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

I've tried a lot to do it, but can't get it exactly right.
 A: We have
$f(x) = Ax^2 + Bx + C; \tag{1}$
set $x = 0$, an integer.  Then we have
$f(0) = C, \tag{2}$
so $C$ is an integer.  Set $x = 1$; then
$f(1) - C = A + B, \tag{3}$
showing $A + B$ is an integer.  Set $x = -1$; then
$f(-1) - C = A - B \tag{4}$
is also an integer.  Since
$2A = (A + B) + (A - B), \tag{4}$
we see that $2A$ is an integer as well.  Noting
$2B = (A + B) - (A - B) \tag{5}$
we also see $2B$ is an integer, "for free".  It appears to me that lab bhattacharjee was onto the right idea, though I did work this out for myself, concurrently as it were.  Cute, I must say . . . 
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
A: Hint $\displaystyle\,\ f(x) = ax^2\!+bc+c\, =\,  2a\, \dfrac{x(x\!-\!1)}2 +  (a\!+\!b)\, x + c \,=\,  2a {x \choose 2} + (a\!+\!b) {x\choose 1} + c{x\choose 0}$
Lemma $\ $ If $\displaystyle\ f(x)\, =\,  c_2 {x \choose 2} + c_1 {x \choose 1} + c_0{x\choose 0}\ $ then $\ f(0),f(1),f(2)\in\Bbb Z\,\Rightarrow\, c_i\in \Bbb Z$
Proof $\ f(0) = c_0\in \Bbb Z\,$ so $\, f(1) = c_0\!+\!c_1\in\Bbb Z\Rightarrow c_1\in\Bbb Z\,$ so $\, f(2) = c_0\!+\!2c_1\! +\! c_2\in\Bbb Z\,\Rightarrow\,c_2\in\Bbb Z$
Remark $\ $ The analogous result (and its converse) is true for arbitrary degree, which leads to a famous result of Polya and Ostrowski on integer-valued polynomials
