Question

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

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

Answer 1

Answer: (A), (B)

The correct options are
A $2^m$
B $2^{m+1}$

${(\sqrt{3}+1)}^{2m}=$ 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 $F=(\sqrt{3}-1{)}^{2m}$.
Let $I+f+F=(\sqrt{3}+1{)}^{2m}+(\sqrt{3}-1{)}^{2m}$
$={\left\{\right(\sqrt{3}+1{)}^2\}}^m+\{{(\sqrt{3}-1{)}^2\}}^m=(4+2\sqrt{3}{)}^m+(4-2\sqrt{3}{)}^m$
$=2^m(2+\sqrt{3}{)}^m+2^m(2-\sqrt{3}{)}^m=2^m\left\{\right(2+\sqrt{3}{)}^m+(2-\sqrt{3}{)}^m\}=2^m.2\left\{{}^m{C}_0{2}^m{+}^m{C}_2\right(2^{m-2}\left)\right(3\left){+}^m{C}_4\right(2^{m-4}\left)\right(3^2)+\cdots \}$
$=2^{m+1}$ k where k is some integer
$\therefore$ 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, $N=\left[\right(\sqrt{3}+1{)}^{2m}]+1=I+f+F=2^{m+1}k$
$\Rightarrow$ $2^{m+1}$ and hence $2^m$ divides N.