IMC 2008 · Day 1 · P3

The theorem is obvious if $p(a_i) = 0$ for some $i$, so assume that all $p(a_i)$ are nonzero and pairwise different.

There exist numbers $s, t$ such that $s \mid p(a_1)$, $t \mid p(a_2)$, $st = \operatorname{lcm}(p(a_1), p(a_2))$ and $\gcd(s, t) = 1$.

As $s, t$ are relatively prime numbers, there exist $m, n \in \mathbb{Z}$ such that $a_1 + sn = a_2 + tm =: b_2$. Obviously $s \mid p(a_1 + sn) - p(a_1)$ and $t \mid p(a_2 + tm) - p(a_2)$, so $st \mid p(b_2)$.

Similarly one obtains $b_3$ such that $p(a_3) \mid p(b_3)$ and $p(b_2) \mid p(b_3)$ thus also $p(a_1) \mid p(b_3)$ and $p(a_2) \mid p(b_3)$.

Reasoning inductively we obtain the existence of $a = b_k$ as required.

The polynomial $p(x) = 2x^2 + 2$ shows that the second part of the problem is not true, as $p(0) = 2$, $p(1) = 4$ but no value of $p(a)$ is divisible by 8 for integer $a$.

Remark. One can assume that the $p(a_i)$ are nonzero and ask for $a$ such that $p(a)$ is a nonzero multiple of all $p(a_i)$. In the solution above, it can happen that $p(a) = 0$. But every number $p(a + n p(a_1) p(a_2) \dots p(a_k))$ is also divisible by every $p(a_i)$, since the polynomial is nonzero, there exists $n$ such that $p(a + n p(a_1) p(a_2) \dots p(a_k))$ satisfies the modified thesis.