Proving or disproving $c-d|p(c)-p(d)$ where $p$ is a polynomial. I have $p(X)=\sum_{i=0}^{n}{a_iX^i}$, where $a_i\in\Bbb{Z}$. Let $c,d\in\Bbb{Z}$. 
Prove or disprove: $c-d|p(c)-p(d)$. 
I did some algebra but I can't think of a way to divide high power parts by $c-d$. I can't on the other hand find a counter example, and it does feel like a true statement. 
I would really appreciate your help here. 

Note that
$p(c) - p(d) = \sum_0^n a_i c^i - \sum_0^n a_i d^i = \sum_0^n a_i (c^i - d^i); \tag{1}$
so much is evident from the definition of $p(X)$; but next, note that that in fact the $i = 0$ term of the rightmost sum is in fact zero:
$a_0(c^0 - d^0) = a_0 (1 - 1) = a_0(0) = 0; \tag{2}$
thus we have
$p(c) - p(d) = \sum_1^n a_i (c^i - d^i); \tag{3}$
we have that
$c^i - d^i = (c - d) (c^{i-1}+c^{i-2}d + c^{i-3}d^2+...+c^{i-1}), \tag{4}$
for $i \ge 1$. which shows that $c - d$ divides every term of (3); hence, $c - d \mid p(c) - p(d)$. QED.
Note:  an inductive proof that $c - d \mid c^i - d^i$ may be found in the answers to this question.
