Question

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

• A. $2^m+1$
• 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 $0\le f<1$
Assuming $F={(\sqrt{3}-1)}^{2m}$
Now,
$I+f+F={(\sqrt{3}+1)}^{2m}+{(\sqrt{3}-1)}^{2m}={(4+2\sqrt{3})}^m+{(4-2\sqrt{3})}^m=2^m\left[\right(2+\sqrt{3}{)}^m+(2-\sqrt{3}{)}^m]=2^m\cdot 2[{}^m{C}_0\times 2^m+{}^m{C}_2\times 2^{m-2}\times 3+\cdots ]=2^{m+1}\cdot [{}^m{C}_0\times 2^m+{}^m{C}_2\times 2^{m-2}\times 3+\cdots ]\Rightarrow I+f+F=2^{m+1}k,k\in N\Rightarrow f+F=2^{m+1}k-I$

$\therefore f+F$ is an integer, so
$\Rightarrow f+F=1$
As $\left[\right(\sqrt{3}+1{)}^{2m}]=I$, so
$\left[\right(\sqrt{3}+1{)}^{2m}]+1=I+1\Rightarrow \left[\right(\sqrt{3}+1{)}^{2m}]+1=I+f+F\Rightarrow \left[\right(\sqrt{3}+1{)}^{2m}]+1=2^{m+1}k$

Hence $\left[\right(\sqrt{3}+1{)}^{2m}]+1$ is divisible by $2^{m+1}.$