brianchung11 wrote: Let $ f(x) = c_m x^m + c_{m - 1} x^{m - 1} + ... + c_1 x + c_0$, where each $ c_i$ is a non-zero integer. Define a sequence $ \{ a_n \}$ by $ a_1 = 0$ and $ a_{n + 1} = f(a_n)$ for all positive integers $ n$. (a) Let $ i$ and $ j$ be positive integers with $ i < j$. Show that $ a_{j + 1} - a_j$ is a multiple of $ a_{i + 1} - a_i$. (b) Show that $ a_{2008} \neq 0$ a) From an elementary of an integer polynomial $ a - b|f(a) - f(b)$ $ P(P(x))\equiv P(x) (\mod P(x) - x)$ Thus $ a_{i + 1} - a_{i}|a_{i + 2} - a_{i + 1},\forall i > 0$ . By induction it is easy to prove that for all $ j > i$ then $ a_{i + 1} - a_{i}$ is a divisor of $ a_{j + 1} - a_{j}$ .Statement (a) has been proved . b) If exist $ n > 0$ such that $ a_n = 0$ then this sequence is periodic (i.e for every $ k > 0$ then $ a_{k + n} = a_k$ Similarity (a) we can prove that for all $ t > 0$ and $ j > i$ then $ a_{i + t} - a_i$ is a divisor of $ a_{j + t} - a_j$ From have : $ a_{j + t} - a_j | a_{kn + 1} - a_{kn + t + 1},\forall j\leq kn$ But $ a_{t + 1} - a_1|a_{j + t} - a_j$ and $ a_{t + 1} - a_1 = a_{kn + t + 1} - a_{kn + 1}$ ,we must have for each t then $ |a_{k + t} - a_t|$ is a constant . Only need to consider the positive integer $ m$ such that $ a_m = 0$ then it gives contradiction .