Page:Annales de mathématiques pures et appliquées, 1830-1831, Tome 21.djvu/117

ce qui donnera (9) et (11),

${\begin{aligned}&x=1,\qquad \qquad \qquad x=-{\frac {1}{2}}\left(1\pm {\sqrt {3}}.{\sqrt {-1}}\right),\\&x=-{\frac {1}{4}}\left\{\left(1+{\sqrt {5}}\right)\pm {\sqrt {2\left(5-{\sqrt {5}}\right)}}\right\},\\\\&x=-{\frac {1}{4}}\left\{\left(1-{\sqrt {5}}\right)\pm {\sqrt {2\left(5+{\sqrt {5}}\right)}}\right\},\end{aligned}}$

${\begin{aligned}&x={\frac {1}{8}}\left\{\left[1+{\sqrt {5}}+{\sqrt {6\left(5-{\sqrt {5}}\right)}}\right]\pm \left[\left(1+{\sqrt {5}}\right){\sqrt {3}}-{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]{\sqrt {-1}}\right\},\\\\&x={\frac {1}{8}}\left\{\left[1+{\sqrt {5}}-{\sqrt {6\left(5-{\sqrt {5}}\right)}}\right]\pm \left[\left(1+{\sqrt {5}}\right){\sqrt {3}}+{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]{\sqrt {-1}}\right\},\\\\&x={\frac {1}{8}}\left\{\left[1-{\sqrt {5}}+{\sqrt {6\left(5+{\sqrt {5}}\right)}}\right]\pm \left[\left(1-{\sqrt {5}}\right){\sqrt {3}}-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]{\sqrt {-1}}\right\},\\\\&x={\frac {1}{8}}\left\{\left[1-{\sqrt {5}}-{\sqrt {6\left(5+{\sqrt {5}}\right)}}\right]\pm \left[\left(1-{\sqrt {5}}\right){\sqrt {3}}+{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]{\sqrt {-1}}\right\}.\end{aligned}}$

Quant aux racines de l’autre équation, comme elles ne différent (4) de celles-ci que par le signe, nous nous dispenserons de les écrire.

17. Soit encore $m=16$ ; les équations à résoudre seront

$x^{16}-1=0,\qquad x^{16}+1=0\,;$

la première revient à

$\left(x^{8}-1\right)\left(x^{8}+1\right)=0,$

dont les racines sont (13)

${\begin{array}{ll}x=\pm 1,&\ \ x=+{\frac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\pm {\sqrt {2-{\sqrt {2}}}}.{\sqrt {-1}}\right),\\x=\pm {\sqrt {-1}},&\ \ x=-{\frac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\pm {\sqrt {2-{\sqrt {2}}}}.{\sqrt {-1}}\right),\\x=+{\frac {1}{2}}\left({\sqrt {2}}\pm {\sqrt {2}}.{\sqrt {-1}}\right),&\ \ x=+{\frac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\pm {\sqrt {2+{\sqrt {2}}}}.{\sqrt {-1}}\right),\\x=-{\frac {1}{2}}\left({\sqrt {2}}\pm {\sqrt {2}}.{\sqrt {-1}}\right),&\ \ x=-{\frac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\pm {\sqrt {2+{\sqrt {2}}}}.{\sqrt {-1}}\right),\end{array}}$