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arXiv 1012.3898math.NT
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关于勒让德多项式II的同余式

Congruences concerning Legendre polynomials II

Zhi-Hong Sun

摘要:

令$p>3$是素数,令$m$是具有$p\nmid m$的整数。本文解决了Z.W. Sun关于$\sum_{k=0}^{p-1}\binom{2k}k^3/m^k\pmod{p^2}$,$\sum_{k=0}^{p-1}\binom{2k}k\b{4k}{2k}/m^k\pmod p$和$\sum_{k=0}^{p-1}\binom{2k}k^2\b{4k}{2k}/m^k\pmod {p^2}.$的研究。特别地,我们证明了$\sum_{k=0}^{\frac{p-1}{2}}\binom{2k}k^3\equiv 0\pmod {p^2}$对于$p\equiv 3,5,6\pmod 7$。令$P_n(x)$为勒让德多项式。本文还证明了$ P_{[\frac {p}{4}]}(t)\equiv -\big(\frac{-6}{p}\big)\sum_{x=0}^{p-1} \big(\frac{x^3-3/2(3t+5)x-9t-7}{p}\big)\pmod p$,并确定了$P_{\frac{p-1}{2}}(\sqrt 2), P_{\frac{p-1}{2}}(\frac{3\sqrt 2}{4}), P_{\frac{p-1}{2}}(\sqrt {-3}),P_{\frac{p-1}{2}}(\frac{\sqrt 3}{2}), P_{\frac{p-1}{2}}(\sqrt {-63}), P_{\frac{p-1}{2}}(\frac {3\sqrt 7}{8}) \pmod p$,其中$t$是$p-$的有理整数,$[x]$是不超过$x$的最大整数,$(\frac {a}{p})$是Legendre符号.作为结果,我们确定了$t=-5/3,-7/9,-65/63$病例中的$P_{[\frac {p}{4}]}(t)\pmod p$,并证实了Z.W.太阳

英文摘要:

Let $p>3$ be a prime, and let $m$ be an integer with $p\nmid m$. In the paper we solve some conjectures of Z.W. Sun concerning $\sum_{k=0}^{p-1}\binom{2k}k^3/m^k\pmod{p^2}$, $\sum_{k=0}^{p-1}\binom{2k}k\b{4k}{2k}/m^k\pmod p$ and $\sum_{k=0}^{p-1}\binom{2k}k^2\b{4k}{2k}/m^k\pmod {p^2}.$ In particular, we show that $\sum_{k=0}^{\frac{p-1}{2}}\binom{2k}k^3\equiv 0\pmod {p^2}$ for $p\equiv 3,5,6\pmod 7$. Let $P_n(x)$ be the Legendre polynomials. In the paper we also show that $ P_{[\frac {p}{4}]}(t)\equiv -\big(\frac{-6}{p}\big)\sum_{x=0}^{p-1} \big(\frac{x^3-3/2(3t+5)x-9t-7}{p}\big)\pmod p$ and determine $P_{\frac{p-1}{2}}(\sqrt 2), P_{\frac{p-1}{2}}(\frac{3\sqrt 2}{4}), P_{\frac{p-1}{2}}(\sqrt {-3}),P_{\frac{p-1}{2}}(\frac{\sqrt 3}{2}), P_{\frac{p-1}{2}}(\sqrt {-63}), P_{\frac{p-1}{2}}(\frac {3\sqrt 7}{8}) \pmod p$, where $t$ is a rational $p-$integer, $[x]$ is the greatest integer not exceeding $x$ and $(\frac {a}{p})$ is the Legendre symbol. As consequences we determine $P_{[\frac {p}{4}]}(t)\pmod p$ in the cases $t=-5/3,-7/9,-65/63$ and confirm many conjectures of Z.W. Sun.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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