Question

If $m$ is a positive integer, then $\left[{(\sqrt{3}+1)}^{2m}\right]+1$, where $\left[x\right]$ denotes greatest integer $\le n$, is divisible by

• A. $2^m$
• B. $2^{m+1}$
• C. $2^{m+2}$
• D. $2^{2m}$

Answer 1

The correct option is B $2^{m+1}$

Let ${(\sqrt{3}+1)}^{2m}=I+f$ where $I$ is some integer and $0\le f<1$. We have $\sqrt{3}-1=\frac{2}{\sqrt{3}+1}$. Therefore,
$0<\sqrt{3}-1<1$
let $F={(\sqrt{3}+1)}^{2m}$
Now,
$I+f+F={(\sqrt{3}+1)}^{2m}+{(\sqrt{3}-1)}^{2m}$
$={\left\{{(\sqrt{3}+1)}^2\right\}}^m+{\left\{{(\sqrt{3}-1)}^2\right\}}^m={(4+2\sqrt{3})}^m+{(4-2\sqrt{3})}^m$
$=2^m{(2+\sqrt{3})}^m+2^m{(2-\sqrt{3})}^m=2^m\{{(2+\sqrt{3})}^m+{(2-\sqrt{3})}^m\}$
$=2^m\cdot 2\{{{}^{m}C}_0{2}^m+{{}^{m}C}_2{(2}^{m-2})\left(3\right)+{{}^{m}C}_4{2}^{m-4}\left(3^2\right)\}=2^{m+1}k$ where $k$ is some integer
$I+f+F$ is an integer, say $J$.
$\Rightarrow f+F=J-I$ is an integer
Since $0<f+F<2$, therefore, $f+F=1$
Now $N=\left[{(\sqrt{3}+1)}^{2m}\right]+1=I+f+F=2^{m+1}k$
$\Rightarrow 2^{m+1}$