of Q1--Q3, may be sought entirely inside the primitive case.

This is a genuine obstruction to further progress by "decorating" existing primitive constructions with larger multiples: such decorations change counting functions by at most a factor $2$ and reciprocal sums by at most a factor $3/2$.

2) ROUND-3 FOUNDATION USED

I rely on the following previously established result.

• Round-2 Corollary 12.1b (part of the vetted earlier foundation). If $A$ satisfies $(*)$ and $a\in A$, then there is at most one element $b\in A$ with $b>a$ and $a\mid b$. In particular, an infinite such $A$ contains neither $1$ nor $2$.

No other earlier lemma is needed in the new proofs below.

3) NEW INSIGHT / TOOL (ROUND-4)

The new tool is the primitive-kernel decomposition.

Given $A$, define its divisibility-minimal kernel \[ K:=\{a\in A:\text{there is no }d\in A\text{ with }d<a\text{ and }d\mid a\}. \] The key observation is that every nonminimal element of $A$ can be assigned to one of its minimal divisors in $K$, and this assignment is automatically injective because Round-2 Corollary 12.1b forbids two distinct larger multiples of the same kernel element.

This converts the whole set $A$ into "one kernel element plus at most one satellite" at each kernel point. That yields the sharp transfer inequalities stated above.

4) ATTACK PLAN (ROUND-4)

Gap left after Round 3. Round 3 analysed a large primitive construction class, but it did not rule out the possibility that the nonprimitive part of a general solution could carry the real difficulty.

Claims to prove now.

1. Construct a canonical primitive subset $K\subseteq A$.
2. Prove that nonkernel elements inject into $K$.
3. Deduce the pointwise counting bound \[ |A\cap[1,N]|\le |K\cap[1,N]|+|K\cap[1,N/2]|. \]
4. Deduce the reciprocal inequalities \[ \sum_{a\in A}\frac1a\le \frac32\sum_{k\in K}\frac1k, \qquad \sum_{a\in A}\frac1{a\log a}\le \frac32\sum_{k\in K}\frac1{k\log k}. \]
5. Use these to reduce Q1--Q3 to primitive sets.
6. As an unconditional bonus, use a theorem on primitive sets to prove \[ \sum_{a\in A}\frac1{a\log a}<2.4 \] for every infinite $A$ satisfying $(*)$.

Why this overcomes the earlier obstacle. It shows that the nonprimitive part of $A$ is never more than a constant-factor perturbation of a primitive set. So the genuinely hard core of the Erdős--Sárközy problem is already present in the primitive case.

5) WORK (ROUND-4)

Notation.

For real $x\ge 1$ write \[ A(x):=|A\cap[1,x]|. \] Given an infinite $A\subseteq\mathbb N$ satisfying $(*)$, define its primitive kernel \[ K:=\{a\in A:\text{there is no }d\in A\text{ with }d<a\text{ and }d\mid a\}. \] Write similarly \[ K(x):=|K\cap[1,x]|. \] Since property $(*)$ is hereditary under passing to subsets, every subset of $A$ also satisfies $(*)$.

Lemma 12.11 (primitive-kernel injection).

Let $A$ be infinite and satisfy $(*)$, and let $K$ be its primitive kernel. Then:

1. (i) $K$ is an infinite primitive subset of $A$, hence also satisfies $(*)$.
2. (ii) For every $n\in A\setminus K$ there exists at least one element $\kappa(n)\in K$ with $\kappa(n)<n$ and $\kappa(n)\mid n$.
3. (iii) Any such choice of $\kappa(n)$ defines an injective map \[ \kappa:A\setminus K\longrightarrow K. \]

Proof. (i) By definition, no element of $K$ is divisible by a smaller element of $A$; in particular no two distinct elements of $K$ divide one another, so $K$ is primitive. Since $K\subseteq A$, it also satisfies $(*)$.

