RESULTS: Erdős #126, counterexample arm (support_6cdfd2e3)

Hunt, not a result. Nothing here is evidence for or against RH (docs/08). Support run for r_186989, arm label construction-refutation.

Verdict: no counterexample construction, and the tensor lane is closed conditionally. The arm was asked for an algebraic / CRT / recursive / tensor / composition construction with $|A|$ exponential in $|S|$. It found none, and it argues (ordinary argument, unreviewed) why the most promising of those five is blocked: every mixed pair of a tensor construction is a solution of the $S$-unit equation $x+y=1$, so a tensor with $|A|$ exponential in $k$ would require that equation to have exponentially many solutions in $|S|$. Every construction known for that equation is subexponential: the run cited $\exp(c(s/\log s)^{1/2})$ (Erdős–Stewart–Tijdeman), and Konyagin and Soundararajan (arXiv:math/0604453) reach $\exp(s^{2-\sqrt2-\varepsilon})$ for $a+b=c$, still subexponential (found on re-landing from the arXiv abstract; the paper was not re-read). The lane therefore cannot refute #126 without first breaking a separate, well-studied barrier.

Two corrections to the parent go the other way and matter more than the non-result:

1. The refutation bar is far lower than a composition law. $r\_186989$ §4 asks for $g(k_1+k_2)\ge g(k_1)g(k_2)$. A bounded prime cost per doubling already refutes #126. That is a much smaller gadget to hunt for.
2. An anti-composition theorem does not prove #126. The converse of §4's implication fails, by an explicit staircase. This is a live trap for the proof arm, which the parent's closing sentence walks into.

Grading (added on re-landing, 2026-10-10). Lemmas 1, 2 and 6, Propositions 3, 4 and 5, Corollary 5.1 and the identity of §5 are this lab's ordinary arguments, unreviewed and pending external verification; none is kernel-checked. The run called Proposition 5 a theorem; it is relabelled for that reason, in HANDBACK.json and results.json too. The published results used are cited where they enter: the prime number theorem (§2), Evertse 1984 (§0, §5) and the Erdős–Stewart–Tijdeman construction (§5). The tensor-lane closure is conditional on subexponential $S$-unit counts, which is open. Every number below is measured: exact integer arithmetic, one route, from one run of probe.py recorded in results_probe.json (re-run on re-landing and reproduced byte for byte); no test in tests/ pins them.

Overlap with the arms already on main. This arm landed after its sibling Erdős #126 arms, so not all of it is new there. Proposition 3 is Hunt #93's amplification result (support_8ea74995 §5), reached independently. Lemma 1 is contained in Hunt #103's normalization lemma (support_60982bf6 §2), and the "every optimal witness lives below 50" correction is also made by Hunts #105 and #107. Lemma 2 is stated in Hunts #104 and #105, and #105 calls it the 1934 construction. Lemma 6 is a weak form of Hunt #107's counting lemma (support_95bb5cb7 §2), which pins that horn from both sides. The loose thread on $|A|\le p-1$ was refuted on main before this landed (see Loose threads). This arm's own contributions are Proposition 4, Proposition 5 with Corollary 5.1, the rows $g(8)\ge 14$ and $g(9)\ge 15$, and the dilated-union scan of §3.

Reproduce with .venv/bin/python hunts/support_6cdfd2e3/probe.py (stdlib only, seconds, writes results_probe.json). Everything measured below is inside a bounded box and is a lower bound on $g(k)$ and only that.

0. Notation

$S$ is a set of $k$ primes; $A$ a finite set of distinct positive integers is admissible for $S$ when $a+b$ is $S$-smooth for every $a\ne b$ in $A$; $g(k)=\max|A|$ over all $S$ with $|S|=k$. $P(n)$ is the set of prime factors of $n$. #126 is $\log g(k)=o(k)$. $N(S)$ denotes the number of solutions of $x+y=1$ in positive $S$-units $x,y\in\mathbb{Q}^{*}$ (Evertse 1984: $N(S)\le 3\cdot 7^{2|S|+3}$).

