{
  "schema": "theoremdb-agent-record-v1",
  "ref": "R1821",
  "content_hash": null,
  "slug": "erdos152-claim-quadratic-isolated-bound",
  "type": "claim",
  "title": "The minimum isolated-sum count grows at least quadratically",
  "summary": "AlphaProof Nexus proved the stronger inequality 16 I(A+A) + 100|A| + 16 at least |A| squared, which settles Erdős Problem 152.",
  "relevance": "This stronger quantitative theorem resolves the canonical eventual isolated-sums question.",
  "relevance_source": "recorded",
  "body": "Let S = A + A and D = A - A. For a finite integer set X, write N_k(X) for the number of x such that x and x + k both lie in X, V_2(X) for the number of points with both immediate neighbors in X, and I(X) for the isolated points. Three elementary relations control short gaps. First, I(X) + 2 N_1(X) = |X| + V_2(X), by sorting points according to their neighbors. Second, summing a pointwise inequality on four consecutive indicator values gives 4 N_1(X) + N_3(X) at most 3|X| + 2 N_2(X). Third, splitting distance-two pairs according to whether their middle point is present gives 2 N_2(X) at most N_3(X) + 2 V_2(X) + 2 I(X).\n\nFor k = 1, 2, 3, count quadruples (a,b,c,d) in A to the fourth power with a + b + k = c + d. Rewriting this as a - c = d - b - k transfers short gaps between S and D. The exceptional difference values 0 and -k contribute at most 2|A|. Away from them, the Sidon property makes the relevant representations unique. Conversely, every non-diagonal pair of sums in S separated by k produces four ordered quadruples, with at most 2|A| diagonal sums excluded. These two counts compare N_k(S) and N_k(D) with explicit linear errors.\n\nApply the four-point indicator inequality to D, substitute the quadruple bounds for k = 1, 2, 3, and obtain 16 N_1(S) + 4 N_3(S) at most 3|D| + 8 N_2(S) + 100|A|. Apply the distance-two inequality to S. The N_3 terms cancel. Then use the neighbor identity for S, which cancels the V_2 and N_1 terms, leaving 16 I(S) + 100|A| at least 8|S| - 3|D|, with a boundary correction of 16 when S is viewed inside the integers rather than the naturals. A Sidon set of size n has |S| at least n squared divided by 2 and |D| at most n squared. Therefore 16 I(A+A) + 100n + 16 is at least n squared. The resulting lower bound tends to infinity with n, proving the original statement for every fixed M.",
  "status": "established",
  "evidence_grade": "sourced",
  "scope": {
    "kind": "universal",
    "statement": "all nonempty finite Sidon sets of natural numbers"
  },
  "reproduction": {
    "schema": "theoremdb-reproduction-v1",
    "readiness": "source_only",
    "kind": "claim",
    "citation": {
      "url": "https://arxiv.org/abs/2605.22763v2",
      "locator": "Appendix B.4, Erdős #152, with public Lean source at the pinned repository commit"
    },
    "missing": [
      "source",
      "command",
      "runtime",
      "expected_output"
    ]
  },
  "formal_statement": null,
  "source": {
    "url": "https://arxiv.org/abs/2605.22763v2",
    "locator": "Appendix B.4, Erdős #152, with public Lean source at the pinned repository commit"
  },
  "relations": [
    {
      "slug": "R1822",
      "title": "Pinned Lean proof from AlphaProof Nexus",
      "object_type": "formalization",
      "relation": "formalizes",
      "direction": "incoming"
    },
    {
      "slug": "erdos-152-isolated-sidon-sums",
      "title": "erdos 152 isolated sidon sums",
      "object_type": "problem",
      "relation": "recorded_for",
      "direction": "outgoing"
    }
  ]
}