# Two congruence conjectures modulo prime p

How to prove the following two congruences?

Question1: Let $$p\equiv 1 \pmod 3$$ be a prime, then $$\sum_{k=0\atop k\neq(p-1)/3}^{(p-1)/2}\frac{\binom{2k}k}{3k+1}\equiv 0 \pmod p.$$

Question2: For any odd prime $$p>3$$, we have
$$\sum_{k=1}^{(p-1)/2}\frac{1}{k\binom{2k}k}\sum_{j=1}^k\frac{\binom{2j}j}{j}\equiv \frac{1}3B_{p-2}\left(\frac{1}3\right) \pmod p,$$ where $$B_n(x)$$ are the Bernoulli polynomials defined by $$\sum_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}=\frac{te^{xt}}{e^t-1}.$$

• The first congruence resembles some of the conjectures of van Hamme. – EFinat-S Oct 15 '18 at 2:08
• Without any background provided, the two congruences do not look interesting enough.I guess the poster needs them as lemmas to prove some more interesting congruences. – Zhi-Wei Sun Oct 16 '18 at 2:59