All of the coefficients of $P(x)$ are divisible by $p$ 
Find a counterexample to the following: If $P(x)$ is a polynomial with integer coefficients and $p$ a prime, then if $p \mid P(x)$ for all $x$ then all of the coefficients of $P(x)$ are divisible by $p$.

I thought about proving this by a proof by contradiction but wasn't sure how to do that. How should we prove the statement?
 A: I think this is false.
Consider the polynomial $x^p-x$, by Fermat's little theorem $x^p\equiv x \bmod p$ so $x^p
-x\equiv 0 \bmod p$
A: Another counterexample: any sequence of $p$ consecutive integers contains a multiple of $p$, so $P(x)=x(x+1)(x+2)...(x+p-1)$ is always a multiple of $p$ for integer $x$, while the leading coefficient is $1$ which is obviously not divisible by $p$.
A: Previous anwers show that the OP claim is false. Here we find all those counterexamples.
For each $P \in \mathbf{Z}[x]$, let $\hat{P}$ be the corresponding polynomial in $\mathbf{Z}_p[x]/(x^p-x)\mathbf{Z}_p[x]$. In other words, if $P(x)=\sum_i \alpha_ix^i$, then
$$
\hat{P}(x)=\sum_{j=0}^{p-1} \underbrace{\left(\sum_{n\ge 0}a_{j+(p-1)n} \pmod p\right)}_{\beta_j}x^j
$$
We claim that if $p\mid P(x)$ for all $x \in \mathbf{Z}$ if and only if $\hat{P}=0$, that is, $\beta_0=\cdots=\beta_{p-1}=0$.
Proof. The "if" part follows by Fermat's little theorem. About the "only if", the polynomial $\hat{P}$ has degree $p-1$ and $p$ roots $0,1,\ldots,p-1$, which is possible only if $\hat{P}=0$.
A: A very simple counterexample:
For $p=2$,
$2$ divides
$x(x+1)=x^2+x$
for all integer $x$.
