2000 Putnam A6

Let $f(x)$ be a polynomial with integer coefficients. Define a

sequence $a_0,a_1,\ldots$ of integers such that $a_0=0$ and

$a_{n+1}=f(a_n)$

for all $n\geq 0$. Prove that if there exists a positive integer $m$ for

which $a_m=0$ then either $a_1=0$ or $a_2=0$.

令 $f(x)$ 为具有整数系数的多项式。定义一个

整数序列 $a_0,a_1,\ldots$ 使得 $a_0=0$ 并且

$a_{n+1}=f(a_n)$

对于所有 $n\geq 0$。证明如果存在正整数$m$

$a_m=0$ 则 $a_1=0$ 或 $a_2=0$。