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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 ✓
(y−1)(y+1)(y2+1)=x(x+1)(x2+1)⇒y=x⇒(x2+1)(x+1)=0⇒x=y=−1✓
Commented by Rasheed.Sindhi last updated on 16/Feb/24
(0,1) & (0,−1) also satisfy.
(0,1)&(0,−1)alsosatisfy.
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).
x4<x4+x3+x2+x+1=(x2+1)(x2+x)<x4+4x3+6x2+4x+1=(x+1)4whenx>0.Ifx<−1,thenx4>x4+x3+x2+x+1>(x+1)4when0<x<−1,thenx4+x3+x2+x+1>x4>(x+1)4So,xisneveranintegerthenThisimpliesx4+x3+x2+x+1isalwaysin−betweentwoconsecutiveperfectfourthpowersx4and(x+1)4whenx>0orx<−1,soitcan′tbeafourthpowerunlessitisequaltooneofthem.[Oronecanconcludefromherethatthereareonlytwointegersleft{−1,0}forx.].x4+x3+x2+x+1=x4⇒x=−1⇒y4=1⇒y=+−1x4+x3+x2+x+1=(x+1)4⇒x=0⇒y=+−1⇒(x,y)=(−1,1),(−1,−1),(0,1),(0,−1).
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.
y4−1=x(x3+x2+x+1)(y−1)(y+1)(y2+1)=x(x+1))(x2+1)(±1)(y−1)(y+1)(y2+1)=(±1)x(x+1)(x2+1)∙y=1(±1)(x4+x3+x2+1)=0x(x3+x2+x+1)=0x=0(x,y)=(0,1)∙y=−1(±1)(x4+x3+x2+1)=0x(x3+x2+x+1)=0x=0(x,y)=(0,−1)∙x=1(y−1)(y+1)(y2+1)=(1+1)(12+1)y4−1=4(Nointegery)∙x=−1−(y4−1)=0y=±1(x,y)=(−1,−1),(−1,1)∙x=y−1(y+1)(y2+1)=(x+1)(x2+1)y3+y2+y+1=y(y2−2y+2)y3+y2+y+1=y3−2y2+2y3y2−y+1=0Nointegerroots.∙x=y+1(y−1)(y2+1)=(x+1)(x2+1)(y−1)(y2+1)=(y+1+1)(y2+2y+1+1)(y−1)(y2+1)=(y+2)(y2+2y+2)y3−y2+y−1=y3+2y2+2y+2y2+4y+4y3−y2+y−1=y3+4y2+6y+45y2+5y+5=0Nointegerroots.∙x=y2+1(y−1)(y+1)=(x+1)(x2+1)(y−1)(y+1)=(y2+1+1)((y2+1)2+1)y2−1=(y2+2)(y4+2y2+2)y2−1=y6+2y4+2y2+2y4+4y2+4y6+4y4+5y2+5=0Nointegerroots.

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