For any $n\in N,\left[{(\sqrt{3}+1)}^{2n}\right]+1$ is divisible by

Let $x={(\sqrt{3}+1)}^{2n}=\left[x\right]+f(\text{where\hspace{0.17em}}[\cdot ]\text{\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}}g\cdot I\cdot f)\to \left(i\right)$

Where $0\le f<1$, now let ${(\sqrt{3}-1)}^{2n}=f^1\to \left(ii\right)$

Where $0<f^1<1$

Adding (i) and (ii), we get

$\begin{array}{l}\Rightarrow \left[x\right]+f+f^1={(\sqrt{3}+1)}^{2n}+{(\sqrt{3}-1)}^{2n}\\ ={(4+2\sqrt{3})}^n+{(4-2\sqrt{3})}^n\\ =2^n\left\{{(2+\sqrt{3})}^n+{(2-\sqrt{3})}^n\right\}\\ =2^n\cdot 2\left\{{ }^n{C}_0{\left(2\right)}^n+{}^n{C}_2{\left(2\right)}^{n-2}\cdot {\left(\sqrt{3}\right)}^2+{}^n{C}_4{\left(2\right)}^{n-4}\cdot {\left(\sqrt{3}\right)}^4+\cdots \right\}\end{array}$

$\Rightarrow \left[x\right]+f+f^1=2^{n+1}\cdot k\to \left(iii\right)$ (where $k$  is any integer)

Hence $f+f^1$ is an integer

$\Rightarrow i.e.,f+f^1=1[\because 0<f+f^1<2]$

From (iii), $\left[x\right]+1=2^{n+1}\cdot k$

$\left[{(\sqrt{3}+1)}^{2n}\right]+1=2^{n+1}\cdot k$ (from (i))

This shows that $\left[{(\sqrt{3}+1)}^{2n}\right]+1$ divisible by $2^{n+1}$ for all $n\in N$.