If $f(x)=Ax^2+Bx+C$ and $2A,A+B,C$ are integers, prove that $f(x)$ is integer whenever $x$ is an integer 
If $f(x)=Ax^2+Bx+C$ and $2A,A+B,C$ are integers, prove that $f(x)$ is integer whenever $x$ is an integer.

I found a way to prove the reverse statement. That is, I can prove that if $f(x)$ is integer whenever $x$ is integer, $2A,A+B,C$ are integers, by finding $f(0),f(1)$ and $f(-1)$.
How can I prove the actual statement? 

Using repeated differences on the values of $f(n)$ for $n=0,1,2,$ we get
$$
\begin{array}{lll}
C & A+B+C & 4A+2B+C \\
A+B & 3A+B  \\
2A  \\
\end{array}
$$
The last line repeats forever because $f$ is a polynomial of degree $2$.
Newton's interpolation formula (or direct verification) then gives
$$
Ax^2+Bx+C = C \binom{n}{0} + (A+B) \binom{n}{1} + 2A \binom{n}{2}
$$
from which the claims follows at once.
This solution tells us where the hypotheses come from.
