gpt pro 5.4 attempt on Math-642, version 2

Erdős Problem #12 --- Round 3 (Continuation & Gap-Closure Mode)

1) ROUND-3 OBJECTIVE

Path (C): obstruction/correction inside the Round-2 square-based program.

Round 2 proved that whenever \[ S\subseteq \mathbb N_{\ge 2} \] is primitive and every prime factor of every $s\in S$ is congruent to $3\pmod 4$, the square set \[ A(S):=\{s^2:s\in S\} \] satisfies (*). That opened a natural constructive program: make $S$ as large as possible.

Round 2, however, only supplied fixed-$k$ squarefree $k$-prime examples (Corollary 12.4b and Proposition 12.5), giving \[ |A_k\cap[1,N]|\asymp \frac{\sqrt N}{\log N}(\log\log N)^{k-1}, \] and it only ruled out one special "private prime" strategy.

The most promising Round-3 task is therefore to settle the internal size question for the whole Round-2 square-based class:

How large can a primitive set $S$ supported on primes $3\bmod 4$ actually be?

I will prove a sharp answer (for infinitely many scales): inside this entire class, one can reach \[ \gg \frac{\sqrt N}{\sqrt{\log N}} \] along an infinite sequence of $N$, but one can never do better than \[ \ll \frac{\sqrt N}{\sqrt{\log N}} \] in general. Thus the Round-2 square-based strategy has an intrinsic $\sqrt{\log N}$ barrier and cannot by itself settle $Q1$.

2) ROUND-2 FOUNDATION USED

I rely on the following specific Round 2 results.

• Round-2 Lemma 12.3 / Corollary 12.3a: for $p\equiv 3\pmod 4$, \[ v_p(x^2+y^2)=2\min(v_p(x),v_p(y)), \] hence $p^{2e}\mid (x^2+y^2)$ forces $p^e\mid x,y$.

• Round-2 Theorem 12.4: if $S$ is primitive and every prime divisor of every element of $S$ is congruent to $3\pmod 4$, then \[ A(S)=\{s^2:s\in S\} \] satisfies property (*).

• Round-2 Proposition 12.6: a certain "private prime" subclass cannot achieve $\gg \sqrt N$. Round 3 replaces that by an obstruction for the entire square-based class coming from Theorem 12.4.

3) NEW INSIGHT / TOOL (ROUND-3)

The genuinely new ingredient is a block-antichain construction inside the multiplicative semigroup \[ \mathcal M:=\{n\in\mathbb N:\ p\mid n\Rightarrow p\equiv 3\pmod 4\}. \]

There are two parts.

1. Analytic size of $\mathcal M$. By a standard Selberg--Delange theorem (specialised below), \[ M(x):=|\mathcal M\cap[1,x]|\asymp \frac{x}{\sqrt{\log x}}. \]

2. Primitive extraction by thick blocks. Although $\mathcal M$ itself is not primitive, a subset of one interval $(x/3,x]$ is primitive, because the least nontrivial element of $\mathcal M$ is $3$. By choosing a very sparse sequence of scales $x_j$ and taking many elements of $\mathcal M\cap(x_j/3,x_j]$ that avoid divisibility by earlier chosen elements, one can inductively build an infinite primitive \[ S\subseteq \mathcal M \] with \[ |S\cap[1,x_j]|\gg \frac{x_j}{\sqrt{\log x_j}} \] for infinitely many $j$.

This is a strict strengthening of Round 2's fixed-$k$ families, and it also shows that the entire Round-2 square-based method has a built-in $\sqrt{\log N}$ loss.

4) ATTACK PLAN (ROUND-3)

Exact Round-2 gap. Round 2 showed that some primitive sets $S\subseteq \mathcal M$ work, but did not determine the true order of magnitude obtainable from this paradigm.

Claims to prove now.

1. Prove an analytic counting estimate for $\mathcal M$: \[ M(x)\asymp \frac{x}{\sqrt{\log x}}. \]

2. Deduce an upper bound for the whole Round-2 square-based class: for every primitive $S\subseteq \mathcal M$, \[ |A(S)\cap[1,N]|\ll \frac{\sqrt N}{\sqrt{\log N}}. \] Hence no set produced via Round-2 Theorem 12.4 can answer $Q1$.

3. Construct an infinite primitive $S\subseteq \mathcal M$ with \[ |S\cap[1,x_j]|\gg \frac{x_j}{\sqrt{\log x_j}} \] for infinitely many $x_j$, and then invoke Round-2 Theorem 12.4 to obtain \[ |A(S)\cap[1,N_j]|\gg \frac{\sqrt{N_j}}{\sqrt{\log N_j}} \qquad\text{for infinitely many }N_j=x_j^2. \]

Why this overcomes the previous obstacle. Round 2 used only thin fixed-$k$ layers of $\mathcal M$. The new construction instead uses full local density of $\mathcal M$ on carefully separated scales, while preserving primitivity by:

• taking each block inside an interval of ratio $<3$, and
• excluding multiples of earlier chosen elements.

5) WORK (ROUND-3)

Notation.

Let \[ \mathcal M:=\{n\in\mathbb N:\ p\mid n\Longrightarrow p\equiv 3\pmod 4\}, \qquad M(x):=|\mathcal M\cap[1,x]|. \] Note that $1\in \mathcal M$, and the least element of $\mathcal M\setminus\{1\}$ is $3$.

Proposition 12.7 (analytic counting input for $\mathcal M$).

There exists a constant $C_{\mathcal M}>0$ such that \[ M(x)=C_{\mathcal M}\frac{x}{(\log x)^{1/2}}+O\!\left(\frac{x}{(\log x)^{3/4}}\right) \qquad (x\to\infty). \] Consequently there exist positive constants $c_1,c_2,x_0$ such that for all $x\ge x_0$, \begin{equation} \label{wsref-0-65-71-3a-4d-2d-75-70-70-65-72} M(x)\le c_2\frac{x}{\sqrt{\log x}}, \end{equation} and \begin{equation} \label{wsref-0-65-71-3a-4d-2d-69-6e-74-65-72-76-61-6c} M(x)-M(x/3)\ge c_1\frac{x}{\sqrt{\log x}}. \end{equation}

Proof. Apply the Selberg--Delange asymptotic of Chang--Martin, Theorem 3.4, with modulus $q=4$ and residue class set $B=\{3\}$. Then $N_B$ in their notation is exactly the set $\mathcal M$ above, and their exponent is \[ \beta=\frac{|B|}{\varphi(4)}=\frac12. \] Hence \[ M(x)=\#\{n\le x:\ n\in\mathcal M\} = \frac{G_B(1)}{\Gamma(1/2)}\frac{x}{(\log x)^{1/2}} +O\!\left(\frac{x}{(\log x)^{3/4}}\right), \] which is the asserted formula with \[ C_{\mathcal M}=\frac{G_B(1)}{\Gamma(1/2)}>0. \] The upper bound (eq:M-upper) is immediate.

For (eq:M-interval), subtract the same asymptotic at $x$ and at $x/3$: \[ M(x)-M(x/3) = C_{\mathcal M}\frac{x}{\sqrt{\log x}} - C_{\mathcal M}\frac{x/3}{\sqrt{\log(x/3)}} + O\!\left(\frac{x}{(\log x)^{3/4}}\right). \] Since \[ \frac{\sqrt{\log x}}{\sqrt{\log(x/3)}}\to 1 \qquad (x\to\infty), \] the main term equals \[ \left(\frac23+o(1)\right)C_{\mathcal M}\frac{x}{\sqrt{\log x}}, \] which is positive for all sufficiently large $x$. After reducing $c_1$ if necessary and enlarging $x_0$, (eq:M-interval) follows. $\square$

Proposition 12.8 (upper bound for the entire Round-2 square-based class).

Let $S\subseteq \mathcal M\setminus\{1\}$ be primitive, and set \[ A(S):=\{s^2:\ s\in S\}. \] Then for all large $N$, \[ |A(S)\cap[1,N]| = |S\cap[1,\sqrt N]| \ll \frac{\sqrt N}{\sqrt{\log N}}. \] In particular, \[ \liminf_{N\to\infty}\frac{|A(S)\cap[1,N]|}{\sqrt N}=0. \] So no set obtained from Round-2 Theorem 12.4 can solve $Q1$.

