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? 
 A: Is A is an integer, then B is an integer and the result is obvious. So suppose $A=\frac{1}{2}+a$ for some integer $a$, then $B=\frac{1}{2}+b$ for some integer $b$. But now $f(x)=ax^2+bx+C+\frac{1}{2}x(x+1)$ which must be an integer for $x$ an integer, because $x(x+1)$ is even.
A: 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.
A: Hint $2A,A+B$ are integers so their subtraction would also be integer . thus $B-A$ is also integer now addition of $B-A,A+B$ will also be integer . Now take cases where $A=0.5x,B=0.5x,C=integer$ where $x\in (2n+1)$ and then second case that all are integers . You can easily prove now the required thing.
