Determining $x$ or $y$ mod $p^2$ with $p=x^2+d{y}^2$

Abstract

Let $p$ be an odd prime and let $d\in \{2,3,7\}$. When $(\frac{-d}{p})=1$ we can write $p=x^2+d{y}^2$ 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 $p\equiv x\equiv 1(\mathrm{mod}4)$ then $$\sum _{k=0}^{(p-1)/2}(3[3\mid k]-1)(2k+1)\frac{{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}^2}{(-16{)}^k}\equiv \left(\frac{2}{p}\right)2x(\mathrm{mod}{p}^2)$$ where $[3\mid k]$ takes $1$ or $0$ according as $3\mid k$ or not, and that if $-p\equiv y\equiv 1(\mathrm{mod}4)$ then $$\sum _{k=0}^{(p-1)/2}\left(\frac{k}{3}\right)\frac{k{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}^2}{(-16{)}^k}\equiv (-1{)}^{(p+1)/4}y\equiv \sum _{k=0}^{(p-1)/2}(1-3[3\mid k])\frac{k{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}^2}{(-16{)}^k}(\mathrm{mod}{p}^2).$$ 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 $m=1,-8,16,-64,256,-512,4096$.

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References in corpus (2)

• Open Conjectures on Congruences
• On sums involving products of three binomial coefficients

Cited by in corpus (1)

• Supercongruences and Complex Multiplication