RESULTS: Erdős #126, additive-combinatorics arm (support_311c14a5)

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

Support run for r_186989 (Hunt #91). The brief: treat $A+A$ off the diagonal as lying in the multiplicative semigroup generated by $S$; work the additive energy / Freiman / sum-product / multiplicative-group angle; derive quantitative dependence on $k$ rather than citing qualitative finiteness; and name the narrowest theorem upgrade that would give $\log g(k) = o(k)$.

Reproduce with .venv/bin/python hunts/support_311c14a5/probe.py (about three minutes of CPU, stdlib only, writes results.json; on landing it took 19 minutes of wall time on a heavily loaded machine).

Answer, in one line. The narrowest upgrade is a subexponential-in-$|S|$ bound on the number of solutions of the weighted two-term $S$-unit equation $\alpha X + \beta Y = 1$, uniform in $\alpha,\beta$. §2 proves that upgrade suffices, by an elementary three-line reduction with no loss. §3 gives the quantitative dependence the brief asked for: the reduction is currently worth $g(k) \le 2 + e^{4k+10}$ from Evertse, which is far weaker than the 1934 bound $3\cdot 2^{k-1}$, and it has a hard floor at $\exp(c\sqrt{k/\log k})$ from Erdős–Stewart–Tijdeman, which is below the target, so the route is not self-defeating. §4 reports the three standard additive-combinatorics tools as inapplicable, with reasons. §5 audits the parent and corrects one of its loose threads.

Landing note (2026-10-21). Landed on main as Hunt #99 from a local-only branch, support/311c14a5, that was never pushed; PR #144 counted this arm among the seven that "produced no branch at all". The run's files are as it wrote them apart from the dated notes marked as added on landing.

Grading (added on landing, 2026-10-21). The Proposition of §2 and the sharpening of §5 ($g(k)=2^k$ exactly under a composition law) are this lab's ordinary arguments, unreviewed and pending external verification; none is kernel-checked. The §4 verdicts on Freiman, sum-product and additive energy are arguments about why each tool does not apply, not proved obstructions in the sense of ALIGNMENT.md §5. The constants of §3 came from one web search each and the run's HANDBACK.json says not to quote the exponents as pinned; the Evertse row is annotated in place below, because it also conflicts with Győry–Stewart–Tijdeman's primary-text quotation of Evertse. The verification counts (165 pair-cases) and the $g_N(3)=5$ search are exact checks and bounded-box lower bounds. Disposition: the attempt at a bound is unresolved (no new bound on $g(k)$), the reduction names a sufficient target and closes nothing, and r_186989's loose thread 3 is a refuted candidate. probe.py was re-run on landing and reproduced results.json byte for byte; no test in tests/ pins these numbers.

Overlap with arms already on main. The Proposition is the reduction of Hunts #96 (support_d5d5ccae) and #104 (support_f3ab3e34), and of sibling arms landed with this one (Hunts #97, #106, and the Győry–Stewart–Tijdeman form quoted in #143); its "no rank inflation" point is #96's choice of Evertse's coefficient-uniform bound over #104's rank-$(k+1)$ group. The composition sharpening of §5 extends Hunt #105 (support_eccd5f5e), which squeezed a composition law to $[2^k, 1.5\cdot2^k]$ and already observed that a proof of $g(3)\le 7$ retires the programme; Hunt #93 (support_8ea74995) gives the general amplification form of the refutation. The refutation of loose thread 3 was on main before this arm landed (Hunts #93, #95, #101, #103), and the OEIS thread of the last loose thread was closed by the literature arm (Hunt #143, support_fcffa3b9 §4.4). This arm's own contributions are the pinning of the composition consequence to $g(k)=2^k$ exactly, the sub-target $X-Y=d$ uniform in $d$, and the reasons each of the three briefed tools fails, with the energy inequality shown dominated by a one-line pigeonhole.

