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Power of $(\sqrt{2} +1)$ and floor function - elim - 01-07-2011 07:24 PM
Prove that $\lfloor (\sqrt{2} + 1)^{2n} \rfloor + 1 - (\sqrt{2} + 1)^{2n} \geq \frac{1}{2} $ and $ (\sqrt{2} + 1)^{2n-1} - \lfloor (\sqrt{2} + 1)^{2n-1} \rfloor \leq \frac{1}{2} , \forall n \in \mathbb{N} $