if-n-N-gt-2-prove-that-n-1-3-n-2-1-3-3-1-0-mod-8-

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

if n ∈ N^(>2) prove that [((n)^(1/3) + ((n + 2))^(1/3) )^3 ] + 1 = 0 (mod 8)

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

Commented by MJS_new last updated on 20/Sep/21

just a try let n=t−1 ⇒ t≥3 f(t)∈R∧0≤f(t)<1 (((t−1))^(1/3) +((t+1))^(1/3) )^3 +1−f(t)=8t f(t)=1−6t+3((((t−1)^2 (t+1)))^(1/3) +(((t−1)(t+1)^2 ))^(1/3) ) we must show 0≤f(t)<1 ∀ t≥3 f(3)≈.0839 and it′s easy to show that lim_(t→∞) f(t) =1

$$\mathrm{just}\:\mathrm{a}\:\mathrm{try} \\ $$$$\mathrm{let}\:{n}={t}−\mathrm{1}\:\Rightarrow\:{t}\geqslant\mathrm{3} \\ $$$${f}\left({t}\right)\in\mathbb{R}\wedge\mathrm{0}\leqslant{f}\left({t}\right)<\mathrm{1} \\ $$$$\left(\sqrt[{\mathrm{3}}]{{t}−\mathrm{1}}+\sqrt[{\mathrm{3}}]{{t}+\mathrm{1}}\right)^{\mathrm{3}} +\mathrm{1}−{f}\left({t}\right)=\mathrm{8}{t} \\ $$$${f}\left({t}\right)=\mathrm{1}−\mathrm{6}{t}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left({t}+\mathrm{1}\right)}+\sqrt[{\mathrm{3}}]{\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{show}\:\mathrm{0}\leqslant{f}\left({t}\right)<\mathrm{1}\:\forall\:{t}\geqslant\mathrm{3} \\ $$$${f}\left(\mathrm{3}\right)\approx.\mathrm{0839} \\ $$$$\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that} \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:{f}\left({t}\right)\:=\mathrm{1} \\ $$

Commented by mathdanisur last updated on 20/Sep/21

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

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