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.
HINT: Note that the numerators are $(x+1)^2+1$ and $(x+1)^2$. Consider separately the cases $x$ odd and $x$ even.
The only time the two floors are different is when $x^2+2x+2$ is a multiple of $4$ (or, equivalently, looking at the right-hand side, $x^2+2x+1 \equiv 3\mod{4}$.

But $x^2+2x+2 = (x+1)^2+1$, and $(x+1)^2\equiv 0, 1\mod{4}$, so we cannot have $x^2+2x+2\equiv 0\mod{4}$. Almost identically, considering the RHS instead, $x^2+2x+1 = (x+1)^2\equiv 0, 1\mod{4}$, so it cannot be $3\mod{4}$.