x and y any integer satisfy equation (x−2004)(x−2006)=2^y the greatest possible value of x+y


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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

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