1. Setup and notation

$S = \{p_1,\dots,p_k\}$ distinct primes. $A \subset \mathbb{Z}_{>0}$ finite with distinct elements, admissible if every off-diagonal sum $a+b$ ($a \ne b$) has no prime factor outside $S$. $g(k) = \max |A|$ over all such $S$ and $A$. An $S$-unit is a rational whose numerator and denominator are $S$-smooth; they form a group $\Gamma_S \le \mathbb{Q}^*$ of rank $k$ (torsion $\{\pm 1\}$). Admissibility says exactly: all off-diagonal sums are positive $S$-units.

Erdős #126 is $f(n)/\log n \to \infty$, equivalently (parent §1, checked and correct) $\log g(k) = o(k)$. Erdős–Turán 1934: $g(k) < 3 \cdot 2^{k-1}$.

2. The reduction. Proved.

Proposition. Let $A$ be admissible for $S$, $|A| = n \ge 3$. Fix any two distinct $a, b \in A$ and put $d = a - b \ne 0$, $\alpha = 1/d$, $\beta = -1/d$. Then the map $$c \;\longmapsto\; (X, Y) = (a+c,\; b+c)$$ sends $A \setminus \{a,b\}$ injectively into the solution set of $$\alpha X + \beta Y = 1, \qquad X, Y \in \Gamma_S .$$ Consequently $n - 2 \le E(k)$, where $$E(k) \;:=\; \max_{S,\;|S|=k}\;\; \max_{\alpha,\beta \in \mathbb{Q}^*}\; \#\{(X,Y) \in \Gamma_S^2 : \alpha X + \beta Y = 1\},$$ so $\;g(k) \le 2 + E(k)$.

Proof. For $c \in A \setminus \{a,b\}$ both $a+c$ and $b+c$ are off-diagonal sums, hence positive $S$-units. And $(a+c) - (b+c) = a - b = d$, so dividing by $d$ gives $\alpha(a+c) + \beta(b+c) = 1$. Injectivity: $X = a+c$ determines $c$. $\square$

Three things are worth saying about how cheap that is.

No rank inflation. The naive move, dividing the difference $u - v = d$ by $d$ and absorbing $1/d$ into the group, costs a generator and lands in rank $k+1$. Putting the constants in the coefficients instead of the group keeps the rank at exactly $k$, and Evertse's theorem is stated for $\alpha X + \beta Y = 1$ with arbitrary fixed nonzero rationals, so nothing is lost by doing so. (For the asymptotic statement $k$ versus $k+1$ would not have mattered; for the constants in §3 it does.)

Uniformity in $\alpha, \beta$ is load-bearing and must be stated. $d = a-b$ is not under our control and is not an $S$-unit in general. A bound on the number of solutions that degraded with the height of $\alpha,\beta$ would be useless here. Evertse's bound is uniform in $\alpha,\beta$; any replacement must be too.

The three-term route is strictly worse. Using four elements $a,b,c,d$ and $(a+b)+(c+d) = (a+c)+(b+d)$ gives a three-term unit equation $x+y+z=1$, nondegenerate for distinct elements (no vanishing subsum: two of the three are positive, and each two-term subsum vanishing forces two elements of $A$ to coincide). But the map from quadruples to solution triples has no evident injectivity, and normalising the constant term costs the same extra generator anyway. Two-term with an honest injection beats three-term with a fibre problem.

Verified on data. probe.py §B re-derives the reduction on all seven of the parent's witnesses ($k=1..7$): the identity $\alpha X + \beta Y = 1$ and the $S$-unit property hold for all $\binom{n}{2}$ choices of $(a,b)$ on every witness, and $c \mapsto X$ is injective in all 165 cases tested, with exactly $n-2$ solutions produced each time. That is a check of the proof's mechanics, not of the bound.

