Congruences concerning Legendre polynomials II

preprint2012arXivOpen access

Abstract

Let p>3p>3 be a prime, and let mm be an integer with p∤mp\nmid m. In the paper we solve some conjectures of Z.W. Sun concerning ∑k=0p−1(2kk)3/mk(modp2)\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 ∑k=0p−12(2kk)3≡0(modp2)\sum_{k=0}^{\frac{p-1}{2}}\binom{2k}k^3\equiv 0\pmod {p^2} for p≡3,5,6(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=0p−1(x3−3/2(3t+5)x−9t−7p)(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 Pp−12(2),Pp−12(324),Pp−12(−3),Pp−12(32),Pp−12(−63),Pp−12(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 p−p-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.

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