arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 1012.4234math.NTmath.CO
收藏

关于勒让德多项式III的同余式

Congruences concerning Legendre polynomials III

Zhi-Hong Sun

摘要:

设$p>3$为素数,$R_p$为分母与$p$互质的有理数集合。令$\{P_n(x)\}$为勒让德多项式。本文主要证明了对于具有$m\not\e 0\pmod p$的$m,n,t\in R_p$,有$$\align &P_{[\frac p6]}(t) \e -\Big(\frac 3p\Big)\sum_{x=0}^{p-1}\Big(\frac{x^3-3x+2t}p\Big)\pmod p, &\Big(\sum_{x=0}^{p-1}\Big(\frac{x^3+mx+n}p\Big)\Big)^2\equiv \Big(\frac{-3m}p\Big) \sum_{k=0}^{[p/6]}\binom{2k}k\binom{3k}k\binom{6k}{3k} \Big(\frac{4m^3+27n^2}{12^3\cdot 4m^3}\Big)^k\pmod p,$$,其中$(\frac ap)$是Legendre符号,$[x]$是最大整数函数.作为应用,我们解决了Z.W. Sun和作者关于$\sum_{k=0}^{p-1}\binom{2k}k\binom{3k}k\binom{6k}{3k}/m^k\pmod {p^2}$,其中$m$是不能被$p$整除的整数。

英文摘要:

Let $p>3$ be a prime, and let $R_p$ be the set of rational numbers whose denominator is coprime to $p$. Let $\{P_n(x)\}$ be the Legendre polynomials. In this paper we mainly show that for $m,n,t\in R_p$ with $m\not\e 0\pmod p$, $$\align &P_{[\frac p6]}(t) \e -\Big(\frac 3p\Big)\sum_{x=0}^{p-1}\Big(\frac{x^3-3x+2t}p\Big)\pmod p, &\Big(\sum_{x=0}^{p-1}\Big(\frac{x^3+mx+n}p\Big)\Big)^2\equiv \Big(\frac{-3m}p\Big) \sum_{k=0}^{[p/6]}\binom{2k}k\binom{3k}k\binom{6k}{3k} \Big(\frac{4m^3+27n^2}{12^3\cdot 4m^3}\Big)^k\pmod p,$$ where $(\frac ap)$ is the Legendre symbol and $[x]$ is the greatest integer function. As an application we solve some conjectures of Z.W. Sun and the author concerning $\sum_{k=0}^{p-1}\binom{2k}k\binom{3k}k\binom{6k}{3k}/m^k\pmod {p^2}$, where $m$ is an integer not divisible by $p$.

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。

↑