Applying binomial expansion we have: $(2+\sqrt{3})^{2n}=\displaystyle \sum _{i=0}^{2n} \left[\binom{2n}{i} 2^{i}\cdot (\sqrt{3})^{2n-i}\right]=\displaystyle \sum _{i=0}^{n} \left[\binom{2n}{2i} 2^{2i}\cdot 3^{n-i}\right]+\displaystyle \sum _{i=1}^{n} \left[\binom{2n}{2i-1} 2^{2i-1}\cdot \sqrt{3} \cdot 3^{n-i}\right]$ $(2-\sqrt{3})^{2n}=\displaystyle \sum _{i=0}^{2n} \left[\binom{2n}{i} 2^{i}\cdot (-\sqrt{3})^{2n-i}\right]=\displaystyle \sum _{i=0}^{n} \left[\binom{2n}{2i} 2^{2i}\cdot 3^{n-i}\right]-\displaystyle \sum _{i=1}^{n} \left[\binom{2n}{2i-1} 2^{2i-1}\cdot \sqrt{3} \cdot 3^{n-i}\right]$ Therefore we have $(2+\sqrt{3})^{2n}+(2-\sqrt{3})^{2n}=2\displaystyle \sum _{i=0}^{n} \left[\binom{2n}{2i} 2^{2i}\cdot 3^{n-i}\right]$ is an even integer. $(2+\sqrt{3})^{2n}-(2-\sqrt{3})^{2n}=2\displaystyle \sum _{i=1}^{n} \left[\binom{2n}{2i-1} 2^{2i-1}\cdot \sqrt{3} \cdot 3^{n-i}\right]=2\sqrt{3}\displaystyle \sum _{i=1}^{n} \left[\binom{2n}{2i-1} 2^{2i-1}\cdot 3^{n-i}\right]$