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

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The discussion focuses on proving that if the quadratic function f(x) = Ax^2 + Bx + C yields integer values for all integer inputs x, then the expressions 2A, A + B, and C must also be integers. Participants explore various mathematical approaches to establish this relationship and discuss the converse, which states that if 2A, A + B, and C are integers, then f(x) will produce integer outputs for all integers x. The conversation includes different proofs and methods to validate these claims, emphasizing the importance of integer properties in polynomial functions. Overall, the thread highlights key mathematical principles related to quadratic functions and integer outputs. The proofs provided contribute to a deeper understanding of the conditions under which these integer relationships hold.
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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.
 
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kaliprasad said:
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.
$f(0)=C \in Z---(2)$
$f(1)=A+B+C\in Z---(3)\,\,\therefore A+B\in Z$
$f(-1)=A-B+C\in Z---(4)$
$(3)+(4): 2A+2C\in Z\,\,\therefore 2A\in Z$
$(3)-(4):2B\in Z---(5)$
prove the converse :
if $x=2k\in Z$
then $f(x)=4k^2A+2Bk+C\in Z$
if $x=2k+1\in Z$
then $f(x)=4k^2A+4Ak+2Bk+A+B+C\in Z$
 
Albert said:
$f(0)=C \in Z---(2)$
$f(1)=A+B+C\in Z---(3)\,\,\therefore A+B\in Z$
$f(-1)=A-B+C\in Z---(4)$
$(3)+(4): 2A+2C\in Z\,\,\therefore 2A\in Z$
$(3)-(4):2B\in Z---(5)$
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
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
 
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