<s>[INST] Getting rid of a floor function in the next expression:$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $, It is known x is odd. I was wondering if there's a way in which you can get rid of a floor function in the next expression:$$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $$ It is known x is odd.
 [/INST] Since $x$ is odd, $x=2n+1$ for some integer $n$, and therefore $(x-2)^2=(2n-1)^2=4n^2-4n+1$. Thus,
$$\left\lfloor\frac{(x-2)^2}4\right\rfloor=\left\lfloor n^2-n+\frac14\right\rfloor=n^2-n\;.$$
And $n=\dfrac{x-1}2$, so the expression reduces to
$$\left(\frac{x-1}2\right)^2-\frac{x-1}2=\frac{x^2-4x+3}4\;.$$
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