<s>[INST] Prove that for all positive integers $x$, $\left\lfloor \frac{x^2 +2x + 2}{4}\right\rfloor =\left\lfloor \frac{x^2 + 2x + 1}{4}\right\rfloor$. Title says it all, basically. I believe it to be true that
$$\left\lfloor \dfrac{x^2 + 2x + 2}{4} \right\rfloor=\left\lfloor \dfrac{x^2 + 2x + 1}{4} \right\rfloor$$ for all positive integers $x$. I am, however, having a difficult time proving this. My current proof reads along the lines of the fact that, when adding $2$ or $1$, it is impossible to cause a large enough difference in the two numbers that, when divided by four and floored, they evaluate to different numbers. This basically comes down to proving that $x^2 + 2x \neq 4k + 2$ for some integer $k$, but I'm not sure if this is a good way of proving it.
Could anyone shine some light on this? Thanks.
 [/INST] $x$ is either odd or even.  If $x$ is even, let $x=2r+1$.  Then the right hand side is 
$$
\lfloor \frac{(x+1)^2}{4} \rfloor = \lfloor 4 \frac{(r+1)^2}{4} \rfloor 
= \lfloor (r+1)^2 \rfloor = (r+1)^2
$$
and the left hand side is 
$$
\lfloor \frac{(x+1)^2+1}{4} \rfloor = \lfloor 4 \frac{(r+1)^2+1}{4} \rfloor 
= \lfloor (r+1)^2 + \frac{1}{4} \rfloor = (r+1)^2
$$
The other answer covers the even case.
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