Proof. Because $S\subseteq \mathcal M$, \[ |A(S)\cap[1,N]|=|S\cap[1,\sqrt N]|\le M(\sqrt N). \] Applying (eq:M-upper), \[ M(\sqrt N)\le c_2\frac{\sqrt N}{\sqrt{\log(\sqrt N)}} = \sqrt 2\,c_2\frac{\sqrt N}{\sqrt{\log N}}, \] which proves the claim. $\square$

Theorem 12.9 (block-antichain construction inside $\mathcal M$).

There exists an infinite primitive set \[ S\subseteq \mathcal M\setminus\{1\} \] such that for infinitely many $x$, \[ |S\cap[1,x]|\gg \frac{x}{\sqrt{\log x}}. \]

Proof. Fix constants $c_1,c_2,x_0$ as in Proposition 12.7, and set \[ \eta:=\frac{c_1}{32c_2}. \] We will inductively construct:

• a strictly increasing sequence $x_1<x_2<\cdots$ tending to infinity,
• finite sets $T_j\subseteq \mathcal M\cap (x_j/3,x_j]$,

such that if \[ S_j:=T_1\cup\cdots\cup T_j, \] then

1. (a) $S_j$ is primitive;
2. (b) $\displaystyle \sum_{s\in S_j}\frac1s\le \eta\sum_{i=1}^j 2^{-i}<\eta$;
3. (c) $\displaystyle |T_j|\ge \frac{c_1}{5}\frac{x_j}{\sqrt{\log x_j}}$.

Once this is done, the infinite union \[ S:=\bigcup_{j\ge 1}T_j \] is primitive, and for each $j$ we have \[ |S\cap[1,x_j]|\ge |T_j|\ge \frac{c_1}{5}\frac{x_j}{\sqrt{\log x_j}}, \] which proves the theorem.

Induction step. Suppose $S_{j-1}$ has already been constructed (for $j=1$, take $S_0=\varnothing$). Choose $x_j$ so large that all of the following hold:

1. (i) $x_j\ge x_0$;
2. (ii) $x_j>3x_{j-1}$ (for $j=1$ there is no restriction);
3. (iii) $x_j>(\max S_{j-1})^2$ (vacuous if $S_{j-1}=\varnothing$);
4. (iv) $\displaystyle \frac{3c_1}{4\sqrt{\log x_j}}\le \frac{\eta}{2^j}$;
5. (v) $\displaystyle \frac{c_1}{4}\frac{x_j}{\sqrt{\log x_j}}\ge 1$.

This is possible because $x_j$ may be taken arbitrarily large.

Now consider the interval \[ I_j:=(x_j/3,x_j]. \] By (eq:M-interval), \[ |\mathcal M\cap I_j| = M(x_j)-M(x_j/3) \ge c_1\frac{x_j}{\sqrt{\log x_j}}. \] Let $E_j$ be the set of elements of $\mathcal M\cap I_j$ divisible by at least one element of $S_{j-1}$. Then by the union bound, \[ |E_j| \le \sum_{s\in S_{j-1}}\#\{n\in\mathcal M\cap I_j:\ s\mid n\}. \] If $s\mid n$ and $n\in \mathcal M$, then $n=sm$ with $m\in\mathcal M$. Hence \[ \#\{n\in\mathcal M\cap I_j:\ s\mid n\}\le M(x_j/s). \] Therefore \[ |E_j| \le \sum_{s\in S_{j-1}} M(x_j/s). \] Because $s\le \max S_{j-1}<\sqrt{x_j}$ by (iii), we have $x_j/s>\sqrt{x_j}$ and so \[ \log(x_j/s)\ge \frac12\log x_j. \] Using (eq:M-upper), \[ M(x_j/s)\le c_2\frac{x_j/s}{\sqrt{\log(x_j/s)}} \le \sqrt 2\,c_2\frac{x_j}{s\sqrt{\log x_j}} \le 2c_2\frac{x_j}{s\sqrt{\log x_j}}. \] Summing over $s\in S_{j-1}$ and using the induction hypothesis (b), \[ |E_j| \le 2c_2\frac{x_j}{\sqrt{\log x_j}}\sum_{s\in S_{j-1}}\frac1s < 2c_2\eta\frac{x_j}{\sqrt{\log x_j}} = \frac{c_1}{16}\frac{x_j}{\sqrt{\log x_j}}. \] Thus the set \[ U_j:=(\mathcal M\cap I_j)\setminus E_j \] of admissible new elements satisfies \[ |U_j| \ge \left(c_1-\frac{c_1}{16}\right)\frac{x_j}{\sqrt{\log x_j}} = \frac{15c_1}{16}\frac{x_j}{\sqrt{\log x_j}}. \] Choose any subset \[ T_j\subseteq U_j \] of cardinality \[ m_j:=\left\lfloor \frac{c_1}{4}\frac{x_j}{\sqrt{\log x_j}}\right\rfloor. \] This is possible because $|U_j|>m_j$ for large $x_j$, and by (v) we have $m_j\ge 1$. Moreover \[ m_j\ge \frac{c_1}{5}\frac{x_j}{\sqrt{\log x_j}} \] for all sufficiently large $x_j$, yielding (c).

