{"paper":{"title":"Congruences concerning Legendre polynomials III","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.CO"],"primary_cat":"math.NT","authors_text":"Zhi-Hong Sun","submitted_at":"2010-12-20T03:16:57Z","abstract_excerpt":"Let $p>3$ be a prime, and let $R_p$ be the set of rational numbers whose denominator is coprime to $p$. Let $\\{P_n(x)\\}$ be the Legendre polynomials. In this paper we mainly show that for $m,n,t\\in R_p$ with $m\\not\\e 0\\pmod p$, $$\\align &P_{[\\frac p6]}(t) \\e -\\Big(\\frac 3p\\Big)\\sum_{x=0}^{p-1}\\Big(\\frac{x^3-3x+2t}p\\Big)\\pmod p, &\\Big(\\sum_{x=0}^{p-1}\\Big(\\frac{x^3+mx+n}p\\Big)\\Big)^2\\equiv \\Big(\\frac{-3m}p\\Big) \\sum_{k=0}^{[p/6]}\\binom{2k}k\\binom{3k}k\\binom{6k}{3k} \\Big(\\frac{4m^3+27n^2}{12^3\\cdot 4m^3}\\Big)^k\\pmod p,$$ where $(\\frac ap)$ is the Legendre symbol and $[x]$ is the greatest integer"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1012.4234","kind":"arxiv","version":4},"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"}