MHB Integer Challenge: Proving $2A, A+B, C$ integers for $f(x)=Ax^2+Bx+C$

[kaliprasad]

Let $f(x) = Ax^2 + Bx +C$ where A,B,C are real numbers. prove that if $f(x)$ is integer for all integers x then
$2A, A + B, C$ are integers. prove the converse as well.

 

[Albert1]

Let $f(x) = Ax^2 + Bx +C---(1)$ where A,B,C are real numbers. prove that if $f(x)$ is integer for all integers x then
$2A, A + B, C$ are integers. prove the converse as well.

Spoiler

$f(0)=C \in Z---(2)$
$f(1)=A+B+C\in Z---(3)\,\,\therefore A+B\in Z$
$f(-1)=A-B+C\in Z---(4)$
$(3)+(4): 2A+2C\in Z\,\,\therefore 2A\in Z$
$(3)-(4):2B\in Z---(5)$
prove the converse :
if $x=2k\in Z$
then $f(x)=4k^2A+2Bk+C\in Z$
if $x=2k+1\in Z$
then $f(x)=4k^2A+4Ak+2Bk+A+B+C\in Z$

 

[kaliprasad]

Spoiler

$f(0)=C \in Z---(2)$
$f(1)=A+B+C\in Z---(3)\,\,\therefore A+B\in Z$
$f(-1)=A-B+C\in Z---(4)$
$(3)+(4): 2A+2C\in Z\,\,\therefore 2A\in Z$
$(3)-(4):2B\in Z---(5)$
prove the converse :
if $x=2k\in Z$
then $f(x)=4k^2A+2Bk+C\in Z$
if $x=2k+1\in Z$
then $f(x)=4k^2A+4Ak+2Bk+A+B+C\in Z$

Above is a good solution different from mine which is as below

Spoiler

we have $f(x) = A x^2 + Bx + C = A (x^2-x) + (A+B) x + C=2A\frac{x(x-1)}{2} + (A+B) x + C$
now x and $\frac{x(x-1)}{2}$ are integers for integer x. so if (A+B),2A and C are integers the $f(x)$ is integer for integer x
if f(x) is integer for all x then $f(0) = C$ is integer.
$f(1) = (A+B) 1 + C$ is integer so $A+B$ is integer
$f(2) = 2A + 2(A +B) + C$ is integer so $2A$ is integer