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Let-P-C-X-p-x-2-1-p-2-x-1-p-0-0-determin-all-polynom-

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Question Number 203975 by witcher3 last updated on 03/Feb/24
Let P∈C[X] p(x^2 +1)=p^2 (x)+1 p(0)=0 determin all polynom
$$\mathrm{Let}\:\mathrm{P}\in\mathbb{C}\left[\mathrm{X}\right] \\ $$$$\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{p}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{1} \\ $$$$\mathrm{p}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{determin}\:\mathrm{all}\:\mathrm{polynom}\: \\ $$
Commented by Rasheed.Sindhi last updated on 03/Feb/24
Let p(x)=ax+b p(0)=b=0 p(x)=ax p(x^2 +1)=a(x^2 +1)=ax^2 +a p^2 (x)+1=(ax)^2 +1=a^2 x^2 +1 ax^2 +a=a^2 x^2 +1 a^2 x^2 −ax^2 =a−1 ax^2 (a−1)=a−1 ax^2 (a−1)−(a−1)=0 (a−1)(ax^2 −1)=0 a=1 ∣ ax^2 =1 One of the solutions is: p(x)=x
$${Let}\:{p}\left({x}\right)={ax}+{b} \\ $$$${p}\left(\mathrm{0}\right)={b}=\mathrm{0} \\ $$$${p}\left({x}\right)={ax} \\ $$$${p}\left({x}^{\mathrm{2}} +\mathrm{1}\right)={a}\left({x}^{\mathrm{2}} +\mathrm{1}\right)={ax}^{\mathrm{2}} +{a} \\ $$$${p}^{\mathrm{2}} \left({x}\right)+\mathrm{1}=\left({ax}\right)^{\mathrm{2}} +\mathrm{1}={a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1} \\ $$$${ax}^{\mathrm{2}} +{a}={a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} −{ax}^{\mathrm{2}} ={a}−\mathrm{1} \\ $$$${ax}^{\mathrm{2}} \left({a}−\mathrm{1}\right)={a}−\mathrm{1} \\ $$$${ax}^{\mathrm{2}} \left({a}−\mathrm{1}\right)−\left({a}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)\left({ax}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${a}=\mathrm{1}\:\mid\:{ax}^{\mathrm{2}} =\mathrm{1} \\ $$$${One}\:{of}\:{the}\:{solutions}\:{is}: \\ $$$${p}\left({x}\right)={x} \\ $$

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