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.
 A: HINT: Note that the numerators are $(x+1)^2+1$ and $(x+1)^2$. Consider separately the cases $x$ odd and $x$ even.
A: 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 
$$\left\lfloor\frac{x^2+2x+2}{4}\right\rfloor=\left\lfloor\frac{(2k-1)^2+2(2k-1)+2}{4}\right\rfloor=\left\lfloor k^2+\frac 14\right\rfloor=k^2.$$
$$\left\lfloor\frac{x^2+2x+1}{4}\right\rfloor=\left\lfloor\frac{(2k-1)^2+2(2k-1)+1}{4}\right\rfloor=\left\lfloor k^2\right\rfloor=k^2.$$
A: $x$ 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.
A: 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 $x^2+2x+3$
A: 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 $3\mod{4}$.
