MHB Can All Outputs of a Quadratic Function Be Integers?

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The discussion revolves around a problem of determining whether all outputs of a quadratic function can be integers given that specific outputs are integers. The function is defined as f(x) = ax^2 + bx + c, with a, b, and c as real numbers. It is established that if f(0), f(1), and f(2) are integers, then f(2010) must also be an integer. However, the integer status of f(2011) remains uncertain and is left for further exploration. The thread encourages engagement with the problem, highlighting a lack of responses to previous challenges.
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Here is this week's POTW:

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Let $f(x)=ax^2+bx+c$ where $a,\,b$ and $c$ are real numbers. Assume that $f(0),\,f(1)$ and $f(2)$ are all integers. Prove that $f(2010)$ is also an integer and decide if $f(2011)$ is an integer.

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No one answered last week's POTW.(Sadface) However, you can find the suggested solution below:

Since $f(0),\,f(1)$ and $f(2)$ are all integers, $f(2)-2f(1)+f(0)$ is also an integer. Now,

$f(2)-2f(1)+f(0)=(4a+2b+c)-2(a+b+c)+c=2a$

So $2a$ is an integer. Since $f(0)$ and $f(1)$ are both integers, $f(1)-f(0)$ is also an integer. Now,

$f(1)-f(0)=(a+b+c)-c=a+b$

So $a+b$ is an integer. Finally, $f(0)=c$, so $c$ is an integer.

Therefore

$\begin{align*}f(2010)&=2010^2a+2010b+c\\&=4038090a+2010a+2010b+c\\&=2019045(2a)+2010(1+b)+c\end{align*}$

which is an integer because $2a,\,a+b$ and $c$ are all integers. Also,

$\begin{align*}f(2011)&=2011^2a+2011b+c\\&=4042110a+2011a+2011b+c\\&=2021055(2a)+2011(1+b)+c\end{align*}$

which is an integer because $2a,\,a+b$ and $c$ are all integers.
 
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