𝛂_1 <𝛂_2 <𝛂_3 <…<𝛂_k ((2^(289) +1)/(2^(17) +1))=2^𝛂_1 +2^𝛂_2 +…+2^𝛂_k k=? 𝛂_1 , 𝛂_2 ,𝛂_3 ....𝛂


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Question Number 162552 by amin96 last updated on 30/Dec/21

$α_{1} < α_{2} < α_{3} < \dots < α_{k}$ $\frac{2^{289} + 1}{2^{17} + 1} = 2^{α_{1}} + 2^{α_{2}} + \dots + 2^{α_{k}} k = ?$ $α_{1}, α_{2}, α_{3} . . . . α_{k}$ positive increasing integers\n\n
Commented bymr W last updated on 30/Dec/21

$c h e c k t h e q u e s t i o n a g a i n!$ $i f α_{1}, . . ., α_{k} a r e a l l p o s i t i v e i n t e g e r s,$ $t h e n n o s o l u t i o n! s i n c e L H S = o d d$ $a n d R H D = e v e n .$ $t h e r e f o r e α_{1}, . . ., α_{k} s h o u l d o n l y b e$ $n o n - n e g a t i v e i n t e g e r s, i . e . z e r o i s$ $a l s o a l l o w e d .$
	Answered by mindispower last updated on 30/Dec/21

	$289 = 17^{2}$ $X^{17} + 1 = (X + 1) (X^{16} - X^{15} + . . . . . . . . . . . . . + 1)$ $\Leftrightarrow X = 2^{17}$ $\frac{{(2^{17})}^{17} + 1}{2^{17} + 1} = \underset{k = 0}{\sum^{16}} ({(- 1)}^{k} 2^{a_{k}})$ $a_{k} = 17 k, \forall k \in [0, 16]$
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