Question

Show that the integer just above ${(\sqrt{3}+1)}^{2n}$ is divisible by $2^{n+1}$ for all $nϵN.$

Answer 1

Let ${(\sqrt{3}+1)}^{2n}={(4+2\sqrt{3})}^n=2^n{(2+\sqrt{3})}^n=I+f$ ...... $\left(i\right)$
where $I$ and $f$ are its integral & fractional parts respectively
$\Rightarrow 0<f<1$
Now, $0<\sqrt{3}-1<1$
$0<{(\sqrt{3}-1)}^{2n}<1$
Let ${(\sqrt{3}+1)}^{2n}={(4-2\sqrt{3})}^n=2^n{(2-\sqrt{3})}^n=f^{\prime }$ ...... $\left(ii\right)$
$0<f^{\prime }<1$
Adding $\left(i\right)$ and $\left(ii\right)$
$I+f+f^{\prime }={(\sqrt{3}+1)}^{2n}+{(\sqrt{3}-1)}^{2n}$
$=2^n[{(2+\sqrt{3})}^n+{(2-\sqrt{3})}^n]={2.2}^n[{}^n{\text{C}}_0{2}^n{+}^n{\text{C}}_2{2}^{n-2}{\left(\sqrt{3}\right)}^2+...]$
$I+f+f^{\prime }=2^{n+1}k$ (where k is a positive integer)
$0<f+f^{\prime }<2\Rightarrow f+f^{\prime }=1$
$I+1=2^{n+1}k.$
$I+1$ is the integer just above ${(\sqrt{3}+1)}^{2n}$ and which is divisible by $2^{n+1}.$