{"paper":{"title":"Fibonacci numbers modulo cubes of primes","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.CO"],"primary_cat":"math.NT","authors_text":"Zhi-Wei Sun","submitted_at":"2009-11-16T19:06:17Z","abstract_excerpt":"Let $p$ be an odd prime. It is well known that $F_{p-(\\frac p5)}\\equiv 0\\pmod{p}$, where $\\{F_n\\}_{n\\ge0}$ is the Fibonacci sequence and $(-)$ is the Jacobi symbol. In this paper we show that if $p\\not=5$ then we may determine $F_{p-(\\frac p5)}$ mod $p^3$ in the following way: $$\\sum_{k=0}^{(p-1)/2}\\frac{\\binom{2k}k}{(-16)^k}\\equiv\\left(\\frac{p}5\\right)\\left(1+\\frac{F_{p-(\\frac {p}5)}}2\\right)\\pmod{p^3}.$$ We also use Lucas quotients to determine $\\sum_{k=0}^{(p-1)/2}\\binom{2k}k/m^k$ modulo $p^2$ for any integer $m\\not\\equiv0\\pmod{p}$; in particular, we obtain $$\\sum_{k=0}^{(p-1)/2}\\frac{\\bino"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"0911.3060","kind":"arxiv","version":9},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"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"}