Proof of a recent conjecture of Z.-W. Sun
arXiv:1604.05019
Abstract
The polynomials are defined by \begin{align*} d_n(x) &= \sum_{k=0}^n{n\choose k}{x\choose k}2^k. \end{align*} We prove that, for any prime , the following congruences hold modulo : \begin{align*} \sum_{k=0}^{p-1}\frac{2k\choose k}{4^k} d_k\left(-\frac{1}{4}\right)^2 &\equiv \begin{cases} 2(-1)^{\frac{p-1}{4}}x,&\text{if with ,} 0,&\text{if ,} \end{cases} [5pt] \sum_{k=0}^{p-1}\frac{2k\choose k}{4^k} d_k\left(-\frac{1}{6}\right)^2 &\equiv 0, \quad\text{if ,} [5pt] \sum_{k=0}^{p-1}\frac{2k\choose k}{4^k} d_k\left(\frac{1}{4}\right)^2 &\equiv \begin{cases} 0,&\text{if ,} (-1)^{\frac{p+1}{4}}{\frac{p-1}{2}\choose \frac{p-3}{4}},&\text{if .} \end{cases} \sum_{k=0}^{p-1}\frac{2k\choose k}{4^k} d_k\left(\frac{1}{6}\right)^2 &\equiv 0, \quad\text{if .} \end{align*} The case of the first one confirms a conjecture of Z.-W. Sun, while the second one confirms a special case of another conjecture of Z.-W. Sun.
4 pages