gpt pro 5.2 attempt on Math-650, version 2

Round 2: Continuation & Gap-Closure (Erdős Problem #32)

1) ROUND-2 OBJECTIVE

Path (C): obstruction / correction.

Round-1 identified that (i) elementary finite obstructions (CRT and factorial blocks) do not by themselves yield the asymptotic lower bounds needed to settle the $o((\log N)^2)$ / $O(\log N)$ questions for the actual primes, and (ii) any successful improvement must exploit structure beyond mere density. In this round we prove a rigorous strategy barrier: there exist prime-like sets $\mathcal P'\subseteq\mathbb N$ with counting function $\mathcal P'(x)\sim x/\log x$ for which every additive complement $B$ must satisfy \[ \liminf_{x\to\infty}\frac{B(x)}{(\log x)^2}\ge 1. \] Consequently, any approach that uses only "PNT-scale" information (growth like $x/\log x$) cannot prove an $o((\log N)^2)$ complement for the actual primes; it would falsely apply to such $\mathcal P'$.

2) ROUND-1 FOUNDATION USED

We rely on the following Round-1 items (used only as foundations, not reproved here):

1. Definition: $A$ is an additive complement to primes if $\exists N_0$ such that every $n\ge N_0$ satisfies $n=p+a$ with $p\in\mathcal P$, $a\in A$.
2. Finite truncation principle (implicit in the Round-1 set-cover viewpoint): to cover $[M,x]$ by a sumset $P'+B$, it suffices to consider $B\cap[0,x]$, since if $p'+b\le x$ then $b\le x$.

3) NEW INSIGHT / TOOL (ROUND-2)

New input is a hard-to-complement theorem (Kolountzakis, 1996): for a suitable random prime-like set $\mathcal P'$, every additive complement $B$ must have size $\gtrsim(\log x)^2$. The proof uses:

1. a packing/greedy construction producing many disjoint truncated translates of $B$;
2. independence of membership events for $\mathcal P'$ on disjoint sets;
3. a union bound over all small $B\cap[0,x]$ and a Borel--Cantelli argument.

4) ATTACK PLAN (ROUND-2)

Round-1 gap addressed: show why arguments based only on coarse density information cannot improve the $(\log N)^2$ scale.

Plan: define a random set $\mathcal P'$ with inclusion probabilities $\mathbb P(n\in\mathcal P')=1/\log n$ (for $n\ge4$). Show that almost surely:

1. $\mathcal P'(x)\sim x/\log x$ (so $\mathcal P'$ is prime-like in growth);
2. if $B(x)\le C(\log x)^2$ infinitely often with some $C<1$, then $B$ cannot be an additive complement of $\mathcal P'$.

Thus, almost surely every additive complement satisfies $\liminf B(x)/(\log x)^2\ge1$.

5) WORK (ROUND-2)

5.1 Setup

Let $\mathbb N=\{0,1,2,\dots\}$. Define a random set $\mathcal P'\subseteq\{4,5,6,\dots\}$ by choosing each integer $n\ge4$ independently with probability \[ \mathbb P(n\in\mathcal P')=\frac1{\log n}. \] For $B\subseteq\mathbb N$ define $B(x):=|B\cap[0,x]|$.

We say $B$ is an additive complement of $\mathcal P'$ if $\exists N_0$ such that every $n\ge N_0$ can be written $n=p'+b$ with $p'\in\mathcal P'$, $b\in B$.

5.2 Main theorem (barrier)

Theorem. With probability $1$ (over the choice of $\mathcal P'$), every additive complement $B$ of $\mathcal P'$ satisfies \[ \liminf_{x\to\infty}\frac{B(x)}{(\log x)^2}\ge 1. \] In particular, for every fixed $C<1$ there is almost surely no additive complement $B$ such that $B(x)\le C(\log x)^2$ for infinitely many $x$.

5.3 Combinatorial packing lemma

Fix $x\in\mathbb N$ and finite $B\subseteq[0,x]$ with $m:=|B|=B(x)$. For $a\in[0,x]$ define $B_{\le a}:=B\cap[0,a]$ and \[ a-B_{\le a}:=\{a-b:\ b\in B_{\le a}\}. \]

Fix $\lambda\in(0,1)$ and set \[ s:=\left\lfloor \frac{\lambda x}{m}\right\rfloor. \]

Lemma 1 (many disjoint truncated translates). There exist integers $a_1,\dots,a_n\in[0,x]$ such that:

1. the sets $a_j-B_{\le a_j}$ are pairwise disjoint;
2. each $(a_j-B_{\le a_j})\cap[1,s]=\emptyset$;
3. and \[ n\ge \frac{(1-\lambda)x}{m^2}. \]

Proof. Choose a maximal family $a_1,\dots,a_n$ satisfying $1$--$2$. Let \[ G:=[1,s]\ \cup\ \bigcup_{j=1}^n (a_j-B_{\le a_j}). \] By maximality, for every $y\in[0,x]$ the set $y-B_{\le y}$ intersects $G$, so $y\in G+B$. Thus $[0,x]\subseteq G+B$ and hence \[ x+1 \le |G+B|\le |G|\cdot|B|\le (s+nm)m. \] Rearranging gives $n\ge (x+1)/m^2 - s/m \ge x/m^2 - s/m \ge (1-\lambda)x/m^2$ since $s\le \lambda x/m$. $\square$

5.4 A single-point coverage bound

Fix $x$, $B\subseteq[0,x]$ with $|B|=m$, and $s\ge4$. If $a\in[0,x]$ satisfies $(a-B_{\le a})\cap[1,s]=\emptyset$, then every positive element of $a-B_{\le a}$ is $>s$.

Lemma 2 (coverage probability bound). For any fixed $\varepsilon>0$, for all sufficiently large $s$ (depending on $\varepsilon$), \[ \mathbb P(a\in \mathcal P'+B)\ \le\ 1-\exp\!\left(-\frac{(1+\varepsilon)m}{\log s}\right). \]

Proof. The event $a\in \mathcal P'+B$ is equivalent to $(a-B_{\le a})\cap \mathcal P'\neq\emptyset$. Independence yields \[ \mathbb P(a\notin \mathcal P'+B)=\prod_{k\in a-B_{\le a}}\left(1-\frac1{\log k}\right). \] For positive $k\in a-B_{\le a}$ we have $k>s$ hence $\log k\ge\log s$ and \[ 1-\frac1{\log k}\ge 1-\frac1{\log s}. \] Also $|a-B_{\le a}|=|B_{\le a}|\le m$, so \[ $\mathbb P$$a\notin \mathcal P'+B$$\ge$

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