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Question Number 209385 by efronzo1 last updated on 08/Jul/24

Answered by Rasheed.Sindhi last updated on 08/Jul/24

p(x)=ax^2 +bx+1⇒p(1)=a+b+1 q(x)=bx^2 +ax+1⇒q(1)=b+a+1 p( q(1) )=q( p(1) ) p(a+b+1)=q(a+b+1) a(a+b+1)^2 +b(a+b+1)+1 =b(a+b+1)^2 +a(a+b+1)+1 a(a^2 +b^2 +1+2ab+2b+2a)+ab+b^2 +b =b(a^2 +b^2 +1+2ab+2b+2a)+a^2 +ab+a a^3 +ab^2 +a+2a^2 b+2ab+2a^2 +b^2 +b =a^2 b+b^3 +b+2ab^2 +2b^2 +2ab+a^2 +a a^3 −b^3 −ab^2 +a^2 b+a−b+a^2 −b^2 +a−b=0 a^3 −b^3 −ab^2 +a^2 b+a^2 −b^2 =0 (a−b)(a^2 +ab+b^2 )+ab(a−b)+(a−b)(a+b)=0 (a−b)(a^2 +ab+b^2 +ab+a+b)=0 a−b≠0⇒(a+b)^2 +(a+b)=0 (a+b)(a+b+1)=0 a+b≠0⇒a+b=−1 p(1)=0 , q(1)=0 p(0)=1, q(0)=1 ▶q(q(q(0)))+p(p(p(1))) =q(q(1))+p(p(0)) =q(0)+p(1) =1+0=1

$${p}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+\mathrm{1}\Rightarrow{p}\left(\mathrm{1}\right)={a}+{b}+\mathrm{1} \\ $$$${q}\left({x}\right)={bx}^{\mathrm{2}} +{ax}+\mathrm{1}\Rightarrow{q}\left(\mathrm{1}\right)={b}+{a}+\mathrm{1} \\ $$$${p}\left(\:{q}\left(\mathrm{1}\right)\:\right)={q}\left(\:{p}\left(\mathrm{1}\right)\:\right) \\ $$$${p}\left({a}+{b}+\mathrm{1}\right)={q}\left({a}+{b}+\mathrm{1}\right) \\ $$$${a}\left({a}+{b}+\mathrm{1}\right)^{\mathrm{2}} +{b}\left({a}+{b}+\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:={b}\left({a}+{b}+\mathrm{1}\right)^{\mathrm{2}} +{a}\left({a}+{b}+\mathrm{1}\right)+\mathrm{1} \\ $$$${a}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{ab}+\mathrm{2}{b}+\mathrm{2}{a}\right)+{ab}+{b}^{\mathrm{2}} +{b} \\ $$$$={b}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{ab}+\mathrm{2}{b}+\mathrm{2}{a}\right)+{a}^{\mathrm{2}} +{ab}+{a} \\ $$$${a}^{\mathrm{3}} +{ab}^{\mathrm{2}} +{a}+\mathrm{2}{a}^{\mathrm{2}} {b}+\mathrm{2}{ab}+\mathrm{2}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{b} \\ $$$$={a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} +{b}+\mathrm{2}{ab}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ab}+{a}^{\mathrm{2}} +{a} \\ $$$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} −{ab}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{a}−{b}+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}−{b}=\mathrm{0} \\ $$$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} −{ab}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)+{ab}\left({a}−{b}\right)+\left({a}−{b}\right)\left({a}+{b}\right)=\mathrm{0} \\ $$$$\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} +{ab}+{a}+{b}\right)=\mathrm{0} \\ $$$${a}−{b}\neq\mathrm{0}\Rightarrow\left({a}+{b}\right)^{\mathrm{2}} +\left({a}+{b}\right)=\mathrm{0} \\ $$$$\left({a}+{b}\right)\left({a}+{b}+\mathrm{1}\right)=\mathrm{0} \\ $$$${a}+{b}\neq\mathrm{0}\Rightarrow{a}+{b}=−\mathrm{1} \\ $$$${p}\left(\mathrm{1}\right)=\mathrm{0}\:,\:{q}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${p}\left(\mathrm{0}\right)=\mathrm{1},\:{q}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\blacktriangleright{q}\left({q}\left({q}\left(\mathrm{0}\right)\right)\right)+{p}\left({p}\left({p}\left(\mathrm{1}\right)\right)\right) \\ $$$$={q}\left({q}\left(\mathrm{1}\right)\right)+{p}\left({p}\left(\mathrm{0}\right)\right) \\ $$$$={q}\left(\mathrm{0}\right)+{p}\left(\mathrm{1}\right) \\ $$$$=\mathrm{1}+\mathrm{0}=\mathrm{1} \\ $$

Commented by Frix last updated on 08/Jul/24

Yes but take a closer look: a(a+b+1)^2 +b(a+b+1)+1 =b(a+b+1)^2 +a(a+b+1)+1 it directly follows: a(a+b+1)+b=b(a+b+1)+a (a−b)(a+b+1)=a−b a+b+1=0

$$\mathrm{Yes}\:\mathrm{but}\:\mathrm{take}\:\mathrm{a}\:\mathrm{closer}\:\mathrm{look}: \\ $$$${a}\left({a}+{b}+\mathrm{1}\right)^{\mathrm{2}} +{b}\left({a}+{b}+\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:={b}\left({a}+{b}+\mathrm{1}\right)^{\mathrm{2}} +{a}\left({a}+{b}+\mathrm{1}\right)+\mathrm{1} \\ $$$$\mathrm{it}\:\mathrm{directly}\:\mathrm{follows}: \\ $$$${a}\left({a}+{b}+\mathrm{1}\right)+{b}={b}\left({a}+{b}+\mathrm{1}\right)+{a} \\ $$$$\left({a}−{b}\right)\left({a}+{b}+\mathrm{1}\right)={a}−{b} \\ $$$${a}+{b}+\mathrm{1}=\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 09/Jul/24

Yes sir, thanks!

$${Yes}\:\boldsymbol{{sir}},\:{thanks}! \\ $$

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