If $p(x)$ maps $\Bbb Z$ to $\Bbb Z$ and $\deg(p)=n$, show $n!\cdot p(x)$ has integer coefficients Given a single variable real polynomial $p(x)$ of degree $n$ that maps integers to integers, show that $n!\cdot p(x)$ has integer coefficients.
I prefer an elementary solution if possible. It seems like some big machinery to take care of this should exist as well but I don't know any names.
For example, if $n=4$, some polynomial like $p(x)=\frac{(x-3)(x-8)(x-50)(x-13)}{24}$ works since $3,8,50,13$ are congruent to $3,0,2,1$ mod $4$. Any integer linear combinations of polynomials of this form works. But I can't say that these are the only possibilities.
What I typed earlier in the preceding paragraph is not correct. I meant polynomials such as $p(x)=\frac{(x-4)(x-5)(x-6)(x-7)}{24}$ works since the product of any $k$ consecutive integers is divisible by $k!$.
 A: Two words: Finite differences.
If $q$ is a polynomial, define another polynomial $\Delta q$ via: $(\Delta q)(x) = q(x+1)-q(x)$. Clearly, $\Delta^k p(0)$ is an integer for all $k$.
Then we know that $$p(x)=\sum_{k=0}^n \frac{\Delta^k p(0)}{k!}(x)_k$$
Where:
$$(x)_k = x(x-1)\cdots(x-k+1)$$

A similar proof, but with some proof of the finite difference result, is to note that the polynomial $(x)_k$ is of degree $k$, and we have $(x+1)_k-(x)_k=k(x)_{k-1}$. So every polynomial can be written uniquely as:
$$p(x)=\sum_{k=0}^{n} a_k(x)_k$$ where $n=\deg p(x)$. We then prove by induction that $a_k \cdot k!$ must be an integer.
If $p:\mathbb Z\to\mathbb Z$, we always have $a_0=p(0)\in\mathbb Z.$ 
Now $\Delta p:\mathbb Z\to\mathbb Z$, and:
$$\Delta p(x)=p(x+1)-p(x)=\sum_{k=0}^{n} ka_k(x)_{k-1}.$$
Since $p$ is smaller degree, by induction, you have that $k!a_k=(k-1)!ka_k\in\mathbb Z$ for each $k>0.$
A: I can prove the $n=2$ case with elementary math.
We're saying that $p$ maps integers to integers. That means that $f=2!\cdot p$ maps integers to multiples of $2$. Let's write $f(n)=\frac{r_1}{s_1}n^2 + \frac{r_2}{s_2}n + \frac{r_3}{s_3}$, and we can assume that each fraction has been written in lowest terms. We want to show that $s_1, s_2, s_3$ all equal $1$.
Since $f(0)$ is an integer, we immediately have that $s_3=1$.
Since $f(1)$ is an integer, we have that $\frac{r_1}{s_1} + \frac{r_2}{s_2}$ is an integer, so $s_1=s_2$. Let's just call it $s$.
Since $f(1)$ and $f(2)$ are both even integers, we have that $r_1 + r_2$ and $4r_1 + 2r_2$ are both multiples of $2s$. Subtracting twice the first expression from the second one, we obtain that $2s$ also divides $2r_1$, which implies that $s|r_1$. However, $r_1$ and $s$ are relatively prime, so $s=1$, q.e.d..
I don't immediately see how this type of argument would gracefully extend to higher degree polynomials, but it might.
