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Question Number 100951 by bemath last updated on 29/Jun/20

Commented by Rasheed.Sindhi last updated on 29/Jun/20

$B y j o h n s a n t u (a n s w e r o f$ $You can't use 'macro parameter character #' in math mode$ $2^{x} (1 + 2^{x + 1}) = (y - 1) (y + 1)$ $s h o w t h a t t h e f a c t o r s y - 1 a n d y + 1$ $a r e e v e n, exactly one of them$ $divisible by 4 . Hence x ⩾ 3 and one$ $of these factors is divisible by 2^{x - 1}$ $but not by 2^{x} . so y = 2^{x - 1} m + ϵ, m odd, ϵ = \pm 1$ $plugging this into the original$ $equation we obtain 2^{x} (1 + 2^{x + 1}) = {(2^{x - 1} m + ϵ)}^{2} - 1$ $= 2^{2 x - 2} m^{2} + 2^{x} m ϵ$ $or equivalently 1 + 2^{x + 1} = 2^{x - 2} m^{2} + m ϵ$ $\Leftrightarrow thus we have the complete$ $list solutions (x, y) : (0, 2), (0, - 2),$ $(4, 23), (4, - 23) .$
Commented by Rasheed.Sindhi last updated on 30/Jun/20

$S i r b e m a t h,$ $Y o u r s o l u t i o n (w h i c h y o u h a v e$ $g i v e n a s a q u e s t i o n) i s s o l u t i o n$ $You can't use 'macro parameter character #' in math mode$ $a n s w e r e d t h e r e b y s i r j o h n s a n t u$ $a n d y o u r t h i s s o l u t i o n i s s a m e$ $a s h i s . W h y {y o u}^{'} v e r e p e a t e d i t$ $a n d g a v e a n s w e r w i t h o u t q u e s t i o n!!!$ $(O n l y t o s a t i s f y m y c u r o s i t y) :)$
Commented by 1549442205 last updated on 01/Jul/20

$I guess perhaps he want everyone know above$ $his question with a good solution for it$
Commented by Rasheed.Sindhi last updated on 01/Jul/20

$M a y b e .$
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