Functional equation $f(x)f(f(x))=x^2$

Question

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.

Answer 1

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 $j$).

Answer 2

Or solutions of the form : $f(x)=ax^b$

Substitutin gives $f(x)=x,\frac{1}{x^2}$