<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] If $x=2k$, then
$$\left\lfloor\frac{x^2+2x+2}{4}\right\rfloor=\left\lfloor\frac{(2k)^2+2\cdot 2k+2}{4}\right\rfloor=\left\lfloor k^2+k+\frac 12\right\rfloor=k^2+k.$$
$$\left\lfloor\frac{x^2+2x+1}{4}\right\rfloor=\left\lfloor\frac{(2k)^2+2\cdot 2k+1}{4}\right\rfloor=\left\lfloor k^2+k+\frac 14\right\rfloor=k^2+k.$$
If $x=2k-1$, then 
$$\left\lfloor\frac{x^2+2x+2}{4}\right\rfloor=\left\lfloor\frac{(2k-1)^2+2(2k-1)+2}{4}\right\rfloor=\left\lfloor k^2+\frac 14\right\rfloor=k^2.$$
$$\left\lfloor\frac{x^2+2x+1}{4}\right\rfloor=\left\lfloor\frac{(2k-1)^2+2(2k-1)+1}{4}\right\rfloor=\left\lfloor k^2\right\rfloor=k^2.$$
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