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

$$Let\xi beapositiveRootof{x}^2-2023x-1$$$$Defineasequence{\phi }_{i}suchThat{\phi }_0=1$$$${\phi }_{n+1}=\lfloor {\phi }_n\xi \rfloor ,findTheRemainderWhen{\phi }_{2023}isdividedby\sqrt{{\phi }_2}$$

Answered by York12 last updated on 02/Sep/23

$$$$$$x^2-2023x-1=0\wedge \xi isapositiveroot$$$$sincetheproductofRoots=-1$$$$\Rightarrow TheotherRoot=\frac{1}{\xi }$$$$\xi -\frac{1}{\xi }=2023\Rightarrow \xi =2023+\frac{1}{\xi }$$$$\Rightarrow {\phi }_n=\lfloor {\phi }_{n-1}\xi \rfloor =\lfloor {\phi }_{n-1}\times 2023+{\phi }_{n-1}\times \frac{1}{\xi }\rfloor$$$$2023{\phi }_{n-1}\in {\mathbb{Z}}^{+}\Rightarrow \lfloor {\phi }_{n-1}\times 2023+{\phi }_{n-1}\times \frac{1}{\xi }\rfloor =2023{\phi }_{n-1}+\lfloor \frac{{\phi }_{n-1}}{2023}\rfloor$$$$Wehave$$$${\phi }_n=\lfloor {\phi }_{n-1}\xi \rfloor \iff {\phi }_n⩽{\phi }_{n-1}\xi <{\phi }_n+1$$$$\Rightarrow \frac{{\phi }_n}{\xi }⩽{\phi }_{n-1}<\frac{{\phi }_n}{\xi }+\frac{1}{2023+\frac{1}{\xi }},\wedge \frac{1}{2023+\frac{1}{\xi }}<1$$$$\Rightarrow \lfloor \frac{{\phi }_n}{\xi }\rfloor \in \{{\phi }_{n-1},{\phi }_{n-1}-1\}$$$$\lfloor \frac{{\phi }_n}{\xi }\rfloor ={\phi }_{n-1}\iff \frac{{\phi }_n}{\xi }={\phi }_{n-1},but{\phi }_n,{\phi }_{n-1}\in {\mathbb{Z}}^{+}\wedge \xi \in {\mathbb{Q}}^{{}^{\prime }}$$$$\Rightarrow \frac{{\phi }_n}{\xi }\notin {\mathbb{Z}}^{+}\Rightarrow \frac{{\phi }_n}{\xi }\ne {\phi }_{n-1}\Rightarrow \lfloor \frac{{\phi }_n}{\xi }\rfloor ={\phi }_{n-1}-1$$$$\Rightarrow {\phi }_n=2023{\phi }_{n-1}+{\phi }_{n-2}-1$$$$\Rightarrow {\phi }_2={2023}^2\Rightarrow \sqrt{{\phi }_2}=2023$$$$\Rightarrow \frac{{\phi }_n}{2023}={\phi }_{n-1}+\frac{{\phi }_{n-2}-1}{2023}$$$$\Rightarrow {\phi }_n\equiv {\phi }_{n-2}-1mod\left(2023\right)$$$$\Rightarrow {\phi }_{2023}\equiv {\phi }_{2021}-1mod\left(2023\right)\equiv {\phi }_{2019}-1mod\left(2023\right)$$$$....\equiv {\phi }_1-1011mod\left(2023\right)\equiv 1012mod\left(2023\right)$$

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