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Question-204397

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Question Number 204397 by Abdullahrussell last updated on 16/Feb/24
Answered by MM42 last updated on 16/Feb/24
(y−1)(y+1)(y^2 +1)=x(x+1)(x^2 +1) ⇒y=x⇒(x^2 +1)(x+1)=0 ⇒x=y=−1 ✓
$$\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)={x}\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow{y}={x}\Rightarrow\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}={y}=−\mathrm{1}\:\checkmark \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 16/Feb/24
(0,1) & (0,−1) also satisfy.
$$\left(\mathrm{0},\mathrm{1}\right)\:\&\:\left(\mathrm{0},−\mathrm{1}\right)\:{also}\:{satisfy}. \\ $$
Answered by AST last updated on 16/Feb/24
x^4 <x^4 +x^3 +x^2 +x+1=(x^2 +1)(x^2 +x) <x^4 +4x^3 +6x^2 +4x+1=(x+1)^4 when x>0. If x<−1, then x^4 >x^4 +x^3 +x^2 +x+1 >(x+1)^4 when 0<x<−1,then x^4 +x^3 +x^2 +x+1>x^4 >(x+1)^4 So,x is never an integer then This implies x^4 +x^3 +x^2 +x+1 is always in-between two consecutive perfect fourth powers x^4 and (x+1)^4 when x>0 or x<−1, so it can′t be a fourth power unless it is equal to one of them. [Or one can conclude from here that there are only two integers left {−1,0} for x.]. x^4 +x^3 +x^2 +x+1=x^4 ⇒x=−1⇒y^4 =1⇒y=+_− 1 x^4 +x^3 +x^2 +x+1=(x+1)^4 ⇒x=0⇒y=+_− 1 ⇒(x,y)=(−1,1),(−1,−1),(0,1),(0,−1).
$${x}^{\mathrm{4}} <{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}\right) \\ $$$$<{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=\left({x}+\mathrm{1}\right)^{\mathrm{4}} \\ $$$${when}\:{x}>\mathrm{0}.\:{If}\:{x}<−\mathrm{1},\:{then}\:{x}^{\mathrm{4}} >{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$$>\left({x}+\mathrm{1}\right)^{\mathrm{4}} \\ $$$${when}\:\mathrm{0}<{x}<−\mathrm{1},{then}\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}>{x}^{\mathrm{4}} >\left({x}+\mathrm{1}\right)^{\mathrm{4}} \\ $$$${So},{x}\:{is}\:{never}\:{an}\:{integer}\:{then} \\ $$$${This}\:{implies}\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$${is}\:{always}\:{in}-{between}\:{two}\:{consecutive}\:{perfect} \\ $$$${fourth}\:{powers}\:{x}^{\mathrm{4}} \:{and}\:\left({x}+\mathrm{1}\right)^{\mathrm{4}} \:{when}\:{x}>\mathrm{0}\:{or}\:{x}<−\mathrm{1},\: \\ $$$${so}\:{it}\:{can}'{t}\:{be}\:{a}\:{fourth}\:{power}\:{unless}\:{it}\:{is}\:{equal}\:{to}\: \\ $$$${one}\:{of}\:{them}.\:\left[{Or}\:{one}\:{can}\:{conclude}\:{from}\:{here}\right. \\ $$$${that}\:{there}\:{are}\:{only}\:{two}\:{integers}\:{left}\:\left\{−\mathrm{1},\mathrm{0}\right\}\: \\ $$$$\left.{for}\:{x}.\right]. \\ $$$$\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}={x}^{\mathrm{4}} \Rightarrow{x}=−\mathrm{1}\Rightarrow{y}^{\mathrm{4}} =\mathrm{1}\Rightarrow{y}=\underset{−} {+}\mathrm{1} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\left({x}+\mathrm{1}\right)^{\mathrm{4}} \Rightarrow{x}=\mathrm{0}\Rightarrow{y}=\underset{−} {+}\mathrm{1} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(−\mathrm{1},\mathrm{1}\right),\left(−\mathrm{1},−\mathrm{1}\right),\left(\mathrm{0},\mathrm{1}\right),\left(\mathrm{0},−\mathrm{1}\right). \\ $$
Answered by Rasheed.Sindhi last updated on 16/Feb/24
y^4 −1=x(x^3 +x^2 +x+1) (y−1)(y+1)(y^2 +1)=x(x+1))(x^2 +1) (±1)(y−1)(y+1)(y^2 +1)=(±1)x(x+1)(x^2 +1) •y=1 (±1)(x^4 +x^3 +x^2 +1)=0 x(x^3 +x^2 +x+1)=0 x=0 (x,y)=(0,1) •y=−1 (±1)(x^4 +x^3 +x^2 +1)=0 x(x^3 +x^2 +x+1)=0 x=0 (x,y)=(0,−1) •x=1 (y−1)(y+1)(y^2 +1)=(1+1)(1^2 +1) y^4 −1=4 (No integer y) •x=−1 −(y^4 −1)=0 y=±1 (x,y)=(−1,−1),(−1,1) •x=y−1 (y+1)(y^2 +1)=(x+1)(x^2 +1) y^3 +y^2 +y+1=y(y^2 −2y+2) y^3 +y^2 +y+1=y^3 −2y^2 +2y 3y^2 −y+1=0 No integer roots. •x=y+1 (y−1)(y^2 +1)=(x+1)(x^2 +1) (y−1)(y^2 +1)=(y+1+1)(y^2 +2y+1+1) (y−1)(y^2 +1)=(y+2)(y^2 +2y+2) y^3 −y^2 +y−1=y^3 +2y^2 +2y+2y^2 +4y+4 y^3 −y^2 +y−1=y^3 +4y^2 +6y+4 5y^2 +5y+5=0 No integer roots. •x=y^2 +1 (y−1)(y+1)=(x+1)(x^2 +1) (y−1)(y+1)=(y^2 +1+1)((y^2 +1)^2 +1) y^2 −1=(y^2 +2)(y^4 +2y^2 +2) y^2 −1=y^6 +2y^4 +2y^2 +2y^4 +4y^2 +4 y^6 +4y^4 +5y^2 +5=0 No integer roots.