3. Quantitative dependence on $k$: what the route is worth today

Annotated on landing (2026-10-21), not corrected. The Evertse row's $e^{4s+6}$ came from a web search, and the run's HANDBACK.json says not to quote the exponents as pinned. Győry–Stewart–Tijdeman 1986 quote Evertse's bound as $3\cdot7^{2w+3}$, read in their primary text by the sibling literature arm (Hunt #143, support_fcffa3b9 T4), and Hunts #96 and #106 use the same $3\cdot7^{d+2s}$. With $s=k+1$ that is $3\cdot7^{2k+3} \approx e^{3.89k+6.94}$, smaller than $e^{4k+10}$ for every $k \ge 0$, so on that reading the route's base is $49$ rather than $e^4 \approx 54.6$. Either way it is far above $2$, and fact (a) below is unchanged. The Beukers–Schlickewei row's $2^{8(r+2)}$ is likewise from a search; main quotes $2^{8r+8}$ (Hunt #104; Hunt #143's T6, secondary), and neither form was checked against the paper here.

Read the table honestly and two facts fall out, and they are the substance of this arm.

(a) The unit-equation route is currently much weaker than 1934. Base $e^4 \approx 54.6$ against base $2$. Any write-up that reaches for "$S$-unit equations have finitely many solutions" as if it were progress on #126 is citing qualitative finiteness where the quantity is the whole problem. The parent's doors section ranks the $S$-unit machinery as the top door; it is the right shape of door, and it is presently on the wrong side of the 1934 bound.

(b) The route's own floor is below the target, so it is not self-defeating. Erdős–Stewart–Tijdeman construct $\alpha,\beta,S$ with $\exp\big((4+o(1))\sqrt{s/\log s}\big)$ solutions, so $E(k)$ is genuinely at least $\exp(c\sqrt{k/\log k})$ and no argument will ever make the Proposition yield a polynomial bound. But $\exp(c\sqrt{k/\log k}) = \exp(o(k))$. The known extremal examples for the unit equation do not obstruct the bound Erdős #126 needs. That is the reason to believe this door is open rather than walled.

So the narrowest upgrade, stated to be checkable:

Target. Prove $E(k) = \exp(o(k))$: for every $\varepsilon > 0$ there is $k_0$ such that for all $k \ge k_0$, all $k$-element prime sets $S$ and all nonzero rationals $\alpha,\beta$, the equation $\alpha X + \beta Y = 1$ has at most $e^{\varepsilon k}$ solutions in $S$-units.

Then $g(k) \le 2 + e^{\varepsilon k}$ for large $k$, hence $\log g(k) = o(k)$, hence $f(n)/\log n \to \infty$: Erdős #126.

Implication direction, explicitly. Target $\Rightarrow$ #126. The converse is false as an implication and should not be asserted: $g$ is a maximum over a very special configuration (all of $A$'s pairwise sums smooth simultaneously), while $E$ counts solutions of one equation with no such coupling, so #126 could hold while $E(k)$ stays exponential. The Proposition is a sufficiency statement only.

Sub-target, if the full one is too strong. The Proposition only ever needs $\alpha = 1/d$, $\beta = -1/d$. That is, it only needs to count pairs of positive $S$-units differing by a fixed $d$, i.e. $X - Y = d$ with $X,Y$ $S$-smooth. A bound $\exp(o(k))$ on that one-parameter family, uniform in $d$, is formally weaker than the full weighted equation and suffices verbatim. This is the narrowest form we could find, and it is where a proof attempt should start. Note it is not weaker than ESD's floor: ESD's construction is of exactly this $X - Y = d$ shape, so $\exp(c\sqrt{k/\log k})$ is a lower bound here too.

4. Additive energy, Freiman, sum–product: three tools, all inapplicable

The brief named these; the honest report is that none of them bites, and the reasons are structural rather than technical.

Freiman / small doubling: no hypothesis to feed it. Freiman-type structure theorems need $|A+A| \le K|A|$. Admissible sets have no such property: on all seven of the parent's witnesses the off-diagonal sumset has size $\binom{n}{2}$ or close to it, i.e. doubling is maximal, not small. Nothing in #126 constrains $|A+A|$; it constrains where $A+A$ lives.