1. Audit of r_186989

Sound. Every witness in its §3 table re-verified from scratch by trial division (parent_witness_audit in results_probe.json, 7/7 admissible). Its §2 sandwich $f(n-1)\le f_0(n)\le f(n)$ is correct as written. Its $g(1)=2$ proof is correct, and so is the Fekete step in §4.

Independently reproduced, in a much smaller box. A separate branch-and-bound (bitset, greedy-colouring bound) exhaustive inside $[1,N]$:

Rows 1–7 agree with the parent's, from a box 20–500× smaller. Rows 8 and 9 are new: $g(8)\ge 14$, $g(9)\ge 15$. The $k$-th root continues to fall.

One claim of the parent is false as stated, and its falsity is a one-line argument rather than a search artifact.

Lemma 1 (dilation invariance). If $A$ is admissible for $S$ and $c\ge 1$ has $P(c)\subseteq S$, then $cA$ is admissible for $S$ and $|cA|=|A|$. Proof. $ca+cb=c(a+b)$ and $P(c(a+b))=P(c)\cup P(a+b)\subseteq S$. $\square$

So optimal witnesses are never unique and are unbounded: from any optimum, $2^t A$ is another for every $t$ when $2\in S$. The parent's door-table entry "every optimal witness has all elements $<50$" and the loose thread "the optimal witnesses are all tiny, and nobody knows why" are both artifacts of ascending branch order. The witnesses recovered here are $\{113,399\}$ at $k=1$ and $\{16,80,112,176\}$ at $k=2$: same sizes, not tiny. The correct normalisation is: divide out $\gcd(A)$, or state the claim as "some optimum is small". The parent's follow-up suggestion (re-run with $N=2^{k+2}$ and compare) would have measured nothing, since Lemma 1 already predicts the answer.

2. An elementary superlinear lower bound

Lemma 2 (interval construction). Let $S=\{p_1,\dots,p_k\}$ be the first $k$ primes and $m=\lfloor (p_{k+1}-1)/2\rfloor$. Then $A=\{1,\dots,m\}$ is admissible for $S$, so $$g(k)\;\ge\;\Big\lfloor \tfrac{p_{k+1}-1}{2}\Big\rfloor\;=\;\big(\tfrac12+o(1)\big)\,k\log k .$$ Proof. Off-diagonal sums lie in $[3,2m-1]\subseteq[3,p_{k+1}-2]$. Any integer $n<p_{k+1}$ has every prime factor $\le n<p_{k+1}$, hence $\le p_k$, hence in $S$. The asymptotic is the prime number theorem. $\square$

This is elementary and is very likely folklore, but the parent's file contains no lower bound on $g$ at all beyond finite search, and this one is asymptotic: it gives $g(k)/k\to\infty$, which no bounded search can. It is also far from the target: superlinear, not exponential.

It is weaker than the measured table for every $k$ tested ($m=11$ at $k=8$ against a measured 14), so it is not sharp either. Both facts are worth having: the sequence 2, 4, 5, 6, 8, 10, 11, 14, 15 is bracketed below by $\sim\frac12 k\log k$ and above by nothing.

3. The refutation bar, corrected downward

$r\_186989$ §4 shows that a full composition law $g(k_1+k_2)\ge g(k_1)g(k_2)$ refutes #126. True, and much more than is needed.

Proposition 3 (bounded-cost doubling suffices). Suppose there are integers $k_0\ge 1$, a real $c>1$ and a $K$ with $g(k+k_0)\ge c\,g(k)$ for all $k\ge K$. Then $g(k)\ge g(K)\,c^{\lfloor (k-K)/k_0\rfloor}$, so $\liminf_k \log g(k)/k \ge (\log c)/k_0>0$ and #126 is false. Proof. Induction along the arithmetic progression $K,K+k_0,K+2k_0,\dots$, and $g$ is non-decreasing between the steps. $\square$

The gadget the counterexample arm should hunt is therefore not a tensor law but the much weaker object: a repeatable way to multiply $|A|$ by any fixed $c>1$ at a fixed prime cost $k_0$. Doubling for ten extra primes, forever, refutes #126. That reframing is the most usable thing this run produces, and Hunt #93 (support_8ea74995 §5) reached the same statement independently.

