{"paper":{"title":"Constructing $x^2$ for primes $p=ax^2+by^2$","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.NT","authors_text":"Zhi-Hong Sun","submitted_at":"2010-12-17T16:38:00Z","abstract_excerpt":"Let $a$ and $b$ be positive integers and let $p$ be an odd prime such that $p=ax^2+by^2$ for some integers $x$ and $y$. Let $\\lambda(a,b;n)$ be given by $q\\prod_{k=1}^\\infty (1-q^{ak})^3(1-q^{bk})^3 = \\sum_{n=1}^\\infty \\lambda(a,b;n)q^n$. In the paper, using Jacobi's identity $\\prod_{n=1}^\\infty (1-q^n)^3 = \\sum_{k=0}^\\infty (-1)^k(2k+1)q^{\\frac{k(k+1)}2}$ we construct $x^2$ in terms of $\\lambda(a,b;n)$. For example, if $2\\nmid ab$ and $p\\nmid ab(ab+1)$, then $(-1)^{\\frac{a+b}2x+\\frac{b+1}2}(4ax^2-2p) = \\lambda(a,b;((ab+1)p-a-b)/8+1)$. We also give formulas for $\\lambda(1,3;n+1),\\lambda(1,7;2n"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1012.3919","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"}