Let-be-a-positive-Root-of-x-2-2023x-1-Define-a-sequence-i-such-That-0-1-n-1-n-find-The-Remainder-When-2023-is-divided-by-2-

Question Number 196872 by York12 last updated on 02/Sep/23

L e t ξ b e a p o s i t i v e R o o t o f x^{2} - 2023 x - 1

D e f i n e a s e q u e n c e φ_{i} s u c h T h a t φ_{0} = 1

φ_{n + 1} = ⌊ φ_{n} ξ ⌋, f i n d T h e R e m a i n d e r W h e n φ_{2023} i s d i v i d e d b y \sqrt{φ_{2}}

Answered by York12 last updated on 02/Sep/23

x^{2} - 2023 x - 1 = 0 \land ξ i s a p o s i t i v e r o o t

s i n c e t h e p r o d u c t o f R o o t s = - 1

\Rightarrow T h e o t h e r R o o t = \frac{1}{ξ}

ξ - \frac{1}{ξ} = 2023 \Rightarrow ξ = 2023 + \frac{1}{ξ}

\Rightarrow φ_{n} = ⌊ φ_{n - 1} ξ ⌋ = ⌊ φ_{n - 1} \times 2023 + φ_{n - 1} \times \frac{1}{ξ} ⌋

2023 φ_{n - 1} \in Z^{+} \Rightarrow ⌊ φ_{n - 1} \times 2023 + φ_{n - 1} \times \frac{1}{ξ} ⌋ = 2023 φ_{n - 1} + ⌊ \frac{φ_{n - 1}}{2023} ⌋

W e h a v e

φ_{n} = ⌊ φ_{n - 1} ξ ⌋ \Leftrightarrow φ_{n} ⩽ φ_{n - 1} ξ < φ_{n} + 1

\Rightarrow \frac{φ_{n}}{ξ} ⩽ φ_{n - 1} < \frac{φ_{n}}{ξ} + \frac{1}{2023 + \frac{1}{ξ}}, \land \frac{1}{2023 + \frac{1}{ξ}} < 1

\Rightarrow ⌊ \frac{φ_{n}}{ξ} ⌋ \in {φ_{n - 1}, φ_{n - 1} - 1}

⌊ \frac{φ_{n}}{ξ} ⌋ = φ_{n - 1} \Leftrightarrow \frac{φ_{n}}{ξ} = φ_{n - 1}, b u t φ_{n}, φ_{n - 1} \in Z^{+} \land ξ \in Q^{^{'}}

\Rightarrow \frac{φ_{n}}{ξ} \notin Z^{+} \Rightarrow \frac{φ_{n}}{ξ} \neq φ_{n - 1} \Rightarrow ⌊ \frac{φ_{n}}{ξ} ⌋ = φ_{n - 1} - 1

\Rightarrow φ_{n} = 2023 φ_{n - 1} + φ_{n - 2} - 1

\Rightarrow φ_{2} = 2023^{2} \Rightarrow \sqrt{φ_{2}} = 2023

\Rightarrow \frac{φ_{n}}{2023} = φ_{n - 1} + \frac{φ_{n - 2} - 1}{2023}

\Rightarrow φ_{n} \equiv φ_{n - 2} - 1 m o d (2023)

\Rightarrow φ_{2023} \equiv φ_{2021} - 1 m o d (2023) \equiv φ_{2019} - 1 m o d (2023)

\dots . \equiv φ_{1} - 1011 m o d (2023) \equiv 1012 m o d (2023)

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