{"paper":{"title":"Proof of a congruence on sums of powers of $q$-binomial coefficients","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.CO"],"primary_cat":"math.NT","authors_text":"Ji-Cai Liu, Victor J. W. Guo","submitted_at":"2015-04-20T16:02:07Z","abstract_excerpt":"We prove that, if $m,n\\geqslant 1$ and $a_1,\\ldots,a_m$ are nonnegative integers, then \\begin{align*} \\frac{[a_1+\\cdots+a_m+1]!}{[a_1]!\\ldots[a_m]!}\\sum^{n-1}_{h=0}q^h\\prod_{i=1}^m{h\\brack a_i} \\equiv 0\\pmod{[n]}, \\end{align*} where $[n]=\\frac{1-q^n}{1-q}$, $[n]!=[n][n-1]\\cdots[1]$, and ${a\\brack b}=\\prod_{k=1}^b\\frac{1-q^{a-k+1}}{1-q^k}$. The $a_1=\\cdots=a_m$ case confirms a recent conjecture of Z.-W. Sun. We also show that, if $p>\\max\\{a,b\\}$ is a prime, then \\begin{align*} \\frac{[a+b+1]!}{[a]![b]!}\\sum_{h=0}^{p-1}q^h{h\\brack a}{h\\brack b} \\equiv (-1)^{a-b} q^{ab-{a\\choose 2}-{b\\choose 2}}[p"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1504.05482","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"}