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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.