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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

$$\boldsymbol{\alpha}_{\mathrm{1}} <\boldsymbol{\alpha}_{\mathrm{2}} <\boldsymbol{\alpha}_{\mathrm{3}} <\ldots<\boldsymbol{\alpha}_{{k}} \\ $$ $$\frac{\mathrm{2}^{\mathrm{289}} +\mathrm{1}}{\mathrm{2}^{\mathrm{17}} +\mathrm{1}}=\mathrm{2}^{\boldsymbol{\alpha}_{\mathrm{1}} } +\mathrm{2}^{\boldsymbol{\alpha}_{\mathrm{2}} } +\ldots+\mathrm{2}^{\boldsymbol{\alpha}_{{k}} } \:\:\:\:\:\:\:\boldsymbol{\mathrm{k}}=? \\ $$ $$ \\ $$ $$\boldsymbol{\alpha}_{\mathrm{1}} ,\:\boldsymbol{\alpha}_{\mathrm{2}} ,\boldsymbol{\alpha}_{\mathrm{3}} ....\boldsymbol{\alpha}_{{k}} \\ $$ positive increasing integers\\n\\n

Commented bymr W last updated on 30/Dec/21

check the question again! if α_1 ,...,α_k are all positive integers, then no solution! since LHS=odd and RHD=even. therefore α_1 ,...,α_k should only be non−negative integers, i.e. zero is also allowed.

$${check}\:{the}\:{question}\:{again}! \\ $$ $${if}\:\alpha_{\mathrm{1}} ,...,\alpha_{{k}} \:{are}\:{all}\:{\color{mathred}{p}\color{mathred}{o}\color{mathred}{s}\color{mathred}{i}\color{mathred}{t}\color{mathred}{i}\color{mathred}{v}\color{mathred}{e}}\color{mathred}{\:}{integers}, \\ $$ $${then}\:{no}\:{solution}!\:{since}\:{LHS}={odd} \\ $$ $${and}\:{RHD}={even}. \\ $$ $${therefore}\:\alpha_{\mathrm{1}} ,...,\alpha_{{k}} \:{should}\:{only}\:{be} \\ $$ $${\color{mathred}{n}\color{mathred}{o}\color{mathred}{n}}\color{mathred}{−}{\color{mathred}{n}\color{mathred}{e}\color{mathred}{g}\color{mathred}{a}\color{mathred}{t}\color{mathred}{i}\color{mathred}{v}\color{mathred}{e}}\:{integers},\:{i}.{e}.\:{zero}\:{is} \\ $$ $${also}\:{allowed}. \\ $$

Answered by mindispower last updated on 30/Dec/21

289=17^2 X^(17) +1=(X+1)(X^(16) −X^(15) +.............+1) ⇔X=2^(17) (((2^(17) )^(17) +1)/(2^(17) +1))=Σ_(k=0) ^(16) ((−1)^k 2^a_k ) a_k =17k,∀k∈[0,16]

$$\mathrm{289}=\mathrm{17}^{\mathrm{2}} \\ $$ $${X}^{\mathrm{17}} +\mathrm{1}=\left({X}+\mathrm{1}\right)\left({X}^{\mathrm{16}} −{X}^{\mathrm{15}} +.............+\mathrm{1}\right) \\ $$ $$\Leftrightarrow{X}=\mathrm{2}^{\mathrm{17}} \\ $$ $$\frac{\left(\mathrm{2}^{\mathrm{17}} \right)^{\mathrm{17}} +\mathrm{1}}{\mathrm{2}^{\mathrm{17}} +\mathrm{1}}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{16}} {\sum}}\left(\left(−\mathrm{1}\right)^{{k}} \mathrm{2}^{{a}_{{k}} } \right) \\ $$ $${a}_{{k}} =\mathrm{17}{k},\forall{k}\in\left[\mathrm{0},\mathrm{16}\right] \\ $$

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