gpt pro 5.2 attempt on Math-642, version 2

Erdős Problem #12 --- Round 2 (Continuation & Gap-Closure Mode)

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction (strategy-level), plus a provable strengthened construction. Round 1 verified the classical "prime-square" example \[ A_1=\{p^2:\ p\equiv 3\!\!\pmod 4\} \] satisfies (*), but it only achieves size $\asymp \sqrt{N}/\log N$. The most promising next step is to amplify the construction while also isolating why the most naive amplification strategies cannot reach \[ \liminf_{N\to\infty}\frac{|A\cap[1,N]|}{\sqrt N}>0 \qquad\text{(Q1).} \] In this round I do two things:

• I prove a Gaussian-$p$-adic valuation lemma (new tool) and use it to build a broad family of infinite sets of squares satisfying (*), strictly extending Round 1's construction.
• I prove an explicit barrier for a whole class of "private prime" (distinguishing-prime) constructions: this strategy can never reach $\gg\sqrt N$ elements up to $N$.

Questions $Q1$--$Q3$ remain open in full generality, but the results below strictly advance the construction side and rule out a natural approach to $Q1$.

2) ROUND-1 FOUNDATION USED

I rely on the following Round 1 results, without re-proving them:

• Lemma 12.1 (Round 1). For fixed $a\in A$, among $A_{>a}$ there are no distinct $b,c$ with $b\equiv -c\pmod a$ (hence at most one class from each pair $\{r,-r\}$).
• Corollary 12.1a (Round 1). If $A$ is infinite and satisfies (*), then $1\notin A$.
• Lemma 12.2 (Round 1). $A_1=\{p^2:\ p\equiv 3\pmod 4\}$ satisfies (*).

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is an exact $p$-adic valuation formula for sums of two squares at primes $p\equiv 3\pmod 4$: \[ v_p(x^2+y^2)=2\min\left(v_p(x),v_p(y)\right)\qquad (p\equiv 3\!\!\!\pmod 4). \] This upgrades the "$-1$ is a nonresidue" argument used in Round 1's Lemma 12.2 from a mod-$p$ contradiction to a prime-power statement, and it lets us build large families $A=\{n^2:\ n\in S\}$ satisfying (*) provided $S$ is chosen to avoid a simple divisibility pattern (an antichain/primitive condition).

4) ATTACK PLAN (ROUND-2)

Round 1 gap (constructive). The only fully verified infinite example in Round 1 was $A_1=\{p^2:\ p\equiv 3\pmod 4\}$. The gap is to systematically enlarge such constructions and to understand what prevents a straightforward enlargement from achieving $\gg \sqrt N$ elements.

Claims to prove in Round 2.

1. Prove the $p$-adic valuation lemma for primes $p\equiv 3\pmod 4$.
2. Use it to prove a general construction theorem: if $S$ is a suitable divisibility-antichain of integers supported on primes $\equiv 3\pmod 4$, then $A=\{n^2:\ n\in S\}$ satisfies (*).
3. Exhibit explicit infinite choices of $S$ (hence $A$) that strictly generalize $A_1$.
4. Prove an obstruction: any "private prime" strategy (each element has a new $p\equiv 3\pmod 4$ that never appears later) cannot yield $\gg \sqrt N$ elements up to $N$.

5) WORK (ROUND-2)

Corollary 12.1b (new, immediate from Round 1 Lemma 12.1).

If $A$ is infinite and satisfies (*), then $2\notin A$. More generally, for any $a\in A$, there is at most one $b\in A$ with $b>a$ and $a\mid b$.

Justification. In Lemma 12.1 take the residue class $0\bmod a$, which is self-opposite; then among $A_{>a}$ there cannot be two distinct multiples of $a$. For $a=2$, both residue classes $0,1$ are self-opposite mod $2$, so among $A_{>2}$ there can be at most one even and at most one odd element; hence $A$ cannot be infinite.

Lemma 12.3 (new: $p$-adic valuation of a sum of two squares).

