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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.

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    $\begingroup$ You have do define domain set and range set of f $\endgroup$ – arberavdullahu Oct 15 '16 at 8:34
  • $\begingroup$ The domain and range are both the nonnegative reals $\endgroup$ – Miles Johnson Oct 15 '16 at 22:49
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$f(x)=1/x^2$ is another one with domain $\mathbb{R^*}$.

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    $\begingroup$ And so is $f(x)=-x^2$. $\endgroup$ – celtschk Oct 15 '16 at 8:47
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    $\begingroup$ Also $f(x) = -1/x^2$. $\endgroup$ – Math Student Oct 15 '16 at 8:48
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    $\begingroup$ For those functions we get $f(f(x))=-x^4$ $\endgroup$ – arberavdullahu Oct 15 '16 at 8:56
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    $\begingroup$ @arberavdullahu you are right. $\endgroup$ – Jean Marie Oct 15 '16 at 9:02
  • $\begingroup$ $f_1(x)=\begin{cases} -x^2 &\text{for } x \ge 0\\x^2 &\text{for } x<0 \end{cases}, f_2(x)=\begin{cases} -1/x^2 &\text{for } x \ge 0\\1/x^2 &\text{for } x<0 \end{cases}$ are another solutions. $\endgroup$ – Rafał Oct 15 '16 at 11:23
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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.

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