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Question Number 175639 by Linton last updated on 04/Sep/22

x^(99) +y^(99) = x^(100)   Interger solutions?

$${x}^{\mathrm{99}} +{y}^{\mathrm{99}} =\:{x}^{\mathrm{100}} \\ $$$${Interger}\:{solutions}?\: \\ $$

Answered by BaliramKumar last updated on 04/Sep/22

x^(99) [1 + ((y/x))^(99) ] = x^(100)    1 + ((y/x))^(99)  = x   ((y/x))^(99)  = x−1  (y/x) = ((x−1))^(1/(99))   y  = x(((x−1))^(1/(99)) )   (0, 0)       &    (1, 0)

$${x}^{\mathrm{99}} \left[\mathrm{1}\:+\:\left(\frac{{y}}{{x}}\right)^{\mathrm{99}} \right]\:=\:{x}^{\mathrm{100}} \\ $$$$\:\mathrm{1}\:+\:\left(\frac{{y}}{{x}}\right)^{\mathrm{99}} \:=\:{x} \\ $$$$\:\left(\frac{{y}}{{x}}\right)^{\mathrm{99}} \:=\:{x}−\mathrm{1} \\ $$$$\frac{{y}}{{x}}\:=\:\sqrt[{\mathrm{99}}]{{x}−\mathrm{1}} \\ $$$${y}\:\:=\:{x}\left(\sqrt[{\mathrm{99}}]{{x}−\mathrm{1}}\right) \\ $$$$\:\left(\mathrm{0},\:\mathrm{0}\right)\:\:\:\:\:\:\:\&\:\:\:\:\left(\mathrm{1},\:\mathrm{0}\right) \\ $$

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