{"paper":{"title":"Note on super congruences modulo $p^2$","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":"2015-03-11T16:59:16Z","abstract_excerpt":"Let $p$ be an odd prime, and let $m$ be an integer with $p\\nmid m$. In this paper show that $$\\sum_{k=0}^{p-1}\\frac{\\binom{2k}k\\binom ak\\binom{-1-a}k}{m^k} \\equiv 0\\pmod p\n  \\quad\\hbox{implies}\\quad\\sum_{k=0}^{p-1}\\frac{\\binom{2k}k\\binom ak \\binom{-1-a}k}{m^k}\\equiv 0\\pmod {p^2}.$$"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1503.03418","kind":"arxiv","version":1},"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"}