Functional equation $f(x)f(f(x))=x^2$ Find all real functions such that
$f(x)f(f(x))=x^2$ and $f(x)=x$ for some $x$.
Obviously $f(x)=x$ is a solution, but I have no clue how to find other solutions.
 A: Here's a whole family of solutions. Let $g(x)$ be any function from the reals to $\{-1,1\}$ such that $g(g(x)x)=1$. Then define $f(x)=g(x)x$. We get
$$f(x)f(f(x))=\left[g(x)x\right]\left[g(g(x)x)\right]\left[g(x)x\right]=g(x)^2x^2\cdot 1=x^2$$
Now, here are some examples of such $g(x)$:
$$g_1(x)=1$$
$$g_2(x)=\begin{cases} 
      1 & x\geq  0 \\
      -1 & x<0
   \end{cases}$$
$$g_3(x)=\begin{cases} 
      1 & x\leq   0 \\
      -1 & x>0
   \end{cases}$$
(it is easy to prove that these all work). Now, let $A$ and $B$ be any sets such that
$$A\cup B=\mathbb{R}$$
$$A\cap B=\emptyset$$
$$a\in A\Rightarrow -a\in A$$
$$b\in B\Rightarrow -b\in B$$
We may finally get to the punchline: For such $A$ and $B$, the function
$$g(x)=\begin{cases} 
      g_i(x) & x\in A \\
      g_j(x) & x\in B
   \end{cases}$$
(where $i$ and $j$ are selected from $\{1,2,3\}$) is another valid $g(x)$. The proof is simple: note that for $x\in A$ we have
$$g_i(x)x\in\{x,-x\}$$
This implies $g_i(x)x\in A$. Therefore
$$g(g_i(x)x)=g_i(g_i(x)x))=1$$
(the same logic applies to $B$). We conclude if $A$ and $B$ follow the conditions above, then $f(x)f(f(x))=x^2$ where $f(x)=g(x)x$ and $g(x)$ is as defined above (for any choice of $i$ and $j$).
A: Or solutions of the form : $f(x)=ax^b$
Substitutin gives $f(x)=x,\frac{1}{x^2}$
