Problem 1555

$$|P(a_n)| \neq 1 \text{ if } 6a(n, P(a_0)) = 1. \quad (*)$$
Let $p = |P(a_n)|$. Then for any natural number $s$ the congruence $P(a_{n+ps}) = P(a_n + psd) \equiv 0 \pmod p$ holds. Therefore from $a_{n+ps}^{n+ps} + 1 \equiv 0 \pmod p$ follows $a_n^{n+ps} + 1 \equiv 0 \pmod p$. Let's choose $s$ such that $ps \equiv 1 \pmod n$. Then we have $a_n^n + 1 \equiv 0 \pmod p$ and
$a_n \pm 1 \equiv 0 \pmod{p}$
, 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 $a_{\varphi(c)}^{\varphi(c)} + 1 \equiv 2 \pmod c$. 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 $P(x) = \pm 1$.