slug: obryant-2024-sum-product-seven-point-cell bibkey: obryant2024visualizing doi: 10.48550/arXiv.2411.08139 url: https://arxiv.org/abs/2411.08139v2 triage: theorem motivation_gids:
- D5/S3/Arith/SumProductSevenPointCell.result
O’Bryant’s seven-point positive-real sum-product cell
Problem
Kevin O’Bryant, Visualizing the Sum-Product Conjecture, arXiv:2411.08139v2, §5.2:
We believe that (23, 13) is not in SPP_{R+}(7).
For a domain X, the spectrum is
Here R+ denotes the strictly positive reals, and A+A and AA denote the pointwise sumset and product set. The exact assertion is that every finite set A of positive reals with seven elements satisfies
The Lean claim retains these quantifiers and hypotheses. Its theorem
result : claim follows from the stronger statement
Motivation
The least possible product cardinality is 2|A|−1. At equality, positivity forces a geometric progression, so a specific corner of the sum-product spectrum becomes a problem about collisions between sums of powers of one real parameter. The bound excludes the proposed cell and identifies the sharp minimum of the sum cardinality along this product-minimal boundary.
Gap
Tier 1, externally stated in arXiv:2411.08139v2 §5.2 and preregistered in
issue #14745. The supplied literature scope lists the integer small-set
papers arXiv:2307.06874 and arXiv:2601.21828, the asymptotic paper
arXiv:2605.28781, and ErdosProblems/52.lean in formal-conjectures, which
states the main sum-product conjecture. These references do not themselves
supply a proof of this seven-element positive-real cell exclusion.
The claimed absence of a subsequent settlement within that scope is
ASSUMED-UNVERIFIED here: the literature information was supplied with the
task and was not independently browsed.
Route
For B={b₀<⋯<bₘ₋₁} of positive reals, the chain
has 2m−1 distinct products. Equality forces a geometric progression. Every increasing path in the ordered product grid from (0,0) to (m−1,m−1) has 2m−1 vertices. At equality each path enumerates the full product set in increasing order. Compare the path that first follows row zero with the path that changes to row one at column j−1. The entries at position j give b₀bⱼ=b₁bⱼ₋₁. Induction yields bⱼ=b₀(b₁/b₀)ʲ. Thus the seven-element case reduces to A={b,br,…,br⁶}, with b>0 and r>1.
There are 28 unordered index pairs. Two different pairs with equal sums can be oriented as i<k≤l<j. Dividing by rⁱ produces
If a+b≤d, then 1+rᵈ−rᵃ−rᵇ≥(rᵃ−1)(rᵇ−1)>0. Hence a+b>d. The thirteen possible triples have the following exact factorizations, where 1+Xᵈ−Xᵃ−Xᵇ=(X−1)EHⱼ.
| d | a | b | j | E |
|---|---|---|---|---|
| 3 | 2 | 2 | 1 | 1 |
| 4 | 2 | 3 | 2 | 1 |
| 4 | 3 | 3 | 3 | 1 |
| 5 | 2 | 4 | 4 | 1 |
| 5 | 3 | 3 | 5 | 1 |
| 5 | 3 | 4 | 6 | X+1 |
| 5 | 4 | 4 | 7 | 1 |
| 6 | 2 | 5 | 8 | 1 |
| 6 | 3 | 4 | 9 | 1 |
| 6 | 3 | 5 | 2 | X²+1 |
| 6 | 4 | 4 | 10 | X+1 |
| 6 | 4 | 5 | 11 | 1 |
| 6 | 5 | 5 | 12 | 1 |
The factors are
The omitted factors E are positive for r>1. Distinct Hⱼ cannot vanish at the same positive real. Polynomial Bézout identities certify this separation in the formal proof. The following independent rational data also separates their positive roots: k is the largest negative-coefficient degree, and u=100^(deg Hⱼ)Hⱼ(L/100), v=100^(deg Hⱼ)Hⱼ((L+1)/100).
