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Question Number 195364 by Rodier97 last updated on 01/Aug/23

$$$$$$\mathrm{show}\mathrm{that}\mathrm{for}\mathrm{any}\mathrm{natural}\mathrm{number}n,$$$$\mathrm{the}\mathrm{natural}\mathrm{number}{(3-\sqrt{5})}^n+{(3+\sqrt{5})}^n\mathrm{is}\mathrm{divisible}$$$$\mathrm{by}{2}^n.$$

Answered by Frix last updated on 31/Jul/23

$$\mathrm{Obviously}\mathrm{some}\mathrm{factors}\mathrm{cancel}\mathrm{out}\mathrm{and}$$$$\mathrm{others}\mathrm{add}.2\mathrm{simple}\mathrm{examples}:$$$${(a-b)}^2+{(a+b)}^2=$$$$=(a^2-2ab+b^2)+(a^2+2ab+b^2)=$$$$=2(a^2+b^2)$$$${(a-b)}^3+{(a+b)}^3=$$$$=(a^3-3a^{2}b+3{ab}^2-b^3)+(a^3+3a^{2}b+3{ab}^2+b^3)=$$$$=2a(a^2+3b^2)$$$$\mathrm{This}\mathrm{should}\mathrm{be}\mathrm{enough}\mathrm{to}\mathrm{see}\mathrm{to}\mathrm{prove}.$$$$(\mathrm{Also}\mathrm{if}b=\sqrt{\beta }\wedge \beta \in \mathbb{N}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{only}\mathrm{factors}$$$$b^{2k}\in \mathbb{N}‘‘{\mathrm{survive}}^{″}\Rightarrow {(a-\sqrt{\beta })}^n+{(a+\sqrt{\beta })}^n\in \mathbb{N})$$

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