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Question Number 80004 by jagoll last updated on 30/Jan/20

x a n d y a n y i n t e g e r s a t i s f y

e q u a t i o n (x - 2004) (x - 2006) = 2^{y}

t h e g r e a t e s t p o s s i b l e v a l u e

o f x + y

Commented by jagoll last updated on 30/Jan/20

t h a n k y o u m i s t e r

Commented by mr W last updated on 30/Jan/20

l e t u = x - 2004

u (u - 2) = 2^{y}

u^{2} - 2 u + 1 = 2^{y} + 1

{(u - 1)}^{2} = (2^{y} + 1)

u = 1 \pm \sqrt{2^{y} + 1}

2^{y} + 1 = {(n + 1)}^{2} s a y

2^{y} = {(n + 1)}^{2} - 1 = n (n + 2)

\Rightarrow n = 2^{p}, n + 2 = 2^{q} \Rightarrow (2^{p - 1} + 1) = 2^{q - 1}

\Rightarrow p = 1 \Rightarrow n = 2 \Rightarrow 2^{y} = 2 \times 4 = 8 \Rightarrow y = 3

\Rightarrow u = x - 2004 = 1 \pm 3

\Rightarrow x = 2008 o r 2002

{(x + y)}_{m a x} = 2008 + 3 = 2011

Answered by mind is power last updated on 30/Jan/20

(x - 2004) (x - 2006) = 2^{a + b}

\Rightarrow x - 2004 = 2^{a}

y - 2006 = 2^{b}

\Rightarrow 2^{a} - 2^{b} = 2^{b} (2^{a - b} - 1) = 2 \Rightarrow b = 1 \Rightarrow a - b = 1 \Rightarrow a = 2

\Rightarrow y = a + b = 3

x = 2^{2} + 2004 = 2008

x + y = 2008 + 3 = 2011 i t s e x a m e s o f 2011 ?

Commented by jagoll last updated on 30/Jan/20

i f x = 2010

(x - 2004) (x - 2006) = 6 \times 4 = 24 \neq 2^{y}

s i r

Commented by mind is power last updated on 30/Jan/20

m i s t a c k s o o r y 2^{y}, y \neq 2^{y}