Mongolian Mathematical Olympiad

$$|P(a_n)|\ne 1\text{ if }6a(n,P(a_0))=1.(\ast )$$ Let $p=|P(a_n)|$. Then for any natural number $s$ the congruence $P(a_{n+ps})=P(a_n+psd)\equiv 0(\mathrm{mod}p)$ holds. Therefore from $a_{n+ps}^{n+ps}+1\equiv 0(\mathrm{mod}p)$ follows $a_n^{n+ps}+1\equiv 0(\mathrm{mod}p)$. Let's choose $s$ such that $ps\equiv 1(\mathrm{mod}n)$. Then we have $a_n^n+1\equiv 0(\mathrm{mod}p)$ and $a_n\pm 1\equiv 0(\mathrm{mod}p)$ , and it implies $|a_n\pm 1|\ge p$. If $|P(a_n)-P(0)|\ne 0$ then there exist infinitely many $n$ with property (*). This contradicts given condition that the fraction is integer and above proved implication. So $P(a_n)=c$ constant. Since there exists $n$ such that $(a_n,c)=1$, we can write $a_{\phi (c)}^{\phi (c)}+1\equiv 2(\mathrm{mod}c)$. Thus starting from a number $n$ inequality $|P(a_n)|>|a_n|+1$ always holds. In case all $a_n$ odd then $P(x)=\pm 1,\pm 2$. In other cases $P(x)=\pm 1$.