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.



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 $0<(\sqrt{5}-\sqrt{3})^{2n}\leq 1$.


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

  • $\begingroup$ Pretty neat and nice. +1 $\endgroup$ – DonAntonio Mar 30 '14 at 10:15

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