51st Ukrainian National Mathematical Olympiad, 3rd Round

Define a sequence $(a_i)$ in the following way: $a_0=a$, $a_i=P(a_{i-1})$ for all natural $i$. Consider minimal number $m$, for which there exists $j<m$, such that $a_j=a_m$. From the condition of the problem it follows that this number does exist. Observe, that $a_0,a_1,...,a_{m-1}$ are distinct. We show that $j=0$. If this is not the case, then our sequence has the following form: $$a_0,a_1,...,a_{j-1},a_j,a_{j+1},...,a_{m-1},a_m=a_j,a_{j+1},...,a_{m-1},a_j,a_{j+1},...,a_{m-1},a_j,a_{j+1},...$$ and $a_0$ is not going to show up in this sequence, which contradicts to our assumption. Thus, $a_0=a_m$, and our sequence is $m$-periodic, in other words $a_p=a_q$ if $p-q:m$. Let $a_l$ be a maximum, $a_k$ be a minimum among $a_0,a_1,...,a_{m-1}$. If $a_l=a_k$, then $P(a_l)=a$ and $P(P(a_l))=a$. Suppose that $a_l>a_k$. Note, that $P(a_l)-P(a_k)$ is divisible by $a_l-a_k$ and $P(a_l),P(a_k)$ are members of our sequence, thus $|P(a_l)-P(a_k)|\le |a_l-a_k|$, and the difference $P(a_l)-P(a_k)$ is 0, or $\pm (a_l-a_k)$.

If $P(a_l)-P(a_k)=a_l-a_k$, then $P(a_k)-a_l$ and $m=1$ with $a_l=a_k$, and the result follows from previous observation. If $P(a_l)-P(a_k)=0$, namely $a_{l+1}=P(a_l)=P(a_k)=a_{k+1}$, and we must have that $(l+1)-(k+1)=l-k$ is divisible by $m$, which implies that $l=k$ and $m=1$. If $P(a_l)-P(a_k)=-(a_l-a_k)$, then $P(a_l)=a_k$, $P(a_k)=a_l$ and $m=2$, which implies that $P(P(a))=a$.