Question Number 175639 by Linton last updated on 04/Sep/22
$${x}^{\mathrm{99}} +{y}^{\mathrm{99}} =\:{x}^{\mathrm{100}} \\ $$$${Interger}\:{solutions}?\: \\ $$
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)](https://www.tinkutara.com/question/Q175642.png)
$${x}^{\mathrm{99}} \left[\mathrm{1}\:+\:\left(\frac{{y}}{{x}}\right)^{\mathrm{99}} \right]\:=\:{x}^{\mathrm{100}} \\ $$$$\:\mathrm{1}\:+\:\left(\frac{{y}}{{x}}\right)^{\mathrm{99}} \:=\:{x} \\ $$$$\:\left(\frac{{y}}{{x}}\right)^{\mathrm{99}} \:=\:{x}−\mathrm{1} \\ $$$$\frac{{y}}{{x}}\:=\:\sqrt[{\mathrm{99}}]{{x}−\mathrm{1}} \\ $$$${y}\:\:=\:{x}\left(\sqrt[{\mathrm{99}}]{{x}−\mathrm{1}}\right) \\ $$$$\:\left(\mathrm{0},\:\mathrm{0}\right)\:\:\:\:\:\:\:\&\:\:\:\:\left(\mathrm{1},\:\mathrm{0}\right) \\ $$