Integer polynomial olympiad practice question Question
I have this problem from Math Olympiad practice.

Let $p$ be a polynomial with real coefficients and degree $n$. Suppose that $\frac{p(b)−p(a)}
{b−a}$ is an integer for all $0 ≤ a < b ≤ n$. Prove that $\frac{p(b)−p(a)}{b−a}$ is an integer for all pairs of distinct integers $a < b$.

Below is a theorem for Divisibility.
If a and b are integers, and p(x) is a polynomial with integer coefficients, then p(a) − p(b) is always divisible by a − b.
Attempt
My initial thinking is to show that p(x) is a polynomial with integer coefficients, then apply the theorem for Disivibility above to prove that $\frac{p(b)−p(a)}{b−a}$ is an integer.
I have not figured out how to do it yet and would welcome suggestions.
 A: Lets define  $P_k(x) = \frac{x(x-1)... (x-k+1)}{k!}$ with $P_0(x)=1$.
Note that $\forall n \in \mathbb{Z}$, $P_k(n)\in \mathbb{Z}$.
If $P\in \mathbb{R}[x]$, $\deg P=n$, verif $P(\{ m,m+1,...,m+n\} ) \subset \mathbb{Z}$, with $m\in \mathbb{Z}$, then there exist integers $a_k$ such that $P=\sum_{k=0}^n a_kP_k$ and so $P(\mathbb{Z}) \subset \mathbb{Z}$.
You can show it by induction, using $P(x+1)-P(x)$, for this problem we only need the second part which is very easy to demonstrate separately by the same way.
The first part is an interesting result that can be useful, a characterisation of polynomials with integer values.
So, we deduce that $\forall x \in \mathbb{Z}$, $\frac{p(x)-p(0)}{x} \in \mathbb{Z}$.
$t(x)=p(x+1)-p(x)$, $\deg t=n-1$
$\frac{t(b)-t(a)}{b-a}=\frac{p(b+1)-p(a+1)}{b-a}-\frac{p(b)-p(a)}{b-a}$
Proceding by induction, because we have $\frac{t(b)-t(a)}{b-a}$ integer for $0\le a\lt b\le{n-1}$,
that is an integer for all $a\lt b$.
Then using:
$\frac{t(b)-t(a)}{b-a}+\frac{p(b)-p(a)}{b-a}=\frac{p(b+1)-p(a+1)}{b-a}$
$\frac{p(b+1)-p(a+1)}{b-a}-\frac{t(b)-t(a)}{b-a}=\frac{p(b)-p(a)}{b-a}$
And that $\forall x \in \mathbb{Z}$, $\frac{p(x)-p(0)}{x} \in \mathbb{Z}$, we can be sure that $\frac{p(b)-p(a)}{b-a}$ always is an integer (taking $b=n$ in the first one, $a=-1$ in the second etc..).
