RESULTS: Erdős #126, ORIGINAL-PROOF arm (support run 517b887f)

Hunt, not a result. Nothing here is evidence for or against RH (docs/08).

Answer to the question asked: no. Processing several primes jointly, or reordering them, cannot replace $2^k$ by $(2-\delta)^k$, because the quantity that the whole Erdős–Turán architecture reduces to is order-blind by definition, and bounding it below by $|A|/(2-\delta)^k$ is not a refinement of the 1934 bookkeeping: it is logically equivalent to the conjectured bound itself. Details in §4. Two further findings that were not asked for and that change how the classical step must be written down are in §3.

Reproduce with python3 hunts/support_517b887f/probe.py (~30 s, stdlib only, writes results.json).

1. The proof, reconstructed

Notation. $A \subset \mathbb{Z}^+$ finite, $|A| = n$, and $S = \{p : p \mid a+b \text{ for some } a \ne b \in A\}$, $|S| = k$. Write $s$ for the number of odd primes in $S$. Define

$$c(A) \;=\; \max\{|A'| : A' \subseteq A,\ a+b \text{ is a power of } 2 \text{ for all } a \ne b \in A'\}.$$

Step 1 (base case, archimedean). $c(A) \le 2$. Proof. Suppose $a<b<c$ positive with $a+b = 2^x$, $a+c = 2^y$, $b+c = 2^z$. Then $x<y<z$, so $z \ge y+1$ and $2a = (a+b)+(a+c)-(b+c) = 2^x + 2^y - 2^z \le 2^x + 2^y - 2^{y+1} = 2^x - 2^y < 0$, impossible. $\square$

This is the only step that uses the ordering of $\mathbb{Z}$. Every other step is local (residues mod $p$).

Step 2 (selection, one binary choice per odd prime). For an odd prime $q$ and a finite $B$, the residue classes mod $q$ pair up as $\{r,-q+r\}$ under $r \mapsto -r$, with $0$ its own partner. Two elements in the same nonzero class sum to $2r \not\equiv 0$; two elements in opposite classes sum to $0$; two elements of class $0$ sum to $0$. So keeping the larger side of each $\{r,-r\}$ pair and at most one element of class $0$ gives $B' \subseteq B$ with $q \nmid a+b$ throughout and

$$|B'| \;\ge\; \tfrac{1}{2}\bigl(|B| - n_0(q)\bigr) + \min(n_0(q),1), \qquad n_0(q) = \#\{a \in B : q \mid a\}. \tag{$\star$}$$

Step 3 (assembly). Iterating Step 2 over the $s$ odd primes and assuming the clean form $|B'| \ge |B|/2$ of ($\star$) gives $c(A) \ge n/2^s$. With Step 1, $n \le 2 \cdot 2^{s} = 2^{k}$ when $2 \in S$ (and $n \le 2$ when $2 \notin S$, by parity). That is the Erdős–Surányi form $g(k) \le 2^k$; the 1934 paper's $3 \cdot 2^{k-1}$ is the same architecture with the base case used as "$\ge 3$ elements is impossible" and a weaker selection.

So the factor 2 per prime enters in exactly one place: the $\pm$ choice in Step 2. There is no second leak. The "3" of $3\cdot 2^{k-1}$ is the base case, and Erdős–Surányi already collected it.

Tightness of the endpoints, against the measured lower bounds for $g$ (hunts/r_186989, audited below): $g(1) = 2 = 2^1$ and $g(2) \ge 4 = 2^2$, so the Erdős–Surányi bound is attained at $k=1,2$. Any $(2-\delta)^k$ must therefore carry a constant $> 1$ and can only bite for $k \ge 3$.

2. Audit of hunts/r_186989/RESULTS.md

Checked, not trusted. Its arm-3 composition argument is correct as written (a superadditive $\log g$ plus Fekete forces $g(k) \ge g(1)^k = 2^k$, refuting the conjecture), and its $g(1) \le 2$ proof is the Step 1 argument above. Its $g_N$ witnesses re-verify: this probe recomputed the extremal sets independently in smaller boxes and reproduced $n = 2,4,5,6,8$ for $k = 1,\dots,5$ with the same or equivalent witnesses. Its claim that the optima live below 50 held in every box used here. One thing it does not say and should: at $k = 1, 2$ its lower bounds meet the classical upper bound $2^k$, which is the sharpest single fact available about how much room the 1934 argument has.