| j | k | L | u | v |
|---|---|---|---|---|
| 1 | 1 | 161 | −179 | 44 |
| 2 | 1 | 132 | −20032 | 22637 |
| 3 | 2 | 183 | −50413 | 3904 |
| 4 | 1 | 122 | −466544 | 5886641 |
| 5 | 2 | 117 | −6339979 | 940976 |
| 6 | 2 | 146 | −19464 | 15623 |
| 7 | 3 | 192 | −9474304 | 3092301 |
| 8 | 1 | 116 | −596583424 | 224480357 |
| 9 | 2 | 112 | −385389568 | 660087893 |
| 10 | 2 | 127 | −1145359 | 4595456 |
| 11 | 3 | 153 | −683634007 | 978453024 |
| 12 | 4 | 196 | −1635610624 | 1132702657 |
Every positive-coefficient degree exceeds k, so Hⱼ(x)/xᵏ is strictly increasing for x>0. The rational signs place its root in (L/100,(L+1)/100); these twelve intervals are disjoint.
For a fixed root, discard the outer index pairs (i,i+d) associated with its rows. Their counts are respectively (4,4,3,2,2,2,2,1,1,1,1,1). At most four pairs are discarded. Every remaining pair sum is distinct, because a collision would have required one of the discarded outer pairs. At least 28−4=24 sums remain. Positive scaling preserves both sum and product cardinalities.
Falsifier
The conclusion would fail if minimum product cardinality did not force a geometric progression, if an unordered-pair collision escaped the thirteen triples, if two distinct factors had a common positive root, or if more than four outer index pairs had to be removed for one factor. The proof must retain strict positivity, seven distinct elements, and the exact product cardinality thirteen throughout these reductions.
Evidence
The designated closed statement is
D5/S3/Arith/SumProductSevenPointCell.result : claim.
The stronger statement is
D5/S3/Arith/SumProductSevenPointCell.stronger_result.
The formal argument uses ordered positive-real products, exact polynomial
identities, root separation, and an injective restriction of the finite
pair-sum map. The rational sign table above was independently recomputed
by integer evaluation
Σᵢ cᵢLⁱ100^(deg H−i); its 24 entries agree with the supplied proof.
These arithmetic readouts support the alternative root-isolation argument;
the retained Lean route uses polynomial separation identities.
Triage
- [proved: D5/S3/Arith/SumProductSevenPointCell.stronger_result] For seven positive reals, |AA|=13 implies |A+A|≥24. Consequently (23,13) is absent from the positive-real spectrum.
- [computed: exact polynomial reduction of the 28 unordered pair sums modulo X²−X−1 and X³−X−1] The bound is sharp for geometric progressions with ratio the golden ratio φ=(1+√5)/2 and the plastic number, the real root greater than one of X³−X−1, approximately 1.3247. Both reductions give 24 distinct sum residues. The positive geometric progressions have thirteen products. These sharpness instances are not separate retained Lean theorems.
- [computed: exact polynomial reduction modulo H₁,…,H₁₂, together with the supplied collision classification] The corresponding sum residue counts are (24,24,25,26,26,26,26,27,27,27,27,27). The supplied calculation identifies {24,25,26,27,28} as the full range of sum cardinalities on seven-term geometric progressions with ratio greater than one; generic ratios give 28. This full-range statement is not separately kernel-checked here.
- [open] Determine the remaining cells of SPP_{R+}(7). The minimum-product boundary does not classify sets with larger product cardinality.
- [open] The paper’s proposed union formula for n≥7 remains a separate question. This exclusion does not establish that formula.
ASSUMED-UNVERIFIED
The source quotation, latest-version designation, and literature scope
are supplied by the task. No independent network verification of
arXiv:2411.08139v2, arXiv:2307.06874, arXiv:2601.21828,
arXiv:2605.28781, or formal-conjectures ErdosProblems/52.lean was performed.
The literature scope is not an exhaustive priority determination.
The full geometric-progression range is attributed to the supplied
classification and computations rather than an additional Lean theorem.