Cubic congruences and sums involving $\binom{3k}k$

Zhi-Hong Sun

Published 2013-10-24, updated 2013-11-20Version 7

Let $p$ be a prime greater than $3$ and let $a$ be a rational p-adic integer. In this paper we try to determine $\sum_{k=1}^{[p/3]}\binom{3k}ka^k\pmod p$, and real the connection between cubic congruences and the sum $\sum_{k=1}^{[p/3]}\binom{3k}ka^k$, where $[x]$ is the greatest integer not exceeding $x$. Suppose that $a_1,a_2,a_3$ are rational p-adic integers, $P=-2a_1^3+9a_1a_2-27a_3$, $Q=(a_1^2-3a_2)^3$ and $PQ(P^2-Q)(P^2-3Q)(P^2-4Q)\not\equiv 0\pmod p$. In this paper we show that the number of solutions of the congruence $x^3+a_1x^2+a_2x+a_3\equiv 0\pmod p$ depends only on $\sum_{k=1}^{[p/3]}\binom{3k}k(\frac{4Q-P^2}{27Q})^k\pmod p$. Let $q$ be a prime of the form $3k+1$ and so $4q=L^2+27M^2$ with $L,M\in\Bbb Z$. When $p\not=q$ and $p\nmid L$, we establish congruences for $\sum_{k=1}^{[p/3]}\binom{3k}k(\frac{M^2}q)^k$ and $\sum_{k=1}^{[p/3]}\binom{3k}k(\frac{L^2}{27q})^k$ modulo p. As a consequence, we show that $x^3-qx-qM\equiv 0\pmod p$ has three solutions if and only if $p$ is a cubic residue of $q$.