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

show that for any natural number n, the natural number (3−(√5))^n +(3+(√5))^n is divisible by 2^n .

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

Answered by Frix last updated on 31/Jul/23

Obviously some factors cancel out and others add. 2 simple 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+3ab^2 −b^3 )+(a^3 +3a^2 b+3ab^2 +b^3 )= =2a(a^2 +3b^2 ) This should be enough to see to prove. (Also if b=(√β) ∧β∈N it′s easy to see only factors b^(2k) ∈N “survive” ⇒ (a−(√β))^n +(a+(√β))^n ∈N)

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

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