To see that $K$ is infinite, suppose instead that $K$ were finite. By Round-2 Corollary 12.1b, each $k\in K$ has at most one larger multiple in $A$. Hence $A$ would contain at most the elements of $K$ together with at most one further element above each $k\in K$, so \[ |A|\le 2|K|<\infty, \] contradicting that $A$ is infinite. Thus $K$ is infinite.

(ii) Fix $n\in A\setminus K$. Consider the finite set \[ D_n:=\{d\in A:d\mid n\}. \] This set is nonempty because $n\in D_n$. Choose $d_0\in D_n$ minimal with respect to the usual order on $\mathbb N$. Since $n\notin K$, we have $d_0<n$. By minimality of $d_0$ inside $D_n$, there is no element $e\in A$ with $e<d_0$ and $e\mid d_0$. Hence $d_0\in K$. Define $\kappa(n):=d_0$. Then $\kappa(n)\in K$, $\kappa(n)<n$, and $\kappa(n)\mid n$.

(iii) Suppose $n_1,n_2\in A\setminus K$ are distinct and satisfy \[ \kappa(n_1)=\kappa(n_2)=k. \] Then $k\in K$, and $k\mid n_1$, $k\mid n_2$, with both $n_1,n_2>k$. This contradicts Round-2 Corollary 12.1b, which says that a fixed element of $A$ can have at most one larger multiple in $A$. Therefore $\kappa$ is injective. $\square$

Proposition 12.12 (counting and reciprocal transfer to the primitive kernel).

Let $A$ and $K$ be as in Lemma 12.11. Then for every $N\ge 1$, \begin{equation} \label{wsref-0-65-71-3a-63-6f-75-6e-74-2d-74-72-61-6e-73-66-65-72} A(N)\le K(N)+K(N/2)\le 2K(N). \end{equation} Moreover, \begin{equation} \label{wsref-0-65-71-3a-72-65-63-69-70-72-6f-63-61-6c-2d-74-72-61-6e-73-66-65-72} \sum_{a\in A}\frac1a\le \frac32\sum_{k\in K}\frac1k, \end{equation} and \begin{equation} \label{wsref-0-65-71-3a-77-65-69-67-68-74-65-64-2d-74-72-61-6e-73-66-65-72} \sum_{a\in A}\frac1{a\log a}\le \frac32\sum_{k\in K}\frac1{k\log k}. \end{equation}

Proof. Split \[ A=K\sqcup B, \qquad B:=A\setminus K. \] By Lemma 12.11 there is an injective map $\kappa:B\to K$ with $\kappa(b)\mid b$ and $\kappa(b)<b$ for every $b\in B$. Since $\kappa(b)$ is a proper divisor of $b$, we have \[ b\ge 2\kappa(b). \]

Now fix $N\ge 1$ and let $b\in B\cap[1,N]$. Then \[ \kappa(b)\le b/2\le N/2. \] Since $\kappa$ is injective, the map $b\mapsto \kappa(b)$ injects $B\cap[1,N]$ into $K\cap[1,N/2]$. Therefore \[ |B\cap[1,N]|\le K(N/2). \] Adding the kernel elements gives \[ A(N)=K(N)+|B\cap[1,N]|\le K(N)+K(N/2)\le 2K(N), \] which proves (eq:count-transfer).

For the unweighted reciprocal sum, use $b\ge 2\kappa(b)$: \[ \sum_{b\in B}\frac1b \le \sum_{b\in B}\frac1{2\kappa(b)} \le \frac12\sum_{k\in K}\frac1k, \] where the last inequality again uses injectivity of $\kappa$. Hence \[ \sum_{a\in A}\frac1a = \sum_{k\in K}\frac1k+\sum_{b\in B}\frac1b \le \frac32\sum_{k\in K}\frac1k, \] proving (eq:reciprocal-transfer).

