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-

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 ?