Next, because every element of $T_j$ is $>x_j/3$, its reciprocal is at most $3/x_j$, so \[ \sum_{t\in T_j}\frac1t \le \frac{3m_j}{x_j} \le \frac{3c_1}{4\sqrt{\log x_j}} \le \frac{\eta}{2^j} \] by (iv). Hence \[ \sum_{s\in S_j}\frac1s = \sum_{s\in S_{j-1}}\frac1s+\sum_{t\in T_j}\frac1t \le \eta\sum_{i=1}^{j-1}2^{-i}+\frac{\eta}{2^j} = \eta\sum_{i=1}^{j}2^{-i}, \] so the induction hypothesis (b) is preserved.

Finally we verify primitivity.

• No old element divides a new element: this is exactly how $U_j$ was defined.
• No new element divides an old element: by (ii), every new element exceeds $x_j/3>x_{j-1}$, while every old element is at most $x_{j-1}$.
• No two distinct new elements divide one another: if $u,v\in T_j$ with $u<v$ and $u\mid v$, then \[ v/u\in \mathcal M\setminus\{1\}, \] so $v/u\ge 3$. But since $u>x_j/3$ and $v\le x_j$, \[ v/u<3, \] a contradiction.

Thus $S_j$ is primitive, completing the induction and the proof. $\square$

Corollary 12.9a (new valid construction for the original divisibility problem).

There exists an infinite set $A\subseteq \mathbb N$ satisfying (*) such that for infinitely many $N$, \[ |A\cap[1,N]|\gg \frac{\sqrt N}{\sqrt{\log N}}. \]

Proof. Let $S$ be as in Theorem 12.9, and define \[ A:=A(S)=\{s^2:\ s\in S\}. \] Since $S$ is primitive and every prime factor of every $s\in S$ is congruent to $3\pmod 4$, Round-2 Theorem 12.4 implies that $A$ satisfies (*).

Now take $N_j:=x_j^2$. Then \[ |A\cap[1,N_j]| = |S\cap[1,x_j]| \gg \frac{x_j}{\sqrt{\log x_j}} = \sqrt 2\,\frac{\sqrt{N_j}}{\sqrt{\log N_j}}. \] Hence the claimed lower bound holds for infinitely many $N=N_j$. $\square$

Theorem 12.10 (sharp size scale inside the Round-2 square-based class).

Among all sets of the form \[ A(S)=\{s^2:\ s\in S\}, \] where $S\subseteq \mathcal M\setminus\{1\}$ is primitive, the correct order of magnitude along infinitely many scales is \[ \frac{\sqrt N}{\sqrt{\log N}}. \] More precisely:

1. every such $A(S)$ satisfies \[ |A(S)\cap[1,N]|\ll \frac{\sqrt N}{\sqrt{\log N}} \qquad (N\to\infty), \] by Proposition 12.8;
2. there exists such an $A(S)$ with \[ |A(S)\cap[1,N_j]|\gg \frac{\sqrt{N_j}}{\sqrt{\log N_j}} \] for infinitely many $N_j$, by Corollary 12.9a.

