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.
It is true because $c-d \mid c^k-d^k, k \in \mathbb {N}$
Alternatively, consider the polynomial $$ p(x) - p(d). $$ Clearly this has $d$ as a root. Therefore $$ x - d \mid p(x) - p(d). $$ Now substitute $x = c$.
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.
The most straightforward approach would be to note that $$ c^n - d^n = (c-d)(c^{n-1} + c^{n-2}d + \dotsb + cd^{n-2} + d^{n-1}) $$ and then use the fact that $p(c)-p(d)$ is a sum of multiples of such terms.
Here is another, more complicated proof which I happen to like:
Consider the polynomial $a(x) = p(x) - p(d)$. Since $q(d) = 0$, $x-d$ divides the polynomial $a(x)$. Since $x-d$ is monic with integer coefficients, and $a(x)$ has integer coefficients, the quotient $q(x)$ when $a(x)$ is divided by $x-d$ is also a polynomial with integer coefficients.
Thus $p(x) - p(d) = a(x) = (x - d)q(x)$. Substituting $x = c$ gives the desired result.