Sum–product: the multiplicative structure is on the wrong set. Erdős–Szemerédi and its descendants lower-bound $\max(|A+A|, |A\cdot A|)$ by exploiting that $A$ cannot be both additively and multiplicatively structured. Here $A$ itself is unconstrained; it is $A+A$ that is multiplicatively structured. There is no sum–product statement in the standard catalogue whose hypothesis is "the sumset lies in a group of bounded rank", and inventing one is the problem, not a tool for it.

Additive energy: gives nothing beyond the trivial count. Let $A \subset [1,N]$ and $\Psi(x,p_k)$ count $S$-smooth integers $\le x$. Cauchy– Schwarz on the off-diagonal sumset gives $E(A) \ge (n^2-n)^2 / \Psi(2N,p_k)$, and $E(A) \le n^3$ always, so $n \le 2\Psi(2N,p_k)$ up to the diagonal correction. But the trivial observation that $a+b$ for fixed $a$ and varying $b$ are distinct $S$-smooth numbers already gives $n - 1 \le \Psi(2N,p_k)$, which is stronger. The energy inequality is dominated by the one-line pigeonhole, and both are $N$-dependent and so cannot bound $g(k)$ at all. We report this because "use additive energy" is a plausible-sounding instruction that measurably buys nothing here.

5. Audit of r_186989 (Hunt #91)

Everything checkable was checked. Two corrections, one sharpening.

Witnesses: all seven confirmed. probe.py §A re-verifies every row of the parent's table by independent trial division. All admissible, no violations. Our own branch-and-bound solver, written from scratch, independently reproduces $g_N(3) = 5$ with the same optimal witness $\{1,3,7,17,47\}$ at both $N=5000$ and $N=60\,000$. The parent's lower bounds stand.

Sharpening of the parent's §4 (composition law). The parent shows a composition law $g(k_1+k_2) \ge g(k_1)g(k_2)$ would give $g(k) \ge 2^k$ and so refute #126. That is correct, and it can be pushed to something falsifiable. Fekete's lemma for the superadditive sequence $\log g$ gives $\lim_k g(k)^{1/k} = \sup_k g(k)^{1/k} \ge g(1) = 2$. Erdős–Turán's $g(k) < 3\cdot 2^{k-1}$ forces $\lim_k g(k)^{1/k} \le 2$. Hence the limit and the supremum equal $2$, so $g(k)^{1/k} \le 2$ for every $k$, while superadditivity gives $g(k) \ge g(1)^k = 2^k$. Therefore:

A composition law does not merely refute #126. It forces $$g(k) = 2^k \quad\text{for every } k \ge 1 .$$

That is a sharp prediction and it is testable at $k=3$: it needs an 8-element admissible set for some 3-element $S$. Our exhaustive search finds $5$ for $S=\{2,3,5\}$ inside $[1,60\,000]$. This is not a proof (the true $g(3)$ could be attained outside the box) but it converts the parent's "measured data is consistent with no such gadget" into a single named integer whose determination would settle the composition law outright. Proving $g(3) < 8$ kills the composition arm permanently. That is a finite, self-contained sub-problem and the cheapest disposal of that branch we can see. (Landing note, 2026-10-21: Hunt #105 reached the same disposal from the weaker corridor $[2^k, 1.5\cdot2^k]$; what this section adds is that the corridor collapses to the single value $2^k$. Hunt #107 found no 6-element set for $S=\{2,3,5\}$ with all pairwise sums $\le 10^5$, and Hunt #103 never saw the first $k$ primes beaten over all subsets of the first nine primes for $k \le 5$; both are bounded searches and neither proves $g(3) \le 7$.)

Correction: the parent's loose thread #3 is false. The parent conjectures that its parity cap generalises: "the same argument mod small primes may cap sets whose $S$ omits 3, mod 3 residue classes", proposing $|A| \le p-1$ when $p \notin S$. It does not generalise, and the reason is that $p=2$ is the only prime for which every residue class is self-annihilating. For $p \notin S$ we need $a + b \not\equiv 0 \pmod p$ for all distinct $a,b \in A$. For $p = 2$ the classes are $0$ and $1$ with $0+0 \equiv 0$ and $1+1 \equiv 0$, so each class holds at most one element and $|A| \le 2$. For any odd $p$, all of $A$ may sit in a single class $r \ne 0$: sums are $\equiv 2r \not\equiv 0$, and the pigeonhole is vacuous, giving no bound at all, let alone $p-1$.

