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Question Number 171402 by mnjuly1970 last updated on 14/Jun/22

$$solve...$$$$\lfloor x^2\rfloor +\lfloor x{\rfloor }^2=x^2+x+1$$$$x=?$$

Answered by floor(10²Eta[1]) last updated on 14/Jun/22

$$\lfloor {\mathrm{x}}^2\rfloor =\mathrm{a}\in \mathbb{Z}\Rightarrow {\mathrm{x}}^2=\mathrm{a}+\alpha ,0⩽\alpha <1$$$$\lfloor \mathrm{x}\rfloor =\mathrm{b}\in \mathbb{Z}\Rightarrow \mathrm{x}=\mathrm{b}+\beta ,0⩽\beta <1$$$$$$$$\mathrm{a}+{\mathrm{b}}^2=\mathrm{a}+\alpha +\mathrm{b}+\beta +1$$$$\Rightarrow {\mathrm{b}}^2-\mathrm{b}=\alpha +\beta +1\Rightarrow 1⩽{\mathrm{b}}^2-\mathrm{b}<3$$$$\Rightarrow {\mathrm{b}}^2-\mathrm{b}\in \{1,2\}\Rightarrow {\mathrm{b}}^2-\mathrm{b}=2\Rightarrow \mathrm{b}=2\mathrm{or}\mathrm{b}=-1$$$$1\mathrm{case}:\mathrm{b}=2$$$$\lfloor \mathrm{x}\rfloor =2,\mathrm{x}=2+\alpha \Rightarrow {\mathrm{x}}^2=4+4\alpha +{\alpha }^2\Rightarrow 4⩽{\mathrm{x}}^2<9$$$${\mathrm{x}}^2\in [4,5)\therefore \lfloor {\mathrm{x}}^2\rfloor =4$$$$4+2^2={\mathrm{x}}^2+\mathrm{x}+1\Rightarrow {\mathrm{x}}^2+\mathrm{x}-7=0$$$$\Rightarrow \mathrm{x}=\frac{-1+\sqrt{29}}{2}$$$${\mathrm{x}}^2\in [5,6)\therefore \lfloor {\mathrm{x}}^2\rfloor =5$$$$5+2^2={\mathrm{x}}^2+\mathrm{x}+1\Rightarrow {\mathrm{x}}^2+\mathrm{x}-8=0$$$$\Rightarrow \mathrm{x}=\frac{-1+\sqrt{33}}{2}$$$${\mathrm{x}}^2\in [6,7)\therefore \lfloor {\mathrm{x}}^2\rfloor =6$$$$6+2^2={\mathrm{x}}^2+\mathrm{x}+1\Rightarrow {\mathrm{x}}^2+\mathrm{x}-9=0$$$$\mathrm{x}=\frac{-1+\sqrt{37}}{2}$$$${\mathrm{x}}^2\in [7,8)\therefore \lfloor {\mathrm{x}}^2\rfloor =7$$$$7+2^2={\mathrm{x}}^2+\mathrm{x}+1\Rightarrow {\mathrm{x}}^2+\mathrm{x}-10=0$$$$\Rightarrow \mathrm{x}=\frac{-1+\sqrt{41}}{2}$$$${\mathrm{x}}^2\in [7,8)\therefore \lfloor {\mathrm{x}}^2\rfloor =8$$$$\Rightarrow \mathrm{x}=\frac{-1+\sqrt{45}}{2}$$$$2\mathrm{case}:\mathrm{b}=-1$$$$\lfloor \mathrm{x}\rfloor =-1,\mathrm{x}=-1+\alpha \Rightarrow {\mathrm{x}}^2=1-2\alpha +{\alpha }^2\Rightarrow 0<{\mathrm{x}}^2⩽1$$$${\mathrm{x}}^2\in (0,1)\therefore \lfloor {\mathrm{x}}^2\rfloor =0$$$$0+{(-1)}^2={\mathrm{x}}^2+\mathrm{x}+1\Rightarrow {\mathrm{x}}^2+\mathrm{x}=0$$$$\Rightarrow \mathrm{x}=-1\Rightarrow {\mathrm{x}}^2\notin (0,1)\left(\mathrm{dont}\mathrm{work}\right)$$$${\mathrm{x}}^2=1\therefore \lfloor {\mathrm{x}}^2\rfloor =1$$$$1+{(-1)}^2={\mathrm{x}}^2+\mathrm{x}+1\Rightarrow {\mathrm{x}}^2+\mathrm{x}-1=0$$$$\mathrm{x}=\frac{-1-\sqrt{5}}{2}⩽-1\left(\mathrm{dont}\mathrm{work}\right)$$$$$$$$\mathrm{solutions}:$$$$\mathrm{x}=\frac{-1+\sqrt{25+4\mathrm{n}}}{2},1⩽\mathrm{n}⩽5,\mathrm{n}\in \mathbb{N}$$$$$$

Answered by mr W last updated on 14/Jun/22

$$x=n+f$$$$\lfloor x{\rfloor }^2=n^2$$$$\lfloor x^2\rfloor =n^2+\lfloor (2n+1)f\rfloor$$$$x^2+x+1=n^2+n+1+(2n+1)f+f^2$$$$2n^2+\lfloor (2n+1)f\rfloor =n^2+n+1+(2n+1)f+f^2$$$$n^2-n-1=(2n+1)f+f^2-\lfloor (2n+1)f\rfloor$$$$n^2-n-1=\left\{(2n+1)f\right\}+f^2⩾0$$$$\Rightarrow n⩽-1orn⩾2$$$$n^2-n-1=\left\{(2n+1)f\right\}+f^2<2$$$$\Rightarrow -1⩽n⩽2$$$$\Rightarrow n=-1orn=2$$$$withn=-1,i.e.-1⩽x<0:$$$$1+1=x^2+x+1$$$$x^2+x-1=0$$$$\Rightarrow x=\frac{-1-\sqrt{5}}{2}<-1rejected.$$$$withn=2,i.e.2⩽x<3:$$$$\lfloor x^2\rfloor =kwith4⩽k⩽8$$$$k+4=x^2+x+1$$$$x^2+x-(k+3)=0$$$$\Rightarrow x=\frac{-1+\sqrt{13+4k}}{2}✓$$$$i.e.x=\frac{-1+\sqrt{29}}{2},\frac{-1+\sqrt{33}}{2},\frac{-1+\sqrt{37}}{2},\frac{-1+\sqrt{41}}{2},\frac{-1+\sqrt{45}}{2}$$

Commented by floor(10²Eta[1]) last updated on 14/Jun/22

$$\mathrm{if}\mathrm{x}=-1\Rightarrow {\mathrm{x}}^2=1$$$$\lfloor {\mathrm{x}}^2\rfloor +\lfloor \mathrm{x}{\rfloor }^2=2\ne {\mathrm{x}}^2+\mathrm{x}+1=1$$

Commented by mr W last updated on 14/Jun/22

$$ihaverejectedx=-1.$$

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