Hong Kong Team Selection Test 1

Let $b_1, b_2, \dots, b_n$ be the integer roots. By Vieta's theorem, we have
$$b_1 b_2 \cdots b_n = (-1)^n a_0, \quad (1)$$
$$b_1 b_2 \cdots b_{n-1} + b_1 b_2 \cdots b_{n-2} b_n + \cdots + b_2 b_3 \cdots b_n = (-1)^{n-1} a_1. \quad (2)$$
Suppose on the contrary that $a_0$ and $a_1$ share a common prime factor $p$. By equation (1), we have $p \mid b_1 b_2 \cdots b_n$. WLOG assume $p \mid b_1$. Then in equation (2), $p$ divides each of the first $n-1$ terms on the left and divides the term on the right. Therefore, we must have $p \mid b_2 b_3 \cdots b_n$. This implies one of $b_2, b_3, \dots, b_n$ is divisible by $p$, contradicting the fact that each of them is relatively prime to $b_1$. Therefore, $a_0$ and $a_1$ must be relatively prime.