Posons $P=X^4-4X^3-36X^2+\lambda X+\mu$ .

$\begin{align*}\ensuremath (\lambda,\mu)\;\mbox{solution}&\Leftrightarrow\exists(z,r)\in\Cc^2/\;\mbox{les racines de}\;P\;\mbox{soient}\;z,z+r,z+2r,z+3r\\ &\Leftrightarrow\exists(z,r)\in\Cc^2/\;\left\{ \begin{array}{l} \sigma_1=4\\ \sigma_2=-36\\ \sigma_3=-\lambda\\ \sigma_4=\mu \end{array} \right. \\ &\Leftrightarrow\exists(z,r)\in\Cc^2/\;\left\{ \begin{array}{l} 4z+6r=4\\ z(3z+6r)+(z+r)(2z+5r)+(z+2r)(z+3r)=-36\\ \sigma_3=-\lambda\\ \sigma_4=\mu \end{array} \right. \\ &\Leftrightarrow\exists(z,r)\in\Cc^2/\;\left\{ \begin{array}{l} 2z+3r=2\\ 6z^2+18rz+11r^2=-36\\ \sigma_3=-\lambda\\ \sigma_4=\mu \end{array} \right. \\ &\Leftrightarrow\exists(z,r)\in\Cc^2/\;\left\{ \begin{array}{l} z=1-\frac{3}{2}r\\ 6(1-\frac{3}{2}r)^2+18(1-\frac{3}{2}r)r+11r^2=-36\\ \sigma_3=-\lambda\\ \sigma_4=\mu \end{array} \right. \\ &\Leftrightarrow\exists(z,r)\in\Cc^2/\;\left\{ \begin{array}{l} -\frac{5}{2}r^2+42=0\\ z=1-\frac{3}{2}r\\ \sigma_3=-\lambda\\ \sigma_4=\mu \end{array} \right. \end{align*}$

D’où la solution (les deux valeurs opposées de $r$ fournissent évidemment la même progression arithmétique) $r=2\sqrt{\frac{21}{5}}$ puis $z=1-3\sqrt{\frac{21}{5}}$ puis les racines $z_1=1-3\sqrt{\frac{21}{5}}$ , $z_2=1-\sqrt{\frac{21}{5}}$ , $z_3=1+\sqrt{\frac{21}{5}}$ et $z_4=1+3\sqrt{\frac{21}{5}}$ , obtenues pour

$\begin{align*} \lambda &= z_1z_2z_3z_4 \\ &= (1-3\sqrt{\frac{21}{5}})(1-\sqrt{\frac{21}{5}})(1+\sqrt{\frac{21}{5}}) (1+3\sqrt{\frac{21}{5}}) \\ &= (1-9\frac{21}{5})(1-\frac{21}{5}) \\ &= \frac{2994}{25}, \end{align*}$

et

$\begin{align*}\ensuremath \mu&=(1-3\sqrt{\frac{21}{5}})(1-\frac{21}{5})+(1-9\frac{21}{5})(1-\sqrt{\frac{21}{5}})+(1-9\frac{21}{5})(1+\sqrt{\frac{21}{5}})+(1-\frac{21}{5})(1+3\sqrt{\frac{21}{5}})\\ &=2(1-\frac{21}{5})+2(1-9\frac{21}{5})=2(2-10\frac{21}{5})=-80 \end{align*}$

exo7 5349