Let $p$ be an odd prime with $p\equiv 3\pmod 4$. For any integers $x,y$, not both $0$, one has \[ v_p(x^2+y^2)=2\min\left(v_p(x),v_p(y)\right). \]

Proof. Write $x=p^\alpha u$ and $y=p^\beta v$ with $\alpha,\beta\ge 0$ and $p\nmid uv$. Without loss of generality assume $\alpha\le \beta$. Then \[ x^2+y^2=p^{2\alpha}\left(u^2+p^{2(\beta-\alpha)}v^2\right). \] If $\beta>\alpha$, then the bracket is congruent to $u^2\not\equiv 0\pmod p$, so $v_p(x^2+y^2)=2\alpha$.

If $\beta=\alpha$, then we must show $u^2+v^2\not\equiv 0\pmod p$. Indeed, if $u^2\equiv -v^2\pmod p$ and $p\nmid v$, then $(uv^{-1})^2\equiv -1\pmod p$, so $-1$ would be a quadratic residue modulo $p$, contradicting $p\equiv 3\pmod 4$. Thus again the bracket is nonzero mod $p$ and $v_p(x^2+y^2)=2\alpha=2\min(\alpha,\beta)$. $\square$

Corollary 12.3a.

If $p\equiv 3\pmod 4$ is prime and $p^{2e}\mid (x^2+y^2)$, then $p^e\mid x$ and $p^e\mid y$.

Proof. Lemma 12.3 gives $2\min(v_p(x),v_p(y))=v_p(x^2+y^2)\ge 2e$. Hence $\min(v_p(x),v_p(y))\ge e$.

Theorem 12.4 (new: Gaussian-antichain square construction).

Let $S\subset\mathbb N_{\ge 2}$ be an infinite set such that:

1. (i) every prime divisor of every $s\in S$ satisfies $p\equiv 3\pmod 4$;
2. (ii) $S$ is primitive: for distinct $s,t\in S$ one does not have $s\mid t$.

Define \[ A(S):=\{s^2:\ s\in S\}. \] Then $A(S)$ satisfies property (*): there do not exist distinct $a,b,c\in A(S)$ with $b,c>a$ and $a\mid (b+c)$.

Proof. Assume for contradiction that $a,b,c\in A(S)$ are distinct with $b,c>a$ and $a\mid (b+c)$. Write $a=s^2$, $b=t^2$, $c=u^2$ with $s,t,u\in S$, and note $t,u>s$ since $t^2,u^2>s^2$. The condition $a\mid(b+c)$ becomes \[ s^2\mid (t^2+u^2). \] Let $p$ be any prime divisor of $s$; by (i) we have $p\equiv 3\pmod 4$. If $v_p(s)=e\ge 1$, then $p^{2e}\mid s^2\mid (t^2+u^2)$, so by Corollary 12.3a we get $p^e\mid t$ and $p^e\mid u$. Since this holds for every prime $p\mid s$, we conclude $s\mid t$ and $s\mid u$.

But $s\mid t$ with $s\neq t$ contradicts primitivity (ii) (and likewise for $u$). Hence no such triple exists. $\square$

Corollary 12.4b (explicit infinite families extending Round 1).

Fix an integer $k\ge 1$, and let \[ S_k:=\{p_1p_2\cdots p_k:\ p_i\ \text{distinct primes with }p_i\equiv 3\pmod 4\}. \] Then $S_k$ is primitive (indeed an antichain under divisibility), so \[ A_k:=A(S_k)=\{(p_1\cdots p_k)^2:\ p_i\equiv 3\pmod 4\ \text{distinct}\} \] satisfies (*). For $k=1$ this recovers the Round 1 example $A_1=\{p^2:\ p\equiv 3\pmod 4\}$.

Proof. If $s,t\in S_k$ and $s\mid t$, then the (squarefree) prime support of $s$ is a subset of that of $t$. Since both have exactly $k$ distinct primes, this forces $s=t$. Thus $S_k$ is primitive, so Theorem 12.4 applies.

Proposition 12.5 (size of $A_k$; analytic input).

Let $k\ge 1$ be fixed. Then, as $N\to\infty$, \[ |A_k\cap[1,N]|\;=\;|S_k\cap[1,\sqrt N]| \;\asymp_k\; \frac{\sqrt N}{\log N}\,(\log\log N)^{k-1}. \] In particular, for every fixed $k$, \[ \liminf_{N\to\infty} \frac{|A_k\cap[1,N]|}{\sqrt N}\,\frac{\log N}{(\log\log N)^{k-1}} \;>\;0. \]

Justification. This is a standard Landau--Sathe--Selberg type asymptotic for squarefree $k$-almost primes, together with the fact that primes $\equiv 3\pmod 4$ have relative density $1/2$ among the primes. (Only the order of magnitude $\asymp_k$ is used here.)

