What solutions are there to $f(f(x))=x^4$?

Question

I know $f(x)=x^2$ is a solution, but I can't seem to find any others and I have no idea how to approach this.

Answer 1

$f(x)=1/x^2$ is another one with domain $\mathbb{R^*}$.

Answer 2

This equation leaves us with immense freedom of choice. There is a continuum of continuous solutions, and even more discontinuous ones.

To find a few, draw a freehand increasing graph from $(2,4)$ to $(4,16)$. Then every point in $(4,16)$ is $f(x)$ for some $x\in(2,4)$, so you may define $f(f(x))$ as $x^4$, then continue in a similar manner to define $f$ on $(16,256)$, and so on. Mind you, this does not even cover all continuous solutions, as illustrated by $1\over x^2$.

To find more, split all $\mathbb R_+$ into countable sets $A_x=\{\dots,\sqrt[4]x,x,x^4,x^{16},\dots\}$, then join the sets in pairs $(A_x,A_y)$, select a representative from each set in each pair, and define $f(x)=y$. Then $f(y)=x^4,\;f(x^4)=y^4$, and so on.