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x-2-x-y-4-y-3-y-2-y-x-4-x-1-4-y-2-y-1-2-find-x-and-y-of-is-

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Question Number 43391 by gunawan last updated on 10/Sep/18
x^2 +x=y^4 +y^3 +y^2 +y x^4 +(x+1)^4 =y^2 +(y+1)^2 find x and y of is…
$${x}^{\mathrm{2}} +{x}={y}^{\mathrm{4}} +{y}^{\mathrm{3}} +{y}^{\mathrm{2}} +{y} \\ $$$${x}^{\mathrm{4}} +\left({x}+\mathrm{1}\right)^{\mathrm{4}} ={y}^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{find}\:{x}\:\mathrm{and}\:{y}\:\:\mathrm{of}\:\mathrm{is}\ldots \\ $$$$ \\ $$
Answered by ajfour last updated on 10/Sep/18
eq.(2) ⇒ 2x^4 +4x^3 +6x^2 +4x=2y^2 +2y ⇒ x^4 +2x^3 +3x^2 +2x=y(y+1) ⇒ x(x+1)(x^2 +1)+x(x+1)^2 =y(y+1) ....(i) eq.(1) ⇒ x(x+1)=y(y+1)(y^2 +1) ....(ii) if (x,y) ≠ (0,0), (−1,0), (0,−1) and (−1,−1) then (ii) ÷(i): x^2 +1 +x+1 = (1/(y^2 +1)) ⇒ x(x+1)+2 = (1/(y^2 +1)) substituting in (ii): (1/(y^2 +1))−2 = y(y+1)(y^2 +1) ⇒ [y(y+1)+2](y^2 +1)^2 = 1 ⇒ (y^2 +y+2)(y^4 +2y^2 +1)=1 ⇒ y^6 +y^5 +4y^4 +2y^3 +5y^2 +y+1=0 No real solutions to the above eq. Hence (x,y) ∈ [ (0,0), (0,−1), (−1,0), (−1,−1) ] .
$${eq}.\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\mathrm{2}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}=\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{y} \\ $$$$\Rightarrow\:{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}={y}\left({y}+\mathrm{1}\right) \\ $$$$\Rightarrow\:\:{x}\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} ={y}\left({y}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$${eq}.\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${x}\left({x}+\mathrm{1}\right)={y}\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)\:\:\:\:….\left({ii}\right) \\ $$$${if}\:\:\left({x},{y}\right)\:\neq\:\left(\mathrm{0},\mathrm{0}\right),\:\left(−\mathrm{1},\mathrm{0}\right),\:\left(\mathrm{0},−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{and}\:\:\left(−\mathrm{1},−\mathrm{1}\right)\:\:{then} \\ $$$$\:\left({ii}\right)\:\boldsymbol{\div}\left({i}\right):\:\:\:\: \\ $$$${x}^{\mathrm{2}} +\mathrm{1}\:+{x}+\mathrm{1}\:=\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\:\:{x}\left({x}+\mathrm{1}\right)+\mathrm{2}\:=\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$${substituting}\:{in}\:\left({ii}\right): \\ $$$$\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}−\mathrm{2}\:=\:{y}\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\left[{y}\left({y}+\mathrm{1}\right)+\mathrm{2}\right]\left({y}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{1} \\ $$$$\Rightarrow\:\:\left({y}^{\mathrm{2}} +{y}+\mathrm{2}\right)\left({y}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow\:\:{y}^{\mathrm{6}} +{y}^{\mathrm{5}} +\mathrm{4}{y}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{3}} +\mathrm{5}{y}^{\mathrm{2}} +{y}+\mathrm{1}=\mathrm{0} \\ $$$${No}\:{real}\:{solutions}\:{to}\:{the}\:{above}\:{eq}. \\ $$$${Hence}\:\:\:\left({x},{y}\right)\:\in\:\left[\:\left(\mathrm{0},\mathrm{0}\right),\:\left(\mathrm{0},−\mathrm{1}\right),\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(−\mathrm{1},\mathrm{0}\right),\:\left(−\mathrm{1},−\mathrm{1}\right)\:\right]\:. \\ $$

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