Let's see what we can learn about $f$ from the condition. Let $a=a_{0}$ be arbitrary, $a_{1}=f\left(a_{0}\right), a_{2}=f\left(a_{1}\right)=a_{0}^{2}-2$, $a_{3}=f\left(a_{2}\right)=a_{1}^{2}-2$ etc.

If $a_{4}=a_{0}$, then naturally $a_{1}=a_{5}, a_{2}=a_{6}, a_{3}=a_{7}$, and thus $a_{0}, a_{1}, a_{2}$ and $a_{3}$ (not necessarily distinct) are roots of the equation $\left(x^{2}-2\right)^{2}-2=x$. Let's set this to zero and factorize it:

$0=x^{4}-4 x^{2}-x+2=\left(x^{2}-x-2\right)\left(x^{2}+x-1\right)$, from which the roots are $\alpha=\frac{\sqrt{5}-1}{2}, \beta=\frac{-\sqrt{5}-1}{2}, \gamma=-1, \delta=2$, and furthermore $\alpha^{2}-2=\beta, \beta^{2}-2=\alpha, \gamma^{2}-2=\gamma$ and $\delta^{2}-2=\delta$.

According to the above considerations, the value of $f(\alpha)$ can only be $\alpha, \beta, \gamma$ and $\delta$. Let's consider each of the four cases separately.
I. $\quad f(\alpha)=\alpha$. Then $f(\alpha)=f(f(\alpha))=\alpha^{2}-2=\beta \neq \alpha$, which is impossible.

II. $f(\alpha)=\beta$. Then $f(\beta)=f(f(\alpha))=\alpha^{2}-2=\beta$, and thus

$$f(\beta)=f(f(\beta))=\beta^{2}-2=\alpha \neq \beta$$

III. $\quad f(\alpha)=\gamma . \quad$ Then $f(\gamma)=f(f(\alpha))=\alpha^{2}-2=\beta$,

$$\begin{aligned} & f(\beta)=f(f(\gamma))=\gamma^{2}-2=\gamma \text { and thus } \\ & f(\gamma)=f(f(\beta))=\beta^{2}-2=\alpha \neq \beta \end{aligned}$$

IV. Similarly, we get a contradiction in the case of $f(\alpha)=\delta$ as well. Since we have no more options for the value of $f(\alpha)$, we cannot define the function $f$ at $\alpha$, so the sought function does not exist.

Patrik Szegedy (Budapest, Fazekas M. Gyak. Gimn., IV. o. t.)