Super congruences involving Bernoulli and Euler polynomials

preprint2014arXivOpen access

Abstract

Let p>3p>3 be a prime, and let aa be a rational p-adic integer. Let {Bn(x)}\{B_n(x)\} and {En(x)}\{E_n(x)\} denote the Bernoulli polynomials and Euler polynomials, respectively. In this paper we show that ∑k=0p−1(ak)(−1−ak)≡(−1)⟨a⟩p+p2t(t+1)Ep−3(−a)(modp3)\sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\equiv (-1)^{\langle a\rangle_p}+ p^2t(t+1)E_{p-3}(-a)\pmod{p^3} and for a≢−12(modp)a\not\equiv -\frac 12\pmod p, ∑k=0p−1(ak)(−1−ak)12k+1≡1+2t1+2a+p2t(t+1)1+2aBp−2(−a)(modp3),\sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\frac 1{2k+1}\equiv \frac{1+2t}{1+2a} +p^2\frac{t(t+1)}{1+2a}B_{p-2}(-a)\pmod{p^3}, where ⟨a⟩p∈{0,1,…,p−1}\langle a\rangle_p\in\{0,1,\ldots,p-1\} satisfying a≡⟨a⟩p(modp)a\equiv \langle a\rangle_p\pmod p and t=(a−⟨a⟩p)/pt=(a-\langle a\rangle_p)/p. Taking a=−13,−14,−16a=-\frac 13,-\frac 14,-\frac 16 in the above congruences we solve some conjectures of Z.W. Sun. In this paper we also establish congruences for ∑k=0p−1k(ak)(−1−ak), ∑k=0p−1(ak)(−1−ak)12k−1, ∑k=1p−11k(ak)(−1−ak)(modp3)\sum_{k=0}^{p-1}k\binom ak\binom{-1-a}k,\ \sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\frac 1{2k-1},\ \sum_{k=1}^{p-1}\frac 1k\binom ak\binom{-1-a}k\pmod{p^3} and ∑k=1p−1(−1)kk(ak), ∑k=0p−1(ak)(−2)k(modp2).\sum_{k=1}^{p-1}\frac {(-1)^k}k\binom ak,\ \sum_{k=0}^{p-1}\binom ak(-2)^k\pmod{p^2}.

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