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Guo","submitted_at":"2014-12-10T00:36:18Z","abstract_excerpt":"The Delannoy polynomial $D_n(x)$ is defined by $$ D_n(x)=\\sum_{k=0}^{n}{n\\choose k}{n+k\\choose k}x^k. $$ We prove that, if $x$ is an integer and $p$ is a prime not dividing $x(x+1)$, then \\begin{align*} \\sum_{k=0}^{p-1}(2k+1)D_k(x)^3 &\\equiv p\\left(\\frac{-4x-3}{p}\\right) \\pmod{p^2}, \\\\ \\sum_{k=0}^{p-1}(2k+1)D_k(x)^4 &\\equiv p \\pmod{p^2}, \\\\ \\sum_{k=0}^{p-1}(-1)^k(2k+1)D_k(x)^3 &\\equiv p\\left(\\frac{4x+1}{p}\\right) \\pmod{p^2}, \\end{align*} where $\\big(\\frac{\\cdot}{p}\\big)$ denotes the Legendre symbol. The first two congruences confirm a conjecture of Z.-W. Sun [Sci. 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W. Guo","submitted_at":"2014-12-10T00:36:18Z","abstract_excerpt":"The Delannoy polynomial $D_n(x)$ is defined by $$ D_n(x)=\\sum_{k=0}^{n}{n\\choose k}{n+k\\choose k}x^k. $$ We prove that, if $x$ is an integer and $p$ is a prime not dividing $x(x+1)$, then \\begin{align*} \\sum_{k=0}^{p-1}(2k+1)D_k(x)^3 &\\equiv p\\left(\\frac{-4x-3}{p}\\right) \\pmod{p^2}, \\\\ \\sum_{k=0}^{p-1}(2k+1)D_k(x)^4 &\\equiv p \\pmod{p^2}, \\\\ \\sum_{k=0}^{p-1}(-1)^k(2k+1)D_k(x)^3 &\\equiv p\\left(\\frac{4x+1}{p}\\right) \\pmod{p^2}, \\end{align*} where $\\big(\\frac{\\cdot}{p}\\big)$ denotes the Legendre symbol. The first two congruences confirm a conjecture of Z.-W. Sun [Sci. 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