Measured, against exactly that. The cheapest repeatable doubling to try is the dilated union $A\cup cA$, which by Lemma 1 is free only when $c$ is $S$-smooth (and then it is not a doubling: $cA$ may overlap, and the cross sums $a+ca'$ are new conditions). Exhaustive over $c\le 200$, with the extra primes drawn from the first twelve primes (2 to 37):

Doubling cost 6 and 7 extra primes at $k=2$ and $k=4$; at $k=3$ no dilation works at all inside the scan. The measured prime cost of doubling is growing with $k$, not bounded, which is what Proposition 3 says #126 requires. This is three data points and a bounded scan; it is evidence about the gadget's shape, not a theorem.

4. Anti-composition is not sufficient for #126

The parent closes: "Erdős #126 … is asking for an anti-composition theorem". That overstates the implication, and the overstatement would misdirect the proof arm.

Proposition 4. There exist non-decreasing $G:\mathbb{N}\to\mathbb{N}$ with $\limsup_k \log G(k)/k>0$ (i.e. #126 false for $G$) such that for every fixed $k_0$ and $c>1$, the law $G(k+k_0)\ge c\,G(k)$ fails for all but a density-zero set of $k$. Proof. Take a rapidly increasing $k_1<k_2<\cdots$ with $k_{j+1}/k_j\to\infty$ and set $G(k)=2^{k_j}$ for $k_j\le k<k_{j+1}$. Then $\log G(k_j)/k_j=\log 2$, so $\limsup>0$. But $G(k+k_0)=G(k)$ unless $k\in[k_{j+1}-k_0,k_{j+1})$ for some $j$, a set of density zero. $\square$

$\log g$ is non-decreasing, so refutation of #126 is exactly $\limsup_k \log g(k)/k>0$: a sparse sequence of jumps refutes it while every uniform composition law fails. Killing all composition gadgets leaves #126 open. The honest statement of the parent's §4 is the one direction it established: composition ⟹ refutation. Not the converse.

5. The tensor lane, closed conditionally

This is the arm's main object. Let $A$ be admissible for $S$, $B$ for $T$, and $C=A\cdot B=\{ab\}$ (assume the $|A||B|$ products are distinct). The "grid lines" of $C$ are free: $ab+ab'=a(b+b')$ and $ab+a'b=b(a+a')$ are $(S\cup T)$-smooth automatically. All the content is in the mixed pairs $ab+a'b'$ with $a\ne a'$, $b\ne b'$. Write

$$u=ab+a'b',\qquad v=ab'+a'b,\qquad M=(a+a')(b+b').$$

Identity. $u+v=M$ and $u-v=(a-a')(b-b')$.

$M$ is $(S\cup T)$-smooth for free. So the tensor is admissible exactly when the smooth number $M$ splits as $u+v$ with both parts smooth, and dividing by $M$:

Proposition 5 (tensor ⟹ $S$-unit equation). If $C=A\cdot B$ is admissible for $S\cup T$, then for every $a\ne a'\in A$, $b\ne b'\in B$ the pair $$x=\frac{ab+a'b'}{(a+a')(b+b')},\qquad y=\frac{ab'+a'b}{(a+a')(b+b')}$$ is a solution of $x+y=1$ in positive $(S\cup T)$-units of $\mathbb{Q}^{*}$. Proof. $x+y=1$ by the identity. Numerators are $(S\cup T)$-smooth by admissibility of $C$; the denominator is $(a+a')(b+b')$ with $a+a'$ $S$-smooth and $b+b'$ $T$-smooth. $\square$

