$p|h(n)$ for some n is equivalent to $h$ mod $p$ having a root In Conrad's paper on Euclidean proofs of Dirichlet's theorem he states (on page 2), 

note that the divisibility condition $p|h(n)$, for some $n$, is equivalent to $h \pmod p$ having a root.

This is not clear to me, maybe I am missing something?
Here $h(x)\in\mathbb{Z}[x]$, so I assume $h \pmod p$ is attained by reducing the coefficients modulo $p$? 
He also doesn't state what kind of root $h \pmod p$ should have; of course it will have complex roots, so does $p|h(n)$ imply that $h \pmod p$ will have real roots/roots in $\mathbb{Z}$/roots in $\mathbb{Z_p}$?
 A: By definition $\ p\mid h(n)\iff \color{#c00}{h(n)\equiv 0}\pmod p$
The latter says that, $\bmod p\!:$ $\ \color{#c00}{n\ \text{is a root of }\ h(x)}.\,$  By Congruence Sum and Product Rules this remains true for any $\,n'\equiv n,\,$ and any $\,h'\equiv h\pmod{p},\,$ where the latter means they have congruent coefficients $\,h_i'\equiv h_i$ for all $i.\, $ In particular it's true when $\,h'_i := h_i\bmod p,\,$ i.e. for $\,h' := h\bmod p,\,$  the polynomial obtained by reducing $\!\bmod p$ all coefficients of $h.$
A: A friend and I think we may have come up with a solution to this question, if we assume that Conrad meant that $h(n)$ as a root in $\mathbb{Z_p}$;
$p|h(n)$ $\Rightarrow$ $h\mod p$ has root in $\mathbb{Z_p}$ :
For $h\in\mathbb{Z}[x]$, write $h'$ for $h$ reduced $\mod p$ (so $h' \in \mathbb{Z_p}$). 
Then $h(n)=pd$, some $d\in\mathbb{Z}$, so $h'(n)=0$, i.e. $h'(x)=0$ for some $x \in \mathbb{Z_p}$, which is equivalent to saying $h'(x)$ has a root in $\mathbb{Z_p}$.
$h\mod p$ has root $\mathbb{Z_p}$ $\Rightarrow$ $p|h(n)$:
As $h'(x)$ has a root in  $\mathbb{Z_p}$ we can write
$h'(x)=(x-a)(a_s x^s +...+a_1 x+a_0)$, then for $n\in \mathbb{Z_p}$, $h'(a)=0$.
Now, $h(x)=(x-a)(a_s x^s +...+a_1 x+a_0)+(b_k x^k +...+b_1 x+b_0)$, where $p|b_i$ $\forall$ $0\leq i \leq k$.
So $h(n)=0+(b_k x^k +...+b_1 x+b_0)$ which is divisible by $p$.
Does this seem correct? 
