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In this paper we give a survey of conjectures and results on this topic and point out the connection between this problem and series for $1/\\pi$."},"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":"1103.4325","kind":"arxiv","version":11},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.NT","submitted_at":"2011-03-22T17:46:57Z","cross_cats_sorted":["math.CO"],"title_canon_sha256":"b47e23fe938a20203ece134d9212927fe4cda79fd07f5b855a9d3039c60ab2e7","abstract_canon_sha256":"8ba6684d74f43ab080478dfe81dcd77b0d29d9f01263af86f076faf5090bfc99"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T02:58:32.997620Z","signature_b64":"nMfWz4OMgg5hvxdkWFUN4rtEufJeMBhart23b3wmzHrc6WCZFjkp2wmthE33alx++Ko8GbVIUsGKo8mW/1rGDQ==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"59d3a566c45432f717a1022e1c9fe3dc0b5aaedc2588946bb2b2d09b5b97804b","last_reissued_at":"2026-05-18T02:58:32.996976Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T02:58:32.996976Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"Conjectures and results on $x^2$ mod $p^2$ with $4p=x^2+dy^2$","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":"2011-03-22T17:46:57Z","abstract_excerpt":"Given a squarefree positive integer $d$, we want to find integers (or rational numbers with denominators not divisible by large primes) $a_0,a_1,a_2,\\ldots$ such that for sufficiently large primes $p$ we have $\\sum_{k=0}^{p-1}a_k\\equiv x^2-2p$ (mod $p^2$) if $4p=x^2+dy^2$ (and $4\\nmid x$ if $d=1$), and $\\sum_{k=0}^{p-1}a_k\\equiv 0$ (mod $p^2$) if $(\\frac{-d}p)=-1$. 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