Corollary 5.1 (the factors are bounded by the unit count). Fix $a\ne a'\in A$ and put $\beta=b/b'$. Then $x=\dfrac{a\beta+a'}{(a+a')(\beta+1)}$ is a non-constant Möbius function of $\beta$ (non-constant because $a\ne a'$), hence injective in $\beta$. The ordered ratios $b_1/b_j$, $j=2,\dots,|B|$, are $|B|-1$ distinct values, so $$|B|\;\le\;N(S\cup T)+1,\qquad\text{and symmetrically}\qquad |A|\le N(S\cup T)+1 .$$

Unconditionally this is weaker than Erdős–Turán: Evertse gives $N\le 3\cdot 7^{2s+3}$, far above $3\cdot 2^{k-1}$. Say that plainly, Proposition 5 buys no new upper bound on $g$. Its value is directional:

• Conditional obstruction. If $N(S)=\exp(o(|S|))$, the expected truth (the run calls it the standard conjecture in this area and cites no source), then any tensor construction has $|A|,|B|=\exp(o(k))$ and so $|C|=\exp(o(k))$. No tensor construction can refute #126.
• What a tensor counterexample would cost. Conversely, an exponential tensor construction forces $N(S)\ge\exp(ck)$, which would beat every known lower bound for $S$-unit solution counts: the run cited $\exp\big((4+o(1))(s/\log s)^{1/2}\big)$ (Erdős–Stewart–Tijdeman 1988), and Konyagin and Soundararajan (arXiv:math/0604453) reach $\exp(s^{2-\sqrt2-\varepsilon})$ for $a+b=c$, larger but still subexponential. It would be a significant independent result. The counterexample lane is not cheap; it is at least as hard as an open problem in a different field.

Why the argument does not extend to general $A$, which is exactly why this closes the tensor lane and no more. For an arbitrary admissible $A$ and four elements $a,b,c,d$ one has $(a+b)+(c+d)=(a+c)+(b+d)=\sigma$, four smooth numbers in two smooth-summing pairs, but $\sigma=a+b+c+d$ need not be $S$-smooth, so the ratios are not $S$-units and Proposition 5 has nothing to divide by. In the tensor case the denominator $M$ factors as (an $A$-sum)$\times$(a $B$-sum) and is smooth for free. That factorisation is the tensor hypothesis, and it is the whole of what the argument uses.

Measured, consistently. Small tensors fail on their mixed pairs by wide margins: $\{1,3\}\otimes\{1,2\}$ over $S\cup T=\{2,3,5\}$ gives $C=\{1,2,3,6\}$ with $2+3=5$ fine but $1+6=7$ outside; $\{1,5,7,11\}\otimes\{1,4\}$ over $\{2,3,5\}$ fails on 10 of its 28 pairs (tensor in results_probe.json). Finite, and not evidence; the argument is the finding.

6. A lemma that turned out too weak to use

Recorded because the proof arm may otherwise spend a run rediscovering it.

Lemma 6 (sumset counting). If $A$ is admissible for $S$, $|A|=n\ge 2$, $M=\max A$, then $2n-3\le\Psi(2M,S)$, the number of $S$-smooth integers $\le 2M$. Proof. $a_1+a_2<a_1+a_3<\cdots<a_1+a_n<a_2+a_n<\cdots<a_{n-1}+a_n$ are $2n-3$ distinct $S$-smooth integers in $[1,2M]$. $\square$ Sharper: the $n-1$ sums $a_n+a_i$ all lie in $(M,2M]$, so $n-1$ is at most the count of $S$-smooth integers in that dyadic window.

It does not obstruct anything. $\Psi(2M,S)\le\prod_{i\le k}(1+\log 2M/\log p_i)$, so forcing $\Psi\ge 2^k$ needs only $\log M\gtrsim\log k$: the lemma places any exponential construction outside a box of size polynomial in $k$, and no further. The parent's box ($N\le 2\cdot 10^5$) is already past that threshold, so the lemma explains nothing about why the searches stall. Smooth numbers are too plentiful in exponent space for counting alone to bite. Stated as a negative so it is not re-attempted.

