# Problem Of The Week #460 March 22nd 2021

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#### anemone

##### MHB POTW Director
Staff member
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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#### anemone

##### MHB POTW Director
Staff member
No one answered last week's POTW. 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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