Does $\left\lfloor\frac{x^2+x}{i}\right\rfloor - \left\lfloor\frac{x^2}{i}\right\rfloor = \left\lfloor\frac{x}{i}\right\rfloor$? Does $\left\lfloor\dfrac{x^2+x}{i}\right\rfloor - \left\lfloor\dfrac{x^2}{i}\right\rfloor = \left\lfloor\dfrac{x}{i}\right\rfloor$?
where $i \le x^2$ and $x$ are any positive integer.
Intuitively, this doesn't seem correct to me but here's my argument which appears valid:
(1) There exists an integer $a$ such that:  $x \equiv a \pmod i$ where $0 \le a < i$
(2) $\left\lfloor\dfrac{x^2+x}{i}\right\rfloor - \left\lfloor\dfrac{x^2}{i}\right\rfloor = \dfrac{x^2 + x - a^2 - a}{i} - \dfrac{x^2 - a^2}{i} = \dfrac{x-a}{i} = \left\lfloor\dfrac{x}{i}\right\rfloor$
Is my argument wrong?  Is my intuition wrong? 

Consider x = 7 and i = 5. 
you get 
$\left\lfloor\dfrac{49+7}{5}\right\rfloor = 11$ 
$\left\lfloor\dfrac{49}{5}\right\rfloor = 9$
$\left\lfloor\dfrac{7}{5}\right\rfloor = 1$
as you see the equation doesn't hold. 
Your intuition is right and the argument is wrong. The same argument should follow the following logic
$\dfrac{x}{i} = k + \dfrac{a}{i}$ where $k$ is highest possible positive integer without making term a negative
$\dfrac{x^2}{i} = l + \dfrac{a^2}{i}$ 
$\left\lfloor\dfrac{x^2+x}{ i}\right\rfloor = k + l + \left\lfloor\dfrac{a+a^2}{i}\right\rfloor$
$\left\lfloor\dfrac{x^2}{i}\right\rfloor = l + \left\lfloor\dfrac{a^2}{i}\right\rfloor$
$\left\lfloor\dfrac{x}{i}\right\rfloor = k + \left\lfloor\dfrac{a}{i}\right\rfloor$
as you see the floor terms are not necessarily equal. There could be cases when $a+a^2$ 
is higher than $i$ when both $a$ and $a^2$ are less than $i$. 
Hope it helps
