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

${\left(\sqrt{3}+1\right)}^{2m}=I+f$ is some integer $\&0\le f<1$

Let $F={\left(\sqrt{3}-1\right)}^{2m}$ 

$\therefore I+f+F=2^{m+1}K$ , where K is some integer

$I+f+F$, is some integer, J(say)

$\therefore f+F=J-I$ is an integer

$\therefore 0<f+F<2\Rightarrow f+F=1$

Now, $\left[{\left(\sqrt{3}+1\right)}^{2m}\right]+1=I+f+F=2^{m+1}.K$