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Question Number 168244 by mnjuly1970 last updated on 07/Apr/22

Answered by MJS_new last updated on 07/Apr/22

$$x⩾0$$$$\mathrm{rhs}\mathrm{is}\mathrm{integer}\Rightarrow \mathrm{lhs}\mathrm{must}\mathrm{be}\mathrm{integer}$$$$x^2⩾0\Rightarrow \lfloor \sqrt{x}{\rfloor }^2+2\lfloor \sqrt{x}\rfloor ⩾2\Rightarrow x⩾1$$$$\mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{is}x=1$$

Commented by mnjuly1970 last updated on 07/Apr/22

$$thxalot..$$

Answered by floor(10²Eta[1]) last updated on 07/Apr/22

$$\lfloor \sqrt{\mathrm{x}}\rfloor =\mathrm{n}⩾0\in \mathbb{Z}\Rightarrow \sqrt{\mathrm{x}}=\mathrm{n}+\alpha ,0⩽\alpha <1$$$$\Rightarrow {\mathrm{x}}^2={(\mathrm{n}+\alpha )}^4\mathrm{and}{\mathrm{x}}^2={\mathrm{n}}^2+2\mathrm{n}-2$$$$0⩽\alpha <1\Rightarrow {\mathrm{n}}^4⩽{(\mathrm{n}+\alpha )}^4<{(\mathrm{n}+1)}^4$$$$\Rightarrow {\mathrm{n}}^4⩽{\mathrm{x}}^2<{(\mathrm{n}+1)}^4\Rightarrow {\mathrm{n}}^4⩽{\mathrm{n}}^2+2\mathrm{n}-2<{(\mathrm{n}+1)}^4$$$$\mathrm{let}\mathrm{f}\left(\mathrm{n}\right)={\mathrm{n}}^4-{\mathrm{n}}^2-2\mathrm{n}+2,\mathrm{we}\mathrm{want}\mathrm{f}\left(\mathrm{n}\right)⩽0$$$$\mathrm{f}\left(0\right)=2,\mathrm{f}\left(1\right)=-1\Rightarrow \mathrm{n}⩾1$$$$\mathrm{if}\mathrm{we}\mathrm{show}\mathrm{that}\mathrm{f}\left(\mathrm{n}\right)\mathrm{is}\mathrm{decreasing}\mathrm{for}\mathrm{n}⩾1{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{done}$$$${\mathrm{f}}^{\prime }\left(\mathrm{n}\right)={4\mathrm{n}}^3-2\mathrm{n}-2⩽0\Rightarrow {2\mathrm{n}}^3⩽\mathrm{n}+1$$$$\mathrm{this}\mathrm{inequality}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=1,$$$$\mathrm{but}\mathrm{for}\mathrm{n}⩾2\mathrm{it}\mathrm{seems}\mathrm{to}\mathrm{be}\mathrm{false}$$$$\mathrm{lets}\mathrm{show}\mathrm{that}{2\mathrm{n}}^3⩾\mathrm{n}+1,\mathrm{\forall }\mathrm{n}⩾2\left(\mathrm{induction}\right)$$$$\mathrm{suppose}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{true}\mathrm{for}\mathrm{any}\mathrm{k}⩾2$$$${2\mathrm{k}}^3⩾\mathrm{k}+1,{\mathrm{let}}^{\prime }\mathrm{s}\mathrm{show}\mathrm{its}\mathrm{true}\mathrm{for}\mathrm{k}+1$$$$2{(\mathrm{k}+1)}^3={2\mathrm{k}}^3+{6\mathrm{k}}^2+6\mathrm{k}+2⩾{6\mathrm{k}}^2+7\mathrm{k}+3⩾7\mathrm{k}+3⩾\mathrm{k}+2$$$$\mathrm{so},\mathrm{in}\mathrm{fact}\mathrm{n}=1\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{value}\mathrm{that}\mathrm{makes}\mathrm{f}\left(\mathrm{n}\right)⩽0$$$$\Rightarrow \sqrt{\mathrm{x}}=\alpha +1\Rightarrow 1⩽\sqrt{\mathrm{x}}<2\Rightarrow \lfloor \sqrt{\mathrm{x}}\rfloor =1$$$$\Rightarrow {\mathrm{x}}^2=1\Rightarrow \mathrm{x}=1$$

Answered by mr W last updated on 07/Apr/22

$$lett=\sqrt{x}$$$$t^4={\left[t\right]}^2+2\left[t\right]-2=integer$$$$\Rightarrow tisalsointeger\Rightarrow \left[t\right]=t$$$$t^4=t^2+2t-2=0$$$$(t^2+2t+2){(t-1)}^2=0$$$$\Rightarrow t=1istheonlyintegersolution$$$$\Rightarrow \sqrt{x}=t=1\Rightarrow x=1$$

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