Bulgarian Mathematical Olympiad

Solution:

a.
Let $n \in A$ and $O_{n}=\{n, f(n),-n, f(-n)\}$. Since $f(f(n))=-n$ and $f(f(-n))=n$, it follows easily that if $k \in A$ then either $O_{k}=O_{n}$ or $O_{n} \cap O_{k}=\varnothing$. Moreover, we obtain $f(n) \neq f(-n)$ for $n \neq 0$.

Further, if $f( \pm n)= \pm n$, then $\mp n=f(f( \pm n))=f( \pm n)= \pm n$, i.e. $n=0$. Also, $f( \pm n)=\mp n$ gives $\mp n=f(f( \pm n))=f(\mp n)$ and then $n=0$. Therefore $|O_{n}|=4$ for $n \neq 0$ which means that $A \setminus \{0\}$ splits into disjoint quadruples. In particular, the number $m$ is even.

b.
Let $m=2k$ and $f: A \rightarrow A$ be a function with the desired property. Set $A_{+}=\{1,2, \ldots, m\}$. We note that $f(-n)=f(f(f(n)))=-f(n)$ and, in particular, $f(0)=0$. Hence for $n \neq 0$ either $f(n)>0$ or $f(-n)<0$. This means that the quadruple $O_{n}$ is uniquely determined by a pair $(n', f(n'))$ of distinct numbers from $A_{+}$—$(n, f(n))$ or $(f(-n), n)$. Therefore $f$ induces a pairing of $A_{+}$ into ordered pairs.

Conversely, any pairing of $A_{+}$ into ordered pairs $(n, k)$ defines a function with the required properties by setting
$$f(0)=0, \quad f(n)=k, \quad f(k)=-n, \quad f(-n)=-k, \quad f(-k)=n$$
It remains to count the number of the pairings of $A_{+}$ into ordered pairs. Ordering all pairs of a given pairing one after another (this can be done in $k!$ ways) we obtain a permutation of the numbers $1,2, \ldots, m$. This gives classes of "equivalent" permutations of $k!$ elements. Therefore the required number is equal to $\frac{m!}{k!}$.