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Question Number 204397 by Abdullahrussell last updated on 16/Feb/24

	Answered by MM42 last updated on 16/Feb/24

	$(y - 1) (y + 1) (y^{2} + 1) = x (x + 1) (x^{2} + 1)$ $\Rightarrow y = x \Rightarrow (x^{2} + 1) (x + 1) = 0$ $\Rightarrow x = y = - 1 ✓$
	Commented by Rasheed.Sindhi last updated on 16/Feb/24

	$(0, 1) & (0, - 1) a l s o s a t i s f y .$
	Answered by deleteduser1 last updated on 16/Feb/24
	$x^4 <x^4 +x^3 +x^2 +x+1=(x^2 +1)(x^2 +x) <x^4 +4x^3 +6x^2 +4x+1=(x+1)^4 when x>0. If x<−1, then x^4 >x^4 +x^3 +x^2 +x+1 >(x+1)^4 when 0<x<−1,then x^4 +x^3 +x^2 +x+1>x^4 >(x+1)^4 So,x is never an integer then This implies x^4 +x^3 +x^2 +x+1 is always in-between two consecutive perfect fourth powers x^4 and (x+1)^4 when x>0 or x<−1, so it can′t be a fourth power unless it is equal to one of them. [Or one can conclude from here that there are only two integers left {−1,0} for x.]. x^4 +x^3 +x^2 +x+1=x^4 ⇒x=−1⇒y^4 =1⇒y=+_− 1 x^4 +x^3 +x^2 +x+1=(x+1)^4 ⇒x=0⇒y=+_− 1 ⇒(x,y)=(−1,1),(−1,−1),(0,1),(0,−1).$
	$x^{4} < x^{4} + x^{3} + x^{2} + x + 1 = (x^{2} + 1) (x^{2} + x)$ $< x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 = {(x + 1)}^{4}$ $w h e n x > 0 . I f x < - 1, t h e n x^{4} > x^{4} + x^{3} + x^{2} + x + 1$ $> {(x + 1)}^{4}$ $w h e n 0 < x < - 1, t h e n x^{4} + x^{3} + x^{2} + x + 1 > x^{4} > {(x + 1)}^{4}$ $S o, x i s n e v e r a n i n t e g e r t h e n$ $T h i s i m p l i e s x^{4} + x^{3} + x^{2} + x + 1$ $i s a l w a y s i n - b e t w e e n t w o c o n s e c u t i v e p e r f e c t$ $f o u r t h p o w e r s x^{4} a n d {(x + 1)}^{4} w h e n x > 0 o r x < - 1,$ $s o i t {c a n}^{'} t b e a f o u r t h p o w e r u n l e s s i t i s e q u a l t o$ $o n e o f t h e m . [O r o n e c a n c o n c l u d e f r o m h e r e$ $t h a t t h e r e a r e o n l y t w o i n t e g e r s l e f t {- 1, 0}$ $f o r x .] .$ $x^{4} + x^{3} + x^{2} + x + 1 = x^{4} \Rightarrow x = - 1 \Rightarrow y^{4} = 1 \Rightarrow y = \underline{+} 1$ $x^{4} + x^{3} + x^{2} + x + 1 = {(x + 1)}^{4} \Rightarrow x = 0 \Rightarrow y = \underline{+} 1$ $\Rightarrow (x, y) = (- 1, 1), (- 1, - 1), (0, 1), (0, - 1) .$
	Answered by Rasheed.Sindhi last updated on 16/Feb/24

	$y^{4} - 1 = x (x^{3} + x^{2} + x + 1)$ $(y - 1) (y + 1) (y^{2} + 1) = x (x + 1)) (x^{2} + 1)$ $(\pm 1) (y - 1) (y + 1) (y^{2} + 1) = (\pm 1) x (x + 1) (x^{2} + 1)$ $∙ y = 1$ $(\pm 1) (x^{4} + x^{3} + x^{2} + 1) = 0$ $x (x^{3} + x^{2} + x + 1) = 0$ $x = 0$ $(x, y) = (0, 1)$ $∙ y = - 1$ $(\pm 1) (x^{4} + x^{3} + x^{2} + 1) = 0$ $x (x^{3} + x^{2} + x + 1) = 0$ $x = 0$ $(x, y) = (0, - 1)$ $∙ x = 1$ $(y - 1) (y + 1) (y^{2} + 1) = (1 + 1) (1^{2} + 1)$ $y^{4} - 1 = 4 (N o i n t e g e r y)$ $∙ x = - 1$ $- (y^{4} - 1) = 0$ $y = \pm 1$ $(x, y) = (- 1, - 1), (- 1, 1)$ $∙ x = y - 1$ $(y + 1) (y^{2} + 1) = (x + 1) (x^{2} + 1)$ $y^{3} + y^{2} + y + 1 = y (y^{2} - 2 y + 2)$ $y^{3} + y^{2} + y + 1 = y^{3} - 2 y^{2} + 2 y$ $3 y^{2} - y + 1 = 0$ $N o i n t e g e r r o o t s .$ $∙ x = y + 1$ $(y - 1) (y^{2} + 1) = (x + 1) (x^{2} + 1)$ $(y - 1) (y^{2} + 1) = (y + 1 + 1) (y^{2} + 2 y + 1 + 1)$ $(y - 1) (y^{2} + 1) = (y + 2) (y^{2} + 2 y + 2)$ $y^{3} - y^{2} + y - 1 = y^{3} + 2 y^{2} + 2 y + 2 y^{2} + 4 y + 4$ $y^{3} - y^{2} + y - 1 = y^{3} + 4 y^{2} + 6 y + 4$ $5 y^{2} + 5 y + 5 = 0$ $N o i n t e g e r r o o t s .$ $∙ x = y^{2} + 1$ $(y - 1) (y + 1) = (x + 1) (x^{2} + 1)$ $(y - 1) (y + 1) = (y^{2} + 1 + 1) ({(y^{2} + 1)}^{2} + 1)$ $y^{2} - 1 = (y^{2} + 2) (y^{4} + 2 y^{2} + 2)$ $y^{2} - 1 = y^{6} + 2 y^{4} + 2 y^{2} + 2 y^{4} + 4 y^{2} + 4$ $y^{6} + 4 y^{4} + 5 y^{2} + 5 = 0$ $N o i n t e g e r r o o t s .$
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