{"paper":{"title":"Determining $x$ or $y$ mod $p^2$ with $p=x^2+dy^2$","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":"2012-10-18T19:58:29Z","abstract_excerpt":"Let $p$ be an odd prime and let $d\\in\\{2,3,7\\}$. When $(\\frac{-d}p)=1$ we can write $p=x^2+dy^2$ with $x,y\\in\\mathbb Z$; in this paper we aim at determining $x$ or $y$ modulo $p^2$. For example, when $p=x^2+3y^2$, we show that if $p\\equiv x\\equiv 1\\pmod 4$ then $$\\sum_{k=0}^{(p-1)/2}(3[3\\mid k]-1)(2k+1)\\frac{\\binom{2k}k^2}{(-16)^k}\\equiv\\left(\\frac2p\\right)2x\\pmod{p^2}$$ where $[3\\mid k]$ takes $1$ or $0$ according as $3\\mid k$ or not, and that if $-p\\equiv y\\equiv 1\\pmod4$ then $$\\sum_{k=0}^{(p-1)/2}\\left(\\frac k3\\right)\\frac{k\\binom{2k}k^2}{(-16)^k} \\equiv(-1)^{(p+1)/4}y\\equiv\\sum_{k=0}^{(p-"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1210.5237","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"}