3. Two defects in the standard statement of Step 2

Both are stated as sketches in the literature summaries with the clean $|B'| \ge |B|/2$; both are false as stated. Witnesses are exact and re-verified by trial division in results.json.

3.1 The per-prime halving is false. Take $A = \{1, 3, 15, 21, 33\}$. Every off-diagonal sum is $\{2,3,11,17\}$-smooth ($4, 16, 22, 34, 18, 24, 36, 36, 48, 54$), so $A$ is admissible with $k=4$. For $q = 3$: four of the five elements are divisible by $3$, so any subset in which $3$ divides no sum contains at most one of them. The maximum is $\{1,3\}$, of size $2 < 5/2$. ($\star$) is what actually holds; the class-0 term is not removable. The unbounded version is $B = q\cdot C \cup \{x\}$: the $q$-clean subset of $B$ has size $2$ however large $C$ is, so there is no constant halving lemma for a single odd prime.

3.2 The assembled bound $c(A) \ge n/2^s$ is false without primitivity. $A = \{3, 9, 21, 51, 141\} = 3 \cdot \{1,3,7,17,47\}$ has all off-diagonal sums $\{2,3,5\}$-smooth, $n = 5$, $s = 2$, and no pairwise sum is a power of $2$, so $c(A) = 1 < 5/4$. The repair is the scaling descent that the sketches leave implicit: divide by $\gcd(A)$ first. Here that returns $\{1,3,7,17,47\}$, for which $1+3 = 4$ and $c = 2$.

Both defects are about the route, not the theorem: $n = 5 \le 2^4$ and $n = 5 \le 2^3$ respectively, so nothing published is contradicted. What is contradicted is the line-by-line reconstruction the brief asked for. A correct write-up needs primitivity plus ($\star$), not "halve once per prime".

Order matters for the intermediate step, and only there. In 3.1, taking $q = 11$ first instead of $q = 3$ gives a clean subset of size $4 = 0.8n$, so reordering rescues that instance. This is the strongest form of the brief's "changing the order" idea that survives, and §4 says why it buys nothing.

Exhaustive check of the repaired form. Over all $104{,}183$ admissible sets of size $\ge 2$ inside the boxes $[1,300]$, $[1,200]$, $[1,150]$ for $S = \{2,3,5\}, \{2,3,5,7\}, \{2,3,5,7,11\}$: zero primitive violations of $c(A) \ge n/2^s$, and $11$ non-primitive violations, all of them scaled copies. The repaired lemma is measured-true in that range and is exactly tight at $A = \{1,2,4,8\}$, where $n = 4$, $s = 2$, $c = 1 = n/2^s$.

4. The barrier

Theorem (order-blindness). Everything the architecture can deliver is contained in the single quantity $c(A)$. Step 1 gives $c(A) \le 2$ unconditionally, and $c(A) \ge 1$ trivially, so $c(A) \in \{1,2\}$ for every admissible $A$. Consequently, for any function $\varphi$:

$$\bigl[\,\forall A \text{ admissible}: c(A) \ge |A|/\varphi(k)\,\bigr] \iff \bigl[\,g(k) \le 2\varphi(k)\,\bigr] \text{ up to a factor } 2 .$$

$c(A)$ is defined by a maximum over subsets and does not mention any prime ordering, any grouping of primes, or any valuation. Hence no reordering and no joint processing of primes can change what this architecture proves; they can only change whether a particular greedy selection attains the maximum. Proving the selection lemma with $\varphi(k) = (2-\delta)^k$ is precisely as hard as proving $g(k) \le 2(2-\delta)^k$. It is not a sharpening of the 1934 bookkeeping; it is the target restated.

Corollary (why "joint" was the natural guess, and why it fails). At the level of residues alone, joint processing is provably worthless: for any distinct odd primes $q_1,\dots,q_s$, the CRT configuration $x_\varepsilon \equiv \varepsilon_i \pmod{q_i}$, $\varepsilon \in \{\pm1\}^s$, has $2^s$ points, and $x_\varepsilon + x_\delta \equiv 0 \pmod{q_i}$ exactly when $\varepsilon_i = -\delta_i$, so a subset clean for every $q_i$ must have $\varepsilon = \delta$: the maximum jointly clean subset is $1$ out of $2^s$. Verified by brute force for $s = 2,3,4$. Worst-case residue data realises the $2^s$ loss exactly, whether the primes are taken one at a time or all at once.

