Problem 1024

. We show that no such function $f$ exists. Suppose $f$ satisfies the condition. Let $g(x)=x^{2}-2=f(f(x))$ and $h(x)=g(g(x))=x^{4}-4 x^{2}+2$. The key is to involve the fixed points of $g$ and $h$. A simple calculation shows that the fixed points of $g$ (i.e., the solutions of $g(x)=x$) are -1 and $2$; the set $S$ of fixed points of $h$ is $\{-1,2, a, b\}$ with
$a=(-1+\sqrt{5}) / 2$ and $b=(-1-\sqrt{5}) / 2$. It is easy to see that $\{-1,2\}$ and $S$ are stable under $f$: indeed, a fixed point $u$ of $g$ satisfies $g(u)=u$ and thus $g(f(u))=f(f(f(u)))=f(g(u))=f(u)$, so $f(u)$ is a fixed point of $g$. The same argument applies to $h$. Since $S$ is stable under $f$ and the restriction of $f^{4}$ is the identity on $S$, we see that $f$ induces a bijection $f^{\prime}$ of $S$ onto $S$. Let $g^{\prime}$ be the bijection induced by $g$ on $S$. We see that $g^{\prime}(-1)=-1, g^{\prime}(2)=2, g^{\prime}(a)=b$ and $g^{\prime}(b)=a$. It is easy to observe that the relation $f^{\prime} \circ f^{\prime}=g^{\prime}$ leads to a contradiction: since $\{-1,2\}$ is stable under $f^{\prime}$, the same is true for $\{a, b\}$; then $f^{\prime}(a)=a$ and $f^{\prime}(b)=b$, or $f^{\prime}(a)=b$ and $f^{\prime}(b)=a$. In both cases, $f^{\prime}\left(f^{\prime}(a)\right)=a$, contradicting $g^{\prime}(a)=b$.