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I have tried a little bit to solve the problem which goes as follows:

My intuition says that there exist no $f:\Bbb R\to\Bbb R$ such that $$f(g(x))=x^{2018}\text{ and }g(f(x))=x^{2019}.$$

Note that $$f(g(f(x)))=f(x)^{2018}\implies f(x^{2019})=f(x)^{2018}$$

Similarly, $$g(x^{2018})=g(x)^{2019}$$ Putting $x=1$ in $f(x^{2019})=f(x)^{2018}$, we get $f(1)=f(1)^{2018}$ and thus $f(1)=0$ or $f(1)=1$.

Similarly, putting $x=1$ in $g(x^{2018})=g(x)^{2019}$ we get $g(1)=g(1)^{2019}$ and thus $g(1)=0,1,-1$.

Now, I can't proceed further. Can anybody solve it? Thanks for assistance in advance.

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2 Answers 2

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Here is a simple proof (to avoid confusion I will write $f^{y}(x)$ instead of $f(x)^{y}$):

Suppose that there are two such functions $f,g$ as in your question.

Note that $\forall\space i \in\{-1, 0, 1\}$, we have $f(i^{2019}) = f(i) = f^{2018}(i)$ and thus $\label{*}\tag{*} f(i) \in \{0, 1\} \quad\forall\space i \in\{-1, 0, 1\}.$

On the other hand, since $g(f(x)) = x^{2019} \space\forall x\in\mathbb{R}$,

  • $g(f(1)) = 1$,
  • $g(f(0)) = 0$,
  • $g(f(-1)) = -1$.

This is impossible since $f$ (and thus also $g\circ f$) only takes two (or fewer) values on $\{-1, 0, 1\}$ after \eqref{*}. $\Longrightarrow\Longleftarrow\quad\square$

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Another classic trick that can be used here is that there is a bijection between the set of fixed points of $fg$ and fixed points of $gf$. If $x$ is such that $f(g(x))=x$ then $g(f(g(x)))=g(x)$ so $g(x)$ is a fixed point of $gf$. Similarly whenever $y$ is a fixed point of $gf$, we can show that $f(y)$ is a fixed point of $fg$. Convince yourself that on the sets of fixed points, this gives a bijection.

Can you see why the $fg$ and $gf$ you've been given fail this condition?

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