And the worst case is realisable by genuine admissible sets, not only by residue patterns. The sweep found, for each $S$ tested, admissible sets with $c(A) = 1$ and $|A|$ equal to the box maximum:

$\{1,2,4,8\}$ is the sharp one: a primitive admissible set on which the repaired selection lemma holds with equality. So the constant $2$ cannot be lowered at $s=2$ by any argument that proves the selection lemma, and any $(2-\delta)^k$ statement must be asymptotic with a constant in front.

5. The smallest missing lemma

The only place where new information could enter is the class-0 term of ($\star$), i.e. the interaction between $A_0 = \{a \in A : q \mid a\}$ and $A_* = A \setminus A_0$. Everything the architecture currently knows about $A_0$ is that it is a scaled copy problem, and the descent handles it at zero cardinality cost only when $A_0 = A$. The missing lemma is:

(L) There is a function $\psi$ with $\log\psi(k) = o(k)$ such that for every primitive admissible $A$ and every odd $q \in S$, $\#\{a \in A: q \mid a\} \le \psi(k) \cdot \#\{a \in A : q \nmid a\}$.

(L) plus ($\star$) gives a selection loss of $2(1+\psi)$ per prime, which is still exponential, so (L) alone does not suffice and I want to be explicit about that. Its interest is that it is the only sub-question in the architecture that is not already equivalent to the full conjecture, and it is a two-set smoothness question: the cross sums $x + qc$ ($x \in A_*$, $qc \in A_0$) must all be $S$-smooth, which is exactly the regime of Győry–Stewart–Tijdeman 1986 ($\omega\bigl(\prod_{a\in A, b\in B}(a+b)\bigr) \gg \log|A|$ for two sets). That is the effective tool the 1934 argument does not use, and it is where an original contribution would have to sit.

6. What this arm could not settle

• No improvement was proved. No $(2-\delta)^k$ bound, and §4 argues that none is reachable inside this architecture without solving the problem.
• (L) is untested. It could be false; the probe did not search for counterexamples to it, and one primitive admissible set with $A_0$ dominating would kill it. That is one cheap follow-up sweep.
• The barrier is against the architecture, not against the problem. A proof that abandons "pass to a subset with power-of-2 sums" is untouched by §4. The $S$-unit route (Evertse–Győry counting for $x+y=1$) is such a proof and is where the prior hunt also pointed.
• All witnesses are exhaustive only inside small boxes. The measured $g_N(k)$ are lower bounds on $g(k)$; nothing here bounds $g$ from above beyond the classical $2^k$.

The doors

This arm measured a ceiling (the $2$ per prime), so it owes the list.

1. Active constraints at the optimum.

2. Frozen-constant inventory.

3. Information class. The whole architecture reads only $\bigl(a \bmod q\bigr)_{q \in S}$ plus one archimedean comparison in the base case. It discards the valuations $v_q(a+b)$ entirely. Any door that could move $g(k)$ below $2^k$ must read more: either the valuation vectors (the $S$-unit equation regime) or the two-set structure of (L). Both are outside this family's configuration ceiling.

Threads

• (L) is one sweep away from a verdict. Search the admissible sets already enumerated for a primitive one whose class-0 part exceeds its class-$*$ part for some odd $q$. If one exists, (L) is dead and §5 loses its only non-circular sub-question.
• $\{1,2,4,8\}$ generalises to $\{1,2,4,\dots,2^m\}$, whose sums are $2^i(1+2^{j-i})$ and whose prime support is $\{2\} \cup \{$primes dividing $2^t+1, t \le m\}$. This is a cheap explicit family relating $g(k)$ to the factorisations of Fermat-type numbers $2^t+1$, and it is where the selection lemma is tight at small $s$. Worth checking whether it, or a variant, gives a better lower bound on $g(k)$ than the clique search for large $k$, since it needs no search at all.