English

Congruences concerning Legendre polynomials II

Number Theory 2012-08-06 v2

Abstract

Let p>3p>3 be a prime, and let mm be an integer with pmp\nmid m. In the paper we solve some conjectures of Z.W. Sun concerning k=0p1(2kk)3/mk(modp2)\sum_{k=0}^{p-1}\binom{2k}k^3/m^k\pmod{p^2}, k=0p1(2kk)\b4k2k/mk(modp)\sum_{k=0}^{p-1}\binom{2k}k\b{4k}{2k}/m^k\pmod p and k=0p1(2kk)2\b4k2k/mk(modp2).\sum_{k=0}^{p-1}\binom{2k}k^2\b{4k}{2k}/m^k\pmod {p^2}. In particular, we show that k=0p12(2kk)30(modp2)\sum_{k=0}^{\frac{p-1}{2}}\binom{2k}k^3\equiv 0\pmod {p^2} for p3,5,6(mod7)p\equiv 3,5,6\pmod 7. Let Pn(x)P_n(x) be the Legendre polynomials. In the paper we also show that P[p4](t)(6p)x=0p1(x33/2(3t+5)x9t7p)(modp) 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 Pp12(2),Pp12(324),Pp12(3),Pp12(32),Pp12(63),Pp12(378)(modp)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 tt is a rational pp-integer, [x][x] is the greatest integer not exceeding xx and (ap)(\frac {a}{p}) is the Legendre symbol. As consequences we determine P[p4](t)(modp)P_{[\frac {p}{4}]}(t)\pmod p in the cases t=5/3,7/9,65/63t=-5/3,-7/9,-65/63 and confirm many conjectures of Z.W. Sun.

Keywords

Cite

@article{arxiv.1012.3898,
  title  = {Congruences concerning Legendre polynomials II},
  author = {Zhi-Hong Sun},
  journal= {arXiv preprint arXiv:1012.3898},
  year   = {2012}
}

Comments

25 pages. arXiv admin note: text overlap with arXiv:1104.3047, arXiv:1104.2789

R2 v1 2026-06-21T17:00:32.914Z