For the logarithmically weighted sum, note first that $K\subseteq A$ and, by the earlier foundation, $2\notin A$ for infinite sets satisfying $(*)$. Hence every $k\in K$ satisfies $k\ge 3$, so $\log k>0$. Also $b\ge 2\kappa(b)$ implies \[ \log b\ge \log \kappa(b). \] Therefore \[ \frac1{b\log b} \le \frac1{2\kappa(b)\log \kappa(b)}. \] Summing over $b\in B$ and using injectivity of $\kappa$ again yields \[ \sum_{b\in B}\frac1{b\log b} \le \frac12\sum_{k\in K}\frac1{k\log k}. \] Adding the kernel contribution gives (eq:weighted-transfer). $\square$

Theorem 12.13 (reduction of Q1, Q2, Q3 to primitive sets).

Let (Q1$_{\mathrm{prim}}$), (Q2$_{\mathrm{prim}}$), (Q3$_{\mathrm{prim}}$) denote the versions of $Q1$, $Q2$, $Q3$ in which one additionally assumes that $A$ is primitive. Then:

1. (a) $Q1$ is true if and only if (Q1$_{\mathrm{prim}}$) is true.
2. (b) $Q3$ is true if and only if (Q3$_{\mathrm{prim}}$) is true.
3. (c) At the yes/no level, $Q2$ is true if and only if (Q2$_{\mathrm{prim}}$) is true. More precisely: if (Q2$_{\mathrm{prim}}$) holds with some exponent $c>0$, then for every $0<c'<c$, $Q2$ holds with exponent $c'$.

Proof. The implications from the general statements to the primitive statements are immediate, since primitive sets are a special case.

(a) Suppose $Q1$ holds, and let $A$ be an infinite set satisfying $(*)$ with \[ \liminf_{N\to\infty}\frac{A(N)}{\sqrt N}>0. \] Let $K$ be its primitive kernel. By (eq:count-transfer), \[ K(N)\ge \frac12 A(N) \qquad\text{for every }N. \] Hence \[ \liminf_{N\to\infty}\frac{K(N)}{\sqrt N} \ge \frac12 \liminf_{N\to\infty}\frac{A(N)}{\sqrt N}>0. \] Since $K$ is infinite, primitive, and satisfies $(*)$, this proves (Q1$_{\mathrm{prim}}$). The converse is trivial.

(b) If (Q3$_{\mathrm{prim}}$) holds and $A$ is arbitrary satisfying $(*)$, let $K$ be its primitive kernel. Then by (eq:reciprocal-transfer), \[ \sum_{a\in A}\frac1a\le \frac32\sum_{k\in K}\frac1k. \] The right-hand side converges by (Q3$_{\mathrm{prim}}$), so $Q3$ holds. The converse is again trivial because primitive sets are a special case.

(c) First, if $Q2$ holds for all sets, then it certainly holds for primitive ones. Conversely, assume (Q2$_{\mathrm{prim}}$) holds with exponent $c>0$. Fix any $c'$ with $0<c'<c$. Let $A$ be arbitrary satisfying $(*)$, and let $K$ be its primitive kernel. By assumption, there are infinitely many $N$ with \[ K(N)<N^{1-c}. \] For such $N$, Proposition 12.12 gives \[ A(N)\le 2K(N)<2N^{1-c}. \] Since $c-c'>0$, for all sufficiently large $N$ one has $2<N^{c-c'}$, hence \[ 2N^{1-c}<N^{1-c'}. \] Therefore there are infinitely many $N$ such that \[ A(N)<N^{1-c'}. \] Thus $Q2$ holds with exponent $c'$. Since the original question only asks for the existence of some positive exponent, the yes/no status of $Q2$ is unchanged by restricting to primitive sets. $\square$

Corollary 12.13a (unconditional weighted reciprocal theorem).

For every infinite set $A\subseteq\mathbb N$ satisfying $(*)$, \[ \sum_{a\in A}\frac1{a\log a}<2.4. \] In particular, \[ \sum_{a\in A}\frac1{a\log a}<\infty. \]

Proof. Let $K$ be the primitive kernel of $A$. By Proposition 12.12, \[ \sum_{a\in A}\frac1{a\log a} \le \frac32\sum_{k\in K}\frac1{k\log k}. \] Now $K$ is primitive and $2\notin K$ because $K\subseteq A$ and infinite such sets $A$ avoid $2$ by Round-2 Corollary 12.1b. Therefore Theorem 4.4 of Lichtman applies and gives \[ \sum_{k\in K}\frac1{k\log k}<1.60. \] Multiplying by $3/2$ yields \[ \sum_{a\in A}\frac1{a\log a}<2.40. \] $\square$

Interpretation relative to Round 3.