$${y}^{\mathrm{4}} −\mathrm{1}={x}\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$$\left.\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)={x}\left({x}+\mathrm{1}\right)\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left(\pm\mathrm{1}\right)\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)=\left(\pm\mathrm{1}\right){x}\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\bullet{y}=\mathrm{1} \\ $$$$\left(\pm\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${x}\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0} \\ $$$$\left({x},{y}\right)=\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\bullet{y}=−\mathrm{1} \\ $$$$\left(\pm\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${x}\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0} \\ $$$$\left({x},{y}\right)=\left(\mathrm{0},−\mathrm{1}\right) \\ $$$$\bullet{x}=\mathrm{1} \\ $$$$\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)=\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{1}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${y}^{\mathrm{4}} −\mathrm{1}=\mathrm{4}\:\left({No}\:{integer}\:{y}\right) \\ $$$$\bullet{x}=−\mathrm{1} \\ $$$$−\left({y}^{\mathrm{4}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${y}=\pm\mathrm{1} \\ $$$$\left({x},{y}\right)=\left(−\mathrm{1},−\mathrm{1}\right),\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$\bullet{x}={y}−\mathrm{1} \\ $$$$\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${y}^{\mathrm{3}} +{y}^{\mathrm{2}} +{y}+\mathrm{1}={y}\left({y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{2}\right) \\ $$$${y}^{\mathrm{3}} +{y}^{\mathrm{2}} +{y}+\mathrm{1}={y}^{\mathrm{3}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{y} \\ $$$$\mathrm{3}{y}^{\mathrm{2}} −{y}+\mathrm{1}=\mathrm{0} \\ $$$${No}\:{integer}\:{roots}. \\ $$$$\bullet{x}={y}+\mathrm{1} \\ $$$$\left({y}−\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left({y}−\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)=\left({y}+\mathrm{1}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{1}+\mathrm{1}\right) \\ $$$$\left({y}−\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)=\left({y}+\mathrm{2}\right)\left({y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{2}\right) \\ $$$${y}^{\mathrm{3}} −{y}^{\mathrm{2}} +{y}−\mathrm{1}={y}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{4} \\ $$$${y}^{\mathrm{3}} −{y}^{\mathrm{2}} +{y}−\mathrm{1}={y}^{\mathrm{3}} +\mathrm{4}{y}^{\mathrm{2}} +\mathrm{6}{y}+\mathrm{4} \\ $$$$\mathrm{5}{y}^{\mathrm{2}} +\mathrm{5}{y}+\mathrm{5}=\mathrm{0} \\ $$$${No}\:{integer}\:{roots}. \\ $$$$\bullet{x}={y}^{\mathrm{2}} +\mathrm{1} \\ $$$$\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)=\left({y}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}\right)\left(\left({y}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${y}^{\mathrm{2}} −\mathrm{1}=\left({y}^{\mathrm{2}} +\mathrm{2}\right)\left({y}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}\right) \\ $$$${y}^{\mathrm{2}} −\mathrm{1}={y}^{\mathrm{6}} +\mathrm{2}{y}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{4}} +\mathrm{4}{y}^{\mathrm{2}} +\mathrm{4} \\ $$$${y}^{\mathrm{6}} +\mathrm{4}{y}^{\mathrm{4}} +\mathrm{5}{y}^{\mathrm{2}} +\mathrm{5}=\mathrm{0} \\ $$$${No}\:{integer}\:{roots}. \\ $$

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