20th Turkish Mathematical Olympiad

The answer is $P(x)=1$, $P(x)=2$ and $P(x)=x$. First recall that for two polynomials $P(x)$ and $Q(x)$ if we have $P(x)=Q(x)$ for infinitely many $x$, then $P(x)=Q(x)$ for every $x$. Plugging in $n=1,2$ gives $P(1)=|P(1)|!$ and $P(2)=|P(2)|!$ which implies that $P(1),P(2)\in \{1,2\}$.

Case 1: $P(2)=1$. Note that $P(n!)>0$ for every positive integer $n$. Then letting $n=m!$ for some positive integer $m$ gives $P(n!)=P(n)!$. Bezout's Theorem implies $n!-2|P(n!)-P(2)|=P(n)!-1$. When $m\ge 2$, $n!-2$ is even and hence so is $P(n)!-1$. In other words $P(n)!$ is odd which implies that $P(n)\in \{0,1\}$. Thus, $P(x)=c$ for some $c\in \{0,1\}$ for infinitely many $x$. Therefore $P(x)$ is constant. As $P(2)=1$, we obtain $P(x)=1$ for every $x$.

Case 2: $P(2)=2$ and $P(1)=1$. Again by Bezout's Theorem we have $5=3!-1|P(3!)-P(1)|=|P(3)|!-1$ and $4=3!-2|P(3!)-P(2)|=|P(3)|!-2$, that is 5 divides $|P(3)|!-1$ and 4 divides $|P(3)|!-2$. Then $|P(3)|=3$ and hence $P(6)=P(3!)=|P(3)|!=6$. Therefore $P(6!)=6!$, $P((6!)!)=(6!)!$ and so on. In other words $P(x)=x$ for infinitely many $x$ and hence $P(x)=x$ for every $x$.

Case 3: $P(2)=2$ and $P(1)=2$. Bezout's Theorem gives $5=3!-1|P(3!)-P(1)|=|P(3)|!-2$, that is 5 divides $|P(3)|!-2$. Then $|P(3)|=2$ and hence $P(6)=P(3!)=|P(3)|!=2$. Thus, $P(6!)=2$, $P((6!)!)=2$ and so on. In other words, $P(x)=2$ for infinitely many $x$ and hence $P(x)=2$ for every $x$. That is easy to verify the solutions.