Baltic Way competition problems, 2013

Notice that $a_{n+1} + a_n = 3(a_n + a_{n-1})$, and let $b_n = a_n + a_{n-1}$, $b > 1$. Then $b_2 = 11$ and $b_n = 11 \cdot 3^{n-2}$, $n > 1$. Using this gives
$$\begin{align*} (1+x)P(x) &= a_1 + \sum_{i=2}^{\infty} a_i x^i + \sum_{i=2}^{\infty} a_{i-1} x^i = \\ &= a_1 x + \sum_{i=2}^{\infty} b_i x^i = \\ &= x + \frac{11}{9} \sum_{i=2}^{\infty} (3x)^i = \\ &= x + \frac{11}{9} \cdot \frac{9x^2}{1-3x} = \\ &= \frac{8x^2 + x}{1-3x} \end{align*}$$
when $x \in ] -\frac{1}{3}, \frac{1}{3}[$. Hence
$$P(x) = \frac{8x^2 + x}{(1 - 3x)(1 + x)} \quad \text{for all } x \in ] -\frac{1}{3}, \frac{1}{3}[,$$
and
$$P\left(\frac{1}{y}\right) = \frac{8+y}{(y-3)(y+1)} = \frac{(y+1)+7}{(y-3)(y+1)} = \frac{(y-3)+11}{(y-3)(y+1)}$$
for all $y \in \mathbb{Z}$ and $|y| > 3$. Assume $P(\frac{1}{y}) \in \mathbb{Z}$ for some $y \in \mathbb{Z}$ and $|y| > 3$. Then $y+1 \mid 7$ and $y-3 \mid 11$, and it is easy to see that $y = -8$ is the only possibility. Since $P(-\frac{1}{8}) = 0$, $y = -8$ is indeed a solution.