Composing F with F

The answer is that there are no such functions $f$ . More generally, applying the below assertion to $g(x) = x^2-2$ to get a solution to the problem.

A more general assertion: If $g$ is a function from $\mathbb{R}$ to itself such that $g$ has exactly two fixed points $\{s,t\}$ and if $g(g(x))$ has exactly four fixed points $\{s,t,u,v\}$ then there is no function $f$ such that $f(f(x))= g(x)$ .

Proof by contradiction: (Assuming there is $f$ such that $f(f(x)) = g(x)$ :-

Step 1: $g(u)=v,g(v)=u$ .

Step 2: $g(f(x))=f(g(x))$ .

Step 3: Using step 2 and the fact that $s,t$ are the fixed points of $g$ , we can show that $f$ takes $\{s,t\}$ to $\{s,t\}$ and $\{s,t,u,v\}$ to $\{s,t,u,v\}$ .

Step 4: by further solving we can show that each possibility $f(v) = s$ or $t$ or $u$ or $v$ leads to a contradiction. so there is no function exist so that $f(f(x))=x^2-2$ .

[Latex Edits - Calvin]

Let $g(x) = x^2 -2$ . Notice that if $g(x) =x$ , then $x^2-2=x$ which has 2 distinct solutions, so $g(x)$ has 2 fixed points. If $g( g(x) )=x$ , then $(x^2-2)^2-2=x \Rightarrow 0=x^4-4x^2-x+2 = (x^2-x-2)(x^2+x-1)$ has 4 distinct solutions, so $g(g(x))$ has 4 fixed points. We will show, by contradiction, that there are no functions $f$ that satisfy $f \circ f = g$ .

Suppose that $f$ exists. Let $a$ and $b$ be the distinct fixed points of $g$ and let $a, b, c$ and $d$ be the distinct fixed points of $g \circ g$ . Let $m=g(c)$ . Then $g(m) = g(g(c)) = c$ , so $g(g(m)) = g(c) = m$ , so $m$ is a fixed point of $g\circ g$ . If $m = a$ , then $c=g(m) = g(a)=a$ is a contradiction. Similarly, if $m=b$ , then $c = g(m) = g(b) = b$ is a contradiction. If $m=c$ , then $c=g(c)$ so $c$ is a fixed point of $g$ , which is a contradiction. Thus, $m = d$ . As such, $g(c)=d$ , and we similarly have $g(d) = c$ .

Now consider $f(x_1)$ for $x_1 \in \{a, b\}$ . Since $g( f(x_1) ) = f( f( f(x_1) ) ) = f( g(x_1) ) = f(x_1)$ , so $f(x_1)$ is a fixed point of $g$ , thus $f(x_1) \in \{a, b\}$ . Similarly, consider $f(x_2)$ for $x_2 \in \{ a, b, c, d\}$ . Since $g( g( f(x_2) ) = f(f(f(f(f(x_2))))) = f( g ( g( x_2) ) ) = f(x_2)$ , so $f(x_2)$ is a fixed point of $g$ , thus $f(x_2) \in \{a, b, c, d\}$ . Consider $f(c)$ .

Case 1: $f(c) = a$ . Then, $f(a) = f( f(c) ) = g(c) = d$ which is a contradiction.

Case 2: $f(c) = b$ . Similarily, $f(b) = f(f(c))=g(c) =d$ which is a contradiction.

Case 3: $f(c) = c$ . Then $g(c) = f( f(c)) = f(c) = c$ which is a contradiction.

Case 4: $f( c) = d$ . Then $d = g(c) = f( f(c) ) = f( d)$ $\Rightarrow g(d) = f( f(d) ) = f(d) = d$ which is a contradiction.

Hence, there are 0 functions $f$ that satisfies the conditions.