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Does there exists two real valued functions $f$ and $g$ on $R$ such that $f \circ g = x^{2018}$ and $ g\circ f = x^{2019}$ ?

My attempt : since $g \circ f $ is bijective thus $f$ is one one and $g$ is onto. Now $f \circ g(-x) = f \circ g(x) $ imply $ g(-x) = g(x) $ (because $ f $ is one one) thus g is even function. Now i dont know how to proceed from here any hint will be helpfull for me.... (I know no such map exists as answer)

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$$g(f(x))=\ x^{2019}$$ $$=> f(g(f(x)))=f(\ x^{2019})$$ $$=> (f(x))^{2018}=f(x^{2019})$$ (by associativity of function composition)

there exist no such f and g because$ f(1)=1, f(-1)=1 => g(1)=1$ and $ g({(-1)}^{2018})=-1$

contradiction hence proved

well one may argue f(1)=0

in that case consider $f(0),f(1),f(-1) $will have either 1 or 0 value and by pigeon hole principle two of them will have atleast same value hence we are done

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  • $\begingroup$ What if $f(1) = 0$? $\endgroup$
    – fleablood
    Jan 8, 2019 at 18:00
  • $\begingroup$ hmm you are correct $\endgroup$
    – Bijayan Ray
    Jan 8, 2019 at 18:02
  • $\begingroup$ consider then f(0),f(-1),f(1) they can be 0 or 1 $\endgroup$
    – Bijayan Ray
    Jan 8, 2019 at 18:04
  • $\begingroup$ by pigeon hole principle atleast two of them will have same value so we are done $\endgroup$
    – Bijayan Ray
    Jan 8, 2019 at 18:05

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