Witness, found by exhaustive search and re-verified (probe.py §D): $S = \{2,5,7,11\}$ omits $3$, and $$A = \{1, 9, 13, 19, 31\}, \qquad |A| = 5 > 2,$$ with all ten off-diagonal sums $\{10,14,20,32,22,28,40,32,44,50\}$ free of the prime $3$. So the mod-2 cap is a coincidence of $p=2$, not the head of a family. (Landing note, 2026-10-21: this thread was refuted on main before this arm landed, in Hunts #93, #95, #101 and #103; the witness here is an independent one.)

Consequence for the parent's doors table. Its rank-1 door ("parity: $2 \in S$ is worth more than every other structural choice combined") survives as a measurement but loses its proposed mechanism-as-family. The correct statement is narrower: $2 \notin S$ caps $|A|$ at $2$; for every other prime, omission costs a constant at most.

6. What we could not settle

• No new upper bound on $g(k)$. §2 hands the problem to $E(k)$; §3 shows the best available $E(k)$ is worse than 1934. This arm improves no constant.
• Whether $E(k) = \exp(o(k))$. That is the open problem, and it is open in the unit-equation literature independently of #126. We did not attempt it.
• $g(3)$ exactly. Our search is bounded, like the parent's. An unbounded argument that $g(3) \le 7$ is missing and would be worth having (§5).
• Whether the coupling between the $\binom{n}{2}$ sums can be exploited. The Proposition throws away everything except one pair $(a,b)$ and the two families $a+c$, $b+c$. All the other smooth sums are discarded. A reduction that uses more of them could in principle beat $E(k)$. We found none.

The doors

This arm measured a route's ceiling, so it owes the list.

1. Active constraints at the optimum. What binds, for the §2 route:

2. Frozen-constant inventory.

3. Information class. The reduction and its sub-target stay inside the data the problem already gives (smoothness of pairwise sums) but move to a different literature: Diophantine counting rather than enumeration. Door 1 in the frozen table, the coupled system over all pairs, is the only one that reads strictly more of the structure than the current family does, and it is therefore the one worth funding if this arm is continued.

Loose threads

• The coupled system. For fixed $A$, every pair $(a,b)$ gives an equation $\alpha_{ab}X + \beta_{ab}Y = 1$ and the solutions are forced to come from one common set of $c$'s. Why it might matter: it is the only place where the problem's real hypothesis (all sums smooth at once) is still unused, and any bound better than $E(k)$ must come from it. First step: for $n=4$, ask whether the four equations can be combined into a single Diophantine constraint on $(a,b,c,d)$ with a rank-independent solution count.
• $g(3) \le 7$, unconditionally. Why it might matter: it disposes of the composition arm permanently (§5) at finite cost. First step: prove admissible sets for $|S|=3$ can be normalised into a bounded box: a $\gcd$-and-scaling normalisation plus a height bound on solutions of $X - Y = d$ in $\{2,3,5\}$-units, which is a computable finite check. (Landing note, 2026-10-21: Hunt #107 measured that primitive admissible sets of sub-extremal size have unbounded height at $S=\{2,3\}$, so a gcd-and-scaling normalisation alone does not reach a bounded box; whether sets of extremal size have bounded height is open there, and Hunt #103 states it as a conjecture. The step needs a height bound that uses the extremal size.)
• Is $g(k)$ in OEIS? The parent raised this and did not run it. Our independent solver reproduces $2,4,5,6,8,10,11$. First step: one OEIS query via ontology/knownness.py. Unrun here; we did not have the budget and say so rather than reporting the check as done. (Closed on landing, 2026-10-21: the sibling literature arm ran it. The sequence matches twenty OEIS entries, all coincidental, and a keyword search for the concept returns nothing, so $g(k)$ is not tabulated there; Hunt #143, support_fcffa3b9 §4.4.)