x^4 +x^3 +x^2 +x+1=y^2 where y is positive integer number then find the positive integal values of (x) for which that holds


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Question Number 192793 by York12 last updated on 27/May/23

$x^{4} + x^{3} + x^{2} + x + 1 = y^{2} w h e r e y i s p o s i t i v e i n t e g e r n u m b e r$ $t h e n f i n d t h e p o s i t i v e i n t e g a l v a l u e s o f (x)$ $f o r w h i c h t h a t h o l d s$
	Answered by deleteduser1 last updated on 27/May/23

	$4 (x^{4} + x^{3} + x^{2} + x + 1) = {(2 y)}^{2}$ ${(2 x^{2} + x)}^{2} < {(2 y)}^{2} = 4 y^{2} < {(2 x^{2} + x + 1)}^{2} w h e n x > 3$ $\Rightarrow 4 y^{2} c a n n o t b e a p e r f e c t s q u a r e w h e n x > 3$ $T h i s i m p l i e s y^{2} c a n n o t a l s o b e$ $S o, w e c h e c k x ⩽ 3$ $\Rightarrow (x, y) = (3, 11)$
	Commented by York12 last updated on 27/May/23
	$(x^2 +(x/2))^2 <(x^4 +x^3 +x^2 +x+1)<(x^2 +(x/2)+1)^2 for x ∈Even numbers then it is impossible to find a perfect square number Between (x^2 +(x/2))^2 & (x^2 +(x/2)+1)^2 becsuse They are consecuative integers. Then x ∈ odd numbers → (x^2 +(x/2))^2 & (x^2 +(x/2)+1)^2 are not perfect squares but there one perfect square between them : (x^2 +(x/2)+(1/2))^2 and since : (x^2 +(x/2))^2 <(x^4 +x^3 +x^2 +x+1)<(x^2 +(x/2)+1)^2 ∴ x^4 +x^3 +x^2 +x+1=(x^2 +(x/2)+(1/2))^2 x^4 +x^3 +x^2 +x+1=x^4 +x^3 +x^2 +(x^2 /4)+(x/2)+(1/4) x^2 −2x−3=0 → x ∈ {3 ,−1} ∴ x = 3 ( That ′s it )$
	${(x^{2} + \frac{x}{2})}^{2} < (x^{4} + x^{3} + x^{2} + x + 1) < {(x^{2} + \frac{x}{2} + 1)}^{2}$ $f o r x \in E v e n n u m b e r s$ $t h e n i t i s i m p o s s i b l e t o f i n d a p e r f e c t s q u a r e n u m b e r$ $B e t w e e n {(x^{2} + \frac{x}{2})}^{2} & {(x^{2} + \frac{x}{2} + 1)}^{2} b e c s u s e$ $T h e y a r e c o n s e c u a t i v e i n t e g e r s .$ $T h e n x \in o d d n u m b e r s \to {(x^{2} + \frac{x}{2})}^{2} & {(x^{2} + \frac{x}{2} + 1)}^{2}$ $a r e n o t p e r f e c t s q u a r e s b u t t h e r e o n e p e r f e c t s q u a r e$ $b e t w e e n t h e m :$ ${(x^{2} + \frac{x}{2} + \frac{1}{2})}^{2}$ $a n d s i n c e :$ ${(x^{2} + \frac{x}{2})}^{2} < (x^{4} + x^{3} + x^{2} + x + 1) < {(x^{2} + \frac{x}{2} + 1)}^{2}$ $∴ x^{4} + x^{3} + x^{2} + x + 1 = {(x^{2} + \frac{x}{2} + \frac{1}{2})}^{2}$ $x^{4} + x^{3} + x^{2} + x + 1 = x^{4} + x^{3} + x^{2} + \frac{x^{2}}{4} + \frac{x}{2} + \frac{1}{4}$ $x^{2} - 2 x - 3 = 0 \to x \in {3, - 1}$ $∴ x = 3 (T h a t^{'} s i t)$
	Answered by MM42 last updated on 28/May/23

	$x (x + 1) (x^{2} + 1) = (y - 1) (y + 1) *$ $t h e l r f t s i d e i s a l w a y s e v e n, s o t h e r i g h t s i d e$ $m u s t a l s o b e e v e n .$ $(y - 1) (y + 1), t w o e v e n n u m b e r s a r e c o n s e c u t i v e$ $w e c a n a r r a n g t h e l e f t s i d e o f t h e a b o v e r e l a t i o n s h i p o f t h e f o l l o w s$ $(x^{2} + x) (x^{2} + 1) o r (x + 1) (x^{3} + x) o r x (x^{3} + x^{2} + x + 1)$ $c a s e 1$ $∣ x^{2} + x - x^{2} - 1 ∣= 2 \Rightarrow x = - 1 \times & x = 3 \Rightarrow y = 11 ✓$ $c a s e 2$ $∣ x^{3} + x - x - 1 ∣= 2 \Rightarrow x = - 1 & \sqrt[3]{3} \times$ $c a s e 3$ $∣ x^{3} + x^{2} + x + 1 - x ∣= 2 \Rightarrow∣ x^{3} + x^{2} + 1 ∣= 2 \times (b e c a u s e x > 0)$ $t h e r e f o r e, t h e o n l y a n s w e r i s; x = 3 & y = 11$
	Commented by York12 last updated on 27/May/23

	$y o u c a n s u b t r a c t 2 a l s o a n d y o u w i l l c o m e u p$ $w i t h t h e s a m e r e s u l t s$
	Commented by MM42 last updated on 27/May/23

	$t h a n k y o u$
	Commented by York12 last updated on 28/May/23

	$y o u a r e w e l c o m e s i r$
	Commented by MM42 last updated on 28/May/23

	$I a m a r e t i r e d t e a c h e r . a n d i a m$ $v e r y h a p p y t o b e m e m b e r o f t h i s c o m m u n i t y$ $g o o d l u c k$
	Commented by York12 last updated on 28/May/23

	$N i c e t o m e e t y o u s i r$ $I a m a j u n i o r h i g h s c h o l a r i n g r a d e 10$ $a n d I a m p r e p a r i n g f o r T h e I M O$
	Commented by MM42 last updated on 28/May/23

	$i h p o e y o u h a v e e g o o d f u t u r e$
	Commented by York12 last updated on 28/May/23

	$t h a n k s r e a l l y a p p r e c i a t e i t$
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