Let $p$ be an odd prime and $P(x)$ a polynomial of degree $p-2$. If $p\mid P(n)+P(n+1)+...+P(n+p-1) \forall n$, must $P$ have integer coefficients? I am stuck on this question from Zeit's book The Art and Craft of Problem Solving.
Question
Let $p$ be an odd prime and $P(x)$ a polynomial of degree at most $p-2$.
If $P(n)+P(n+1)+...+P(n+p-1)$ is an integer divisible by $p$ for every integer $n$, must $P$ have integer coefficients?
Attempt
I think $P$ must have integer coefficients. Suppose $P(x)=\sum_{k=0}^{p-2} a_k \cdot x^k$. Define $H(n)=P(n)+P(n+1)+...+P(n+p-1)$. Then we can expand $H(n)$ into a polynomial $H(n)=\sum_{k=0}^{p-2} c_k \cdot n^k$, where each of the coefficients $c_k$ are some linear combination of $a_k$.
If $c_i$ is not an integer for some $i$, then I think by choosing some suitable $n_i$, we can cancel all denominators of $c_j$ where $j \neq i$, while leaving $c_i$ still a fraction. Then $H(n_i)$ will be non-integer. But I don't know how to prove this properly. The $c_i$ are already very complicated to express.
However, if $c_i$ is integer for all $i$, then I am stuck.
I also don't know where to use the fact that $p\mid H(n)$ for all $n$.
Thanks for the help!

The answer is no.  For example, $$P(x):=\frac{x\,(x+1)\,(x+5)}{6}$$
is not in $\mathbb{Z}[x]$.  However, for $p=5$, we see that $P(x)$ satisfies the required conditions.  To see this, we observe that
$$\sum_{r=0}^4\,P(n+r)=5\,\left(\frac{n(n-1)(n+1)}{6}+2n^2+8n+11\right)$$
for all $n\in\mathbb{Z}$.