Round 3 showed that a large square-based primitive class cannot reach the target in $Q1$. Round 4 now shows that this is not a defect of having insisted on primitiveness: adding nonprimitive multiples can only change the counting function by a factor at most $2$. Thus any genuine progress on Q1 or Q3 must come from finding a qualitatively new primitive construction, or from proving a theorem for primitive sets.

6) ADVERSARIAL VERIFICATION

(i) Could a divisibility component have long chains and break the reduction? No. In fact, Round-2 Corollary 12.1b is much stronger: a fixed $a\in A$ has at most one larger multiple in $A$ in total. So chains of length $\ge 3$ are impossible automatically, because if $a\mid b\mid c$ with $a<b<c$, then $a$ would have two larger multiples $b$ and $c$.

(ii) Is the map $\kappa$ well-defined when a nonkernel element has several minimal divisors? Yes. One may choose any minimal divisor in $K$. Injectivity does not depend on uniqueness: if two distinct elements were assigned the same $k\in K$, then $k$ would divide two distinct larger elements of $A$, contradicting Round-2 Corollary 12.1b.

(iii) Could the counting bound fail near the endpoint $N/2$? No. If $b\in A\setminus K$ and $\kappa(b)=k$, then $b\ge 2k$. Hence $b\le N$ implies $k\le N/2$. This is the only place where $N/2$ enters.

(iv) Is there any issue with logarithms at small numbers? No. Infinite sets satisfying $(*)$ contain neither $1$ nor $2$, by the earlier foundation. Hence every $k\in K$ satisfies $k\ge 3$, so $\log k>0$.

(v) Does the reduction really preserve the open questions? For Q1 and Q3, yes exactly: primitive is a special case in one direction, and Proposition 12.12 gives the reverse transfer. For Q2, the exponent may shrink from $c$ to any $c'<c$ when passing from primitive sets back to general sets, but since Q2 only asks whether some positive exponent exists, the yes/no status is preserved.

(vi) Could the primitive kernel $K$ be finite even when $A$ is infinite? No. If $K$ were finite, each $k\in K$ could support at most one larger multiple in $A$, so $A$ would have at most $2|K|$ elements.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

This round does not solve the original Erdős--Sárközy questions outright, but it isolates the true core of the problem.

The three open questions may be reduced to primitive sets. In particular, any eventual counterexample to Q1 or Q3 must already be primitive, and any proof of Q1--Q3 may be carried out after restricting to primitive sets.

This is a strict advance beyond Round 3 because it proves that the nonprimitive part of a set satisfying $(*)$ is asymptotically inessential:

• it can increase counting functions by at most a factor $2$;
• it can increase reciprocal sums by at most a factor $3/2$;
• it yields an unconditional explicit theorem \[ \sum_{a\in A}\frac1{a\log a}<2.4. \]

So the unresolved core has now been sharply reformulated as follows:

Does there exist an infinite primitive set $A$ satisfying $(*)$ with \[ \liminf_{N\to\infty}\frac{|A\cap[1,N]|}{\sqrt N}>0? \] Must every primitive such $A$ satisfy \[ \sum_{a\in A}\frac1a<\infty? \] And does some positive-exponent gap in Q2 already hold in the primitive case?

That primitive reformulation is the minimal corrected statement isolated by this round.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 70%

9) REFERENCES

• T. F. Bloom, Erdős Problem #12, accessed 2026-03-07.
• J. D. Lichtman, A proof of the Erdős primitive set conjecture, Forum of Mathematics, Pi 11 $2023$, e18, 1--21.

10) OUTPUT FORMAT

Here $x=12$, so the requested export filename is \[ \texttt{12\_solutions.tex}. \]