Let (\sqrt{3}+1)^{2n}= I + f
where I is an integer and 0≤f<1
We have (\sqrt{3}-1)^{2n}= f_{1}
where f1<1
we have
(\sqrt{3}+1)^{2n}+ (\sqrt{3}-1)^{2n}= 2[(\sqrt{3})^{2n} + \begin{pmatrix} 2n\\ 2 \end{pmatrix}(\sqrt{3})^{2n-2}+....+1 ]
= an even integer
so I + f + f1 is an even integer \Leftrightarrow f + f_{1} is an integer
Since 0<f+ f1<2 we must have f + f1 = 1
So, f = 1- f1
But,\lim_{n\rightarrow \propto } (\sqrt{3}-1)^{2n} = 0 = f_{1}
So, f = 1- f_{1} = 1
\lim_{n\rightarrow \propto }\left\{(\sqrt{3}+1)^{2n} \right\} = 1