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.

Let $x^2+2x+1=(x+1)^2$ but $(x+1)^2\equiv0,1\pmod4$
i.e., $x^2+2x+1=4a$ or $4a+1$ for any integer $a\ge0$ 
So, $x^2+2x+2=4a+1$ or $4a+2$
In either case, the floor by $\pmod4$ remains same even for $x^2+2x+3$
