Question Number 218656 by Mamadi last updated on 14/Apr/25
$${resolve}\:{the}\:{equation}\:{with}\:{unknow}\:{p} \\ $$$${P}\:\:{is}\:{polynom}\: \\ $$$$\left.\mathrm{1}\right)\:{P}\left({X}^{\mathrm{2}} \right)=\left({X}^{\mathrm{2}} +\mathrm{1}\right){P}\left({X}\right) \\ $$$$\left.\mathrm{2}\right)\:{P}\:\mathrm{0}{P}\:={P} \\ $$
Answered by MrGaster last updated on 14/Apr/25
$${P}\left({X}\right)=\mathrm{0} \\ $$
Commented by Mamadi last updated on 14/Apr/25
$${ok}\:{but}\:{can}\:{you}\:{prove}\:{for}\:{me}?? \\ $$
Answered by maths2 last updated on 14/Apr/25
$$\left.\mathrm{1}\right);{let}\:{n}={deg}\left({P}\right) \\ $$$${deg}\left({p}\left({x}^{\mathrm{2}} \right)\right)=\mathrm{2}{n}\Rightarrow\mathrm{2}{n}=\mathrm{2}+{n}\Rightarrow{n}=\mathrm{2} \\ $$$${p}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\Rightarrow{ax}^{\mathrm{4}} +{bx}^{\mathrm{2}} +{c}=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({ax}^{\mathrm{2}} +{bx}+{c}\right) \\ $$$$\Rightarrow{b}=\mathrm{0};\left({c}+{a}\right)=\mathrm{0};{c}=−{a} \\ $$$${p}\left({x}\right)={a}\left({x}^{\mathrm{2}} −\mathrm{1}\right); \\ $$$$\left(\mathrm{2}\right){dep}\left({pop}\right)={n}^{\mathrm{2}} \Rightarrow{n}^{\mathrm{2}} ={n}\Rightarrow{n}\in\left\{\mathrm{0},\mathrm{1}\right\} \\ $$$${p}\left({x}\right)={ax}+{b} \\ $$$$\Rightarrow{a}\left({ax}+{b}\right)+{b}={ax}+{b}\Rightarrow{x}\left({a}^{\mathrm{2}} −\mathrm{1}\right)+{b}\left({a}−\mathrm{1}\right)=\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\Rightarrow{a}\in\left\{−\mathrm{1},\mathrm{1}\right\} \\ $$$${ba}=\mathrm{0}\Rightarrow{b}=\mathrm{0} \\ $$$${p}\left({x}\right)={X};{P}\left({X}\right)=−{X} \\ $$
Commented by Mamadi last updated on 14/Apr/25
$${tank}\:{you}\:\:{very}\:{much} \\ $$