show-that-for-any-natural-number-n-the-natural-number-3-5-n-3-5-n-is-divisible-by-2-n-

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

show that for any natural number n,

the natural number {(3 - \sqrt{5})}^{n} + {(3 + \sqrt{5})}^{n} is divisible

by 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} - 2 a b + b^{2}) + (a^{2} + 2 a b + b^{2}) =

= 2 (a^{2} + b^{2})

{(a - b)}^{3} + {(a + b)}^{3} =

= (a^{3} - 3 a^{2} b + 3 {a b}^{2} - b^{3}) + (a^{3} + 3 a^{2} b + 3 {a b}^{2} + b^{3}) =

= 2 a (a^{2} + 3 b^{2})

This should be enough to see to prove .

(Also if b = \sqrt{β} \land β \in N {it}^{'} s easy to see only factors

b^{2 k} \in N “ {survive}^{″} \Rightarrow {(a - \sqrt{β})}^{n} + {(a + \sqrt{β})}^{n} \in N)

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