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Question Number 154649 by mathdanisur last updated on 20/Sep/21

$$\mathrm{if}\mathrm{n}\in {\mathbb{N}}^{>2}\mathrm{prove}\mathrm{that}$$$$\left[{(\sqrt[3]{\mathrm{n}}+\sqrt[3]{\mathrm{n}+2})}^3\right]+1=0\left(\mathrm{mod}8\right)$$

Commented by MJS_new last updated on 20/Sep/21

$$\mathrm{just}\mathrm{a}\mathrm{try}$$$$\mathrm{let}n=t-1\Rightarrow t⩾3$$$$f\left(t\right)\in \mathbb{R}\wedge 0⩽f\left(t\right)<1$$$${(\sqrt[3]{t-1}+\sqrt[3]{t+1})}^3+1-f\left(t\right)=8t$$$$f\left(t\right)=1-6t+3(\sqrt[3]{{(t-1)}^2(t+1)}+\sqrt[3]{(t-1){(t+1)}^2})$$$$\mathrm{we}\mathrm{must}\mathrm{show}0⩽f\left(t\right)<1\mathrm{\forall }t⩾3$$$$f\left(3\right)\approx .0839$$$$\mathrm{and}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{show}\mathrm{that}$$$$\underset{t\to \mathrm{\infty }}{\lim }f\left(t\right)=1$$

Commented by mathdanisur last updated on 20/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathbf{S}\mathrm{er},\mathrm{thank}\mathrm{you}$$

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