{"paper":{"title":"Sum of consecutive powers as a perfect power","license":"http://creativecommons.org/licenses/by/4.0/","headline":"For k ≡ 2 mod 4 the equation x^k + (x+1)^k = y^n (n ≥ 3) has only the solutions x = 0, -1 when 6 ≤ k ≤ 100 or k has an odd prime factor ≡ 3 mod 4.","cross_cats":[],"primary_cat":"math.NT","authors_text":"Angelos Koutsianas, Nikos Tzanakis","submitted_at":"2026-05-18T13:06:04Z","abstract_excerpt":"In this paper we study the equation $$\n  x^k + (x+1)^k = y^n,\\quad n\\geq 3, $$ when $k\\equiv 2\\pmod{4}$. We prove that the only solutions are for $x=0, -1$ when $6\\leq k\\leq 100$ or for a $k$ with odd prime factors congruent to $3\\pmod{4}$. We use linear forms in logarithms, the modular method and the resolution of Thue equations."},"claims":{"count":3,"items":[{"kind":"strongest_claim","text":"We prove that the only solutions are for x=0, -1 when 6≤k≤100 or for a k with odd prime factors congruent to 3 mod 4.","source":"verdict.strongest_claim","status":"machine_extracted","claim_id":"C1","attestation":"unclaimed"},{"kind":"weakest_assumption","text":"The linear forms in logarithms and the modular method together produce effective bounds that cover all possible solutions without exception for the stated range of k (abstract, methods paragraph).","source":"verdict.weakest_assumption","status":"machine_extracted","claim_id":"C2","attestation":"unclaimed"},{"kind":"one_line_summary","text":"For k ≡ 2 mod 4 the equation x^k + (x+1)^k = y^n (n ≥ 3) has only the solutions x = 0, -1 when 6 ≤ k ≤ 100 or k has an odd prime factor ≡ 3 mod 4.","source":"verdict.one_line_summary","status":"machine_extracted","claim_id":"C3","attestation":"unclaimed"}],"snapshot_sha256":"535be2a00dfa9925af4fde2529c4c57c5965fa9217f2641517f943dd3cbe276b"},"source":{"id":"2605.18348","kind":"arxiv","version":1},"verdict":{"id":"eeda31ac-14ca-4d10-ad67-70ad64a9cdf0","model_set":{"reader":"grok-4.3"},"created_at":"2026-05-20T00:02:59.223182Z","strongest_claim":"We prove that the only solutions are for x=0, -1 when 6≤k≤100 or for a k with odd prime factors congruent to 3 mod 4.","one_line_summary":"For k ≡ 2 mod 4 the equation x^k + (x+1)^k = y^n (n ≥ 3) has only the solutions x = 0, -1 when 6 ≤ k ≤ 100 or k has an odd prime factor ≡ 3 mod 4.","pipeline_version":"pith-pipeline@v0.9.0","weakest_assumption":"The linear forms in logarithms and the modular method together produce effective bounds that cover all possible solutions without exception for the stated range of k (abstract, methods paragraph).","pith_extraction_headline":""},"integrity":{"clean":true,"summary":{"advisory":0,"critical":0,"by_detector":{},"informational":0},"endpoint":"/pith/2605.18348/integrity.json","findings":[],"available":true,"detectors_run":[{"name":"ai_meta_artifact","ran_at":"2026-05-19T23:33:35.155125Z","status":"skipped","version":"1.0.0","findings_count":0},{"name":"external_links","ran_at":"2026-05-19T23:32:00.443138Z","status":"completed","version":"1.0.0","findings_count":0},{"name":"claim_evidence","ran_at":"2026-05-19T23:21:58.815804Z","status":"completed","version":"1.0.0","findings_count":0}],"snapshot_sha256":"41533a28d672fb205cb2055eb7f7dc683b7b9269a4177bb7de1466a42ff17703"},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}