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-

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 AST 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} + \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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