Thus the full Round-2 square-based construction method has now been pushed to its natural asymptotic limit (up to constants) on an infinite subsequence.

6) ADVERSARIAL VERIFICATION

(i) Does the block construction really preserve primitivity? Yes. Inside one block $(x_j/3,x_j]$, divisibility would force a quotient in $\mathcal M\setminus\{1\}$, hence at least $3$, but every quotient of two numbers in that interval is $<3$. Across different blocks, earlier elements are explicitly forbidden from dividing later ones, and later elements are too large to divide earlier ones.

(ii) Is the estimate for excluded multiples legitimate? Yes. If $s\in S_{j-1}$ divides $n\in\mathcal M$, then $n=sm$ with $m\in\mathcal M$. Thus the number of multiples of $s$ up to $x_j$ is at most $M(x_j/s)$. No independence assumption is used; only the union bound.

(iii) Did the proof accidentally use more than finite reciprocal mass of old elements? No. The key inequality is \[ |E_j|\ll \frac{x_j}{\sqrt{\log x_j}}\sum_{s\in S_{j-1}}\frac1s. \] Only the finite reciprocal sum of the previously chosen blocks enters. The construction deliberately makes the blockwise reciprocal contributions summable.

(iv) Is the analytic input applied correctly? Yes. Chang--Martin Theorem 3.4 counts integers all of whose prime factors lie in a fixed set of reduced residue classes. With $q=4$ and the single class $3\bmod 4$, this is exactly the semigroup $\mathcal M$. The exponent is $\beta=1/\varphi(4)=1/2$, giving the $\sqrt{\log x}$ denominator.

(v) Does Corollary 12.9a settle $Q1$? No. It only yields the lower bound \[ |A\cap[1,N_j]|\gg \frac{\sqrt{N_j}}{\sqrt{\log N_j}} \] for infinitely many $N_j$. This is far from the required \[ \liminf_{N\to\infty}\frac{|A\cap[1,N]|}{\sqrt N}>0. \] So Q1 remains open.

(vi) Does this contradict the density-zero theorem of Erdős--Sárközy? No. Both \[ \frac{x}{\sqrt{\log x}}=o(x) \qquad\text{and}\qquad \frac{\sqrt N}{\sqrt{\log N}}=o(\sqrt N). \] So the new construction is sparse enough to be consistent with all known upper bounds.

(vii) Is the Round-2 theorem being used in exactly the permitted way? Yes. I do not re-prove Round-2 Theorem 12.4; I invoke it exactly once, at the end, after constructing a larger primitive $S$. The new mathematical work is entirely the analytic counting of $\mathcal M$ plus the primitive extraction argument.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 3 does not solve the original Erdős problem in full generality. However, it gives a sharp obstruction/correction for the whole constructive framework introduced in Round 2:

Within the entire square-based class supplied by Round-2 Theorem 12.4, the best possible size scale is \[ |A\cap[1,N]|\asymp \frac{\sqrt N}{\sqrt{\log N}} \] along infinitely many scales, and in particular this class can never realise the target in $Q1$.

This is a substantive advance beyond Round 2 for two reasons:

1. it replaces the fixed-$k$ families by a much denser valid construction, namely \[ |A\cap[1,N_j]|\gg \frac{\sqrt{N_j}}{\sqrt{\log N_j}} \] for infinitely many $N_j$;
2. it proves that no construction obtainable from Round-2 Theorem 12.4 can exceed this order by more than constants.

So the next unresolved gap is now sharply isolated:

Any eventual positive solution to $Q1$ must leave the Round-2 square-based paradigm entirely (or augment it with new structure not covered by primitive subsets of $\mathcal M$).

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 55%

9) REFERENCES

• B. Chang and G. Martin, The smallest invariant factor of the multiplicative group, especially Theorem 3.4, used here with $q=4$ and residue class set $B=\{3\}$ to count integers all of whose prime factors are congruent to $3\pmod 4$.
• T. F. Bloom, Erdős Problem #12, problem statement and current status page, accessed 2026-03-07.
• Round-2 results from the present investigation: Lemma 12.3, Corollary 12.3a, and Theorem 12.4.

10) OUTPUT FORMAT

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