2

Do there exist functions $f,g : R → R$ such that $f (g(x)) = x^2$ and $g( f (x)) = x^3 \text{ , }\forall x ∈ R$.

Simply applying $g$ on both sides of equation $1$ and $f$ on equation $2$ respectively, we get

$g(x)^3=g(x^2)$ and $f(x)^2=f(x^3)$.

It does seem like there aren't functions that satisfy this. But how do i prove that? Incase there are such functions, what should be the next step now? Pluggin in $0$ or $1$ would give a lot of cases and it doesn't really seem like the right approach. In an exam i might maybe bash through all of them systematically if i cant find a better alternative, but for now please help :).

Aayam Mathur
  • 613
  • Duplicate: https://math.stackexchange.com/questions/2974068/existence-of-two-functions-f-and-g-for-which-f-circ-g-x-x2-g-circ-f-x/2978339#2978339 – Ivan Neretin Jun 05 '20 at 19:43

1 Answers1

2

Notice that $x\mapsto x^3$ is bijective, so from $g(f(x))=x^3$, $f$ must be injective and $g$ must be surjective.

Moreover, from $f(x)^2 = f(x^3)$ you get $$ f(0)^2 = f(0) \qquad f(1)^2 = f(1) \qquad f(-1)^2 = f(-1) $$ so $f(0),f(1),f(-1)$ may only assume the values $0$ or $1$, that is a contradiction since $f$ is injective.

Exodd
  • 10,844