Let $p$ be an odd prime and let $d\in\{ 2,3,7\}$. When $(\frac{ -d}p)=1$ we can write $p=x^2+dy^2$ with $x,y\in\mathbb Z$; in this paper we aim at determining $x$ or $y$ modulo $p^2$. For example, when $p=x^2+3y^2$, we show that if $p\equiv x\equiv 1\pmod 4$ then $$\sum_{ k=0}^{ (p-1)/2}(3[3\mid k]-1)(2k+1)\frac{ \binom{ 2k}k^2}{ (-16)^k}\equiv\left(\frac2p\right)2x\pmod{ p^2}$$ where $[3\mid k]$ takes $1$ or $0$ according as $3\mid k$ or not, and that if $-p\equiv y\equiv 1\pmod4$ then $$\sum_{ k=0}^{ (p-1)/2}\left(\frac k3\right)\frac{ k\binom{ 2k}k^2}{ (-16)^k} \equiv(-1)^{ (p+1)/4}y\equiv\sum_{ k=0}^{ (p-1)/2}(1-3[3\mid k])\frac{ k\binom{ 2k}k^2}{ (-16)^k}\pmod{ p^2}.$$ We also determine $$\sum_{ k=0}^{ p-1}\frac{ k\binom{ 2k}k^3}{ m^k}\sum_{ k\le j<2k}\frac1j\quad \mbox{ mod}\ p$$ for $m=1,-8,16,-64,256,-512,4096$.
Categories: math.NT math.CO