Proof of a recent conjecture of Z.-W. Sun

Song Guo, Victor J. W. Guo

Published 2016-04-18Version 1

The polynomials $d_n(x)$ 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 $p$, the following congruences hold modulo $p$: \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 $p=x^2+y^2$ with $x\equiv 1\pmod{4}$,} 0,&\text{if $p\equiv 3\pmod{4}$,} \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 $p>3$,} [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 $p\equiv 1\pmod{4}$,} (-1)^{\frac{p+1}{4}}{\frac{p-1}{2}\choose \frac{p-3}{4}},&\text{if $p\equiv 3\pmod{4}$.} \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 $p>5$.} \end{align*} The $p\equiv 3\pmod{4}$ 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.