Correction on re-landing (2026-10-10). The threshold above is understated. The product is an upper bound on $\Psi$, so it gives a necessary condition and cannot show that $\Psi$ is large; trivially $\Psi(2M,S)\le 2M$, so an exponential $A$ already needs $\max A$ exponential in $k$, not polynomial. Hunt #107 (support_95bb5cb7 §2) pins $\log\Psi(2M,S)/k$ from both sides: it tends to 0 iff $\log 2M=o(\log\mathrm{rad}\,S)$, so counting forces an exponential $A$ to height $\mathrm{rad}(S)^{c}$, for a constant $c>0$ set by the growth rate, and no further. The negative conclusion of this section stands, in #107's sharper form: counting bounds height, not $|A|$.

7. What this run could not settle

• No counterexample, and no proof that none exists. The tensor lane is closed only conditionally on subexponential $S$-unit counts, and that is itself open. CRT and recursive lanes were considered and not developed: a CRT splicing has the same defect as the dilated union, the cross sums are $|A|^2$ unconstrained conditions against $O(1)$ free parameters, but I have no theorem for them, only the measured scan of §3 and the counting argument, which §6 shows is too weak. Calling those lanes closed would be an overclaim.
• No upper bound on $g(k)$. Corollary 5.1 is one for tensors only, and it is worse than the 1934 bound.
• Whether $g(k)\ge\frac12 k\log k$ is the best known. Lemma 2 is elementary enough that it is very likely in the literature; I did not search. If the published lower bound is larger, the bracket in §2 tightens from the other end.

The doors

This run measured a ceiling (the tensor lane's ceiling, and $g_N(k)$ to $k=9$), so it owes the list.

1. Active constraints at the optimum.

2. Frozen-constant inventory.

3. Information class. Doors 2 and the box stay inside what this family reads (integers and the smoothness of their sums) and are under its ceiling. Door 1 requires reading more: the $S$-unit counting literature (Evertse–Győry, Erdős–Stewart–Tijdeman). That is the same door $r\_186989$ ranked first and could not open. This run's contribution is to show it is not optional for the counterexample side either, Proposition 5 puts the tensor lane underneath it rather than beside it.

Loose threads

• Erdős–Stewart–Tijdeman transfer. Their construction gives $S$ with $\exp(c(s/\log s)^{1/2})$ solutions to $x+y=1$. Why it might matter: if it can be pushed through Proposition 5 backwards it would give $g(k)\ge\exp(c(k/\log k)^{1/2})$, superpolynomial, still consistent with #126, and far above $\frac12 k\log k$. First step: check whether their $S$-unit solutions can be arranged with a common denominator $(a+a')(b+b')$ of the required product shape. (Re-landing note: Konyagin and Soundararajan's $\exp(s^{2-\sqrt2-\varepsilon})$ family, arXiv:math/0604453, is the larger candidate for the same transfer.)
• The parity fact generalises and the parent's version is right. $S$ omitting 2 caps $|A|\le2$. The parent's suggested extension ($|A|\le p-1$ when a small $p\notin S$) is worth doing; it is the only iterable statement in either file. First step: pigeonhole on residues mod $p$ against $r\leftrightarrow -r$. Closed in the negative on main before this arm landed: $|A|\le p-1$ fails for every odd $p\notin S$ ($A=\{1,3,7,13\}$, $S=\{2,5,7\}$), because the pigeonhole bounds occupied residue classes, not $|A|$. See Hunts #93, #95, #101 and #103.
• CRT lane never developed. Why it might matter: it is the one of the five named schemes with no theorem against it here. First step: ask whether $A=\{a: a\equiv r_i \bmod m_i\}$ can force $a+b\equiv 0$ mod a large smooth modulus for all pairs, which is the only way to beat the $|A|^2$-conditions problem.