Menu Close

1-lt-2-lt-3-lt-lt-k-2-289-1-2-17-1-2-1-2-2-2-k-k-1-2-3-k-positive-increasing-integers-




Question Number 162552 by amin96 last updated on 30/Dec/21
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_k   positive increasing integers
$$\boldsymbol{\alpha}_{\mathrm{1}} <\boldsymbol{\alpha}_{\mathrm{2}} <\boldsymbol{\alpha}_{\mathrm{3}} <\ldots<\boldsymbol{\alpha}_{{k}} \\ $$$$\frac{\mathrm{2}^{\mathrm{289}} +\mathrm{1}}{\mathrm{2}^{\mathrm{17}} +\mathrm{1}}=\mathrm{2}^{\boldsymbol{\alpha}_{\mathrm{1}} } +\mathrm{2}^{\boldsymbol{\alpha}_{\mathrm{2}} } +\ldots+\mathrm{2}^{\boldsymbol{\alpha}_{{k}} } \:\:\:\:\:\:\:\boldsymbol{\mathrm{k}}=? \\ $$$$ \\ $$$$\boldsymbol{\alpha}_{\mathrm{1}} ,\:\boldsymbol{\alpha}_{\mathrm{2}} ,\boldsymbol{\alpha}_{\mathrm{3}} ….\boldsymbol{\alpha}_{{k}} \\ $$positive increasing integers

Commented by mr W last updated on 30/Dec/21
check the question again!  if Ξ±_1 ,...,Ξ±_k  are all positive integers,  then no solution! since LHS=odd  and RHD=even.  therefore Ξ±_1 ,...,Ξ±_k  should only be  nonβˆ’negative integers, i.e. zero is  also allowed.
$${check}\:{the}\:{question}\:{again}! \\ $$$${if}\:\alpha_{\mathrm{1}} ,…,\alpha_{{k}} \:{are}\:{all}\:{positive}\:{integers}, \\ $$$${then}\:{no}\:{solution}!\:{since}\:{LHS}={odd} \\ $$$${and}\:{RHD}={even}. \\ $$$${therefore}\:\alpha_{\mathrm{1}} ,…,\alpha_{{k}} \:{should}\:{only}\:{be} \\ $$$${non}βˆ’{negative}\:{integers},\:{i}.{e}.\:{zero}\:{is} \\ $$$${also}\:{allowed}. \\ $$
Answered by mindispower last updated on 30/Dec/21
289=17^2   X^(17) +1=(X+1)(X^(16) βˆ’X^(15) +.............+1)  ⇔X=2^(17)   (((2^(17) )^(17) +1)/(2^(17) +1))=Ξ£_(k=0) ^(16) ((βˆ’1)^k 2^a_k  )  a_k =17k,βˆ€k∈[0,16]
$$\mathrm{289}=\mathrm{17}^{\mathrm{2}} \\ $$$${X}^{\mathrm{17}} +\mathrm{1}=\left({X}+\mathrm{1}\right)\left({X}^{\mathrm{16}} βˆ’{X}^{\mathrm{15}} +………….+\mathrm{1}\right) \\ $$$$\Leftrightarrow{X}=\mathrm{2}^{\mathrm{17}} \\ $$$$\frac{\left(\mathrm{2}^{\mathrm{17}} \right)^{\mathrm{17}} +\mathrm{1}}{\mathrm{2}^{\mathrm{17}} +\mathrm{1}}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{16}} {\sum}}\left(\left(βˆ’\mathrm{1}\right)^{{k}} \mathrm{2}^{{a}_{{k}} } \right) \\ $$$${a}_{{k}} =\mathrm{17}{k},\forall{k}\in\left[\mathrm{0},\mathrm{16}\right] \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *