{"paper":{"title":"Quartic residues and sums involving $\\binom{4k}{2k}$","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.NT","authors_text":"Zhi-Hong Sun","submitted_at":"2013-11-25T16:59:35Z","abstract_excerpt":"Let $p$ be an odd prime and let $m\\not\\equiv 0\\pmod p$ be a rational p-adic integer. In this paper we reveal the connection between quartic residues and the sum $\\sum_{k=0}^{[p/4]}\\binom{4k}{2k}\\frac 1{m^k}$, where $[x]$ is the greatest integer not exceeding $x$. Let $q$ be a prime of the form $4k+1$ and so $q=a^2+b^2$ with $a,b\\in\\Bbb Z$. When $p\\nmid ab(a^2-b^2)q$, we show that for $r=0,1,2,3$, $p^{\\frac{q-1}4}\\equiv (\\frac ab)^r\\pmod q$ if and only if $$\\sum_{k=0}^{[p/4]}\\binom{4k}{2k}\\Big(\\frac{a^2}{16q}\\Big)^k\\equiv (-1)^{\\frac{p^2-1}8a+\\frac{p-1}2\\cdot \\frac{q-1}4}\\Big(\\frac pq\\Big) \\Big"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1311.6364","kind":"arxiv","version":2},"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"}