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We show the formula $h(p) \\equiv h(-p) m(p)$ (mod $16$), where $m(p)$ is an integer defined using the \"negative\" continued fraction expansion of $\\sqrt{p}$. Our result solves a conjecture of Richard Guy."},"verification_status":{"content_addressed":true,"pith_receipt":true,"author_attested":false,"weak_author_claims":0,"strong_author_claims":0,"externally_anchored":false,"storage_verified":false,"citation_signatures":0,"replication_records":0,"graph_snapshot":true,"references_resolved":false,"formal_links_present":false},"canonical_record":{"source":{"id":"1407.3261","kind":"arxiv","version":2},"metadata":{"license":"http://creativecommons.org/licenses/by-nc-sa/3.0/","primary_cat":"math.NT","submitted_at":"2014-07-11T19:45:31Z","cross_cats_sorted":[],"title_canon_sha256":"47e07ac9594dc52bae99d7cadeaa94a58ab9ac15039224e9d6d7d364dfb44d03","abstract_canon_sha256":"f84a31f452791d90d911ebe770d196940821ed70e526f278ca6e5ebd538a6913"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T02:32:30.647372Z","signature_b64":"215EcpCWMqn3AgTBS9L904Yw6gZ5tl8hZWbaamNiiXXsOGdpEaaSWGwcUxuZbeRdfdMVxfEYO58G8ySFi0r+CA==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"8802491fbab87f11c8c20612309676da3ebcb877c1201766970dc56146fae055","last_reissued_at":"2026-05-18T02:32:30.647006Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T02:32:30.647006Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"Proof of a conjecture of Guy on class numbers","license":"http://creativecommons.org/licenses/by-nc-sa/3.0/","headline":"","cross_cats":[],"primary_cat":"math.NT","authors_text":"Allen Yuan, Benjamin Gunby, Lynn Chua, Soohyun Park","submitted_at":"2014-07-11T19:45:31Z","abstract_excerpt":"It is well known that for any prime $p\\equiv 3$ (mod $4$), the class numbers of the quadratic fields $\\mathbb{Q}(\\sqrt{p})$ and $\\mathbb{Q}(\\sqrt{-p})$, $h(p)$ and $h(-p)$ respectively, are odd. It is natural to ask whether there is a formula for $h(p)/h(-p)$ modulo powers of $2$. We show the formula $h(p) \\equiv h(-p) m(p)$ (mod $16$), where $m(p)$ is an integer defined using the \"negative\" continued fraction expansion of $\\sqrt{p}$. 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