Super congruences involving Bernoulli and Euler polynomials

Zhi-Hong Sun

paper

Super congruences involving Bernoulli and Euler polynomials

arXiv:1407.0636

Abstract

Let $p > 3$ be a prime, and let $a$ be a rational p-adic integer. Let ${B_{n} (x)}$ and ${E_{n} (x)}$ denote the Bernoulli polynomials and Euler polynomials, respectively. In this paper we show that $k = 0 \sum p - 1 (k a) (k - 1 - a) \equiv (- 1)^{⟨ a ⟩_{p}} + p^{2} t (t + 1) E_{p - 3} (- a) (mod p^{3})$ and for $a \neq \equiv - \frac{1}{2} (mod p)$ , $k = 0 \sum p - 1 (k a) (k - 1 - a) \frac{1}{2 k + 1} \equiv \frac{1 + 2 t}{1 + 2 a} + p^{2} \frac{t ( t + 1 )}{1 + 2 a} B_{p - 2} (- a) (mod p^{3}),$ where $⟨ a ⟩_{p} \in {0, 1, \dots, p - 1}$ satisfying $a \equiv ⟨ a ⟩_{p} (mod p)$ and $t = (a - ⟨ a ⟩_{p}) / p$ . Taking $a = - \frac{1}{3}, - \frac{1}{4}, - \frac{1}{6}$ in the above congruences we solve some conjectures of Z.W. Sun. In this paper we also establish congruences for $\sum_{k = 0}^{p - 1} k (k a) (k - 1 - a), \sum_{k = 0}^{p - 1} (k a) (k - 1 - a) \frac{1}{2 k - 1}, \sum_{k = 1}^{p - 1} \frac{1}{k} (k a) (k - 1 - a) (mod p^{3})$ and $\sum_{k = 1}^{p - 1} \frac{( - 1 ) ^{k}}{k} (k a), \sum_{k = 0}^{p - 1} (k a) (- 2)^{k} (mod p^{2}) .$

30 pages

Super congruences involving Bernoulli and Euler polynomials

Abstract

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