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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 rhs is integer ⇒ lhs must be integer x^2 ≥0 ⇒ ⌊(√x)⌋^2 +2⌊(√x)⌋≥2 ⇒ x≥1 the only solution is x=1

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

Commented by mnjuly1970 last updated on 07/Apr/22

thx alot..

$$\:{thx}\:{alot}.. \\ $$

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

⌊(√x)⌋=n≥0∈Z⇒(√x)=n+α, 0≤α<1 ⇒x^2 =(n+α)^4 and x^2 =n^2 +2n−2 0≤α<1⇒n^4 ≤(n+α)^4 <(n+1)^4 ⇒n^4 ≤x^2 <(n+1)^4 ⇒n^4 ≤n^2 +2n−2<(n+1)^4 let f(n)=n^4 −n^2 −2n+2, we want f(n)≤0 f(0)=2, f(1)=−1⇒n≥1 if we show that f(n) is decreasing for n≥1 we′re done f′(n)=4n^3 −2n−2≤0⇒2n^3 ≤n+1 this inequality is true for n=1, but for n≥2 it seems to be false lets show that 2n^3 ≥n+1, ∀n≥2 (induction) suppose it′s true for any k≥2 2k^3 ≥k+1, let′s show its true for k+1 2(k+1)^3 =2k^3 +6k^2 +6k+2≥6k^2 +7k+3≥7k+3≥k+2 so, in fact n=1 is the only value that makes f(n)≤0 ⇒(√x)=α+1⇒1≤(√x)<2⇒⌊(√x)⌋=1 ⇒x^2 =1⇒x=1

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

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

let t=(√x) t^4 =[t]^2 +2[t]−2=integer ⇒t is also integer ⇒[t]=t t^4 =t^2 +2t−2=0 (t^2 +2t+2)(t−1)^2 =0 ⇒t=1 is the only integer solution ⇒(√x)=t=1 ⇒x=1

$${let}\:{t}=\sqrt{{x}} \\ $$$${t}^{\mathrm{4}} =\left[{t}\right]^{\mathrm{2}} +\mathrm{2}\left[{t}\right]−\mathrm{2}={integer} \\ $$$$\Rightarrow{t}\:{is}\:{also}\:{integer}\:\Rightarrow\left[{t}\right]={t} \\ $$$${t}^{\mathrm{4}} ={t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{2}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)\left({t}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{1}\:{is}\:{the}\:{only}\:{integer}\:{solution} \\ $$$$\Rightarrow\sqrt{{x}}={t}=\mathrm{1}\:\Rightarrow{x}=\mathrm{1} \\ $$

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