prove that (2+(√3))^(2n) +(2−(√3))^(2n) is an even integer and that (2+(√3))^(2n) −(2−(√3))^(2n) =w(√3) for some integers w,for all integer n≥1.


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Question Number 17867 by Mr easymsn last updated on 11/Jul/17

$p r o v e t h a t {(2 + \sqrt{3})}^{2 n} + {(2 - \sqrt{3})}^{2 n} i s a n$ $e v e n i n t e g e r a n d t h a t {(2 + \sqrt{3})}^{2 n} - {(2 - \sqrt{3})}^{2 n} = w \sqrt{3}$ $f o r s o m e i n t e g e r s w, f o r a l l i n t e g e r n ⩾ 1 .$
	Answered by alex041103 last updated on 11/Jul/17

	$We expand using$ ${(a + b)}^{n} = \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) a^{n - k} b^{k}$ $We get$ ${(2 + \sqrt{3})}^{2 n} + {(2 - \sqrt{3})}^{2 n} =$ $= \underset{k = 0}{\sum^{2 n}} (\begin{matrix} 2 n \\ k \end{matrix}) 2^{2 n - k} {(\sqrt{3})}^{k} [1 + {(- 1)}^{k}]$ $When k \equiv 1 (\mod 2)$ $(\begin{matrix} 2 n \\ k \end{matrix}) 2^{2 n - k} {(\sqrt{3})}^{k} [1 + {(- 1)}^{k}] = 0$ ${That}^{'} s why k \equiv 0 (\mod 2)$ $\Rightarrow {(2 + \sqrt{3})}^{2 n} + {(2 - \sqrt{3})}^{2 n} =$ $= \underset{k = 0}{\sum^{n}} (\begin{matrix} 2 n \\ 2 k \end{matrix}) 2^{2 n - 2 k} {(3^{1 / 2})}^{2 k} \times 2$ $= 2 \underset{k = 0}{\sum^{n}} (\begin{matrix} 2 n \\ 2 k \end{matrix}) 2^{2 n - 2 k} \times 3^{k}$ ${It}^{'} s easy to show that$ $\underset{k = 0}{\sum^{n}} (\begin{matrix} 2 n \\ 2 k \end{matrix}) 2^{2 n - 2 k} \times 3^{k} \in N$ $\Rightarrow \frac{{(2 + \sqrt{3})}^{2 n} + {(2 - \sqrt{3})}^{2 n}}{2} \in N$ $The same procedure works for the next statement$ ${(2 + \sqrt{3})}^{2 n} - {(2 - \sqrt{3})}^{2 n} =$ $= \underset{k = 0}{\sum^{2 n}} (\begin{matrix} 2 n \\ k \end{matrix}) 2^{2 n - k} {(\sqrt{3})}^{k} [1 + {(- 1)}^{k + 1}]$ $Then we substitute k = 2 t + 1 and$ ${(2 + \sqrt{3})}^{2 n} - {(2 - \sqrt{3})}^{2 n} =$ $= \underset{k = 0}{\sum^{n - 1}} (\begin{matrix} 2 n \\ 2 k + 1 \end{matrix}) 2^{2 n - 2 k} (\sqrt{3}) 3^{k}$ $= \sqrt{3} \underset{k = 0}{\sum^{n - 1}} (\begin{matrix} 2 n \\ 2 k + 1 \end{matrix}) 2^{2 n - 2 k} 3^{k}$ $\Rightarrow {(2 + \sqrt{3})}^{2 n} - {(2 - \sqrt{3})}^{2 n} = w \sqrt{3}, w \in N$
	Commented by alex041103 last updated on 12/Jul/17

	$I s t h e r e a p r o b l e m, s i r ?$
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