Proposition 12.6 (new: "private prime" obstruction for square-based constructions).

Consider square-based constructions $A=\{s^2:\ s\in S\}$ where each $s\in S$ has a private prime $p_s\equiv 3\pmod 4$ dividing $s$ such that $p_s\nmid t$ for all $t\in S$ with $t>s$. Then for every $N$, \[ |A\cap[1,N]|=|S\cap[1,\sqrt N]|\ \le\ \pi_{3\bmod 4}(\sqrt N), \] where $\pi_{3\bmod 4}(X)$ counts primes $p\le X$ with $p\equiv 3\pmod 4$. In particular, this strategy can never achieve $|A\cap[1,N]|\gg \sqrt N$.

Proof. Define a map $\varphi:S\cap[1,\sqrt N]\to\{p\le \sqrt N:\ p\equiv 3\pmod 4\}$ by $\varphi(s)=p_s$. If $\varphi(s)=\varphi(t)=p$ with $s<t$, then $p=p_s$ divides $t$, contradicting that $p_s$ is private to $s$. Hence $\varphi$ is injective, so $|S\cap[1,\sqrt N]|\le \pi_{3\bmod 4}(\sqrt N)$. $\square$

6) ADVERSARIAL VERIFICATION

(i) Prime-power subtlety in Lemma 12.3. The key step is excluding $u^2+v^2\equiv 0\pmod p$ when $p\equiv 3\pmod 4$ and $p\nmid uv$; this is exactly the "$-1$ nonresidue" property used in Round 1, now localized to the unit part. No step uses $p=2$, and indeed $p=2$ is excluded.

(ii) Dependence on the "square" structure. Theorem 12.4 crucially uses that $b+c=t^2+u^2$ is a sum of two squares. Without $A$ being a set of squares (or more generally Gaussian norms), Corollary 12.3a is unavailable. So the theorem is not claiming a general resolution of $Q1$--$Q3$, only a principled amplification of the prime-square example.

(iii) Quantifiers in (*). In Theorem 12.4, if $a=s^2\in A(S)$ and $b=t^2,c=u^2\in A(S)$ satisfy $b,c>a$, then indeed $t,u>s$. Thus $s\mid t$ contradicts primitivity, so the forbidden configuration cannot arise. Distinctness of $a,b,c$ is used only at the final step (to rule out the trivial equality case).

(iv) Obstruction proposition sanity. The injection argument in Proposition 12.6 is robust: it uses only the definition of "private prime", not property (*). Hence it truly rules out a wide class of potential constructions for $Q1$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

• The original questions $Q1$--$Q3$ remain open in full generality.
• Round 2 provides a new structural tool (Lemma 12.3) and a broad construction principle (Theorem 12.4) that strictly extends Round 1's verified example.
• Explicit families $A_k$ are produced for every $k\ge 1$ (Corollary 12.4b), giving \[ |A_k\cap[1,N]|\asymp_k \frac{\sqrt N}{\log N}(\log\log N)^{k-1}, \] which is asymptotically larger than the $k=1$ (prime-square) construction for every fixed $k\ge 2$.
• A key obstruction is isolated: any square-based strategy that assigns each element a new "private" $p\equiv 3\pmod 4$ cannot reach $\gg\sqrt N$ (Proposition 12.6). Thus a positive answer to $Q1$, if true, must involve substantial prime re-use without a private-prime certificate.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 35%

9) REFERENCES

• P. Erdős and A. Sárközy, original formulation and density-zero result (see bibliographic pointers collected in the Erdős Problems database entry for Problem #12).
• E. M. Wright, A Simple Proof of a Theorem of Landau, Proc. Camb. Phil. Soc. 50 $1954$, 304--309. (For the classical asymptotic of $k$-almost primes used in Proposition 12.5.)
• Standard algebraic number theory fact: for $p\equiv 3\pmod 4$, $p$ is inert in $\mathbb Z[i]$, giving Lemma 12.3 (proved directly above).