Determining $x$ or $y$ mod $p^2$ with $p=x^2+dy^2$

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 $x, y \in Z$ ; in this paper we aim at determining $x$ or $y$ modulo $p^{2}$ . For example, when $p = x^{2} + 3 y^{2}$ , we show that if $p \equiv x \equiv 1 (m o d 4)$ then $\sum_{k = 0}^{(p - 1) / 2} (3 [3 ∣ k] - 1) (2 k + 1) \frac{{(\binom{2 k}{k})}^{2}}{(- 16)^{k}} \equiv (\frac{2}{p}) 2 x (m o d p^{2})$ where $[3 ∣ k]$ takes $1$ or $0$ according as $3 ∣ k$ or not, and that if $- p \equiv y \equiv 1 (m o d 4)$ then $\sum_{k = 0}^{(p - 1) / 2} (\frac{k}{3}) \frac{k {(\binom{2 k}{k})}^{2}}{(- 16)^{k}} \equiv (- 1)^{(p + 1) / 4} y \equiv \sum_{k = 0}^{(p - 1) / 2} (1 - 3 [3 ∣ k]) \frac{k {(\binom{2 k}{k})}^{2}}{(- 16)^{k}} (m o d p^{2}) .$ We also determine $\sum_{k = 0}^{p - 1} \frac{k {(\binom{2 k}{k})}^{3}}{m^{k}} \sum_{k \leq j < 2 k} \frac{1}{j} m o d p$ for $m = 1, - 8, 16, - 64, 256, - 512, 4096$ .

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

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

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Determining xx or yy mod p2p2 with p=x2+dy2p=x2+dy2

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Determining xx or yy mod p2p2 with p=x2+dy2p=x2+dy2

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

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AI智能校验

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

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AI智能校验