Smallest integer greater than the given number

Question

What is the smallest integer greater than the real number $(\sqrt{5}+\sqrt{3})^{2n}$ (for non-negative integer $n$)?
The answer is $(\sqrt{5}+\sqrt{3})^{2n}+(\sqrt{5}-\sqrt{3})^{2n}$.
I just checked for $n=1,2$ and $3$ and it works. I don't know how to derive the answer or how to prove it. Help.

Answer 1

Hints:

Use the binomial formula to check that $(\sqrt{5}+\sqrt{3})^{2n}+(\sqrt{5}-\sqrt{3})^{2n}$ is an integer.

Show that $0< \sqrt{5}-\sqrt{3}\leq1$, hence $0<(\sqrt{5}-\sqrt{3})^{2n}\leq 1$.

Hence:

$$(\sqrt{5}+\sqrt{3})^{2n}<(\sqrt{5}+\sqrt{3})^{2n}+(\sqrt{5}-\sqrt{3})^{2n}\leq(\sqrt{5}+\sqrt{3})^{2n}+1$$ Thus, it follows that ....